1 Introductionaplustestbank.eu/sample/Solution-Manual-for-Fundamentals... · 2017-04-07 · 1....

1 Introduction

5This text is associated with Themis Matsoukas, Fundamentals of Chemical Engineering Thermodynamics 0-13-269306-2 (978-0-13-269306-6). Copyright © 2013 Pearson Education, Inc. Do not redistribute.

Full file at http://AplusTestbank.eu/Solution-Manual-for-Fundamentals-of-Chemical-Engineering-Thermodynamics-132693062

1. INTRODUCTION

Problem 1.1 Solution

We will need the following unit conversions:

1 ft D 0:3048 m; 1 lb D 0:454 kg; 1 lbmol D 454 mol

(Note on the conversion mol to lbmol: one mol has a mass equal to the molecular weight in g, while onelbmol has a mass equal equal to the molecular weight in lb.) We also need the molar mass of ammoniawhich is

Mm D 17 g/mol D 17 � 10�3 kg/mol

a) Specific Volume:

V D1

41:3 lb/ft3D 0:02421 ft3/lb D 0:00151 m3/kg D 1:51 cm3/g

b) Molar Volume

Vmolar D VMm D 2:567 � 10�5 m3/mol D 25:67 cm3/mol D 0:4116 ft3/lb-mol



http://AplusTestbank.eu/Solution-Manual-for-Fundamentals-of-Chemical-Engineering-Thermodynamics-132693062

Problem 1.2 Solution First we write the given equation as

TK DB

A � .lnPbar/=.ln 10/� C

where TK refers to T in kelvin, Pbar refers to pressure in bar, and the logarithm is natural. Next we use

TF D 1:8.TK � 273:15/C 32 and Pbar DPpsi

14:5

and substitute these values into the above equation. After some manipulation the result is

TF DB.1:8/.ln 10/

A ln 10C ln.14:5/ � lnPpsi� C.1:8/ � .1:8/.273:15/C 32

Doing the algebra,

TF D5062:37

13:2153 � lnPpsi� 302:217

Therefore,A0 D 13:2153; B 0 D 5062:37 C 0 D 302:217




1. INTRODUCTION

Problem 1.3 Solution a) The mean velocity is

Nv D

�8kBT

�m

�1=2where kB D 1:38 � 10�23 J=K, T D 273:16 K. The mass of the water molecule is

m DMm

NAVD

18 � 10�3 kg=mol6:022 � 1023 mol�1 D 2:98904 � 10

�26 kg:

The mean velocity isNv D 566:699 m=s D 2040:12 km=h D 1267 mph

This result depends only on temperature and since all three phases are the same temperature, molecules havethe same mean velocity in all three phases.

b) The mean kinetic energy isNEkin D

12mv2 D 1

2mv2

where v2 is the mean squared velocity,

v2 D3kBT

m:

With this the mean kinetic energy is

NEkin D3

2kBT D 5:65441 � 10

�21 J

This is the mean kinetic energy per molecule in all phases. The number of molecules in 1 kg of water is

N D1 kg

18 � 10�3 kg=mol6=022 � 1023 D 3:34556 � 1025

The total kinetic energy in 1 kg of water at 0.01 ıC (regardless of phase) is

NEkin D�5:65441 � 10�21 J

� �3:34556 � 1025

�D 189; 171 J D 189 kJ

Comment: This is the translational kinetic energy of the molecule, i.e., the kinetic energy due to themotion of the center of mass. A water molecule possesses additional forms of kinetic energy that arisefrom the rotation of the molecule, the bending of bonds, and the vibration of oxygen and hydrogen atomsabout their equilibrium positions. These are not included in this calculation as the Maxwell-Boltzmanndistribution refers specifically to the translational kinetic energy.

c) The above calculation shows that the mean kinetic energy depends only on temperature (it is independentof pressure or of the mass of the molecule). Therefore, oxygen at 0.01 ıC has the same kinetic energy aswater at the same temperature:

NEkin D3

2kBT D 5:65441 � 10

�21 J





3 4 5 6 7 8 9 10

-4.´10-22

-2.´10-22

0

2.´10-22

4.´10-22

r HÞL

FHJ

ou

leL

Potential

a) The potential has a minimum just above 4 A or so. To determine this value accurately we must set thederivative of the potential equal to zero solve for the value of r . To do this easily, we define a new variablex D r=� and rewrite the potential as:

ˆ D a�x�12 � x�6

�By chain rule we now have:

dF

drDdF

dx

dx

drD��12x�13 C 6x�7

� dxdr

Setting this to zero and solving for x we have:

�12x�13 C 6x�7 D 0 ) x D 21=6

Since r D �x, the value of r that minimizes the potential is

r� D 21=6� D .4:24964/.3:786 A/ D 4:25 A

b) If we imagine N molecules to be situated at the center of cubes whose sides are equal to r�,

L

L

the volume occupied isV D N.r�/

3

These N molecules correspond to N=NAv mol and their total mass is

M DN

NAVMm

where Mm is the molar mass of methane (Mm D 16 � 10�3 kg/mol). For the density, therefore, we obtainthe following final formula:

� DMm

NAV.r�/3

By numerical substitution we finally obtain the density:

� D16 � 10�3 kg/mol

.6:024 � 1023 mol�1/.4:25 A/.10�10 m/A/D 346 kg/m3 D 0:346 g/cm3




1. INTRODUCTION

c) Specific volumes of saturated liquid methane are listed in Perry’s Handbook from which we can computethe densities. We notice (as we would have expected) that liquid volumes near the critical point (Tc D190:55 K) vary with pressure, from 162.3 kg/m3 at the critical point to 454 kg/m3 around 90 K. Our valuecorresponds to Perry’s tabulation at about 160 K. Our calculation is approximate and does not incorporatethe effect of pressure and temperature. Notice that if we pick a distance somewhat different from r� theresult will change quite a bit because if the third power to which this distance is raised. But the importantconclusion is that the calculation placed the density right in the correct range between the lowest and highestvalues listed in the tables. This says that our molecular picture of the liquid, however idealized, is fairly closeto reality.





Assuming molecules to be sitting at the center of a cubic lattice with spacing L, the volume occupied by Nmolecules is NL3 and the density is

� DMm

NAVL3

where Mm is the molar mass. Solving for the intermolecular distance,

L D

�Mm

NAV�

�a) For liquid water with � D 1000 kg=m3,

L D

�18 � 10�3 kg=mol

.6:022 � 1023 mol�1/.1000 kg=m3/

�1=3D 3:10 � 10�10 m D 3:01 A

b) For steam with � D 0:4 kg=m3

L D

�18 � 10�3 kg=mol

.6:022 � 1023 mol�1/.0:4 kg=m3/

�1=3D 4:2 � 10�9 m D 42 A

The intermolecular distance is about 10 times larger in the vapor.




1. INTRODUCTION


F

P0

P=0

2 R

The force needed to separate the two haves is equal to the force that is exerted by pressure on one hemisphere:

F D �P A

where �P is the pressure difference between the atmospheric pressure and the contents of the sphere, andA is the cross-sectional area of the sphere (�R2). Assuming the sphere to be fully evacuated, the pressuredifference is equal to the atmospheric pressure (� 1 bar D 105 Pa). The force now is

F D P0.�R2/ D .105 Pa/.�0:52 m2/ D 19; 635 N

This force corresponds to a mass

M DF

gD39269:9 N9:81 N=kg

D 2; 000 kg:

Assuming an average weight of 80 kg/person, it would take the weight of 25 people (!) to separate thespheres.




2 Phase Diagrams of Pure Fluids




2. PHASE DIAGRAMS OF PURE FLUIDS

Problem 2.1 Solution a) 25 ıC, 1 bar: liquid, because the temperature is below the saturation temperatureat 1 bar (99.63ıC).

10 bar, 80ıC: liquid, because the temperature is below the saturation temperature at 10 bar (179.88 ıC).

120 ıC, 50 bar: At 120 ıC the vapor pressure is 198.54 kPa = 1.9884 bar. Since the actual pressure is higher,the state is liquid. (Or, at 50 bar the saturation temperature is 263.91 ıC. Since the actual temperature islower the state is liquid.)

b) Liquid, because the pressure (1 atm = 760 mm Hg) is higher than the vapor pressure of bromobenzene(10 mg H) at the same temperature.

c) Liquid, because the temperature is lower than the boiling point at the same pressure.

Note: All of these statements will make better sense if you plot your information on the PV graph.




Problem 2.2 Solution a) From steam tables at 40 bar we collect the following data:

600 ıC 650 ıC40 bar 0:0989 m3=kg 0:1049 m3=kg

By interpolation at V D 100 g=cm3 D 0:1 m3=kg we find

T D 600C650 � 600

0:1049 � 0:0989.0:1 � 0:0989/ D 609:2 ıC

The system is at 40 bar, 609.2 ıC: the phase is vapor.

b) At 6 bar we find that the desired volume lies between that of the saturated liquid (VL D 0:0011 m3=kg)and saturated vapor (VV D 0:3156): the system is a vapor/liquid mixture. The vapor and liquid fractions areobtained by lever rule:

xL D0:3156 � 0:1

0:3156 � 0:0011D 0:314 D 31:4%

xV D 1 � 0:314 D 0:686 D 68:6%





Problem 2.3 Solution a) At 1 bar, 210 ıC, the specific volume of steam is found by interpolation to be

V D 2219:2 cm3/g D 2:2192 m3/kg

Since the total volume of the vessel is 3.5 m3, the mass is

M DV t

VD

3:5 m3

2:2192 m3/kgD 1:577 kg

b) Since the system is a mixture of liquid and vapor, it is saturated. From the saturated tables at 20 bar wefind by interpolation the following:

T D 212:37 ıC; VL D 1:18 cm3/g; VV D 99:54 cm3/g

The specific volume of the mixture is

V D lVL C vVV D .0:85/.1:18/C .0:15/.99:54/ D 15:93 cm3/g

The total volume required for storage is V t DMV , where V is the specific volume calculated above:

V t DMV D .525 kg/.1000 g/kg/.15:93 cm3/g/ D 8363250 cm3 D 8:36 m3




Problem 2.4 Solution Solution Initial state:

P D 45 bar; T D 257:41 ıC; V L D 1:269 cm3/g; V V D 44:037 cm3/g

The specific volume is obtained from the lever rule with XL D 0:25:

V D xLV L C .1 � xL/V V D .0:25/.1:269/C .1 � 0:25/.44:037/ D 33:345 cm3/g

a) We know P D 80 bar, V D 33:345 cm3/g. From the steam tables we find:

T D 375 ıC; V D 32:222 cm3/g

T D 400 ıC; V D 34:31 cm3/g

By interpolation at V D 33:345 cm3/g we obtain T D 388:5 ıC.

b) From saturated tables we find:

T D 272 ıC; P D 56:802 bar; V D 34:42 cm3/g

T D 274 ıC; P D 58:587 bar; V D 33:29 cm3/g

By interpolation at V D 33:345 cm3/g we find P D 57:76 bar.

c) The volume fraction of the liquid, �L, and the mass fraction of the liquid, xL, are related as follows:

vol. occupied by liq.vol. occupied by liq.+vol. occupied by vap

DxLV L

xLV L C xV V V

from which,

xL DV V �L

V L C .V V � V L/�L

The saturated volumes are those at the initial state and with �L D 0:25 we find

xL D.44:037/.0:25/

.1:269/C .44:037 � 1:269/.0:25/D 0:920429

Therefore, the new specific volume in the tank is

V 0 D .0:920429/.1:269/C .1 � 0:920429/.44:037/ D 4:6721 cm3/g

If we have mass M initially and mass M 0 at the end, we can write for the tank volume:

V t D NV D V tM 0 )M 0

MDV

V 0

orM �M 0

MDV

V 0� 1 D

33:345

4:6721D 6:14

Therefore, we must add 6.14 times the original mass (that’s 614%).





Problem 2.5 SolutionP .bar/ 0:5 15

V .m3=kg/ 1:124 0:031

� .kg=m3/ 0:88968 32:2581

If we interpolate for V we findV D 0:257138 m3=kg:

If we interpolate for � we find

� D 25:7681 kg=m3) V D 1=� D 0:0388077 m3=kg:

These results are quite different. Which one should we pick?

For the molar volume we haveV D

ZRT

P:

Using � D 1=V , the molar density is

� DP

ZRT(2.1)

We conclude that while V is inversely proportional to P , 1=V is proportional, provided thatZ does not varymuch in the given range. Therefore, we would accept the interpolation in �.





a) 198.3 C

b) We first obtain the specific volumes of the saturated liquid and vapor at 15 bar:

V L D 0:0015 m3=kg; V V D 0:1317 m3=kg

Next we calculate the total mass of each phase:

mL D .0:5 m3/.0:0015 m3=kg/ D 434:8 kg

mV D .11:5 m3/.0:1317 m3=kg/ D 87:3 kg

The total mass ism D 434:8C 87:3 D 522:1 kg

c) The quality of the steam is:

xV D87:3

522:1D 0:167

d) The specific volume after 87% of the mass is removed is

V 0 D0:13 � 522:1

12 m3 D 0:1768 m3=kg

The temperature is obtained by interpolation at 15 bar between 250 ıC and 300 ıC. We find

T 0 D 320 ıC





Problem 2.7 Solution a) Since the cooker contains both vapor and liquid, the state is saturated steam.Therefore, T D T sat D 120:23 ıC.

b) From steam tables we obtain the specific volumes of the saturated phasees:

V L D 1:061 cm3/g; V V D 885:44 cm3/g

The total volume of the liquid in the cooker is V L;tot D .0:25/.8/ liter = 2000 cm3. Therefore, the mass ofthe liquid is

mL DV L;tot

V LD2000

1:061D 1885:0 g

The volume of the vapor in the cooker is V L;tot D .0:75/.8/ liter = 6000 cm3 and its mass is given by

mV DV L;tot

V VD

6000

885:44D 6:78 g

The total mass is

m D 1885:0C 6:78 D 1891:78 g D 1:89 kg

c) The mass fractions of the liquid (xL), and of the vapor (xV ) are

xL D1885:0

1885:0C 6:78D 0:9964 D 99:64%; xV D 1 � xL D 0:004 D 0:4%

Even though the vapor occupies 75% of the volume, it only represents 0.4% of the total mass.

d) The quick solution is to take a look at the PV graph. The initial state is at A and the final state, B , isreached by constant-volume cooling. This state is obviously in the two-phase region because the originatingstate was also inside that region as well. We conclude that P D P sat.25 ıC/ D 3:166 kPa. (If, however,state A were in the superheated region, we would be able to tell if B is superheated or vapor/liquid and wewould have to do the solution in more detail as shown below.)

Detailed solution: The total volume of the system as well as the specific volume remain constant. Thespecific volume is

V DVcooker

MtotD8000 cm3

1891:78 gD 4:229 cm3/g

From the saturated steam tables at 25 ıC we find, V L D 1:003 cm3/g, V V D 43400 cm3/g. The specificvolume of the system is between these two values, therefore we still have a saturated system. We concludethat P D P sat.25 ıC/ D 3:166 kPaD 0:03166 bar.




P

V

initial state

const. V

process

�nal state

2 bar 120.21°C

25 °C0.003175

bar

This solution is more general and will work regardless of where the initial state is.

e) When the system has cooled, the outside pressure is 1 bar and the inside pressure is 0.03166 bar. There-fore, the lid remains closed under the action of this pressure difference. The force is

F D �R2.Pout � Pin/ D .�/.0:2 m2/.1 � 0:03166/ bar � 105N/bar D 12166 N

To put this force into perspective we calculate the mass whose weight is 12166 N:

M DF

gD

12166 N9:81 m/s2

D 1240 kg D 2732 lb

If you can lift 2700–2800 lb then you could remove that lid! (Note: Whether you take the outside pressureto be 1 bar or 1 atm or something similar, the conclusion remains that the required force is indeed very large.






It is very helpful to draw the PV graph, shown below:

1 bar A

B 80 C

1 bar

C

D'

C'

D

E

F

G

H

part (a) part (b)

part (c) part (d)

a) The initial state (A) is compressed liquid (80 ıC, 1.013 bar). The process is conducted under constantvolume. Assuming the isotherm to be vertical, the final state, B, is at the same temperature and on thesaturation line. Therefore: T D 80 ıC, P D 47:36 kPa.

Notice that the temperature has not changed. This is a consequence of the fact that we have approximatedthe isotherm with a vertical line. In reality, the isotherm is not vertical and state B should be at a temperaturesomewhat below 80 ıC. However, the steepness of the isotherm means that this temperature is very close to80 ıC. If we had the value of � and ˇ we could calculate this temperature and would verify that it is indeedextremely close to 80 ıC.

b) The initial state is a V/L mixure (state C). The process is under constant volume, therefore, the final stateis located at the intersection of the vertical line through C and the saturation line (state D). From steamtables:

1 bar; saturated:VL D 1:043; V V D 1693:7

The mass fractions of the liquid and the vapor are:

l D0:5=1:043

0:5=1:043C 0:5=1693:7D 0:99938

v D 1 � l D 0:00062




The specific volume at the initial state is (in cm3/g)

V D .0:99938/.1:043/C .0:00062/.1693:7/ D 2:085

At the final state: V D 2:085, and saturated. From steam tables we find, T � 367 ıC, P D 203:13 bar(saturated liquid).

Note 1: Even though the mass fraction of the vapor is very nearly 0, it would not be correct to set it equal to0. While v is small, when multiplied by a large V V it makes a significant contribution to the specificvolume of the mixture. If we had set v �, we would have concluded that the initial state is practicallysaturated liquid which means that the final pressure is almost 1 bar. Clearly, this approximation missesthe right pressure by more than 200 bar!!!

Note 2: In this case the specific volume of the vapor-liquid mixture was very close to the liquid side and forthis reason the final state was liquid. In other words, under heating the vapor condenses and becomesliquid. If, however, the initial volume was much closer the vapor side (state C’), then heating wouldproduce vapor. In this case, heating would cause the liquid to evaporate. That is, after heating thecontents of the vessel the final state might either saturated vapor or saturated liquid. Can you establisha criterion for the initial specific volume to determine whether the final state is vapor or liquid?

c) The final state saturated vapor (state F). The process is cooling under constant volume, therefore, theinitial state must be somewhere on the vertical line through F and above point F (since cooling implies thatthe initial state is at higher T ). We conclude the initial state is superheated vapor.

d) By similar arguments as above, we determine that the initial state is vapor/liquid mixture. Notice thathere we are heating a vapor/liquid mixture and as a result the vapor condenses to produce saturated liquid!





Problem 2.9 Solution The specific volume of water under these conditions in the tank is

V D12 m3

6:2 kgD 1:935 m3=kg:

According to the steam tables, the volume of saturated vapor at 1.4 bar is between 1.694 m3=kg (at 1 bar)and 1.1594 m3=kg (at 1.5 bar). The calculated value is higher, therefore, the state is to the right of thesaturated vapor and it must be superheated.

b) We need an entry in the steam tables such that pressure is 1.4 bar and the specific volume is 1:935m3=kg.To locate this state, we interpolated in the steam tables between 1.0 bar and 1.5 bar at various temperaturesand construct the table below:

T 150 200 250 300 350 400 450

V .at 1:0 bar/ 1:9367 2:1725 2:4062 2:6389 2:8710 3:1027 3:3342

V .at 1:4 bar/ 1:4158 1:5901 1:7623 1:9335 2:1040 2:2742 2:4442

V .at 1:5 bar/ 1:2856 1:4445 1:6013 1:7571 1:9123 2:0671 2:2217

We can now see that the desired value is between 300 ıC and 350 ıC. By interpolation between these twotemperatures we find

T D 300 ıCC.350 � 300/ ıC.1:5 � 1:0/ bar

.1:4 � 1:0/ bar D 300:6 C

c) If we add more steam while keeping temperature constant to 300.6 C, pressure will increase and thespecific volume will decrease. The vapor will become saturated when the specific volume in the tank is thatof saturated vapor at 300.6 C. We obtain this value by interpolation in the saturated steam tables between300 ıC and 302 ıC:

T ıC V m3=kg300 0:02166

300:593 0:0211498

302 0:01994

The specific volume when the tank is saturated is V D 0:0211498 m3=kg. The total mass is

12 m3

0:0211498 m3=kg6:2 kg D 567:4 kg

The amount that must be added is.567:4 � 6:2/ D 561:2 kg




Problem 2.10 Solution Solution

The graphs below show the PV in various combinations of linear and logarithmic coordinates and the ZPgraph.

1.0

0.8

0.6

0.4

0.2

0.0

Com

pre

ssib

ility F

acto

r

250200150100500

Pressure (bar)

100 C

200 C

300 C

400 C

0.01

0.1

1

10

100

1000

Pre

ssu

re (

ba

r)

100

101

102

103

104

105

Specific volume (cm3/g)

100 C

200 C

300 C

400 C

250

200

150

100

50

0

Pre

ssu

re (

ba

r)

12 4 6

102 4 6

1002 4 6

1000


100 C200 C

300 C

400 C

250

200

150

100

50

0

Pre

ssu

re (

ba

r)

100806040200


100 C

200 C

300 C

400 C

Comments:

� The volumes span a very wide range and in order to see the shape of the saturation line, we must plotonly a smaller range. In the above graph, the volume axis ranges from 0 to 100 cm3/mol.

� By doing the V axis in log coordinates we can now look at a very wide range of values withoutsqueezing the graph into nothingness. Notice that in the log plot the volume goes from 1 to 100,000cm3/g.

� The steam tables do not contain data for the compressed liquid region and so our isotherms stop at thesaturated liquid. We could extrapolate them into the liquid by drawing them as vertical lines.

� The ZP graph has the familiar look. Notice that the isotherms are better separated on this graph.






a) We find that for ethane,Tc D 305:3K; Pc D 48:72bar

The reduced conditions are

Tr D298:15 K305:3 K

D 0:976; Pr D10 bar48:72 bar

D 0:21

The desired isotherm is between Tr D 0:95 and Tr D 1:0. By graphical interpolation we findZ � 0:92. theisotherm is fairly linear between Pr D 0 and the value of Pr corresponding to the given pressure, therefore,the truncated virial equation is valid.

b) The reduced conditions are

Tr D244:15 K305:3 K

D 0:80; Pr D10 bar48:72 bar

D 0:21

Z cannot find at Tr D 0:80 under the given pressure. therefore, the truncated virial equation is not valid.

c) Since the temperature is �35ıC, lower than that boiling point, the ethane is liquid. The truncated virialequation is not valid.




Problem 2.12 Solution a) Vapor.

b) The molar mass can be obtained from the relationship between molar volume and density:

� DMm

V) Mm D �V

We know density at two pressures, so we need the molar volume in one of them. We choose the lowestpressure because at 0.01 we are justified (below) to assume ideal-gas state:

Justification: The critical pressure is not known but it must be higher than the saturation pressure at 25 C,which is 64.3 bar. That is, the reduced pressure is at most

0:01 bar64:3 bar

D 0:0002

From generalized graphs is it clear that at such low reduced pressures the state is essentially ideal.

The molar volume is

V DRT

PD.8:314 J=mol=K/.298:15 K/

0:01 � 105 PaD 2:47882 m3=mol

The molar mass is

Mm D�0:177 kg=m3� .2:47882 m3=mol/ D 43:8 � 10�3 kg=mol

c) The second virial coefficient can be calculated from the truncated virial equation

PV

RTD 1C

BP

RT

We justify the use of this equation at 25 ıC, 20 bar as follows:

Justification: The reduced temperature is Tr D 0:97. The reduced pressure is not known but it must be lessthan

20 bar64:3 bar

D 0:31

since the critical pressure must be higher than 64.3 bar. From generalized graphs we see that for Tr >� 1,the isotherm in the pressure range Pr D 0 up to about 0.31 is quite linear. This of course is a judgement callbut is as good as we can do with the information we have.

Solving the truncated virial for B:

B D V �RT

P

where V DMm=�. Using the data at 20 bar with V D 0:00110239 m3=mol we obtain:

B D �1:370 � 10�4 m3=mol

d) We will answer this question using the truncated virial equation

Justification: If the truncated virial is valid at 25 ıC, 20 bar, as assumed above, it is valid for all pressuresless than 20 bar.





Solving the truncated virial for V and using P D 12 bar:

V DRT

P� B D 0:00192866 m3=mol

The number of moles to be stored is

n D20 kg

43:9 � 10�3 kg=molD 456 mol

and the volume of the tank is

V tankD .456 mol/.0:00192866 m3=mol/ D 0:88 m3




Problem 2.13 Solution a) We collect the data for this problem:

Tc D 190:56 KPc D 45:99 bar! D 0:011

We calculate the second virial using the Pitzer equation:

Tr D 1:5646

Pr D 0:434877

B0 D �0:12319

B1 D 0:112756

B D �0:0000420106 m3=mol

The compressibility factor is

Z DBPc

RTD 0:966104:

b) The molar volume in the tank is

V DZRT

PD 0:0011974 m3=mol

The total number of moles is

n DV tank

VD 835:144 mol

c) If we double the number of moles, the new molar volume in the tank is

V2 DV tank

2nDV

2D 0:000598699 m3=mol:

We use the truncate virial to solve for the new pressure P2 (since temperature is the same as before, thesecond virial does not need to be recalculated):

PV

RTD 1C

BP

RT) P D

RT

V � BD 38:7 bar

d) To validate the applicability of the truncated virial we check with the Z0 graph and notice that isothermsaround Tr D 1:6 remain linear up to fairly high pressures. The pressure of this problem, P D 38:7 barcorresponds to Pr D 0:84, which is still within the linear range of the isotherm, as far as we can tell bynaked eye.





Summary of resultsP D 20 barT D 298:15vKVtank 1:0 m3

Tc D 190:56 KPc D 45:99 bar! D 0:011

Tr D 1:5646

Pr D 0:434877

B0 D �0:12319

B1 D 0:112756

B D �0:0000420106 m3=molZ D 0:966104

V D 0:0011974 m3=moln D 835:144 moln2 D 1670:29 molV2 D 0:000598699 m3=molP2 D 38:6886 bar




Problem 2.14 Solution a) The critical constants and acentric factor of nitrogen are:

Pc D 34:0 bar; Tc D 126:2 K; ! D 0:038

The reduced conditions are

Pr D7 bar34:0 bar

D 0:206; Tr D110 K126:2 K

D 0:872

From Eqs (2.28), (2.29) we have

B0 D 0:083 �0:422

0:8721:6D �0:442395

B1 D 0:139 �0:172

0:8724:2D �0:166745

B0 C !B1 D �0:442395 � 0:166745 � 0:038 D �0:448731

The second virial coefficient is calculated form Eq. (2.27)

B DRTc

Pc.B0 C !B1/ D .�0:448731/

.8:314 J/mol K/.126:2 K/34:0 � 105 Pa

D �1:3848 � 10�4 m3/mol

The compressibility factor is

Z D 1CBP

RTD 1 �

.1:3848 � 10�4m3/mol/.7 � 105 Pa/.8:314 J/mol K/.110 K/

D 0:8940

The molar volume is finally calculated to be

V DZRT

PD.0:8940/.8:314 J/mol K/.110 K/

7 � 105 PaD 1:168 � 10�3 m3/mol

The mass of nitrogen in the tank

m D nMw DV tank

VMw D

5 m3

1:168 � 10�3 m3/mol28:014 g/mol D 119:92 kg

b) In this process the specific volume stays constant. For the two unkowns, P and T , we have the followingtwo equations:

P satVRT

D 1CBP sat

RT

P satD e14:9542�

588:72�6:6CT

These should be solved by trial and error. For example, specify T , solve for P sat from the second equation,solve for V from first equation, and if the answer does not match the known volume, try again. The solutionis as following:

T (K) B P sat (mmHg) Z V (m3/mol)100 -1.6710�10�4 5715.79 0.846850 9.2393�10�4

90 -2.0477�10�4 2684.39 0.902059 1.8860�10�3

99 -1.7038�10�4 5338.88 0.852655 9.8597�10�4

98 -1.7377�10�4 4979.38 0.858414 1.0535�10�3

97 -1.7726�10�4 4636.94 0.864119 1.1273�10�3

96 -1.8085�10�4 4311.17 0.869766 1.2078�10�3

96.48 -1.7911�10�4 4465.49 0.867063 1.1682�10�3





The volume solved in part a) is 1.168�10�3 m3/mol. Therefore the tempeture is 96.48 K, and the pressureis P D 4465:49 mmHg D 5:95 � 105 Pa.

c) The reduced state in part (a) is Tr D 0:87 and Pr D 0:206. The desired isotherm is between Tr D 0:80

and Tr D 0:90. From a generalized Z � P graph we see that isotherms in this range are fairly linear,therefore the truncated virial is acceptable.

In part (b) we found Tr D 0:76 and Pr D 0:175. The desired isotherm is between Tr D 0:70 and Tr D 0:80.Again, the isotherm is fairly linear between Pr D 0 and the value of Pr corresponding to the given pressure,therefore, the truncated virial equation is valid.




Problem 2.15 Solution a) The second virial coefficient is directly related to the slope of an isotherm onthe ZP graph. Specifically,

Z D 1CBP

RTC � � � )

BP

RTD

�@Z

@P

�T

ˇPD0

This suggests the following graphical solution: calculate Z from the steam tables at constant T , plot themversus pressure, and obtain the slope of the line near P D 0. To facilitate calculations, suppose that P is inkPa, V

From the steam tables at 200 ıC we have:

P (kPa) V (cm3/g) Z

1 218350 0.999110 21820 0.998420 10900 0.997530 7267.5 0.997640 5447.8 0.997150 4356. 0.996675 2900.2 0.9953100 2172.3 0.9940200 1080.4 0.9887250 861.98 0.9861300 716.35 0.9834350 612.31 0.9806400 534.36 0.9780

Notice that we had to go up to 400 kPa (4 bar) to see enough change in Z so that we can obtain the slope ofthe line. The graph is shown below:

1.00

0.99

0.98

0.97

0.96

0.95

Z

5004003002001000

Pressure (kPa)

200 C

The line shown is tangent to the points at P D 0 and its slope is�5:00554 � 10�5. The easy way to drawthis line is to fit a straight line through the points closest to the origin, say below 200 kPa. A smarter way isto use all the points and a quadratic equation:

f .P / D aP 2 C bP C c

Then, the equation of the tangent line at P D 0 is (why?)

bP C c

and its slope is b. Following this procedure we find

slope D �5:00554 � 10�5 kPa�1 D �5:00554 � 10�8 Pa�1





The second virial coefficient is

B D .slope/.RT / D .�5:00554 � 10�8 Pa�1/.8:314 J/mol K/.473:15 K/ D �1:969 � 10�4 m3/mol

b) Using the above value of B we can calculate the molar volume of water at 14 bar as follows:

PV

RTD 1C

BP

RT)

V DRT

PC B D

.8:314 J/mol K/.473:15 K/14 � 105 Pa

� 1:969 � 10�4 m3/mol

D 2:613 � 10�3 m3/mol

D 145:15 cm3/g

The value from the steam tables is 142.94 cm3/g. This agreement is very good indicating that the truncatedvirial equation is valid at these conditions.




Problem 2.16 Solution We need the density of methane under these conditions. We will calculate it viathe compressibility factor using the Lee-Kesler tables. The critical parameters of methane are:

Tc D 190:6 K; Pc D 45:99 bar; ! D 0:012

The reduced state of methane is

Tr D298:15 K190:6 K

D 1:564; Pr D75 bar45:99 bar

D 1:631

We must perform a double interpolation in the tables since neither value is listed. The calculation is sum-marized below:

Interpolations for Z0

Pr1.5 1.631 2.0

1.5 0.8689 0.8595 0.83281.564 0.88111.6 0.9 0.8931 0.8738

Interpolations for Z1

Pr1.5 1.631 2.0

0.1345 0.1466 0.18060.1433

0.1303 0.1414 0.1729

Here, numbers in regular font are form the Lee-Kesler tables and those in bold are interpolations. First weinterpolate atTr D 1:5 to calculate Z0 at Pr D 1:631 and we obtain 0.8595. Next, we do the same atTr D 0:6 to calculate Z0 at Pr D 1:631 from which we find 0.8931. Finally, we interpolate between thesetwo values to obtain Z0 at Tr D 1:564 to find the value of Z0 at the desired state. We find

Z0 D 0:8811

The procedure is similar for Z1 where we find

Z1 D 0:1433

With these values we calculate the compressibility factor, the molar volume, density, and total volume of thetank:

Z D 0:8828; V D 2:92 � 10�4; m3/mol � D 54:8 kg/m3; V tankD 18:2 m3





Problem 2.17 Solution a) At the given conditions, Tr D 0:963675, Pr D 0:949281. The state is veryclose to the critical, therefore, far removed from the ideal-gas state.

b) Using the Lee Kesler method we find

Z0 D 0:381614;

Z1 D �0:597626;

Z D 0:247148;

V D 0:0000860517 m3=mol

The number of moles in the tank is

n D200 kg

4410�3 kg=molD 4545:45 mol

and the volume of the tank is

V tankD V n D 0:391 m3

c) At 25 ıC, 1 bar, CO2 is essentially in the ideal-gas state and its molar volume is

V2 DRT

P2D 0:0243725 m3

which means that the moles in th tank are

n2 DV tank

V2D 16:048 mol ) m2 D 0:706 kg

We must remove 199:294s kg.





Outline:

a) Calculate Pr and Tr and find the compressibility factor using the Lee-Kesler tables or graphs.

b) With the compressibility factor known, calculate the specific volume and then the total volume of the tank(since the total mass is known).

c) Calculate the new molar volume after 90% (45 kg) is removed. Calculate Z. Since we know Z and T weshould be able to obtain P . If we use the Lee-Kesler graph for Z we must do a trial-and-error procedure:choose P , calculate Z, if it doesn’t match the known Z try another pressure and continue.

Alternatively, use the truncated virial equation: estimate B using the Pitzer correlation and use

PV

RTD 1C

BP

RT

to solve for P . Once you have obtained B , confirm that the use of the virial equation was justified.

Calculations

a) The critical parameters of ethylene are:

! D 0:087

TC D 282:3 K

PC D 50:40 bar

The reduced temperature and pressure are

Tr D293:15 K282:3 K

D 1:05615; Pr D80 bar50:40 bar

D 1:5873

Checking with the generalized Z graph we see that the state is well-removed from the ideal-gas state.

b) We calculate the compressibility factor using the generalized Lee-Kesler graphs (or by interpolation intables):

Z0 D 0:322979; Z1 D 0:0390341 Z D .0:322979/C .0:087/.0:0390341/ D 0:326

As expected, the compressibility factor is quite below its ideal value. The molar volume of the gas is

V D ZRT

PD 0:00010113 m3=mol

The number of moles in the tank is

n D50 kg

28 � 10�3 kg=molD 1785:71 mol

and the total volume is

V totD nV D .1785:71 mol/.0:00010113 m3=mol/ D 0:181 m3

c) The number of moles left in the tank is

n0 D .0:1/n D 178:6 mol





and the new molar volume is

V 0 DV tot

n0D 0:00101128 m3=mol

We will use the truncated virial since it does not require iterations, but we will have to justify its use after-ward.

P 0V 0

RTD 1C

BP 0

RT) P 0 D

RT

V 0 � B

We calculate the second virial coefficient using the Pitzer correlation:

B0 D 0:083 �0:422

D� 0:303682

B1 D 0:139 �0:172

T 4:2rD 0:00226241

B DRTc

Pc.B0 C !B1/ D �0:0001413 m3=mol

Finally, the pressure is

P 0 DRT

V 0 � BD 21:5 bar

Check validity of truncated virial At the final state, Tr D 1:05615, Pr D 0:426709. If the isotherm atthis Tr is sufficiently linear up to Pr D 0:426709, the truncated virial is acceptable. The graph in thebook looks “linear enough.” But we want to be more precise, so we will calculate the compressibilityfactor form the virial equation and from the Lee-Kesler graphs and will compare:

Using the virial equation:

Z D 1CBP 0

RTD 0:877

Using the Lee-Kesler tables we find Z0 D 0:8677, Z1 D �0:007497 and

Z D .0:8677/C .0:018/.�0:007497/ D 0:867

Pretty close.

Note: We could have used the ideal-gas law subject to the same condition: after the calculation weshould check whether the ideal-gas assumption is correct or not:

P 0 DRT

VD 24:5 bar

With Tr D 1:05615, Pr D 24:5=50:40 bar D 0:486, the Lee-Kesler charts give Z D 0:846. This valueis more than 5% way from the ideal-gas state, therefore, we reject the calculation.

Trial and error Here is how to perform a trial-and error solution using the Lee-Kesler tables. We needa starting guess for P - we will use the result fo the ideal-gas calculation, Pguess D 24:5 bar. From theLee-Kesler tables we find

Z0 D 0:846798; Z1 D �0:00762552; Zguess D 0:846135; Vguess D 0:00085568 m3=mol




We compare this to the known value V 0 D 0:00101128 m3=mol by calculating the ratio

Vguess

V 0D 0:8461

The correct pressure must make this ratio equal to 1. Since it is less than 1 (i.e., Vguess < V 0) we mustguess a lower pressure, to allow volume to increase. We choose Pguess D 20 bar and repeat until theratio is sufficiently close to 1. The table below summarizes the results of these iterations.

iteration Pguess Z0 Z1 Zguess Vguess Vguess=V0

1 24:5 0:846881 �0:00762573 0:846217 0:000856171 0:84662

2 20 0:877931 �0:00730925 0:877295 0:00108733 1:0752

3 22 0:864336 �0:00754181 0:86368 0:000973139 0:962283

4 21 0:871173 �0:00744249 0:870525 0:00102756 1:0161

The solution is somewhere between 21 and 22 bar. We continue in the same manner if we want tobracket the solution more precisely.





Problem 2.19 Solution We need to calculate densities, i.e. we need V or Z. The ideal-gas law is out ofthe question because the pressure is too high. Same for the truncated virial equation. We could use either theLee-Kesler tables or an equation of state. Both methods would be appropriate since krypton is a non-polarcompound.a) We need the density of krypton, so we will first find the compressibility factor at the indicated conditions.From tables we find

Pc D 55:02; Tc D 209:4; ! D 0

The given conditions, 20 ıC, 110 bar, correspond to reduced conditions

Pr D110

55:02D 2:0; Tr D

20C 273:15

209:4D 1:4

From the Lee-Kesler tables we find (notice that we don’t need Z1 since the acentric factor is 0)

Z D Z0 D 0:7753

The molar volume of Kr is

V DZRT

PD.0:7753/.8:314 J/mol K/.293:15 K/

110 � 105 PaD 1:72 � 10�4 m3

The number of moles corresponding to 2000 kg of Kr (Mw D 83:8) is

n D2000 � kg

83:8 � 10�3 kg/molD 23866:3 mol

and the required volume of the tank is

V tankD nV D .23866:3 mol/.1:72 � 10�4 m3/ D 4:1 m3

b) We will calculate the mass in the tank when the pressure is the maximum allowable. At 180 bar, 25 ıC,we have

Pr D 3:272; Tr D 1:42 � 1:4

From Lee-Kesler by interpolation:

Z D Z1 D 0:7202C0:7761 � 0:7202

5:0 � 3:0.3:272 � 3:0/ D 0:7278

The specific volume is

ZRT

PD.0:7278/.8:314 J/mol K/.298:15 K/

180 � 105 PaD 1:0 � 10�4 m3

and the number of moles of Kr in the tank is

n DV tank

VD

4:1 m3/mol1:0 � 10�4 m3

D 40907:2 mol

The corresponding mass is

M D nMW D .40907:2 mol/.83:8 � 10�3 kg/mol/ D 3428 kg

That is, 3428 kg is the maximum mass that can be stored at 25 ıC without exceeding the safety limit. It is,therefore, safe to store 2500 kg.




Problem 2.20 We collect the following information for n-butane:

Pc D 37:96 bar; Tc D 425:1 K; ! D 0:2; Vc D 255 cm3/mol; Zc D 0:274

a) We need the molar volume of the liquid. Our options are: Lee-Kesler, and Rackett. We choose the Rackettequation because it is known to be fairly accurate while the accuracy of the Lee-Kesler is not very good inthe liquid side. Still, if you did the problem using L-K I will consider the solution correct.

V D .255 cm3/mol/.0:274/.1�293:15=425:1/0:2857

D 100:9 cm3/mol

The moles is

ML D107 cm3

100:9 cm3/molD 99:1 � 103 mol

Note: This problem could also be done using the Lee-Kesler. The solution requires more calculations andthe final result is very close to the above. This calculation is given at the end of this solution. b) For thevolume of the vapor we use Lee-Kesler. The required interpolation is shown below.

Tr D 0:7

Pr D 0:05 Pr D 0:0545 Pr D 0:1

Z0 0:9504 0.9455 0:8958

Z1 �0:0507 �0:0566 �0:1161

from which we obtain the compressibility factor:

Z D 0:9455C .0:2/.�0:0566/ D 0:9342

The molar volume is

V DZRT

PD.0:9342/.8:314/.293:15/

2:07 � 105D 1:1 � 10�2 m3/mol

The moles of the vapor are

MV D10 m3

1:1 � 10�2 m3/molD 909 mol

c) To answer this question we take a look at the PV graph.

A

A'

B

B'

T

T + d





Both sides of the tank undergo constant-volume processes as indicated by the dashed lines. The graphshows the two states at the initial temperature T , as well as the states at some higher temperature, T C ı.It is obvious that the pressure in the liquid side will always be higher than the pressure of the vapor side.Therefore, the pressure of 40 bar will be reached first in the liquid side, causing that alarm to go off.

d) To calculate the temperature at the state we recall that for liquids with constant ˇ and �, we have

lnV2

V1D ˇ�T � ��P

SInce volume is constant, V1 D V2 and solving for �T we find

�T D��P

ˇD3:4 � 10�4 bar�1.40 � 2:07/ bar

2:54 � 10�3 K�1D 5 ıC

The alarm will sound at T D 20C 5 D 25 ıC. At that point the pressure of the liquid side will be 40 barwhile that of the vapor will be not much higher than 2 bar!

Calculation of liquid V using Lee-Kesler:

If you opted to do the calculation using the Lee-Kesler tables, the correct solution is shown below. First wecalculate the reduced temperature and pressure.

Tr D 0:69 � 0:7; Pr D 0:545

Note: because the phase is liquid, one must extrapolate to Pr D 0:0545 from the listed values for the liquid(shown in the tables in italics):

Tr D 0:7

Pr D 0:0545 Pr D 0:2 Pr D 0:4

Z0 9:45 � 1�3 0:0344 0:0687

Z1 �4:1785 � 10�3 �0:0148 �0:0294

With these values we obtain the following:

Z D .9:45 � 1�3/C .0:2/.�4:1785 � 10�3/ D 8:614 � 10�3

V DZRT

PD.8:614 � 10�3/.8:314 J/mol K/.293:15 K/

2:07 � 105 PaD 1:01 � 10�4 m3/mol

ML D10 m3

1:01 � 10�4 m3/molD 98:5 � 103 mol

The answer is very close to that obtained using the Rackett equation but the Lee-Kesler method requires morecalculations.




Problem 2.21 Solution Solution a) Filled with xenonWe need the volume of the tank which will obtain by first calculating the molar volume of xenon. We willdo this calculation using the Pitzer method and the Lee Kesler tables. For xenon:

Tc D 289:7 KIPc D 58:4 barI! D 0IMw D 131:30 g/mol

The reduced temperature and pressure are

Tr D132C 273:15

289:7D 1:39852 � 1:4; Pr D

82

58:4D 1:4:

Interpolating at Tr D 1:4 between Pr = 1.2 and Pr = 1.5 we find

Z0 D 0:836436

Since ! D 0, Z D Z0 D 0:836436. Using SI units, the molar volume of xenon is

V DZRT

PD .0:836436/.8:314/.405:15/82 � 105 D 3:436 � 104 m3 /mol

Since the tank contains 10,000 kg, or

n D10; 000 kg

131:30 � 10�3 kg molD 76; 161 mol

the volume of the tank isV t D nV D 26:16 m3

Filled with steam The specific volume of steam in the tank is

V D26:16 m3

10000 kgD 2:616 cm3/g

At 200 ıC, the saturated volumes of water are 1.156 and 127.2 cm3 /g. Since the specific volume lies be-tween the two values, the steam is a saturated vapor/liquid mixture and the pressure is equal to the saturationpressure at 200 ıC: P D 15:45 bar.

b) If the mass in the tank is reduced to half, the specific volume doubles:

V D .2/.2:616/ D 5:232 cm3/g

This value is still between that of the saturated vapor and liquid, therefore the pressure remains constant.





Problem 2.22 Solution a) Vapor.

b) At 0.1 bar, 200 C, o-xylene is essentially in the ideal-gas state (why?).

V DRT

PD 0:393377 m3=mol

The volume of the tank is

V tankD .200 mol/.0:393377 m3=mol/ D 39:3 m3: (2.2)

c) At 44.9 bar, 200 C, the reduced temperature an pressure is

Tr D 0:75; Pr D 1:2

From the Lee-Kesler tables we find

Z0 D 0:84435; Z1 D 0:0453;Z D 0:8585

The molar volume isV2 D

ZRT

P2D 0:000752133 m3=mol

and the number of moles

n DV tank

V2D 52301:5 mol




Problem 2.23 Solution a) We first calculate the coefficient of thermal expansion form the empirical equa-tion given above:

ˇ D1

V

�@V

@t

�P

�@t

@T

�where t stands for temperature in celsius and T for temperature in kelvin. Using the polynomial expressiongiven in the problem statement we find

ˇ Da1 C 2a2t C 3a3t

2

1C a1t C a2t2 C a3t3

For constant pressure process,

dV

VD ˇdT )

V2

V1D eˇ.T2�T1/

For this calculation we will use an average value of ˇ between 18 ıC and 40 ıC:

ˇav D 0:5.ˇ18 C ˇ40/ D 0:5.7:09 � 10�4C 7:27 � 10�4/ K�1

D 7:18 � 10�4 K�1

With this value, the change in volume isV2

V1D 1:016

or an increase of 1/6%.

Note: We have treated ˇ as nearly constant. To determine the validity of this assumption we make a graphof ˇ in the temperature range of interest:

20 25 30 35 400.0000

0.0002

0.0004

0.0006

0.0008

0.0010

T H°CL

ΒH1

�KL

The coefficient ˇ changes very slowly with temperature, therefore the use of an average value is justified.

b) We start withdV

VD ˇdT � �dP:

Assuming the contained to be rigid, volume remains constant, i.e., dV D 0. We then integrate the aboveequation and solve for �P :

0 D ˇdT � �dP ) ��P D ˇ�T ) �P Dˇ�T

�

Numerical substitutions:

�P D.7:18 � 10�4 K�1/.22 K/

52 � 10�6 barD 287 bar





Problem 2.24 Solution The numerical results are summarized in the atatched tables:




P=70 barFluid CO2Tc 304.129 KPc 73.74 barw 0.225

T 293.15 KP 70. barPsat 57.2 barPhase Liquid

a 0.381864 Joule Meter3

Mol2

b 0.0000297087 Meter3

Mol

A 0.449994B 0.085326

Z roots 0.186938

Z 0.186938

V 0.0000650879 Meter3

Mol

moles 7681.92 Molmass 338.081 kg

P=60 barT 293.15 KP 60. barPsat 57.2 barPhase Liquid


Mol2

b 0.0000297087 Meter3

Mol

A 0.385709B 0.0731366

Z roots0.1694180.3380980.492484

Z 0.169418

V 0.000068819 Meter3

Mol

moles 7265.44 Molmass 319.752 kg

This text is associated with Themis Matsoukas, Fundamentals of Chemical Engineering Thermodynamics 0-13-269306-2 (978-0-13-269306-6). Copyright © 2013 Pearson Education, Inc. Do not redistribute.



50 barT 293.15 KP 50. barPsat 57.2 barPhase Vapor


Mol2

b 0.0000297087 Meter3

Mol

A 0.321424B 0.0609471

Z roots0.1609090.186580.652512

Z 0.652512

V 0.000318067 Meter3

Mol

moles 1572. Molmass 69.1836 kg

2 solution-2-Table.nb





First we collect the parameters for isobutane:

Tc D 408:1 K; Pc D 36:48 bar; ! D 0:181; Mw D 58:123 � 10�3 kg/mol

We are given T D 294:26 K, P D 4:13793 bar. With this information we find that the compressibilityequation has three real roots:

Z1 D 0:0185567; Z2 D 0:100577; Z3 D 0:880866;

We know that the phase is liquid (since the given temperature is below the saturation temperature at thegiven pressure), therefore the correct compressibility factor is the smallest of the three:

Z D 0:0185567

The corresponding molar volume is

V D ZRT

PD .0:0185567/

.8:314 J/mol K/.294:26 K/4:13793 � 105 Pa

D 1:09713 � 10�4 m3/mol

The amount (moles) of isobutane is

n D5000 kg

58:123 � 10�3 kg/molD 86024:5 mol

Therefore, the volume of the tank is

V tank D .86024:5 mol/.1:09713 � 10�4 m3/mol/ D 9:44 m3

a) Before we solve the problem it is useful to look at the PV graph first.

P1

P2

V

1

2

Since the volume of the tank and the mass of isobutane remain the same, the molar volume also stays thesame. In other words, the new state must be on the vertical line that passes through the initial state 1. Sincetemperature is higher, the final state will be above state 1 (marked as state 2 in the above figure). Thisis somewhat surprising: one might think that some vapor may be generated since temperature increases.Instead, the system moves firther into the compressed liquid region! This is because heating takes placeunder constant volume.1

1If heating were to take place under constant pressure insead, the final state would move to the right of state 1, possibly creatingsome vapor.





b) Since the total volume and mass in the tank remain the same, the molar volume must also stay the same,namely,

V D 1:09713 � 10�4 m3/mol

The pressure can now be calculated directly from the SRK equation:

P DRT

V � b�

a

V.V C b/

Notice, however, that the parameter a must be recalculated because it depends on temperature. With T D308:15 K we find

a D 1:68988 J m3/mol2

Using this value of a, the previous value of b, and V D 1:09713 � 10�4 m3/mol, the SRK equation gives

P D 70:1 bar

This represents an increase of 66 bar even though temperature increased only by 20 ıF! The reason is thatisotherms in the compressed liquid state are very steep, resulting in large pressure change under constant-volume heating.





a) At 30 ıC, 1 bar the SRK equation has the following three real roots. Since the phase is vapor (why?) wepick the largest root:

Z D 0:977286 V DZRT

PD 0:0246314 m3=mol

b) At 30 ıC, 10 bar the SRK equation has the following three real roots. The phase is liquid (why?), thereforewe pick the smallest root:

Z D 0:0413973 V DZRT

PD 0:000104337 m3=mol

b) At 30 ıC, 4.05 bar the SRK equation has the following three real roots. Since the system is saturated, thesmallest root is the liquid and largest is the vapor:

ZL D 0:0413973; VL DZLRT

P sat D 0:000104698 m3=mol

ZV D 0:901582; VV DZVRT

P sat D 0:00561071 m3=mol

The literature values from the NIST Web Book are

VL D 0:00010678 m3=mol VV D 0:0055461 m3=mol

The SRK values are off by �2% (liquid) and 1:2% (vapor). These errors are pretty small.






150 °C150 °C

30 °C

sat. Liq sat. Vap

TcTcT

c

T

c

1´10-4 5´10-40.001 0.005 0.010 0.0500

10

20

30

40

50

V Hm3�molL

PHb

arL

Note: For maximum readability use semilog coordinates (linear axis for pressure, logarithmic axis for vol-ume).




Problem 2.28 Solution The the general form for the differential of V is

dV

VD ˇdT � �dP

Using the given expressions for ˇ and � we have

d lnV DdT

T�dP

PŒA�

Integration is this differential from V0, T0, P0 to V , T , P is very simple in this case because the variableshappen to be separated (each of the three terms contains one variable only). The result is

lnV

V0D ln

T

T0� ln

P

P0ŒB�

The same result is obtained if we adopt an arbitrary integration path, say from T0, P0, under constant T toT0, P , and then under constant P to T , P . As we can easily verify, the differential of the above is indeedequal to Eq. [A]. Equation [B] can be rearranged to write

lnV

V0D ln

�T

T0�P0

P

�or

PV

TDP0V0

T0

In other words we have obtained the ideal-gas law.

Based on the final result we can certainly say that this equation of state is not appropriate for liquids. Evenbefore integration, however, we could reach the same conclusion by looking at the T and P dependence ofˇ and �. The inverse dependence of ˇ on T (and of � on P ) indicates that these parameters vary quite a bitwith pressure and temperature. This is a characteristic of gases. The values of � and ˇ for solids and liquidsare typically small numbers and vary much less with temperature and pressure.





Problem 2.29 Solution Using the hint we write:

� D �1

V

�@V

@P

�T

D �

�1

V

�.�@P@V

�T

Starting with the SRK equation,

P DRT

V � b�

a

V.V C b/

the derivative wrt V is: �@P

@V

�T

Da

V 2b C 2V

.b C V /2�

RT

.b � V /2

Numerical substitutions

T D 303:15 KP D 4:05 bara D 1:60525 J=m3=mol2

b D 0:000075256 m3=mol

ZL D 0:0168238

VL D 0:000104698 m3=mol.@P=@V /T D �1:62039 � 10

12 bar mol=m3

�L D 0:000589445 bar�1

ZV D 0:901582

VV D 0:00561071 m3=mol.@P=@V /T D �6:44372 � 10

7 bar mol=m3

�V D 0:276596 bar�1





a) By definition,

ˇ D1

V

�@V

@T

�P

The partial derivative will be approximated as a finite difference between two states A and B at the samepressure: �

@V

@T

�P

�VA � VB

TA � TB

For V we must use a value between VA and VB . Choosing V D .VA C VB/=2, the final result is

ˇ �2

VA C VB

VA � VB

TA � TB

A B

C

D

const. P

Pressure

Volume

a) At 25ıC, 1 bar, the system is compressed liquid. Assuming the liquid to be incompressible, the requiredvolumes are those of the saturated liquid.

TA D 20 ıC VA D 0:001002 m3=kgTB D 30 ıC VB D 0:001004 m3=kg

The coefficient of isothermal compressibility is

ˇ D 1:99402 � 10�4 K�1

b) The answer at 10 bar is the same because the assumption of incompressibility implies that the isothermsare vertical and the molar volumes the same as in the previous part.

c) In this case the state is superheated vapor. We select two temperatures around 200 ıC and apply the sameprocedure:

TA D 150 ıC VA D 1:9367 m3=kgTB D 250 ıC VB D 2:4062 m3=kg

and we findˇ D 2:16215 � 10�3 K�1





Problem 2.31 Solution Solution

Assuming isotherms in the compressed liquid region to be vertical, ˇ is calculated as

ˇ D1

V

�@V

@T

�P

D1

V1CV2

2

V2 � V1

T2 � T1

where T1, T2 are two temperatures around 24 ıC, and V1, V2, are the volumes of the saturated liquid at thesetemperatures, to be calculated using the Rackett equation. With T1 D 20ıC, T2 D 30 ıC we find

ˇ D 1:475 � 10�3 K�1

Calculation are shown in the attached notebook.




Rackett Equation

Data for ethanolIn[6]:= Tc = 513.9 K;

Pc = 61.48 bar;w = 0.645;Vc = 167 cm3 ê mol;Zc = 0.24;

Rackett Equation

In[5]:= V@t_D := Vc ZcH1-têTcL0.2857

CalculationsIn[14]:= T1 = H20 + 273.15L K;

T2 = H30 + 273.15L K;

In[29]:= V1 = V@T1DV2 = V@T2Db =

1ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅHV1+V2LÅÅÅÅÅÅÅÅÅÅÅÅÅÅ

2

V2 - V1ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅT2 - T1

;

Print@"b = ", bDOut[29]= 54.4331 cm3

ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅmol

Out[30]=55.2419 cm3ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅmol

b =0.00147493ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅK

Rackett-Ethanol.nb 1







3 Energy & the First Law




3. ENERGY & THE FIRST LAW

Problem 3.1 Solution a) Assuming equilibrium, the pressure inside equals the total external pressure,atmospheric plus pressure due to the weight of the piston. The pressure due to the weight of the piston is

Pp D.30 kg/.9:81 m/s2/

�4.0:2/2 m2

D 9367:86 kg /m s2 D 9367:86 Pa D 0:09367 bar

The total pressure isP D 1C 0:09367 D 1:09 bar

Notice that the weight of the piston makes a very small contribution to the total pressure.

b) Work is done by the system as it expands against the surroundings under constant pressure. Assumingthe processes to be reversible, this work is

W D

Z 2

1

PdV D P.V2 � V1/ D P.ah/

where a D �D2=4 is the area of the piston and h D 0:5 m is the distance travelled by the piston. The resultis,

W D 1738:37 J

c) The pressure inside the cylinder is constant throughout this problem.




Problem 3.2 Solution a)

P DRT

V � b

When V !1 at constant T , the right-hand side goes to RT=P , which is the ideal-gas result.

b)

W D �

Z 2

1

PdV D �RT

Z 2

1

dV

V � bD �RT ln

V2 � b

V1 � b:

c)

Z DPV

RTD

V

V � b

Set V D 1=�:

Z D1

1 � b�

expand in powers of � at � D 0Z D 1C b�C b2�2 C � � �

and set � D 1=V :

Z D 1Cb

VCb2

V 2C � � �

We identify the second and third virial as

B D b; C D b2





Problem 3.3 Solution a) For this constant-pressure expansion, the amount of work is

W D �P0.Vs � Vl/ D �P0

�1

�s�1

�l

�where P0 D 1 bar is the atmospheric pressure, V is the specific volume, � is the density, and s, l , in thesubscript denote solid or liquid, respectively. By numerical substitution we obtain

W D �.1 � 105 Pa/�

1

917 kg/m3�

1

1000 kg/m3

�D �9:05 J/kg

Since 1 liter of water has a mass of 1 kg, the work in part (a) is 9.05 J.

b)

v D

�2W

M

�1=2D

�2.9:05 J/kg/

1 kg

�1=2D 4:25 m/s D 15 km/h D 9:6 mph

c)

h DW

MgD

.9:05 J/kg/.1 kg/.9:81 m/s2/

D 0:92 m





The work is

W D �

Z V2

V1

PdV

with the integral to be calculated along a path of constant temperature. We will calculate the integral nu-merically by application of the trapezoidal rule. Given a list of n pressures Pi and volumes Vi , the integralis

W D �

n�1XiD1

�1

2.Pi C PiC1/ .ViC1 � Vi /

�Using the steam tables we construct the following table:

i P .bar/ V .m3=kg/ �12.Pi C PiC1/ .ViC1 � Vi / .kJ=kg/

1 3:0 0:7965 0

2 3:5 0:6815 37:375

3 4:0 0:5952 32:3625

4 4:5 0:5281 28:5175

5 5:0 0:4744 25:5075

6 5:5 0:4305 23:0475

7 6:0 0:3939 21:045

8 6:5 0:3629 19:375

9 7:0 0:3364 17:8875

The work is the sum of the last column:

W D 205:1 kJ=kg

The work is positive, as it should be for compression.

b) The heat is

Q D Uf � Ui �W D 2799:5 � 2718:7 � 205:118 D �215:418 kJ=kg

This is the amount of heat that must be given to the steam in order to maintain constant temperature.

Alternative method for integration: Another way to perform the integration is to fit a function to the dataand integrated the fitted equation. Fitting pressure to a polynomial of second degree in V we find

P D 13:8246 � 25:7837V C 15:4057V 2

with P in bar and V in m3=kg. Integrating this from V1 D 0:7965 m3=kg to V2 D 0:4744 m3=kg we find

�

Z 0:4744

0:7965

�13:8246 � 25:7837V C 15:4057V 2

�D 2:0403 bar m3=kg

Converting into kJ/kg we findW D 204:0 kJ=kg

which compares very favorably with the trapezoidal rule. The fitted line is compared to the data from thesteam tables in the graph below.





0.4 0.5 0.6 0.7 0.83.0

3.5

4.0

4.5

5.0

5.5

V Hm3�kgL

PHb

arL




Problem 3.5 Solution We calculate the work as

W D �

Z Pf

Pi

PdV � �1

2.Pi C PiC1/.VI � ViC1/

where Pi and Vi are tabulated data along the isotherm:

i P .bar/ V .m3=kg/ �12.Pi C PiC1/.VI � ViC1/

1 10 0:3066 0

2 9 0:3411 �0:32775

3 8: 0:3843 �0:3672

4 7:5 0:4102 �0:200725

5 7 0:4398 �0:2146

6 6:5 0:4739 �0:230175

7 6 0:5137 �0:24875

8 5:5 0:5608 �0:270825

9 5 0:6173 �0:296625

�2:15665

The work isW D �2:15665 bar m3=kg D �215:665 kJ=kg

The heat isQ D Uf � Ui �W

where Ui D 2963:6 kL=kg and U2 D 2957:8 kJ=kg are the internal energies in the initial and final states.We find

Q D 221 kJ=kg





Problem 3.6 Solution a) At the initial state:

P1 D 5 bar; T1 D 151:84 C; V1 D 0:3748 m3=kg; U1 D 2560:7 kJ=kg

The mass in the tank ism D V tank=V1 D 2:66809 kg

At the final state we know P2 D 10 bar and V2 D V1 D 0:3748 m3=kg. By interpolation

T2 D 543:856 C; U2 D 3199:89 kJ=kg

b) The amount of heat isQ D U2 � U1 D 639:194 kJ=kg

and since we have a total of 2.668 kg,

QtotD mQ D 1705:43 kJ

The heat is positive, i.e., it is added to the system.





For constant-volume heating,�UAB D Q

At the initial state:

P1 D 1 bar; T1 D 100ıC; V1 D 1:6960 m3=kg; U1 D 2506:2 kJ=kg

At the final state we know P2 D 2:5 bar, V2 D VA D 1:6960 m3=kg. By interpolation at P D 2:5 bar,V D 1:6960 m3=kg we find

T2 D 646:4ıC; U2 D 3384:41 kJ/kg

Therefore,Q D U2 � U1 D 878:2 kJ/kg





Problem 3.8 Solution a) At 5 bar, 200 ıC, we find V1 D 0:425m3=kg. We look for a saturated state withthe same specific volume. By interpolation in the tables we find

T2 D 146:885ıC; P2 D 4:37662 bar; U2 D 2556:2 kJ=kg

b)Q D U2 � U1 D �87:2037 kJ=kg

c) The internal energy of saturated liquid at T2 D 146:885 ıC is 623.13 kJ/kg. For the additional heat wefind

Q0 D �.1937:86 .kJ=kg/




Problem 3.9 Solution a) The specific volume of steam at the given conditions is V1 D 0:2060 m3=kg.With m1 D 1 kg, the volume of the tank is

V tankD m1V1 D 0:206 m3

b) After removing 0.5 kg, m2 D 0:5 kg and the new specific volume in the tank is

V2 DV tank

V 2D 0:412 m3=kg

From the steam tables at 200 ıC, 0.412 m3=kg, we find

P2 D 5:164 bar; U2 D 2642:75 kJ=kg:

c) We know the specific volume and pressure. By interpolation at 10 bar, 0.412 m3=kg we find

T3 D 623:291ıC; U3 D 3338:72 kJ=kg

d) The amount of heat isQ D U2 � U1 D 695:972 kJ=kg





Problem 3.10 Solution At the initial state (2 bar, 120.21 ıC) we have:

VL D 0:00106 m3=kg; VV D 0:8857 m3=kg; UL D 491:94 kJ=kg; UV D 2529:1 kJ=kg

AlsoxL D

mL

mL CmVD 0:903862; m D mL CmV D 12:17 kg:

a)V tank

D mLVL CmV VV D 1:04793 m3

b)U1 D xLUL C xVUV D 687:789 kJ=kg

The total internal energy isU totD mU1 D 8370:39 kJ

c) At 10 bar the system is still in the two-phase region (why?), and its temperature is 179.89ıC, which isless than the temperature alarm. The pressure alarm will sound first.

d) At this state we know VV D V1 D 0:0861076m3=kg. This corresponds to T2 D 220 ıC, P2 D 23:19 bar,U2 D 2601:4 kJ=kg. The amount of heat is

Q D U2 � U1 D 1913:61 kJ=kg:

The total amount of heat isQtotD Qm D 23288:7 kJ

e) Both alarms will sound.

f) See graph below.

200°C

179.89°C10 bar

2 bar

15.55 bar

23.19 bar

120.21°C

220°C




Problem 3.11 Solution a) The specific volume under these conditions is

V D xLVL C xV VV D 0:484854 m3=kg

When the system becomes saturated, it is saturated vapor. By interpolation:

T D 141:792 C; P D 3:80126 bar; V D 0:484854 m3=kg

b) The specific volume in this case is V D 0:00167473 m3=kg. When the system reaches saturation, it issaturated liquid:

T D 343:907 C; P D 153:346 bar; V D 0:00167473 m3=kg

c) By drawing these cases on the PV graph we conclude that if the specific volume of the two phase systemis larger than the critical volume, the saturated system at constant volume is vapor; otherwise it is liquid.Therefore, these questions can be decided by comparing the specific (or molar) volume to that at the criticalpoint.

The critical volume of toluene is 316 cm3=g. The molar volume of the saturated liquid at 1 atm (boilingtemperature 384 K) may be calculated from the Racket equation with Tr D 384=591:75 D 0:648923,ZC D 0:264 and VC D 316 cm3=g:

VL D 0:000117707 m3=kg

For the volume of the vapor, use the ideal gas law:

V DRT

PD 0:000753868 m3=kg

The volume fo the two-phase system is

V D 0:98VL C 0:02VV D 0:000753868 m3=mol

This is larger than Vc , therefore upon heating, the liquid will disappear and the contents will be vapor.





Problem 3.12 Solution a) The initial state is at its boiling point. The specific volume is

V1 D1 m3

1:2 kgD 0:833333 m3=kg

From the steam tables,

VL D 0:00104 kg; VV D 1:694kg

UL D 407:05 kJ=kg; UV D 2505:5 kJ=kg

Using the lever rule,

xL DVV � V1

VV � VLD 0:50838

U1 D xLUL C xVUV D 1438:69 kJ=kg

At P2 D 3 bar, V2 D V1 D 0:833333 m3=kg we find

T2 D 273:2 C; U2 D 2764:71 kJ=kg

b) Before: 49.16%. After: 100%

c) Q D U2 � U1 D 1326:02 kJ=kg




Problem 3.13 Solution a) The final state is saturated liquid at 148 bar. From the steam tables,

P2 D 148 bar; T2 D 341:062 C; V2 D 0:0017 m3=kgU2 D 1546:14 kJ=kg

From the tables at 300 ıC we find:

VL D 0:001404 m3=kg; VV D 0:02166 m3=kg

With V1 D V2 D 0:0017 m3=kg, the lever rule gives

xL DVV � V1

VV � VLD 0:985387

b) By lever rule at the initial state

U1 D xLUL C xVUV D 1350:69 kJ=kg

The heat isQ D U2 � U1 D 195:454 kJ=kg

c) The specific volume in the tank must be no larger than the critical volume, which is Vc D 0:003110m3=kg(why?).This means that the minimum fraction of liquid is

X 0L DVV � Vc

VV � VLD 0:915778

d) Hint: Draw the PV graph and start from a vapor liquid state whose specific volume is less than thecritical volume.

pressure

liquid vapor

volume

AB C

e) When the initial volume is less that the critical, the system is almost entirely liquid with a very smallamount of vapor. Upon heating the expansion of the liquid essentially compresses the vapor and causes it tocondense.





Problem 3.14 Solution a) The specific volume is

V1 DV tank

mD 0:03333 m3=kg (3.1)

For saturated steam at 120ıC we find

L V

V 0:00106 0:8913 m3=kgU 503:57 2528:9 kJ=kg

We calculate the liquid fraction using the lever rule:

xL DVV � V1

VV � VLD 0:963748

and the internal energyU1 D xLUL C .1 � xL/UV D 576:993 kJ=kg

b) We locate the state by interpolation in the saturated tables at saturated vapor volume VV D V1. We find

T2 D 274:314 C;

P2 D 58:8526 bar;

U2 D 2849:63 kJ=kg

c) Q D U2 � U1 D 2272:64 kJ=kg.

d) Not possible.




Problem 3.15 Solution a) The two tanks together constitute a closed system. The PV work is zero be-cause the total volume is constant; there is no shaft work (none is mentioned); and the presence of insulationindicates that no heat is exchanged between the system and the surroundings. Application of the first lawgives

�U12 D QCW D 0 ) U1 D U2 Œ1�

That is, during this process both V and U state constant. We will use this information to specify the finalstate.

From the steam tables we collect information about the initial states in the two tanks:

tank A tank BPA D 5 bar PB D 10 barTA D 151:84

ıC TB D 500ıC

VL D 1:093 cm3/g VB D 353:96 cm3/gVV D 374:68 cm3/g UB D 3124:3 kJ/kgUL D 639:569 kJ/kgUV D 2560:2 kJ/kg

Based on this information and the liquid fraction given in tank A we find in the find:

VA D .0:07/.1:093/C .0:93/.374:68/ D 348:529 cm3/g

UA D .0:07/.639:569/C .0:93/.2560:2/ D 2425:76 kJ/kg

MA D2 m3

348:529 cm3/gD 5:74 kg

The mass in tank B is

MB D1 m3

353:96 cm3/gD 2:83 kg

Combined system of tank A and B The specific volume of the two-tank system is

VC D.2C 1/ m3

.5:74C 2:83/ kgD 350:321 cm3/g

The initial total internal energy is:

MAUA CMBUB D .5:74/.2425:76/C .2:83/.3124:3/ D 22746:7 kJ

This is also equal to the final internal energy, as Eq. [1] shows. Therefore, the specific internal energy at thefinal state is

UC D22746:7 kJ

.17:7C 2:83/ kgD 2656:21 kJ/kg

Therefore, the final state is defined by the conditions:

VC D 350:321 cm3/g; UC D 2656:21 kJ/kg

Determination of P , T We need to locate two states that bracket desired V and U . One such pair is:P1 D 7:5 bar, T1 D 167:76 ıC and P2 D 7:5 bar, T2 D 350 ıC. A second pair is P2 D 7:75 bar,T1 D 169:10 ıC and P2 D 7:75 bar, T2 D 350 ıC. By interpolation between the two states (see attachednotebook) we find

T D 213:2 ıC; P D 5:84 bar





b) If the desired pressure is 15 bar, the state is defined by the conditions P D 15 bar, V D 350:321 cm3/g.Because this state lies outside the range of the steam tables at 15 bar, we extrapolate from the last two entriesin the table (600ıC and 650 ıC). The final result is

T 0 D 866:3 ıC; U 0 D 3766:19 kJ/kg

The amount of heat is obtained by application of the first law. Since there is no work, the first law gives:

�U D Q ) Q D U 0 � UC D .3766:19 � 2656:21/ kJ/kg D C1109:98 kJ/kg

This heat must be added.




Problem 3.16 Solution a) At 1 bar, 80 ıC, the state is compressed liquid. The specific volume is V1 D0:001029 m3=kg, therefore, the mass of water in the tank is

M1 D1 m3

0:001029 m3=kgD 971:8 kg

b) The specific volume in tank 2 is

V2 D1 m3

20 kgD 0:05 m3=kg

At 20 bar, this specific volume falls between saturated liquid and vapor. From the steam tables we find:

P D 20 bar; T D 212:37 ıC; V L D 0:00118 m3=kg; V V D 0:0996 m3=kg;

The liquid mass fraction in the tank is

xL D0:0996 D �:05

0:0996 � 0:00118D 0:504

c) When the valve is open, the specific volume of the mixed system is

V D2 m3

.971:8C 20/ kgD 0:0020165 m3=kg

This value places the state in the vapor/liquid reegion at 175 ıC, therefore, the pressure is equal to thesaturation pressure. By interpolation,

P D P satD 8:93 bar





Problem 3.17 Solution a) First we point out that we know the complete state (P , T , V , n) in eachcompartment initially. Their calculation is left as an exercise.

By first law.n1 C n2/Uf D n1U1 C n2U2 ) n1.Uf � U1/C n2.Uf � U2/ D 0

with internal energy expressed per mole. Using �U D CV�T , the above equation becomes

n1CV .Tf � T1/C n2.Tf � T2/ D 0 ) Tf Dn1T1 C n2T2

n1 C n2D 545 K D 272 ıC

Since total volume is conserved we also have

.n1 C n2/Vf D n1V1 C n2V2 ) Vf D 0:0352508 m3=mol

The pressure is

Pf DRTf

VfD 1:286 bar

b) We collect data for the initial state in each compartment:

m .kg/ T .ıC/ P .bar/ V .m3=kg/ U kJ=kgcompartment 1: 0:0023015 200 1 2:1725 2582:9

compartment 2: 0:0012909 400 2 1:5493 2967:1

By first law the internal energy in the final state is

.m1 Cm2/Uf D m1U1 Cm2U2 ) Uf D 2720:96 kJ=kg

We also know the specific volume at the final state:

.m1 Cm2/Vf D m1V1 Cm2V2 ) Vf D 1:94856 m3=kg

Since we know the internal energy and the specific volume, we have enough information to locate the statein the steam tables. This requires some tedious trial-and-error because the tables are organized by P and T ,not V and U . We find

Tf D 240:23ıC; Pf D 1:26 bar




Problem 3.18 Solution From the steam tables at 6 bar we find (in standard steam table units):

V U H

Liquid 0:0011 669:72 670:5

Vapor 0:3156 2566:8 2756:1

The properties at the initial state are:

V1 D 0:19955 m3=kg; U1 D 1866:78 kJ=kg; H1 D 1986:51 kJ=kg

a) With Q D �1500 kJ=kg,H2 D H1 CQ D 486:5 kJ=kg

From steam tables,

T2 D 116ıC

V2 D 0:001057 m3=kg

U2 D 486:61 kJ=kg

The work isW D P.V2 � V1/ D U2 � U1 �Q D C119:8 kJ=kg

(Confirm that both calculations giver the same answer.)

b) With Q D C1500 kJ=kg,H2 D 3486:51 kJ=kg

By interpolation in the steam tables, the final state is:

P2 D 6 bar

T2 D 501:482 C;

V2 D 0:593153 m3=kg;

U2 D 3130:62 kJ=kg;

H2 D 3486:51 kJ=kg

The work isW D P.V2 � V1/ D U2 � U1 �Q D �236:16 kJ=kg





Problem 3.19 Solution a) Since heating is under constant pressure, the amount of heat is equal to thechange in enthalpy:

Q D H2 �H1 ) H2 D H1 CQ

At the final state we know pressure and enthalpy, therefore, we can obtain all other properties.

From the steam tables at 30 bar, 400 ıC we find:

U1 D 2934:6 kJ.kg; H1 D 3232:5 kJ.kg

from which we obtainH2 D .3232:5C 200/ kJ/kg D 3432:5 kJ/kg

By interpolation in the steam tables at P D 30 bar, H D 3432:5 kJ/kg we find

T2 D 409:4ıC U2 D 3089:51 kJ/kg

b) The change in U is

�U D U2 � U1 D .3089:51 � 2934:6/ kJ/kg D 154:9 kJ/kg

We note that �U and �H are not the same.




Problem 3.20 Solution We collect data in the initial and final states:

state P .bar/ T .ıC/ V .m3=kg/ H .kJ=kg/1 10 200 0:206 2828:3

2 10 179:89 0:00113 762:68

a) 179.89 ıC.

b) Q D H2 �H1 D �2065:62 kJ=kg.

c) W D �P.V2 � V1/ D 204:87 kJ=kg





Problem 3.21 Solution a) We collect the following data from the steam tables:

P1 D 1 bar T1 D 20ıC H1 D 83:92 kJ=kg

P2 D 1 bar T2 D 99:61ıC HL D 417:44 kJ=kg HV D 2674:9 kJ=kg

At the final stateH2 D 0:05HL C 0:95HV D 2562:03 kJ=kg

The heat isQ D H2 �H1 D 2478:11 kJ=kg

b) If the heat is Q0 D 2100 kJ the final enthalpy is

H 02 D H1 CQ D 2183:92 kJ=kg

The state is vapor-liquid mixture at 1 bar with quality

x0V DH 02 �HL

HV �HLD 0:7825





a) The heat is

Q D

Z T2

T1

CPdT D R

"27:0423.T2 � T1/ �

0:0224477.T 22 � T21 /

2C0:000115337.T 32 � T

31 /

3

#D 71051:8 J=mol

b) The work is W D �P.V2 � V1/. For the volume we use the Rackett equation:

V D VcZ.1�Tr /

0:2857

c

This empirical equation gives the volume of saturated liquid. At the final state the state is indeed saturated.At the final it is compressed liquid, but we may assume that the volume is the same as that of the saturatedliquid at the same temeperature. That is,

V1 D 0:000158356 m3=mol V2 D 0:000241088 m3=mol

Using P D 1:1 bar, the work isW D �9:10 J=mol:

c) The enthalpy at the initial state is

H1 D H2 �Q D 38592 � 71051:8 D �32459:8 J=mol

d) The internal energy at the initial state is

U1 D H1 � PV1 D �32477:3 J=mol






a) For constant pressure heating, Q D �H . From steam tables:

initial final

T .ıC/ 300 450

P .bar/ 20 20

V .m3=kg/ 0:1255 ‹

U .kJ=kg/ 2773:2 3031:0

H .kJ=kg/ 3024:3 3358:1

For the heat we findQ D 3358:1 � 3024:3 D 333:8 kJ=kg

The work isW D �U �Q D .3031 � 2773:2/ � 333:8 D �76 kJ=kg

b) For constant V heating, Q D �U . In the final state we know V2 D V1 D 0:1255 m3=kg and T2 D450 ıC. The internal energy (and pressure) of the final state are obtained by interpolation. We find:

initial final

T .ıC/ 300 450

P .bar/ 20 25:9

V .m3=kg/ 0:1255 0:1255

U .kJ=kg/ 2773:2 3025:09

H .kJ=kg/ 3024:3 ‹

The amount of heat isQ D 3027 � 2773:2 D 251:9 kJ=kg

c) For the CP we have

CP �333:8 kJ=kg.450 � 300/ ıC

D 2:23 kJ=kg ıC

and for the CV :

CV �251:9 kJ=kg.450 � 300/ ıC

D 1:68 kJ=kg ıC

These values are calculated by approximating derivatives with finite differences. For more accurate values,the temperature difference should be chosen as small as possible while still being big enough to show adifference in H and U that is above the truncation error of the tables.

Note: Entries with a question mark can be obtained by nterpolation but becasue they are not needed in thisproblem they have not been calculated.





From the steam tables we obtain the following properties of saturated water:

T D 200 C P satD 15:55 bar VL D 0:001157 m3=kg UL D 850:60 kJ=kg HL D 852:39 kJ=kg

VV D 0:1272 m3=kg UV D 2594:3 kJ=kg HV D 2792:1 kJ=kg

The properties at the initial state (xL D 0:8, XV D 0; 2) are:

V1 D 0:0263656 m3=kg U1 D 1199:34 kJ=kg H1 D 1240:33 kJ=kg

a) Constant-pressure cooling: the enthalpy at the final state is

H2 D H1 �Q D 1240:33 � 600 D 640:3 kJ=kg

This in the compressed region at P D 15:55 bar, H2 D 640:3 kJ=kg. Assuming the enthalpy of thecompressed liquid to be the same as that of the saturated liquid at the same temperature, we use the saturatedtables to find T � 152 ıC.

b) Constant-volume cooling: The internal energy in the final state is

U2 D U1 �Q D 1199:34 � 600 D 599:34 kJ=kg:

We know U D 599:34 kJ=kg and V D 0:0263656 m3=kg. This volume is between VL and VV at alltemperatures below 200 ıC, therefore the final state must be a vapor liquid mixture. To locate the precisetemperature, we notice that U and V must be given by

V2 D xL0VL0 C xV 0VV 0

U2 D xL0UL0 C xV 0UV 0

where VL0 , UL0 , V˙V’; UV 0 , are the saturation properties of water at the final state. Solving for the liquidfraction,

xL0 DVV 0 � V2

VV 0 � VL0DUV 0 � U2

UV 0 � UL0

We now construct a trial-and-error procedure as follows: we pick a temperature in the saturated tables andcalculate xL0 using the volume and then using the internal energy. If the two values are not the same, werepeat at another temperature until the two agree:

T ıC VL VV UL UV xL(using V ) xL(using U ) (ratio)

124 0:001064 0:7927 520:56 2533:2 0:968039 0:960857 1:00747

126 0:001066 0:7483 529:07 2535:3 0:966142 0:964974 1:00121

128 0:001068 0:7068 537:58 2537:4 0:964154 0:969117 0:99487

130 0:001070 0:6681 546:10 2539:5 0:962077 0:973292 0:98847

132 0:001072 0:6318 554:63 2541:6 0:959898 0:977498 0:98199

The answer is approximately 126 ıC (the ratio of the two values of xL is shown in the last column – pickthe one closest to 1).






1 bar

50 bar

400 °C

20 °C 30 °C55 bar

450 °C

A

A’

B

B’

C D

E

P

V

The heat capacities will be calculated from their definition by approximating the derivatives with finitedifferences between states that are close to the specified state.

a) At 25 ıC, 1 bar we are in the compressed liquid region. With reference to the attached figure, the CP is

CP �HB �HA

TB � TAD125:75 � 83:92

30 � 20D 4:183 kJ=kg K

and the CV is

CV �UB � UA

TB � TAD125:74 � 83:92

30 � 20D 4:182 kJ=kg K

Comments

For the calculation of CV we need two states at the same V . Here we have assumed that the volume inA is essentially the same as in A because V does not change appreciably from 20 ıC to 30 ıC.

The enthalpy and internal entropy of the compressed liquids (A, B) is taken to be that of the saturatedliquid at the same temperature (A0, B 0).

b) At 50 bar, 400 ıC we are in the superheated region. For CP we use states C and D, which are at thesame pressure and close temperature:

CP �HD �HC

TD � TCD3317 � 3196:6

450 � 400D 2:408 kJ=kg K

For CV , we use states E and C , which are at the same specific volume. State E must be obtained byinterpolation.

CV �HE �HC

TE � TCD3006:39 � 2907:4

456:122 � 400D 1:764 kJ=kg K




Problem 3.26 Solution a) The specific volume is

V D3

2:031L=mol D 0:0014771 m3=mol

This places the system in the V/L region. We find

xL D 0:704

The saturated proeprties at 50 ıC are

V .m3=mol/ U .J=mol/ H .J=mol/L W 0:00010717 18691:8 18745

V W 0:00473934 35796:3 38147

with the internal energy calcualted asU D H�P=�. Using these values and the level rule with xL D 0:704,the enthapy end internal energy of the initial state are

H1 D 24483 kJ=kg; U1 D 23750:4 kJ=kg

b) We are seeking saturated vapor with density

�V D1

0:0014771 m3=molD 0:677 L=mol

From the given tables we see that the final temperature is 100 ıC and P2 D 15:26 bar. We find U2 DH2 � P2=�V D 39668:9 kJ=mol and the heat is

Q D U2 � U1 D 15918:6 J=mol

c) At 0 ıC, P D 4:96 bar, the system is compressed liquid:

H2 � HL.0ıC/ D 11624 J=mol

and the heat isQ D H2 �H1 D �12859: J=mol

Alternative calculation of enthalpy at 0 ıC Start from saturated liquid at 50 ıC and cool under constantpressure to 0 ıC:

H2 D HL.50ıC/C CP;L.0 ıC � 50 ıC/ D 18745 J=molC .153 J=mol K/.�50 ıC/ D 11095 J=mol

and the heat is11095 � 24483 D �13388: J=mol

The small discrepancy (4%) is due to the fact that the given value of CP is an average over the tempera-ture range 0-100 C.





Problem 3.27 Solution The initial state is compressed liquid and the final state is superheated vapor. Thecalculation must be broken into three parts: liquid, liquid/vapor, and vapor:

�H1 D CPL.Tsat� T1/ D 327:2 J=mol

�H2 D �Hvap D 21242 J=mol

�H3 D CPV .T2 � Tsat/ D 9151:75 J=mol

The total amount of heat is

Q D 327:2C 21242C 9151:75 D 30720:9 J=mol




Problem 3.28 Solution The amount of heat from initial temperature to the saturated vapor is

Q1 D CPV .Tsat� T1/ D �5413:67 J

Since the amount of heat that is removed is more than that, the final state is to the left of the saturated vapor(on the PV graph). The additional amount of heat removed is

Q2 D Q �Q1 D .�10; 000/ � .�5413:67/ D �4586:33 J=mol

Since this is less than the heat of vaporization, the final state is vapor/liquid. The fraction of the liquid is(why?)

xL D�Q2

�HvapD�.�4586:33/

21242D 0:216:

The final mixture contains 21.6% liquid.





Problem 3.29 Solution Using the subscript 1 for the liquid and 2 for the vapor, the enthalpy in each partbefore mixing is:

T1 D 50ıC P1 D 10 bar H1 D H

sat;LC CLP .T1 � T

sat/ D �7858:15 J=mol


sat;VC CVP .T2 � T

sat/ D 62400:5 J=mol

For adiabatic mixing, Q D 0, and for constant-pressure process, Q D �H . Therefore,

�H D 0 ) .n1 C n2/H � n1H1 � n2H2 D 0 ) H Dn1H1 C n2H2

n1 C n2D 38980:9 J=mol K

This value palces the system in the liquid/vaopr region: the temeprature is The fraction of the liqud is125.7 ıC and the liquid fraction is

xL DH sat;V �H

H sat;V �H sat;L D 0:21




Problem 3.30 Solution Using subscript 1 for the initial liquid, and 2 from the initial vapor, the enthalpyof each part before mixing is


sat;LC CLP .T1 � T

sat/ D �14449:1 J=mol


sat;VC CVP .T2 � T

sat/ D 46556:7 J=mol

In the final state the enthalpy is

H D .0:25/.22645/C .0:75/.43320/ D 27813:8 J=mol

Suppose we mix n1 moles of liquid with n2 moles of vapor, with n1 C n2 D 1. By energy balance

n1H1 C n2H2 D .n1 C n2/H )n1

n2H1 CH2 D

�n1

n2C 1

�H )

n1

n2DH2 �H

H �H1

By numerical substitution,n1

n2D

46556:7 � 27813:8

27813:8 � .�14449:1/D 0:443

The liquid and the vapor must be mixed at a ratio 0.443 to 1.





Problem 3.31 Solution a) The amount of heat is equal to the amount needed to heat the liquid to thesaturation temperature plus the amount to produce 95% vapor:

Q1 D CLP .T

sat� T1/C xV�Hvap D 40215:6 J=mol

b) The final state is in the two-phase region (why?). By energy balance:

Q2 D CLP .T

sat� T1/C x

0V�Hvap ) x0V D

Q2 � CLP .T

sat � T1/

�HvapD 0:677:

b) The final state is in the superheated vapor region (why?) region. By energy balance:

Q3 D CLP .T

sat� T1/C�Hvap C C

VP .T3 � T

sat/

which is solved for the final temperature T3:

T3 D TsatCQ3 � C

LP .T

sat � T1/ ��Hvap

CVPD 207:1 ıC




Problem 3.32 Solution When a liquid is heated under constant pressure, first its temperature rises untilthe boiling point, then the liquid evaporates under constant temperature, and then, if more heat is added, thetemperature of the vapor increases further. In the present problem we do not know where the final state is.First we will calculate the amount of heat needed to obtain saturated liquid and we will compare this to theknown amount added. This heat is equal to the enthalpy change of the liquid and will be calculated usingthe CP of the liquid:

QAL D

TLZTA

CPdT D

120:23Z40

.72:4C 0:01T � 1:5 � 10�6T 2/dT D 6077 J/mol D 337:6 kJ/kg

The amount of heat added is more than that, therefore, the final state is past the saturated liquid. Next, wecalculate the amount of heat to produce saturated vapor:

QAV D QAL C�Hvap D 337:6C .2706:3 � 504:701/ D 2201:6 kJ/kg

This is more than the amount given. Therefore, the final state is a mixture of liquid and vapor at 2 bar,120.23 ıC. To obtain the mass fractions of the liquid and vapor we must first calculate the enthalpy of themixture. We know that the enthalpy of the saturated liquid is 504.701 kJ/kg. The amount of heat added pastthe saturated liquid is equal to

�HLB D Qtot �QAL D 1200 � 337:6 D 862:4 kJ/kg

where B denotes the final state. Therefore, the enthalpy at the final state is

HB D HL C�HLB D 504:701C 862:4 D 1367:1 kJ/kg

The fraction of the liquid is now calculated by the lever rule:

l DHV �HB

HV �HLD

2706:3 � 1367:1

2706:3 � 504:701D 0:61 D 61%

We repeat this problem assuming the value of the CP is not known.

From the steam tables, the initial enthalpy at 2 bar, 40 ıC, is 167.5 kJ/kg. Upon adding 1200 kJ/kg the finalenthalpy becomes

H2 D H1 CQ D 167:5C 1200 D 1367:5 kJ/kg

This enthalpy is between that of the saturated vapor and saturated liquid at 2 bar, therefore, the system is avapor-liquid mixture. The quality of that mixture is

v D1367:5 � 504:701

2706:3 � 504:701D 0:39; ) l D 0:61

Comment: The solution is simpler if we use the steam tables instead of the heat capacity. The method usingheat capacities, however, is often the only option if we do not have detailed tabulations of properties.

Exercise Determine the final state if the amount of heat added is 3000 kJ/kg instead of 1200 kJ/kg.





Problem 3.33 Solution a) From the steam tables, HL D 1006:37 kJ=kg, HV D 2803 kJ=kg.

�Hvap D HV �HL D 2803 � 1006:37 D 1796:63 kJ=kg

b) Q D xV�Hvap D .0:75/.1796:63/ D 1347:47 kJ=kg

c) The enthalpy at the final state is

H2 D HL CQ D 1347:47C 750 D 1756:37 kJ=kg

This value is between the saturated liquid and saturated vapor, therefore the state is a vapor-liquid mixtureat 30 bar, 233.86 ıC:

xL D2803 � 1756:37

2803 � 1006:37D 0:583

d) The enthalpy in the final state is

H2 D HL CQ D 1347:47C 1200 D 2956:37 kJ=kg

This value is higher than that of the saturated vapor, therefore, the system is superheated vapor. The temper-ature is obtained by interpolation at P D 30 bar; H D 2956:37 kJ=kg. We find

T2 D 286:2ıC




Problem 3.34 Solution From the steam tables we find:

P D 1 bar T sat D 99:61 ıC HL D 417:44 kJ=kg HV D 2674:9 kJ=kgP D 1 bar T D 20 ıC H � 83:92 kJ=kg

a) The heat isQ D H2 �H1 D

The enthalpy at the initial state isH1 D 83:92 kJ=kg

and at the final,

H2 D xLHL C xVHV D .0:05/.417:44/C .0:95/.2674:9/ D 2562:03 kJ=kg

The heat isQ D 2562:03 � 83:92 D 2478:11 kJ=kg

b) From the steam tables, the enthalpy at the initial state is H1 D 3278:5 kJ=kg. At the final state

H2 D H1 �Q D 3278:5 � 3000 D 278:5 kJ=kg

The state is compressed liquid at 1 bar. By interpolation in the saturated tables,

T2 D 66:5ıC





Problem 3.35 Solution We collect the following from the steam tables:

P D 30 bar T1 D 20ıC H1 D 83:92 kJ=kg

P D 30 bar T sat D 233:86 ıC HL D 1008:37 kJ=kg HV D 2803:3 kJ=kg

a) The enthalpy at the final state is

H2 D .1 � xV /HL C xVHV D 2354:57 kJ=kg

The heat isH2 �H1 D 2270:65 kJ=kg

b) The final enthalpy isH2 D H1 C 2000 kJ=kg D 2083:92 kJ=kg

The state is in the V/L region and the quality of the steam is

xV DH2 �HL

HV �HLD 0:599

c) The final enthalpy isH2 D H1 C 3000 kJ=kg D 3083:92 kJ=kg

The state is in the superheated vapor region. By interpolation we find

T2 D 336:8ıC




Problem 3.36 Solution Using subscripts 1 and 2 for the two compartments before the partition was re-moved, and F for he final state, the energy balance gives:

n1U1 C n2U2 D .n1 C n2/UF )

n1.UF � U1/C n2.UF � U2/ D 0 )

n1CV .TF � T1/C n2CV .TF � T2/ D 0 )

TF Dn1T1 C n2T2

n1 C n2[1]

At the initial state each compartment has a volume equal to half of the total:

P1V total

2D n1RT1 [2]

P2V total

2D n2RT2 [3]

from which it follows

n1

n1 C n2D

P1=T1

P1=T1 C P2=T2D 0:727

n1

n1 C n2D 0:273

Substitution into Eq. [1] givesTF D 327:3 K

For the final pressure we have

PF D .n1 C n2/RTF

V total [4]

Solving [2-3] for n1, n2, and adding the result we obtain

n1 C n2 DV total

2R

�P1

T1CP2

T2

�[5]

Substitution into [4] gives

PF D1

2

�P1TF

T1C P2

TF

T2

�D 1:5 bar

Alternative solution (Outline): The actual volume of the tank is not given and actually, that volumedoes not matter, i.e., the solution is the same regardless of the volume. This means that we could chosethis volume arbitrarily to solve the problem. Set, for example V tank D 1m3 and use that to calculate themoles in each compartment. With n1, n2, known, calculate TF from [1] and PF from [4]. The resultsshould be the same.





Problem 3.37 Solution a) At 120 ıC and 10 bar ethanol is compressed liquid. At 140 ıC and 1 bar it isvapor.

b) At 1800 ıC, 20 bar, we find Tr D 4, Pr D 0:325. Under these conditions enthanol can be assumed to bein the ideal-gas state (check with the generalized ZP graph).

c) Since ethanol can be assumed ideal and the molar volume is constant between initial and final state:

P2 DP1T2

T1D 1036:58 K

Q D CV .T2 � T1/ D .CP �R/.T2 � T1/ D 170:1 kJ=mol (3.2)




Problem 3.38 Solution The pressure at state 2 is:

P2 D P1

�T1

T2

�CP =R

D 22:9 bar

The various quantitites are calculated as follows:

�U D .CP �R/�T

�H D CP�T

Q12 D �U12

Q23 D �H12

W12 D �U12 �Q12

W23 D �U23 �Q23

The results are summarized below in J/mol:

�U �H Q W

1! 2 8961:55 12495 0 8961:55

2! 3 �9383:27 �13083 �13083 3699:73

total �13083 12661:3






The results are summarized below (in kJ/kg):

ig steam tables errorQ 803:541 813:8 �1:26062%W �184:926 �187:1 �1:1617%�H 803:541 813:8 �1:26062%�H 618:615 626:7 �1:29015%

Under the conditions of this problem steam is fairly close to the ideal-gas state.




Problem 3.40 Solution a) The final pressure is

P2 D p1

�T2

T1

� QCP =R

wirh QCP given by

QCP D

Z T2

T1

CPdT=T

ln.T2=T1/[1]

We find:

QCP D 37:3841 J=mol K

P2 D 3:79 bar

b) The final tempeature is

T2 D T1

�P2

P1

�R= QCP

The QCP is given by Eq. [1] but the final temperature is not known, so this has to be done by trial and error.Starting with a guess for QCP , calculate T2, then recalculate QCP between T1 and T2 and repeat the processuntil the results don’t change any more. As a first guess, use the CP at T1 D 500 ıC This is shown in thetable below:

iteration QCP .J=mol K/ T .K/1 38:1382 676:414

2 37:6599 675:267

3 37:6541 675:253

4 37:654 675:253

5 37:654 675:253

The answer converges very quickly to a final temeprature of 675 K.







4 Entropy & the Second Law




4. ENTROPY & THE SECOND LAW

Problem 4.1 Solution We will calculate entropy using

�S D

ZdQrev

T;

along the following reversible path: (a) Constant pressure path from P D P1, T1, to P D P1, V D V2,followed by (b) constant-volume path from P D P1, V D V2 to P D P2, V D V2, where V2 is the specificvolume of the final state. This path involves an intermediate state whose pressure is that of the initial stateand its specific volume that of the final state. The properties in the three states involved are:

1 10 2

P 36 36 400 .bar/T 250 803:03 22 .ıC/V 0:0568 0:1371 0:1371 .m3=kg/U 2619:7 3658:04 2943:4 .kJ=kg/H 2824:0 4151:48 3244:9 .kJ=kg/

Constant-pressure path: Along this path, dQ D dH D CPdT , and the entropy change is

�S110 D

Z T10

T1

CPdT

T:

Constant-pressure path: Along this path, dQ D dH D CPdT , and the entropy change is

�S110 D

Z T10

T1

CV dT

T:

The calculation requires the CP and CV as a function of temperature. Here we will simplify the calculationusing an approximation for the two heat capacities. The value of CP will be calculated from the knownenthalpies at states 1 and 1’, both of which are at the same temperature. The CV will be calculated from theknown internal energies in states 10 and 2, which are at the same volume:

CP �4151:48 � 2824:0

803:03 � 250D 2:400 kJ=kg K

CV �3658:04 � 2943:4

803:03 � 400D 1:773 kJ=kg K

The entropy change along each path is

�S110 D CP lnT10

T1D .2:400 kJ=kg K/ ln

803:03C 273:15

250C 273:15D 1:7314 kJ=kg K

�S102 D CV lnT10

T1D .1:773 kJ=kg K/ ln

400C 273:15

803:03C 273:15D �0:831978 kJ=kg K




Problem 4.2 Solution The entropy change of the copper is

�Sc D mcCpc lnT0

TiD .2 kg).0:38 kJ/kg/ ln

298:15

473:15D �0:3510 kJ/K

The entropy change of the air, treated as a bath is

�Sa D �Q

T0(4.1)

where Q D mcCpc.T0 � Ti / is the heat exchanged. Since it is calculated with respect to copper, a minussign is needed when this heat is used to calculate the entropy change of the bath:

�Sa D �mcCpc.T0 � Ti /

T0D �

.2 kg).0:38 kJ/kg/.298:15 � 473:15/473:15

D 0:4461 kJ/K

The entropy change of the universe is

�Suniv D �0:3510C 0:4461 D 0:0951 kJ/K

and it is positive, as it should.





Problem 4.3 Solution Entropy change of ice: The process is conducted under constant pressure (dQ DdH ), therefore, the entropy change will be calculated as

�S D

ZdQ

TD

ZdH

T

The calculation must be split into three parts: cold ice to melting temperature; melting of ice at constantT D 0 ıC; and warming of water from 0 ıC to final temperature 40 ıC. With T1 D 268:15 K, Tf D273:15 K, Tbath D 313:15 K, we have

�Sa D mCP;ice lnTfusion

T1D 0:0378729 kJ=K

�Sb D m�Hfusion

TfusionD 1:24557

�Sc D mCP;liq lnTbath

TfusionD 0:571244 kJ=K

The total entropy change is

�Sice D �Sa C�Sb C�Sc D 1:85469 kJ=kg

Entropy change of bath: The bath rejects heat at constant pressure. Its entropy change is

�Sbath D�Q

Tbath

where the minus sign indicates thatQ is calculated taking the ice to be the system. The heat is also calculatedin three steps:

�Ha D mCP;ice.Tfusion � T1/ D 10:25 kJ

�Hb D m�Hfusion D 334

�Hc D mCP;liq.Tbath � Tfusion/ D 167:2 kJ

The entropy change of the bath is

�Sbath D�.10:25C 334C 167:2/

313:15D �1:63324 kJ=K

The entropy generation isSgen D 1:85469 � 1:63324 D 0:221445 kJ=K




Problem 4.4 Solution By energy balance,

Q D �H D mCP .T2 � T1/

from which we obtain the final temperature of the bath,

T2 D T1 DQ

mCP: [1]

The entropy change of the bath is

�S D

Z T2

T1

CPdT

TD CP ln

T2

T1[2]

with T2 obtained from Eq. [1].

Numerical Substitutions: The results are summarized below:

m .kg/ T2 .ıC/ �S .kJ=K/ jQj=T1 .kJ=K/

1 63:9 0:307725 0:319336

10 42:3 0:318122 0:319336

1000 40:02 0:319324 0:319336

1 40 � 0:319336

Notes

1. Temperature is shown in ıC but it must be in kelvin when calculating entropy.

2. The last column shows the entropy change calculated by the bath formula, jQj=Tbath with Tbath D

T1 D 313:15 K (= 40ıC). As the mass of the bath approaches infinity, the entropy change approachesthe value calculated by the bath formula.

f)

�S D mCP lnT2

T1

When T2 � T1, the above becomes

�S D mCP lnT2

T1� mCp

�1 �

T2

T1

�D�mCP .T2 � T1/

T1D�Q

T1





Problem 4.5 Solution a) The PV graph is shown below. As we see, the initial state is superheated vaporand the final is compressed liquid. Calculations will be done separately in each region. We also need theheat of vaporization at 78ıC. This can be obtained from a handbook or using the Pitzer equation below:

�Hvap

RTcD 7:08.1 � Tr/

0:354C 10:95!.1 � Tr/

0:456) �Hvap D 38111:5 J=mol

2 bar

1 bar

0.5 bar

78 °C

1

2

3

4

b) The enthalpy is calculated as follows:

�H12 D 0

�H23 D ��Hvap D �38111:5 J=mol

�H23 � 0

The final result is�H14 D �38111:5 J=mol:

Note: �H12 is zero because we are assuming ideal gas and the process is isothermal; �H12 is zero be-cause the enthalpy of compressed liquid is essentially the same as that of the saturated liquid at the sametemperature.

c) Entropy is calculated similarly:

�S12 D �5:76 J=mol K

�S23 D ��Hvap

T sat D �108:533 J=mol K

�S23 � 0

Therefore�S14 D �114:296 J=mol K

d) The heat in each part is

Q12 D �U12 �W12 D �W12 D �RT lnP2

P1D �2023:62 J=mol

Q23 D ��Hvap D �38111:5 J=mol

Q34 D 0

The total heat isQ14 D �40135:1 J=mol




Problem 4.6 Solution a) The process is shown schematically in the PV graph above. The initial stateis superheated vapor but we do do not know whether the final state is in the vapor, the vapor/liquid or thecompressed liquid region. We will calculate the amount of heat from A to V and from A to L and willcompare with the given amount.

From A to V :QAV D C

VP .TV � TA/ D .60/.83 � 200/ D �7020 J/mol

Since the amount of heat removed is larger than that, the final state is to the left of V . To go from A to L theheat is

QAL D QAV ��Hvap D �7020 � 36740 D �43760 J/mol

Since the heat removed (25000 J/mol) is less than that, the final state is somewhere between L and V .Therefore, the temperature of the final state is 83 ıC.

b) For entropy we have:�SAB D �SAV C�SVB

For the entropy from A to V we have:

�SAV D CVP ln

TV

TAD .60/ ln

83C 273:15

200C 273:15D �17:0 J/mol K

From V to B , since the process is isothermal

�SVB DQVB

TVD

�17980

83C 273:15D �50:5 J/mol K

Finally, for the entropy change from A to B we find

�SAB D �17:0 � 50:5 D �67:5 J/mol K

Note: It turns out that the heat capacity of the liquid is not needed. It would be required if we had to docalculations in the compressed-liquid region of the phase diagram, i.e. if the amount of heat removed wasmore than 7020C 36740 D 43760 J/mol.

Although the problem does not ask for the fraction of vapor and liquid in the final state, here is how wewould calculate it:

Method 1 We notice the amount of heat removed past V is

QVB D Qtot �QAV D �25000 � .�7020/ D �17980 J/mol

To produce 100% liquid, starting from V , we must remove heat in the amount of�Hvap; since we have onlyremoved 17980 J/mol, the fraction of the vapor that is condensed is

l D17980

36740D 48:9%

Alternative calculation of l: The heat from V to B is QVB D �17980 J/mol. This is also equal to:

QVB D HB �HV Œ1�

whereHB D lH

LC .1 � l/HV

�HV� l�Hvap Œ2�

Combining Eqs. [1] and [2] and solving for l we find

QVB D �l�Hvap ) l D�QVB

�HvapD17980

36740D 49%





Problem 4.7 Solution a) The state is vapor:

SA D SL C�Svap C CVP ln

Ta

T sat D SL C�Hvap

T sat C CVP ln

Ta

T sat

b) Vapor-liquid mixture:

SB D xLSL C xV SV D xLSL C xV

�SL C

�Hvap

T sat

�

c) Compressed liquid:

SC D SL C CLP ln

TC

T sat

d) Same as c.


P .bar/ T .K/ S .J=mol K/

part a 10 573:15 4649:95

part b 10 489:95 4581:43

part c 10 293:15 4451:64

part d 10 293:15 4451:64

NOTE: The entropy of saturated liquid should have been given as 49.582 J=mol K. Below are thecorresponding results if the correct units are used used :

P .bar/ T .K/ S .J=mol K/

part a 10 573:15 131:044

part b 10 489:95 62:5221

part c 10 293:15 �67:2612

part d 10 293:15 �67:2612




Problem 4.8 Solution Outline: For reversible isothermal process,

Q D T�S

The work is then obtained asW D �U �Q

Numerical substitutions (in standard steam-table units) The initial state is obtained directly from the steamtables. The final state is a vapor-liquid mixture at 250 ıC. From the saturated steam tables we find

P satD 39:76; UL D 1080:7; UV D 2601:9;

T D 250 SL D 2:7934; SV D 6:0722

The properties of the final state are obtained by application of the lever rule. The results are summarized inthe table below:

1 2

T 250 250

P 15 39:76

U 2696:0 1567:5

S 6:7111 3:8426

xV � 0:32

We find:

�U D �1128:52 kJ=kg

�S D �2:86848 kJ=kg K

Q D �1500:7 kJ=kg

W D C372:1 kJ=kg

The heat is negative, i.e., it must be removed from the system.





Problem 4.9 Solution At the initial state we have

T1 D 250ıC;

P1 D 15 bar;

U1 D 2696: kJ=kg;

S1 D 6:7111 kJ=kg K

At the final state T2 D T1 D 250 ıC and

S2 D S1 CQ

T1D 6:7111C

�2000

523:15D 2:8881 kJ=kg K

This places the state in the vapor-liquid region at 250 ıC. We find:

xL D 0:971116; P2 D 39:76 bar; U2 D 1124:64 kJ=kg

The work isW D U2 � U1 �Q D 428:638 kJ=kg




Problem 4.10 Solution We have the following to equations:

U2 � U1 D QCW

Q D .S2 � S1/T

These can be solved by trial and error: guess P2, calculate U2 and S2, then check if the above equationsare satisfied; if not, repeat with a new guess. Following this procedure we find that the final state is in thetwo-phase region. In this region the equations to be solved become:

U2 � U1 D QCW

Q D .S2 � S1/T

U2 D xLUL C .1 � xL/UV

S2 D xLUL C .1 � xL/SV

which must now be solved for xL, U2, S2 and Q. We find:

xL D 0:823575

S2 D 3:37186 kJ=kg K

U2 D 1349:08 kJ=kg

Q D �1746:92 kJ=kg





Problem 4.11 Solution At the final state:

V2 D 0:001091 m3=kg

U2 D 631:73 kJ=kg

H2 D 632:25 kJ=kg

S2 D 1:8420 kJ=kg K

The entropy change of steam is

�Ssteam D S2 � S1 D �5:7727 kJ=kg

The entropy change of the bath is (why?)

�Sbath D ��Ssteam D C5:7727 kJ=kg

The heat isQ D

S2 � S1

TD �2442:72 kJ=kg

The work isW D U2 � U1 CQ D 491:548 kJ=kg





At the initial state we have

P1 D 1 bar

T1 D 150ıC

U1 D 2582:9 kJ=kg

S1 D 7:6147 kJ=kg K

At the final state P2 D 20 bar and S2 D S1. By interpolation:

P2 D 20 bar

T2 D 566:073ıC

U2 D 3231:79 kJ=kg

S2 D 7:6147 kJ=kg K

For adiabatic process the work is

W D U2 � U1 D 648:888 kJ=kg





Problem 4.13 Solution We will take the ideal-gas heat capacity of nitrogen to be constant and equal toCP D 29:5 J=mol K.

The entropy change is

�S D CP lnT2

T1�R ln

P2

P1

Solving for T2:

T2 D

�P2

P1e�S=R

�R=CP

D 510 K (4.2)

b) The work isW D CV .T2 � T1/ D .CP �R/.T2 � T1/ D 4910:69 J=mol

For reversible operation, �S D 0 and the exit temperature would be

T 02 D

�P2

P1

�R=CP

D 437:8 K

The corresponding amount of work is

W 0 D .CP �R/.T02 � T1/ D 3382:22 J=mol

The efficiency of the compressor is

� DW

W 0D 69%




Problem 4.14 Solution a) The PV graph is shown below. As we see, the initial state is compressedliquid and in the final superheated vapor. Calculations will be done separately in each region. We also needthe heat of vaporization at 65ıC. This can be obtained from a handbook or using the Pitzer equation below:

�Hvap

RTcD 7:08.1Tr/

0:354C 10:95!.1Tr/

0:456) �Hvap D 36735:7 J=mol

2 bar

1 bar

0.2 bar

65 °C

1

2

3

4

b) The enthalpy is calculated as follows:

�H12 � 0

�H23 D �Hvap D 36735:7 J=mol

�H23 D 0

The final result is�H14 D 36735:7 J=mol:

Note: �H34 is zero because we are assuming ideal gas and the process is isothermal; �H12 is zero be-cause the enthalpy of compressed liquid is essentially the same as that of the saturated liquid at the sametemperature.

c) Entropy is calculated similarly:

�S12 D 0

�S23 D�Hvap

T sat D 108:637 J=mol K

�S23 � 0

Therefore�S14 D 108:637 J=mol K

c) The heat in each part is

Q12 D 0

Q23 D �Hvap D 36735:7 J=mol

Q34 D �U12 �W12 D �W12 D �RT lnP4

P3D 4524:74J=mol

The total heat isQ14 D 41260:4 J=mol





Problem 4.15 Solution Treating steam as an ideal gas, we obtain

�S D �R lnP2

P1D �39:8032 J=mol K D �2:21129 kJ=kg K

The result from steam tables is

�S D 7:1834 � 9:4216 D �2:2382 kJ=kg K

The ideal-gas calculation is off by�1:2%. The good agreement is due the fact that steam at this low pressureis very close to the ideal-gas state.




Problem 4.16 Solution For a reversible process the heat can be calculated as

Q D

BZA

TdS D T

BZA

dS D T .SB � SA/

where we took into consideration the fact that temperature stays constant during this path. From the steamtables we find:

T D 400 C PA D 10 bar UA D 2957:8 kJ/kg SA D 7:4668 kJ/kg KT D 400 C PB D 5 bar UB D 2964:8 kJ/kg SB D 7:7954 kJ/kg K

For the heat we findQ D T .SB � SA/ D 221:2 kJ/kg

The work is calculated as W D �UAB �Q and we find

W D �215:4 kJ=kg





Problem 4.17 Solution The properties in the two compartments initially are:

A B

m 1 1 kgP 10 20 barT 200 800 ıCV 0:1943 0:247 m3=kgU 2582:8 3658:1 kJ=kgS 6:585 8:1791 kJ=kg K

For the overall system initially,

Ui DmAUA CmBUB

mA CmBD 3120:45 kJ=kg

Vi DmAVA CmBVB


Si DmASA CmBSB


In the final state,Uf D Ui D 3120:45 kJ=kg Vf D Vi D 0:22065 kJ=kg

These conditions determine the final state. This requires a double interpolation. We perform this interpola-tion between the following four states:

P1 D 15 bar; P2 D 16 bar; T1 D 450 ıC; T2 D 500 ıC

P T Œij � V U S

15 450 Œ11� 0:2192 3035:9 7:4259

16 450 Œ21� 0:2053 3034:9 7:3948

15 500 Œ12� 0:2352 3120:8 7:5716

16 500 Œ22� 0:2203 3120: 7:6787

Solving the double interpolation formula for P and T at the final state we find,

Pf D 15:98 17bar; Tf D 500:256ıC

and with P and T known the final entropy is

Sf D 7:67818 kJ=kg K

The entropy generation is

Sgen D Sf D Si D 7:67818 � 7:38205 D 0:296126 kJ=kg K




Problem 4.18 Solution The work that could be extracted is equal to the ideal work for the system (brick)to go from 800 ıC to 25 ıC.

Wideal D �U � T0�S

�U � �H D CP .T2 � T1/

�S D CP lnT2

T1

Numerical substitutions: Using T1 D 1073:15 K, T2 D 298:15 K, T0300 K we obtain

�U D �697:5 kJ=kg

�S D �1:153 kJ=kg K

Wideal D �351:7 kJ=kg

Negative works means work produced.





Problem 4.19 Solution a) The efficiency f the first cycle is

� D 1 �298:15

873:15D 0:6585

and the amount of work isW1 D �.0:6585/.1000 kJ/ D �658:5 kJ

This is produced by cycle 1 and absorbed by cycle 2, i.e.,

W2 D �W1 D 658:5 kJ

b) In the second cycle, QC2 D 1000 kJ. By energy balance,

QH2 CQC2 CW2 ) QH2 D 1658:5 kJ

The efficiency of this cycle is

�2 D�W2

QH2

D 39:7%

The high temperature is calculated as follows:

�2 D 1 �TC2

TH2) TH2 D

TC1

1 � �2D 1448:15 K D 1175 ıC

1000 kJ 1658 kJ

600 °C

25 °C 600 °C

TH

658 kJ

341 kJ 1000 kJ

Carnot 1 Carnot 2




Problem 4.20 Solution All results in standard steam table units.

a)1 2 3 4

P 20 20 10 10

T 212:38 212:38 179:89 179:89

V 0:00118 0:0996 0:183622 0:0145363

U 906:27 2599:2 2482:13 887:956

H 908:62 2798:4 2665:75 902:483

S 2:447 6:3392 6:3392 2:447

xL 1 0 0:0552782 0:930599

b)�U Q W

1! 2 1692:93 1889:78 �196:85

2! 3 �117:075 0 �117:075

3! 4 �1594:17 �1763:32 169:154

4! 1 18:3135 0 18:3135

total 0 126:458 �126:458

c)W D �126:458

d) The two ratios are equal:

jW j

jQH jD 0:0669166

1 �tL

tHD 0:0669166






1 2 3 4

T .ıC/ 700 700 500: 500:

P .bar/ 20 10 3:35894 6:72414

U .kJ=kg/ 3471:7 3476:3 3130:21 3127:52

H .kJ=kg/ 3918:2 3924:1 3486:17 3482:58

S .kJ=kg K/ 7:9509 8:2755 8:2755 7:9509

QH D 315:884 kJ=kg

QC D �250:964 kJ=kg

W D �64:92 kJ=kg

� D �W=QH D 20:6%

�Carnot D 1 � TC =TH D 20:6%




5 Calculation of Properties




5. CALCULATION OF PROPERTIES

Problem 5.1 Solution a) We begin with

dH D CPdT C

�V � T

�@V

@T

�P

�dP

Since this represents an exact differential, applying the criterion of exactness we obtain:�@CP

@P

�T

D

�@V

@T

�P

�

�@T

@T

�P

�@V

@T

�P

� T

�@2V

@T 2

�P

)

�@CP

@P

�T

D �T

�@2V

@T 2

�P

b) In the ideal-gas state, �@V

@T

�P

D

�@RT=P

@T

�P

DR

P;

�@2V

@T 2

�P

D 0

from which we conclude @C

igP

@P

!T

D 0

or that the C igP is independent of pressure. Since C ig

V D CigP �R, we also conclude the constant-volume heat

capacity is also independent of pressure in the ideal-gas state.

c) Using the definition of ˇ, �@V

@T

�P

D ˇV

from which�@2V

@T 2

�P

D V

�@ˇ

@T

�P

C ˇ

�@V

@T

�P

D V

�@ˇ

@T

�P

C Vˇ2 D V

��@ˇ

@T

�P

C ˇ2�

Therefore, �@CP

@P

�T

D �T V

��@ˇ

@T

�P

C ˇ2�

For liquid away from the critical point, s well as for solids in general, V and ˇ are small numbers, and ˇ isonly weakly dependent on pressure. Under these assumptions,�

@CP

@P

�T

� 0

that is, the heat capacity of condensed phases is essentially independent of pressure.




Problem 5.2 Solution a) The residual properties can be calculated as follows:

HRD

Z P

0

�V � T

�@V

@T

�P

�dP

SR D

Z P

0

�R

P�

�@V

@T

�P

�dP

with the integrations performed at constant temperature. Solving the equation of state for V we have

V DRT

PC b

from which we obtain the partial derivative, �@V

@T

�P

DR

P

The residual properties are now obtained by integration:

HRD bP

SR D 0

b) For isothermal process from (T; PA) to (T; PB ):

�SAB D �R lnPB

PAC SRB � S

RA D �R ln

PB

PA

Q D T�S

The work isW D �UAB �Q

We need the internal energy, which we calculate as follows:1

�UAB D �HAB ��.PV /AB

�HAB D HRB �H

RA D b.PB � PA/

�.PV /AB D b.PB � PA/

which finally give�UAB D 0

We conclude then that the work isW D �Q

1For the PV term, use the equation of state to write PV D RT � Pb and then calculate the difference between initial andfinal state.






P DRT

V � b�

a

V 2�@P

@T

�V

DR

V � b�@P

@V

�T

D �RT

.V � b/2/C2a

V 3

ˇ D �.@P=@T /V

.@P=@V /T


V D 0:0007882 m3

ˇ D 0:00377314 K�1




Problem 5.4 Solution The amount of heat is

Q D T�S:

The work is practically zero because the liquid is essentially incompressible (� � 0). For isothermal process,

�S D �

ZˇVdP � �ˇV�P

Numerical substitutions

ˇ D 0:00142597 1=K

V D 0:234369 m3=kg

�S D �0:635 kJ=kg K

Q D �186:146 kJ=kg

Heat must be removed from the system in order to maintain temperature constant during compression.






Starting withH D H ig

CHR (5.1)

we differentiate with respect to T at constant P :

CP D CigP C

@HR

@T

!P

:

The partial derivative on the right hand side is the residual heat capacity and represents a correction to theideal-gas heat capacity for deviations form ideality. This term will be calculated from the equation of state.First we will calculate HR, then we will estimate its derivative by numerical differentiation.

The residual enthalpy for the truncated virial is

HRD

�B � T

dB

dT

�P

and B is calculated by the Pitzer method

B DRTc

Pc.B0 C !B1/

B0 D 0:083 �0:422

T 1:6r

B1 D 0:139 �0:172

T 4:2rdB

dTDR

Pc

�0:6752

T 2:6rC0:7224!

T 5:2r

�We now calculate the residual enthalpy at �50 ıC, 38 bar, and �45 ıC, 38 bar:

HR1 D �490:508 J=mol

HR2 D �511:448 J=mol

The residual heat capacity is

CRP DHR2 �H

R1

T2 � T1D 4:189 J=mol K

This value to be added to the ideal-gas heat capacity of oxygen in order to account for deviations fromideality.




Problem 5.6 Solution Using residuals to express enthalpy at two close temperatures T1 and T2 D T1 C

ıT , we have

H2 �H1 D CigP ıT CH

R2 �H

R1 :

Dividing both sides by ıT :H2 �H1

ıTD C

igP C

HR2 �H

R1

ıT

The LHS is the CP of the fluid:

CP D CigP C

HR2 �H

R1

ıT

Using T1 D 15 ıC, T2 D 25 ıC, we find

HR1 D �32208:4 J=mol; HR

2 D �31960:4 J=mol

From handbooks, C igP .20

ıC/ D 141:2 J=mol K. Therefore,

CP D 141:2C�31960:4 � .�32208:4/

10D 190:7 J=mol K





Problem 5.7 Solution Some conversion factors are given in the footnotes of table 2-198. Otherwise youmay use:

1 Btu D 1055 JI 1 lbmol D 454 molI 1 K D .1:8/ ıRankine

and recall that the relationship between temperature differences in ıC and in ıF is the same as between Kand ıRankine.

a) C igp .300 K/ D 35:9 J/mol K D 8:58 Btu/lbmol ıF

b)C

igP;mh D 46:1 J/mol K D 11:02 Btu/lbmol ıF

CigP;ms D 43:8 J/mol K D 10:5 Btu/lbmol ıF

c)�H ig

D 46123 J/mol D 19842 Btu/lbmol

�S igD 42:56 J/mol K D 10:18 Btu/lbmol




Problem 5.8 Solution a) The residual volume is

V R D V �RT

P

With V D 0:00104 m3=kg, T D 372:76 K, we find

V D �0:0309725 m3=mol D 1:7207 m3=kg

b) We write the residual volume as

V R DRT

P.Z � 1/

and use the Lee-Kesler graphs for Z. For ethanol,

Tc D 513:92 K; Pc D 61:48 bar; ! D 0:649

With Tr D 0:726, Pr D 0:0163, we find Z0 D 0:9862, Z1 D �0:0124, and

V R D 0:000678 m3=mol

c) For isothermal process,

�S D �R lnP2

P1C SR2 � S

R1

From the Lee-Kesler graphs we find,

T .K/ P .bar/ Tr Pr s0 s1

state 1 513:92 1 1 0:01627 �0:0109 �0:0136

state 2 513:92 100 1 1:62655 �2:8572 �2:9970

With these values the entropy change is,

�S D �78:05 J=mol K





Problem 5.9 SolutionTc D 543:90 K; Pc D 28:70 bar; ! D 0:326

Using the Lee Kesler method we find:

T .K/ P .bar/ Tr Pr h0 h1 s0 s1 Z0 Z1

initial 543:9 1 1: 0:0348 �0:0351 �0:0316 �0:02347 �0:0293 0:9883 �0:0024

final 543:9 28:7 1: 1: �2:5738 �2:3825 �0:02347 �2:3106 0:9883 �0:07887

�H D HR2 �H

R1 D �14945:6 J=mol

�S D �R lnP2

P1C SR2 � S

R1 D �51:917 J=mol K

�U D �H � .P2V2 � P1V1/ D �18207:9 J=mol

Q D Tc�S D �28237:7 J=mol

W D �U �Q D 10029:8 J=mol

Notes:

1. The temperature of the initial and final states is the same, therefore, the terms involving C igP drop out.

2. The volume is calculate as V D ZRT=P

3. The initial state is sufficiently close to the ideal gas state so that the residual properties could be takento be zero.




Problem 5.10 Solution Outline:

a) The initial state is superheated vapor; the final state is compressed liquid.

b) The entropy change

�S D �R lnP2

P1C SR2 � S

R1

For the residuals, use the Lee-Kesler graphs.

c) The amount of heat isQ D T�S

Numerical substitutions At the initial state, we assume ideal-gas state; at the final state we obtain the residualentropy from the LK graphs.

Tc D 369:83 K; Pc D 42:48 bar; ! D 0:152

Tr Pr s0 s1

0:878 0:0002 0 0

0:878 0:4 �0:0292 �0:07834

�S D �62:18 J=mol K

Q D �20182 J=mol





Problem 5.11 Solution From the SRK we obtain the following residuals:

T .K/ P .bar/ HR .J=mol/ SR .J=mol K/

220 1 �24419:7 �92:1899

300 1 �163:626 �0:352764

Using C igP D 96:5 J=mol K we find

�H D 31976:1 J=mol

�S D 121:767 J=mol

Sgen D 15:1801 J=mol K




Problem 5.12 SolutionTc D 154:6; Pc D 50:43; ! D 0:022

CigP D 30:1798 � 0:0149153T C 0:0000547061T

2� 4:99671 � 10�8T 3 C 1:48821 � 10�11T 4

(with C igP in J/mol/K, T in K).

�H D

Z T

T1

CigP dT CH

R.T / �HR.T1/

�S D

Z T

T1

CigP

TdT �R ln

P2

P2C SR.T / � SR.T1/

The table below is constructed using the above equations an the SRK equation with T1 D 293:15 K, P1 D1 bar, P2 D 10 bar

T .K/ H .J=mol/ S .J=mol K/300 128:677 �18:6667

350 1630:25 �14:0377

400 3143:94 �9:99566

450 4674:94 �6:38958

500 6226:45 �3:12063

550 7800:25 �0:121002

600 9397:05 2:65751

650 11016:8 5:25018

700 12658:6 7:68349

750 14321:4 9:97771

800 16003:5 12:1488

At T D 550 we find �S � 0 and �H D 7800:25 J=mol. If better accuracy is needed, then the tabulationsshould be done at smaller temperature internals.





Problem 5.13 Solution Outline: The enthalpy at P; T using the ideal-gas refence state at T0 D 70 K,P0 D 7:83 bar is

H D CigP .T � T0/CH

RD 29.T � 70/CHR (a)

where HR is the residual enthalpy at P , T , from the above table. The enthalpy of vaporization is (why?)

�Hvap D HRV �H

RL

with HL, HV , calculated from Eq. (a).


a/ � 5938 J=mol

b/ � 4104 J=mol

c/C 4573 J=mol




Problem 5.14 Solution Using the SRK we obtain the following residuals:

T .K/ P .bar/ phase HR .J=mol/ SR .J=mol K/

266 1 sat. liq. �22563:1 �84:5351

266 1 sat. vap. �196:679 �0:470042

300 10 �21034:4 �60:30

a) HR D �21034:4 J=mol, SR D �60:3

b) H D 4809:7 J=mol, S D 16:694 J=mol K

c) H D �17753:4 J=mol, S D �67:8 J:mol K

d) At 1 bar:�Hvap D 22366:4 J=mol; �Svap D 84:065 J=mol K

At 10 bar:�Hvap D 17401:8 J=mol; �Svap D 50:6875 J=mol K





Problem 5.15 Solution Enthalpy and entropy at .T; P / are calculated as follows:

H D CigP .T � T0/CH

R�HR

0 CH0

S D CigP ln

T

T0�R ln

P

P0C SR � SR0 C S0

where HR, SR, are the residual properties at T , P , HR0 , SR0 are the residuals at the reference state, and

H0, S0, are the enthalpy and entropy of the refence state.

1. Reference state: The actual state at P D 20 bar, T D 400 K (H0 D 0, S0 D 0):

H (J/mol) S (J/mol K)state 1 �7251:19 1:43185

state 2 �35137:1 �102:908

�.2 � 1/ �27885:9 �104:34

2. Reference state: The ideal-gas state at P D 20 bar, T D 400 K (H0 D �2562:44 J=mol, S0 D�4:63942 J=mol K):,

H (J/mol) S (J/mol K)state 1 �9813:63 3:20757

state 2 �37699:5 �107:548

�.2 � 1/ �27885:9 �104:34





Tc D 369:83 K; Pc D 42:48 bar; ! D 0:152

a) With T D T0 D Tc , P D P0 D Pc we have

S D SR.Tc ; Pc/

With s0 D �2:1674, s1 D �2:3106 we find

S D �20:9397 J=mol K

b) The entropy at the final state is

S D �R lnP2

P1C SR

The state may be assumed ideal (why?); with SR D 0, we find

S D 126:888 J=mol K

The entropy change of propane is

�S D .126:888/ � .�20:9397/ D 147:828 J=mol K

d) The reference state is chosen arbitrarily. The specification “ideal gas” refers to the details of the calcula-tion, to the actual state of the system.





Problem 5.17 Solution a)To shift the steam tables from their original reference state to a new one at T0,P0, all that is needed to do is subtract the value of H0 from all enthalpies, subtract the value of S0 from allentropies, and recalculate internal energy as U D H � PV using the new value of H (which amounts tosubtracting H0 from all internal energies). This makes the enthalpy and entropy at the new reference statezero while keeping all differences in U , H or S among any two states the same:

original ref state new ref state

V 2:1725 2:1725

U 2658:2 �118:9

H 2875:5 98:4

S 7:8356 1:2506

b) We develop the equation for enthalpy and leave entropy as an exercise. The enthalpy in the steam tablesis calculated as

H D

Z T

Ttriple

CigP dT CH

R�H

R;Ltriple (a)

where H is the enthalpy at P , T , HR is the residual enthalpy at P , T , and HR;Ltriple is the enthalpy at the

reference state (saturated liquids at the triple point). The enthalpy H 0 at P , T , that is calculated using theideal-gas reference state at T0, P0, is

H 0 D

Z T

T0

CigP dT CH

R (b)

Taking the difference, solving for H 0, and combining the two integrals into one, we have

H 0 D H C

Z Ttriple

T0

CigP dT CH

R;Ltriple (c)

The last step is the calculation of the residual enthalpy of saturated liquid water at the triple point.

HRL;triple D H

RV;triple ��Hvap;triple � ��Hvap;triple

Because pressure is very low (Ptriple D 0:006117 bar), we set HRV;triple � 0, from which it follows,

HRL;triple � ��Hvap;triple (d)

Combining this result with Eq. (c) we finally have

H 0 D H C

Z Ttriple

T0

CigP dT ��Hvap;triple

The corresponding result for the entropy is

S 0 D S C

Z Ttriple

T0

CigP

dT

T�R ln

Ptriple

P0��Hvap;triple

Ttriple

In both cases, the new reference state amounts to adding a constant to the tabulated values, therefore, alldifferences will remain unaffected.




6 Balances in Open Systems




6. BALANCES IN OPEN SYSTEMS

Problem 6.1 Solution We recognize the process as a Carnot cycle. To determine whether the processis feasible or not we may either compare the efficiency of the cycle to the theoretical Carnot efficiency, orcalculate the entropy generation to see if it is positive. Both methods are equivalent and we will use themethod of Sgen. Since this is a cycle, the entropy generation is due only to exchanges of heat with thesurroundings:

Sgen D�Q1

T1C�Q2

T2

where Q1 D C3500 kJ=kg, Q2 D �.0:53/Q1 D �1855 kJ=kg, T1 is the temperature of the heatingbath, and T2 that of the cooling bath. The cooling process produces a V=L mixture at 1 bar, therefore, itstemperature must be at most T2 D .99:61C 273:14/ D 372:76 K. The heating bath adds heat to the steamthat is produced by the isentropic compression of the V=L mixture to 20 bar. From the steam tables wefind that the compressed state is in the V=L region and its temperature is 212:38 C 273:15 D 485:53 K.Therefore, the heating bath must be at least that hot, i.e., T2 D 485:53 K. with these values we find

Sgen D �2:23223 kJ=kg K < 0:

Since this is negative, the process is not feasible.





W D �45 J=mol K

�H D CP .T2 � T1/ D �9083 J=mol

Q D �H �W D �4583 J=mol

b)

�S D Cp lnT2

T1�R ln

P2

P1D �1:01778 J=mol K

d)Wideal D �H � T0�S D �8779:55 J=mol





Problem 6.3 Solution a) By first law:

�. PnH/ D PQC PW ) nAcP .TC � TA/C nBCP .TC � TB/ D PQC PW

Solving for TC we findTC D 313:15 K

b) By second law

Sgen D �. PnS/ �PQ

Tbath

The entropy changes of the streams is

�. PnS/ �PQ

TbathD nA

�CP ln

TC

TA�R ln

PC

PA

�C nB

�CP ln

TC

TB�R ln

PC

PB

�D �535:675 J=s K

For Tbath we take the temperature of boiling water at 1 atm (Tbath D 373:15 K). We find

PSgen D 53:9 W=K > 0:

The process is feasible, I recommend further investigation.

c) The boiling temperature of water at 10 bar is 179:89 ıC D 453:04 K. The entropy generation now is

PSgen D �50:7 W=K < 0:

This process is not feasible: the heat produced by the process is not at sufficiently high temperature toproduce steam at 10 bar without additional expenditure of work.




Problem 6.4 Solution a) By first law:

�. PnH/ D PQC PW ) nAcP .TC � TA/C nBCP .TC � TB/ D PQC PW

Solving for TC we findTC D 184:2 K

b) By second law

Sgen D �. PnS/ �PQ

Tbath

The entropy changes of the streams is

�. PnS/ �PQ

TbathD nA

�CP ln

TC

TA�R ln

PC

PA

�C nB

�CP ln

TC

TB�R ln

PC

PB

�D �2922:15 J=s K

For Tbath we take the temperature of boiling water at 1 atm (Tbath D 373:15 K). We find

PSgen D �1582:21 W=K < 0:

The process is not feasible – do not forward to upper management.

c) The boiling temperature of water at 10 bar is 179:89 ıC D 453:04 K. The entropy generation now is

PSgen D �1818:5 W=K < 0:

This process is even less feasible.






30 bar78 C

245 CVL

b)

�S D �Svap C CP lnT

T sat

�Svap D�Hvap

T sat

By numerical substitution�S D 68:57 J=mol K




Problem 6.6 Solution a)1 2

T .ıC/ 300 60

P .bar/ 2 2

H .kJ=kg/ 3072:1 251:15

S .kJ=kg K/ 7:894 0:8312

pressure

12

volume

b)Q D H2 �H1 D �2820:95 kJ=kg

c)�S D S2 � S1 D 7:0628 kJ=kg s

d)

�Ssur D �Q

TsurD 9:46151 kJ=kg

Sgen D �S C�Ssur D 2:39871 kJ=kg k





Problem 6.7 Solution From steam tables we collect the following information:

1 2 3 4

(steam in) (steam out) (water in) (water out)m .kg=s/ 1 1 Pmwater PmwaterT .ıC/ 300 151:84 20 60

P .bar/ 5 5 1 1

H .kJ=kg/ 3064:6 2748:1 83:92 251:15

S .kJ=kg K/ 7:4614 6:8206 0:2965 0:8312

a) Energy balance

Pmsteam.H2 �H1/ D Pmwater.H4 �H3/ ) Pmwater D 1:88 kg=s

b) Entropy Generation

PSgen D Pmsteam.S2 � S1/C Pmwater.H4 �H3/ D 0:3779 kW=K




Problem 6.8 Solution1 2

T .ıC/ 200 60

P .bar/ 1 1

H .kJ=kg/ 2875:5 251:15

S .kJ=kh K/ 7:8356 0:8312

a)Q D H2 �H1 D �2624:35 kJ=kg

b)�S D S2 � S1 D �7:0044 kJ=kg K

c) Taking Tbath D 60ıC D 333:15 K:

Sgen D S2 � S1 �Q

TbathD 1:652 kJ=kg K > 0






[DUPLICATE PROBLEM 6.8]




Problem 6.10 Solution The mass and energy balance equations are:

Pm1 C Pm2 D Pm3

Pm1H1 C Pm2H2 D Pm3H3

The unknowns in these two equations are the mass flow rates of streams 2 and 3. From the steam tables,

1 2 3

P 5 5 5 barT 300 40 151:84 ıCH 3064:6 167:54 1019:6 kJ=kgS 7:4614 0:5724 2:7534 kJ=kg KxV s/h comp. liq. 0:18


Sgen D Pm1S1 C Pm2S2 � Pm3S3 D 0:526 W=K

and the lost work isWlost D T0Sgen157:9 W

We find

Pm1 DH1 �H3

H3 �H2Pm1 D 2:4 kJ=s

Pm3 D Pm1 C Pm2 D 3:4 kJ=s






a)Reversible operation For reversible operation, the exit state is fixed by the condition:

S2rev D S1

Since pressure and entropy are known, all properties of the exit state are known. The results are shownbelow:

1 2 2rev

P 30 8 8 barT 450 261:586 296:318 ıCH 3344:7 2975:15 3049:06 kJ=kgS 7:0853 7:0853 7:2202 kJ=kg K

The reversible work isWrev D H2 �H2rev D �369:5 kJ=kg

b) Actual Operation The actual work is

W D .0:8/Wrev D �295:6 kJ=kg

From this we obtain the actual enthalpy at the exit:

H2 D H1 CW D 3049:1 kJ=kg

Finally, with the enthalpy and pressure of the exit stream known, all other properties are obtained from thesteam tables:

1 2

P 30 8: ıCT 450 296:3 barH 3344:7 3049:1 kJ=kgS 7:0853 7:2202 kJ=kg K

The entropy generation isSgen D S2 � S1 D 0:1349 kJ=kg K

and the lost workWlost D T0Sgen D 40:5 kJ=kg




Problem 6.12 Solution1 2

P .bar/ 35: 3:

T .ıC/ 500 200

H .kJ=kg/ 3451:5 2866

S .kJ=kg K/ 7:1592 7:3132

a)

Pm.H2 �H1/ D PW ) Pm DPW

H2 �H1D 1:28 kg=s

b) The power lost to heat losses is

.60 kJ=kg/.1:28 kg=s/ D 76:9 W

therefore, the power generated isPW 0 D 673 W






1

2

3

5

4

Outline of calculations:

1. With known state 1, calculate exit of reversible turbine

2. With known state 2, calculate exist of turbine with efficiency of 75%

3. With known states 3, 4, and 5, perform energy balance around mixing point:

m3H3 Cm4H4 D .m3 Cm4/H5

and solve for the ration m2=m1.

The result are summarized below on a basis of 1 kg in stream 2:

1 2 3 4 5

T .ıC/ 500 500 99:61 158:295 99:61

P .bar/ 40 40 1 1 1

m .kg/ 1 0:842379 1 0:842379 1:84238

H .kJ=kg/ 3445:8 3445:8 2575:41 2793:01 2674:9

S .kJ=kg K/ 7:0919 7:0919 7:0919 7:65135 7:3588

Entropy generation:Sgen D m5S5 �m1S1 �m2S2 D 0:4917 kJ=kg K





a) From steam tables we determine the states of inlet and outlet:

1 2

P .ıC/ 50 1

T .bar/ 400 99:61

H .kJ=kg/ 3196:6 2629:75

S .kJ=kg K/ 6:6481 7:23768

The work isW D H2 �H1 D �566:8 kJ=kg

and the mass flow is

Pm DPW

WD 1:411 kg=s:

b) If the turbine were operating reversibly, the exit state would be atP2 D 1bar , S 02 D S1 D 6:6481 kJ=kg K.This state is in the V/L region at 1 bar, 99.61 ıC:

L V

H .kJ=kg/ 417:44 2674:9

S .kJ=kg K/ 1:3026 7:3588

By lever rule,

xL DSV � S

02

SV � SLD 0:117351

The enthalpy isH 02 D xLHL C .1 � xL/HV D 2409:99 kJ=kg

The reversible work isWrev D H

02 �H1 D �786:615 kJ=kg

and the efficiency is

� DW

WrevD�566:8

�786:615D 72%

c) The entropy generation isSgen D Pm.S2 � S1/ D 0:832 kW=K






Before we begin we note that the molar flow rate of the gas is

Pn DPV

V1DPV

RT1P1

where PV is the volumetric flow rate at the inlet and V1 is the molar volume at the inlet conditions. Bynumerical substitution (watch the units!)

Pn D 0:580918 mol/s

Absolute enthalpies and entropies are calculated as follows:

H.P; T / D CP .T � Tref/

S.P; T / D CP lnT

Tref�R ln

P

Pref

where Pref is the reference pressure (1 bar) and Tref is the reference temperature (293.15 K).

a) For reversible operation, the compressor exit temperature, T2;rev would be obtained from the condition�S12 D 0:

�S12 D CP lnT2;rev

T1�R ln

P2

P1) T2;rev D T1

�P2

P1

�R=CP

By numerical substitution:T2;rev D 496:736 K

The actual temperature at 2 is obtained from the known efficiency:

� DCP .T2;rev � T1/

CP .T2 � T1/

Solving for T2 we findT2 D 547:633 K

The work in the compressor isPW D Pn.H2 �H1/ D 5366:36 W

b) The heat in the heat exchanger is

PQ D Pn.H3 �H2/ D �5366:36 W

c) The entropy generation is

PSgen D Pn.S3 � S1/C . PSgen of coolant in Heat Exchanger/

Since no information is given about the coolant, we make a best case estimate by assuming the coolant torun at sufficiently high flow rate so that its temperature, T 0 is essentially constant between inlet and outlet.This coolant temperature is unknown but we do know that it cannot be more than 20ıC (why?). We assumeT 0 D 20ıC D 293:15 K because that corresponds a best-case scenario, i.e., any temperature less thanthat would give an even larger entropy generation (it left as an exercise to confirm this with a numericalcalculation). Under these assumptions, the entropy generation of the coolant is (why?)

. PSgen of coolant in Heat Exchanger/ D� PQ

T 0




After numerical substitutions we findPSgen D C7:185 W/K

which is positive, in accordance with the second law.

1 2 3molar rate (mol/s) 0:580918 0:580918 0:580918

P (bar) 1 10 10

T (K) 293:15 547:633 293:15

H (J/mol) 0: 9237:72 0

S (J/mol K) 0: 3:54091 �19:1437

PQ (W) = +5366.36PW (W) = -5366.36PSgen (W/K) =+7.185





Problem 6.16 Solution a) The work in the compressor is

W D CP .T2 � T1/ D 5250 J=mol

b) The entropy generation is

Sgen D �S12 D CP lnT2

T1�R ln

P2

P1D 0:4736 J=mol K

c) For reversible operation �S D 0:

CP lnT 02T1�R ln

P2

P1D 0 ) T 02 D T1

�P2

P1

�R=CP

D 465:7 K

and the corresponding work is

W 0 D CP .T02 � T1/ D 5027:67 J=mol

c) T 02 D 465:7 K




Problem 6.17 Solution a) The power of the compressor is

PW D Pn�H12 D PnCP .T2 � T1/ D 10:3 kW

b) If the compressor operated reversibly, the exit temperature would be (why?)

T 02 D T1

�P2

P1

�R=CP

D 575:6 K

and the corresponding power would be

PW 0 D PnCP .T02 � T1/ D 6:72 kW

The efficiency is

� DPW

PW 0D 65:3%






[DUPLICATE PROBLEM 6.17]




Problem 6.19 Solution a) Reversible operation The exit state is fixed by the known pressure and entropy(S2rev D S1 D 7:8356 kJ=kg K). We find

T2rev D 558ıC;H2rev D 3603:9 kJ=kg

The reversible work isWrev D H2rev �H2 D 728:3 kJ=kg

b) Actual operation The actual work is

W DWrev

0:75D 971:1 kJ=kg

and the enthalpy at the actual exit state is

H2 D H C 1CW D 3846:6 kJ=kg

With known enthalpy and pressure, al properties of the final state are obtained from the steam tables. Theresults results are summarized below:

1 2rev 2

P 1 12 12 barT 200 558:0 666:5 ıCH 2875:5 3603:9 3846:6 kJ=kgS 7:8356 7:8356 8:1099 kJ=kg K


and the lost work isWlost D T0Sgen D 82:3 kJ=kg





Problem 6.20 Solution The properties of the initial state are:

H1 � 83:92 kJ=kgI S1 � 0:2965 kJ=kg=KI

a) Nearly no change in temperature.b) At P D 0:00706 the properties of the saturated phases are

T satD 2 ıC HL D 8:39 kJ=kgI SL D 0:0306 kJ=kg=K

HV D 2504:6k J=kgI SV D 9:1027 kJ=kg=K

The exit state is a mixture of vapor and liquid with xL D 97%.




Problem 6.21 Solution a) From steam tables we collect the following information:

1 sat. L sat. VP .bar/ 20 5 5

T .ıC/ 15 0:008726 0:008726

H .kJ=kg/ 83:92 21:02 2510

S .kJ=kg K/ 0:2965 2510 2510

The final state is in the VLE region. By level rule,

xL DHV �H1

HV �HLD 0:9747

The entropy isS2 D xLSL C .1 � xL/SV D 0:3024 kJ=kg K


b) If the final pressure is reached by isentropic expansion, then

x0L DSV � S1

SV � SLD 0:9754

By lever rule,

H 02 D x0LHL C .1 � x

0L/HL D 83:934 kJ=kg

The work isW 0 D H 02 �H1 D 0:014 kJ=kg






�H D 0 ) CP .T2 � T1/ D 0 ) T2 D T1 D 25ıC

b) The final state is still in the compressed liquid region. The temperature is nearly the same as in the inletof the throttling valve.

c) The final state is in the V/L region at 0.1 bar, 45.81 ıC. We collect the following data:

T .ıC/ HL .kJ=kg/ H .kJ=kg/(inlet) 60 251:15 2608:8

(outlet) 45:81 191:81 2583:9

At the inlet of the valve:H1 D 0:5HL1 C 0:5HV1 D 1429:98 kJ=kg

At the exitxL2 D

HV 2 �H1

HV 2 �HL2D 0:482

The liquid fraction decreases by a small amount during this process.




Problem 6.23 Solution The properties of the inlet are:

V1 D 0:001002 meter3=kg; H1 D 83:92 kJ=kg; S1 D 0:2965 kJ=kg K

The reversible work isWrev D V1.P2 � P1/ D 0:014028 kJ=kg

The actual work isWact D

Wrev

�D 0:0179846 kJ=kg

The enthalpy of the exit stream is

H2 D H1 CWact D 83:938 kJ=kg

The temperature rise is

�T DH2 �H1

CLPD 0:004 ıC

where CLP D 4:18 kJ=kg K.





Problem 6.24 Note: In this problem ˇ is given, therefore, the full form of the equations for �H and �Sshould be used.

a) The reversible work is

�Hrev DP2 � P1

�D 3:125 kJ=kg

The actual work is�H D

�Hrev

�D 5:04032 kJ=kg

The mass flow rate is

Pm DPW

�HD 1:984 kg=s

b) The temperature change is obtained by solving the enthalpy equation

�H D CP�T C.1 � ˇT1/.P2 � P1/

�) �T D 3:4 K

c) The entropy generation is

PSgen D Pm

�CP ln

T2

T1�ˇ.P2 � P1/

�

�D 0:01375 kW=K





4000

3900

3800

3700

3600

3500

3400

3300

3200

3100

3000

2900

2800

2700

2600

2500

2400

2300

2200

2100

2000

Enth

alpy

(kJ/

kg)

9.59.08.58.07.57.06.56.05.55.04.5

Entropy (kJ/kg K)

96

94

92

90

88

86

84

828078

7674

727065

98100%

650 °C

700 °C750 °C

600 °C

550 °C

500 °C

450 °C

400 °C

350 °C

200 °C

250 °C

300 °C

180 °C

160 °C

140 °C

120 °C

100 °C

80 °C

60 °C

40 °C

20 °C

0.01 b

ar0.0

15 ba

r0.0

2 bar

0.03 b

ar0.0

4 bar

0.06 b

ar0.

08 b

ar0.

1 ba

r0.

15 b

ar

0.2

bar

0.3

bar

0.4

bar

0.6

bar

0.8

bar

1.0

bar

1.5

bar

2.0

bar

3.0

bar

4.0

bar

6.0

bar

8.0

bar

10 b

ar

15 b

ar

20 b

ar

30 b

ar

40 b

ar

60 b

ar

80 b

ar

100

bar

150

bar

200

bar30

0 ba

r400

bar600

bar80

0 ba

r

1000

bar

% quali

ty

3190

2665

2490

3910 1

2

3`

3

Q D �720 J=mol K

Wrev D �700 J=mol K

W D �525 J=mol K






a) We collect the following data from the steam tables:

A B C

T .ıC/ 25 212:38 ‹

P .bar/ 20 20 20

Pm .kg=s/ 10 7 17

H .kJ=kg/ 104:84 908:62 ‹

S .kJ=kg K/ 0:3952 2:447 ‹

By energy balance:

PmCHC � PmAHA � PmBHC D PQ ) HC D 1265:22 kJ=kg

State C is vapor/liquid mixture at 20 bar, 212.38 ıC. By lever rule we find:

xL D 0:8113; SC D 3:1815 kJ=kg K

b) The entropy generation is

PSgen D Pm3S3 � PmASA � PmBSB D 33:0 kW=K




Problem 6.27

boiler

condenser

pump turbine

heat in

work in work out

heat out

QB

WTWB

QC

1 2

34

Solution1 2 3 4

Pm .kg=s/ 52:8 52:8 52:8 52:8

P .bar/ 60 60 1 1

T .ıC/ 101:424 700 232:572 99:61

H .kJ=kg/ 425:11 3894:5 2939:82 417:44

S .kJ=kg=K/ 1:32309 7:4248 7:96524 1:3026

phase comp L s/h s/h sat L





Problem 6.28 Solution (Outline)

a) Efficiency:

� Dj PQboilerj � j PQcondj

j PQboilerjD 15%

b) Entropy generation:

PSgen D �j PQboilerj

Tboiler�j PQcondj

TcondD 1:823 kW=K

c) Maximum power

PWmax D

�1 �

Tcond

Tboiler

�j PQboilerj D 693 kW




Problem 6.29 Solution Summary of Results:

1 2 3 4 5 6 7 8

m .kg/ 1 1 1 0:926237 1 1 0:0737626 0:0737626

P .bar/ 60 60 5 1 1 60 5 1

T .ıC/ 145:84 700 372:279 210:836 99:61 101:424 151:84 99:61

H .kJ=kg/ 615: 3894:5 3214:53 2896:96 417:44 425:11 640:19 640:19

S .kJ=kg K/ 1:8011 7:4248 7:70619 7:87873 1:3026 1:32309 1:8606 1:90018

phase c/L s/h s/h s/h sat L c/L sat L 0:901327

Q W Sgen Wlost(kJ/kg) (kJ/kg) (kJ/kg K) (kJ/kg)

Turbine 1 0 �679:971 0:281393 84:4178

Turbine 2 0 �294:148 0:159809 47:9427

Pump 0 7:67 0:0204931 6:14794

Boiler 3279:5 0 2:25372 676:115

Cond �2313:05 0 0:0700662 21:0199

Feedwater heater 0: 0 0:0468206 14:0462

Valve 0: 0 0:00291971 0:875912

TOTAL 966:449 �966:449 2:83522 850:565





Problem 6.30 Solution We take the evaporator to be at -30 ıC (5 ıC below the required temperature) andthe condenser at 30 ıC (5 ıC above the temperature of cooling water). The following are data for propane(R290) from the NIST Web Book:

T .ıC/ P .bar/ H .kJ=kg/ S .kJ=kg K/�30 1:68 127:71 0:7220

�30 1:68 540:44 2:4195

T .ıC/ P .bar/ H .kJ=kg/ S .kJ=kg K/20 10:79 251:86 1:1790

30 10:79 279:09 1:2704

30 10:79 605:95 2:3486

30 10:79 605:97 2:3486

40 10:79 626:61 2:4156

50 10:79 647:03 2:4798

60 10:79 667:44 2:5420

-25°C

+25°C

1

2

4

3

evaporator

condenser

1 2 3 4

T .ıC/ �30 51:3097 30 �30

P .bar/ 1:6785 10:79 10:79 1:6785

H .kJ=kg/ 540:44 649:703 279:09 279:09

S .kJ=kg K/ 2:4195 2:48795 1:2704 1:34461

phase V V L 0:633223




Q (kJ/kg) W(kJ/kg) Sgen (kJ/kg K)evaporator 261:35 0 9:78656

condenser �370:613 0 7:29151

compressor 0 109:263 0:0684463

throttling 0: 0 0:0742108

total �109:263 109:263 17:2207

cop 2:39193 �

Max cop 4:0525 �

mass 20:1856 mol/s





Problem 6.31 Solution Outline

1. Use 1 kg as the basis in stream 1.

2. Calculate the mass flow rates in 2 and 6 by energy balance around the mixing point (2+6!7)

3. Stream 3 is saturated liquid at 0.5 bar. Obtain stream 4 via throttling calculation.

4. All streams are known - complete the energy balances.

1 2 3 4 5 6 7

m .kg/ 1 0:955924 0:0440758 0:0440758 0:0440758 0:0440758 1

P .bar/ 0:5 0:5 0:5 0:008726 0:008726 0:5 0:5

T .ıC/ 81:32 81:32 81:32 5 5 500 100

H .kJ=kg/ 2543:62 2645:2 340:48 340:48 2510:1 3489:2 2682:4

S .kJ=lg K/ 7:30642 7:593 1:091 1:2248 9:0249 9:1565 7:6952

Sgen Wlost

separator 0 0

valave 0:00589755 1:76926

evaporator 0 0

compressor 0:00580038 1:74011

mixing 0:0332874 9:98623




Problem 6.32 Solution a) The ideal work is

Wideal D �H12 � T0�S12

where T0 D 298:15 K is the temperature of the surroundings. The enthalpy change is

�H12 D CVP .T2 � T1/ ��Hvap D �24236 J=mol

The entropy change is

�S12 D CVP ln

T2

T1��Hvap

T2D �89:88 J=mol K

The ideal work is now found to be:Wideal D 2561:67 J=mol

b) The work of the refrigerator is

W DQEV

COPD 6924:57 J=mol

where QEV D ��H12. c) The lost work is

Wlost D T0Sgen


Sgen D �QEV

TEV�QCOND

TCOND

The heat in the condenser of the refrigerator is

QCOND �W �QEV D �31160 J=mol

For the temperature of the evaporator we use TEV D �7 C D 266:15 K and for the condenser TCOND D

T0 D 398:15 K:Sgen D 13:4516 J=mol

The lost work finally isWlost D 4010:61 J=mol

Notes:

1. For practical purposes the temperature of the evaporator must be below �7 ıC. In such case theentropy generation would be even higher.

2. Notice that the lost work is not equal to the difference between the ideal work and the actual work ofthe refrigerator (through they are close). Why not?

3. The assumption of ideality is left as an exercise.






1. Calculate the rate of heat removal in the evaporator:

PQevap D .cop/ PWcomp

2. Calculate the heat in the condenser

PQcond D �Qevap � PWcomp

3. Calculate the entropy generation:

PSgen D �PQevap

Tevap�PQcond

Tcond

4. Obtain the properties of streams 3 and 15. Solve for stream 4 doing a throttling calculation6. Calculate the heat per unit mass in the evaporator:

Qev D H1 �H2

7. Calculate the flow rate of the refrigerant

Pm DPQev

Qev

8. Calculate the enthalpy of stream 2:

H2 D H1 CPWcomp

Pm9. Obtain the temperature of stream 2 from its known enthalpy and pressure.

Numerical Substitutions Using data from the NIST web book (search for R134a and select units to matchthe ones given here) we obtain the following results:

0° F

90° F

1

2

4

3

evaporator

condenser

1 2 3 4

T 0 140:907 89:998 0

P 21:171 119:01 119:01 21:171

H 17028 19538:4 10752 10752

S 42:417 44:025 28:14 28:764

phase sat V V sat L xL D 0:6763




(btu/lbmol) (btu/hr)Qevap 6276 31812:5

Qcond �8786:4 �44537:5

Wcomp 2510:4 12725

PSgen D 11:8 btu=hr R

Pm D 5:07 lbmol=hr





Problem 6.34 Solution We will operate the evaporator at 21 psia (T D �14:7 ıF) and the condenser at166 psi (T D 84:8 ıF), which allow for about 5ıF of temperature difference with the specified temperatures.We use data from the NIST Web Book (notice the mixed units chosen in this solution):

1

2

4

3

evaporator

condenser

1 2 3 4

T .ıF/ �14:741 310:497 84:816 �14:741

P .psi/ 21 166: 166 21

H .kJ=kg/ 1572:4 1957:58 481:73 481:73

S .kJ=kg K/ 6:4699 6:65689 1:9493 2:05754

phase V V L 0:809355

Q W Sgen Wlost.kJ=kg/ .kJ=kg/ .kJ=kg K/ .kJ=kg/

Evaporator 1090:67 0 0:0464765 13:943

Condenser �1475:85 0 9:6301 2889:03

Valve 0 385:181 0:108242 32:4725

Compressor 0 0 0:186986 56:0958





20 C 30 C

T3=200 C T3=473.15 KT3=473.15 K

T2=303.15 K

T1=298.15 K

Q

Q

Qsolar

W

-(Q+W)

house

solar panels

The process may be represented by two Carnot cycles. One is a refrigerator that pumps an amount Q ofheat from the house (T1) to the surroundings (T2); the other one is an engine that power the refrigerator byabsorbing heat at T3 and rejecting it at T2. We assume reversible cycles and set the entropy genration ineach cycle equal to zero:

�Qsolar

T3�QCW

T2D 0; �

QCW

T2�Q

T3D 1

Solving for W=Q and Qsolar=Q we find

W

QD 0:0168;

Qsolar

QD 1:59

The air conditioning requires 0.0168 W of work per W of heat that is pumped; to generate this amount ofwork, the solar panels must absorb 1.59 W of heat.






1 2 3

We take the temperature in 3 to be 100 ıF (10 ıF) above that of the cooling water. This makes the pressureof the stream 212.01 psi (data from the NIST Web Book). The results are summarized in the table below:

1 2 3

T (F) 80 240:203 100

P psi 80 212:01 212:01

H (Btu/lb) 12197: 13570:1 3631:7

S (Btu/lb R) 25:178 25:5908 8:1188

phase V V sat L

W D H2 �H1 D 1373:09 btu=lb

Q D H3 �H2 D �9938:39 btu=lb





1. Set the flow rate of stream 1 to 1 kg (basis).

2. Take the temperature of stream 2 to be 5 ıC above that of the cooling water; this fixes the pressure ofstream 2 since the state is saturated liquid.

3. The temperature in stream 11 is �40 ıC; since the stream is saturated vapor, this fixes P11.

4. Now that we know the low and the high pressure in the process calculate the intermediate pressure as.PlowPhigh/

1=2. All pressures are now known.

5. Perform a throttling calculation for streams 3 and 6.

6. Perform an energy balance around the separator and solve for the enthalpy of stream 5. Obtain theremaining properties of 5 from the known enthalpy and pressure.

7. Perform a throttling calculation to obtain stream 10.

8. Obtain stream 12 by balance around compressor 1.

9. Obtain 13 by balance on the mixing point between the two compressors.

10. Obtain 1 by energy balance around compressor 2.

11. Obtain the mass flow rate bay total balance around

Complete the calculations using data from the NIST Web Book.






1. Set the flow rate of stream 1 to 1 kg(basis).

2. Set T2 D 30 ıC; this fixes P2.

3. Set T5 D �20 ıC; this sets P5. All pressures are known.

4. Obtain stream 3 by balance around the throttling valve.

5. Obtain the mass flow in 6 and 4 by balance around the separator.

6. Obtain stream 1 by balance around the compressor.

Perform the numerical calculations using data from the NIST Web Book.





Note The solution is worked out for m2 D 20m1 and for T13 D 3 ıC; the temperature of stream 14 is notspecified but will be calculated instead.

Data for methane (NIST Web Book)

T .C/ P .bar/ H .kJ=mol/ S .J=mol K/ phase T .C/ P .bar/ H .kJ=mol/ S .J=mol K/ phase�161:64 1 �0:0089481 �0:079738 L �100 30 3:943 26:908 L�161:64 1 8:1908 73:456 V �95:876 30 4:327 29:103 L�160 1 8:249 73:975 V �95:876 30 8:662 53:555 V�150 1 8:599 76:940 V �90 30 9:138 56:200 V�140 1 8:944 79:634 V �80 30 9:741 59:408 V�130 1 9:286 82:110 V �70 30 10:250 61:978 V�120 1 9:626 84:406 V �60 30 10:714 64:207 V�110 1 9:965 86:549 V �50 30 11:151 66:211 V�100 1 10:303 88:560 V �40 30 11:570 68:052 V�90 1 10:641 90:455 V �30 30 11:979 69:765 V�80 1 10:978 92:250 V �20 30 12:379 71:378 V�70 1 11:316 93:955 V �10 30 12:773 72:907 V�60 1 11:654 95:580 V 0 30 13:164 74:365 V�50 1 11:993 97:136 V 8 30 13:475 75:488 V�40 1 12:334 98:628 V 10 30 13:553 75:764 V�30 1 12:676 100:060 V 20 30 13:941 77:111 V�20 1 13:019 101:450 V 30 30 14:329 78:413 V�10 1 13:365 102:790 V0 1 13:714 104:090 V3 1 13:819 104:47 V10 1 14:065 105:350 V20 1 14:42 106:58 V30 1 14:778 107:78 V40 1 15:14 108:96 V

Below we outline the steps of the solution:

1. Ignoring pressure drop in pipes, the process operates between two pressures, 1 bar and 30 bar. Allstreams before the compressor and after the expander and the throttling valve are at 1 bar; the rest areat 30 bar.

2. On a basis of 1 kg in stream 1, the flow rate in 2, 3 and 4 is 10; in 12, 13 and 14 it is 9. Let the flowrate of 10 be V ; then the flow rate in 8 is 1C V and in 11 is 9 � V . In summary, the only unknownflow rate is that of stream 10.

3. With all pressures known and with the information given in the problem, the states of streams 1, 5, 6,9, 10 and 13 are known.

4. Stream 11 is calculated from the known efficiency of the expander. We find:

T11D � 132:964 bar; P11 D 1 bar; H11 D 9:18484 kJ=mol; S11 D 81:3762 J=mol K

5. Perform an energy balance around a box that contains heat exchanger HE2, the throttling valve, theseparator, and the mixing point at the top of the separator. This system receives streams 5 and 11 asinlet, and exits streams 9 and 13 as outlets. There is no heat or work exchanged in this part of theprocess. The energy balance reads

m5H5 Cm11H11 D m9H9 Cm13H13

and the only unknown is the flow rate V .





1 2 3

6

7 8

10

1112

14

4

compressorprecooler

HE1

HE2

expander

V/L

separator5

13

9

11

6. Obtain 12 by mixing 10 and 11.

7. Obtain 7 by energy balance in HE2.

8. Obtain 8 by throttling 7.

9. Obtain 9 and 10 by separating 8.

10. Obtain 14 by balance around HE1.

11. Obtain 2 by mixing 1 and 14.

12. Obtain 3 by solving the compresor.

1 2 3 4 5 6 7

m .kg/ 1 20 20 20 18:3805 1:61951 18:3805

T .ıC/ 20 27:9064 349:792 30 8 8 �95:876

P .bar/ 1 1 30 30 30 30 30

H .kJ=kg/ 14:42 14:7031 29:2565 14:329 13:475 13:475 7:74469

S .J=kg K/ 106:58 107:529 110:572 78:413 75:488 75:488 48:3803

phase V V V V V V xL D 0:212

8 9 10 11 12 13 14

m .kg/ 18:3805 1 17:3805 1:61951 19 19 19

T .ıC/ t8 �161:64 �161:64 �132:964 �159:245 3 28:3226

P .bar/ 1 1 1 1 1 1 1

H .kJ=kg/ 7:74469 �0:008948 8:1908 9:18484 8:27553 13:819 14:7179

S .J=kg K/ 69:4553 �0:079738 73:456 81:3762 74:1988 104:47 107:579

phase xL D 0:0544 L V V V V V





The procedure for solving this problem is outlined in Example 7.23 of the book. The results below wereobtained using data from the NIST Web Book.

1 2 3 4 5 6 7 8 9

m .kg/ 1 14:25 14:25 14:25 14:25 14:25 1 13:25 13:25

P .bar/ 1 1 150 150 150 1 1 1 1

T .ıC/ 25 25 t3 30 �84:1578 �183:09 �183:09 �183:09 25

H .kJ=kg/ 271:01 271:01 h3 242:62 64:6412 64:6412 �133:58 79:6 271:01

S .kJ=kg K/ 6:4106 6:4106 s3 5:0335 4:27301 5:14041 2:9395 5:3065 6:4106

The calculation of intercoolers between compression stages is left as an exercise.






The solution requires trial and error:

1. Set the exit temperature to 200 ıC

2. Pick an exit pressure

3. Calculate the efficiency

4. Repeat the above steps until the efficiency matches the one given in the problem

We demonstrate the calculation of stage 1. The inlet state is

T1 D 25ıC P1 D 1 bar H1 D 271:01 kJ=kh S1 D 6:4106 kJ=kg K

We pick P2 D 2bar . Then,

T2 D 200ıC P2 D 2 bar H2 D 435:26 kJ=kh S2 D 6:6631 kJ=kg K

The corresponding efficiency is found to be 35% (this requires the calculation of the reversible exit atP2 D 2 bar). We try a higher pressure and repeat; after a few trials we find that at P2 D 4 bar the efficiencyis 79.8%. This si close enough, therefore the first stage discharges to 4 bar. The inlet to the second state isat 1 bar, 25 ıC. We then repeat the procedure for the second stage and so on. Since the first stage increasesthe inlet pressure by a factor of 4, we try the same compression ratio as the guess for the second stage. Iturns out this guess is good; we continue the same way. For the last stage we cannot force the efficiency tobe 80% because both the pressure (150 bar) and temperature (200 ıC) are fixed; we will accept any solutionthat gives an efficiency that is less than 80% (why?)

The results are summarized in the table below.

inlet outletstage T .C/ P .bar/ H .kJ=kg/ S .kJ=kg K/ T .C/ P .bar/ H .kJ=kg/ S .kJ=g K/1 25 1 271:01 6:4106 200 4 435:09 6:4826

1 25 4 270:27 6:0484 200 16 434:07 6:1198

1 25 16 267:32 5:6802 200 64 430:22 5:7494

1 25 64 241:29 5:0681 200 150 424:36 5:5113





pressurized line at

P1, T1

tank at

P, T

1

The internal energy in the tank is

U D H1m

M0 C ımC U0

M0

M0 C ım(1)

whereM0 is the mass in the tank initially, U0 is the internal energy in the tank initially,H1 is the enthalpy inthe line, and ım is the amount of steam added. At the end of the process the pressure is P D P1 D 10 bar.We also have

V DM0 C ım

V tank (2)

We now solve for the temperature by trial-and-error as outlined below:

1. Pick a temperature (the pressure is known and equal to 10 bar).

2. Calculate ım from Eq (2).

3. Calculate U from Eq. (1) and compare with the value at the chosen temperature. If they are not equal,pick a new temperature and repeat.

Numerical Substitutions At the initial state we have

P0 D 1 bar; T0 D 150ıC; V0 D 1:9367 kJ=kg; U0 D 2582:9 kJ=kg

The initial mass in the tank is

M0 DV tank

V0D 1:549 kg

We now pick various temperatures at P D 10 bar and perform the calculations outlined above. These are





summarized in the table below:

T .ıC/ U a .kJ=kg/ U b .kJ=kg/ U a=U b ım .kg/200 2622:3 3001:84 0:873566 13:0141

250 2710:5 2995:37 0:904896 11:3431

300 2793:7 2989:25 0:934583 10:0789

350 2875:7 2983:32 0:963927 9:07044

400 2957:8 2977:48 0:993389 8:23571

450 3040:8 2971:72 1:02324 7:53088

500 3124:9 2965:99 1:05358 6:92316

550 3210:4 2960:27 1:08449 6:39379

U a is from steam tables at the indicated temperatureU b is from Eq 1.

The solution is found where U a=U b D 1. From the above table we see that the final temperature is prettyclose to 400 ıC. The amount of steam added to the tank is 8.23 kg.





tank at

P, T

open air at

P0, T0

vented fluid at

P0, T'

The initial state in the tank is

Pi D 10 bar

Ti D 200ıC

Vi D 0:206 m3=kg

Ui D 2622:3 kJ=kg

Hi D 2828:3 kJ=kg

Si D 6:6955 kJ=kg K

The tank contains 24.27 kg of steam. After the tank is vented the pressure is 1 bar. The temperature is definedby the isentropic condition Sf D Si . From steam tables we find that the state is vapor/liquid mixture:

Tf D 99:61ıC; xL D

SV � S2

SV � SLD 0:109524

The mass in the tank is

mf DV tank

xLVL C .1 � xL/VVD 3:31438 =kg

i.e., the amount of steam that was vented is ım D 20:9575 kg.

The enthalpy in the tank is

Hf D �xLHL C .1 � xL/HV D 2427:65 kJ=kg

The temperature at the vent is 99.61 ıC and the steam exits as a vapor/liquid mixture with xL D 0:109524

(why?).







7 VLE of pure Fluid




7. VLE OF PURE FLUID

Problem 7.1 Solution By application of the Clausius-Clapeyron equation,

�Hvap DRT 2

P sat

dP sat

dT

At T D 298:15 K, we find

dP sat

dTD 0:008748 K�1

P satD 0:2017 bar

�Hvap D 411 kJ=kg




Problem 7.2 Solution We pick a pressure at 500 ıC low enough that the state may be considered ideal.Using subscript 2 for this state and 1 for the state given in the problem statement, we have

f1 D P2 exp�G1 �G2

RT

�and

GR1 D RT ln�1 D RT ln.P2=f1/

As state 1 we use P2 D 0:05 bar. The results are summarized below:

1 2

T .K/ 773:15 773:15

P .bar/ 200 0:05

H .kJ=kg/ 3211:8 3489:7

S .kJ=kg K/ 6:0704 10:22

G .kJ=mol/ �26:6675 �79:4141

f .bar/ 183:104 0:05

� .�/ 0:915519 1

GR .kJ=mol/ �567:356Gig .kJ=mol/ �26100:2





Problem 7.3 a) Calculate the fugacity of solid acetylene at its triple point.b) Calculate the change in the chemical potential of the vapor when the pressure is reduced from 1.3 atm to0.1 atm at the constant temperature of �84 ıC.c) State and justify your assumptions.Additional data at the triple point: Ttriple D �84

ıC, Ptriple D 1:3 atm, V solid D 34 cm3=mol, V liq D

42:7 cm3=mol, V vap D 12; 020 cm3=mol.

Solution a) At the triple point the fugacity of all phases is the same:

f solidD f liqu

D f vap� P triple

D 1:3 atm

b) The change of chemical potential in isothermal process is

��12 D

ZPdV D RT ln

P

P triple D �4033 J=mol

(The volumes of the phases are not needed in this problem.)




Problem 7.4 Solution a) At the triple point,

f S D f L D f V

The fugacity of the solid is therefore equal to the fugacity of the vapor, which in turn can be calculated fromthe Pitzer tables. For CO2: Tc D 304:2 K, Pc D 73:8 bar, ! D 0:228. At the given conditions:

Tr D 0:7119 Pr D 0:0701

By interpolation in the � tables:�0 D 0:9368 �1 D 0:9381

so that� D .0:9368/.0:9381/0:228 D 0:9232

The fugacity isf D �P D .0:9232/.5:17/ D 4:77 bar

b) Since ice is incompressible, the Poynting equation applies: At P2 D 70 bar,

f2 D f1 exp�V.P2 � P1/

RT

�where V is the molar volume of dry ice. The molar volume is calculated from the given density:

V DMW�D44 � 10�3 kg/mol1585:25 kg/m3

D 2:776 � 10�5 m3/mol

exp�V.P2 � P1/

RT

�D 0:0999453

Poynting Factor D 1:10511

f2 D f1 � .Poynting Factor/ D .4:77 bar/.1:10511/ D 5:27 bar





Problem 7.5 Solution The properties are calculated as follows:

H D CigP .T � T0/CHR

S D CigP ln

T

T0�R ln

P

P0C SR

G D H � TS

Using the parameters for benzene

Tc D 562:1 K; Pc D 48:9 bar; ! D 0:21; CigP D 82:7 J=mol K

we calculate the data in the tables below:

Tr Pr h0 h1 s0 s11 1: 1: �2:57382 �2:38248 �2:16743 �2:31059

2 0:841754 0:3 �1:13314 �12:7117 �1:22856 �12:3133

3 0:841754 0:920245 �1:8617 �11:095 �1:50131 �11:4189

T .K/ P .bar/ H .J=mol/ S .J=mol K/ G .J=mol/1 562:1 48:9 �14366:4 �22:0541 �1969:74

2 473:15 14:3 �29826:3 �51:2669 �5569:34

3 473:15 45 �26942:2 �45:9688 �5192:13




Problem 7.6 Solution [DUPLICATE PROBLEM WITH 7.5 IN US EDITION]





Problem 7.7 Solution a) At 40 ıC (313.15 K) we find

Pr D 0:0205; Tr D 1:026

From the tables (E.13 and E.14) we find

�0 D 0:9941; �1 D 0:9994

(We have used the entries at Tr D 1:02 because �0 and �1 do not vary much between Tr D 1:02 andTr D 1:05. For Pr we interpolated between the listed values. Pr D 0:01 and Pr D 0:05.) Therefore,

� D .0:9941/.0:9994/0:1 D 0:9940

The fugacity isf D �P D .0:9940/.1/ D 0:994 bar

b) At 50 bar we have Tr D 0:6654, P r D 1:026 � 1. By interpolation in tables we find

�0 D 0:0745; �1 D 0:0715 ) � D .0:0745/.0:0715/0:1 D 0:0572

andf D .0:0572/.50/ D 2:86 bar

Alternatively, we may use the Poynting equation:

f D f sat exp��P � P sat

RTV L

�D �satP sat exp

��P � P sat

RTV

�

The saturation pressure at �70 ıCis 2.58 bar, corresponding to Pr D 0:0515 � 0:05. By interpolation atTr D 0:6654, Pr D 0:1 we find

�0 D 0:9453; �1 D 0:9378; ) �satD .0:9453/.0:9378/0:1 D 0:9392

The molar volume of the liquid at Tr D 0:6654 can be estimated from the Rackett equation:

V L D VcZ.1�Tr /

0:28852

c

With Vc D 0:279, Vc D 0:0001455 m3/mol, we find VL D 0:000128785 m3/mol. Therefore, the Poyntingfactor is

Poynting D exp.50 � 2:58/ � 105 � 0:000128785

8:314 � 203:15D 1:44

Therefore, the fugacity isf D .0:9392/.2:58/.1:44/ D 3:38 bar

and� D

3:38 bar503:38 bar

D 0:06766

The two methods do not agree very well. Generalized graphs can usually be trusted in the vapor region butare not very accurate in the liquid region.

c) We use the Pitzer correlation for �. The fugacity is then calculated as f D �P . At -58.8 the reducedtemperature is 0.7. Using tabulated data at Tr D 0:7 we obtain the table below:




Pr �0 �1 P (bar) � f (bar)0.010 0.9908 0.9908 0.49 0.9899 0.4820.050 0.9528 0.9528 2.44 0.9482 2.3100.100 0.9057 0.9036 4.87 0.8966 4.3680.200 0.4560 0.1182 9.74 0.3683 3.5890.400 0.2360 0.1163 19.49 0.1903 3.7090.600 0.1626 0.1147 29.23 0.1309 3.8280.800 0.1262 0.1131 38.98 0.1015 3.9561.000 0.1045 0.1116 48.72 0.0839 4.089





Problem 7.8 Solution From the data given in the problem we calculate the compressibility factor of thefluid in the tank:

Z DPV tot

nRTD 0:9164

This value is close to the ideal gas state. Assuming the validity of the truncated virial equation,

ln� D Z � 1 D �0:08358 ) � D 0:91981





a) We write the equation of state in volume-explicit form:

P DRT

VC

a

V 2)

�@P

@T

�V

DR

V

We calculate the residual enthalpy and residual entropy as follows:

HRD PV �RT C

Z V

1

�T

�@P

@T

�V

� P

�dV D �

2a

V

SR D lnPV

RTC

Z V

1

��@P

@T

�V

�P

V

�dV D R log

�1 �

a

RT V

�The fugacity coefficient is

ln� DGR

RTDHR � TSR

RT) ln� D �

2a

RT V� log

�1 �

a

RT V

�b) From the given information we find

V DZRT

PD 0:0221794 m3=mol

We then havea D �RTV.�1CZ/ D 6:70809 Jm3=mol2

The fugacity coefficient isln� D �0:112167 ) � D 0:893895

c)GR D RT ln� D �282:704 J=mol






90 K1.321

50

AB V

C

P (atm)

Volume

a) State A is superheated vapor. We calculate the compressibility factor based on the following information:

V D1

0:0406l/mol; P D 1 atm; T D 300 K ) ZA D 1:00035

This is ideal, therefore we may assume � � 1 ) f � 1 atm.

b) State B is saturated liquid. By the equilibrium criterion (see PV graph):

�B D �V

We calculate the compressibility factor at state B based on the following information:

V D1

0:1864l/mol; P D 1:321 atm; T D 90 K ) ZB D 0:959

This value is still close the ideal-gas state; the proper approximation here that for “moderate pressures”according to which

ln� � ZB � 1 D �0:0406 ) �B � 0:960; fB D .0:960/.1:321/ D 1:27 atm

If we assume ideal-gas state, the answer is

� � 1 ) �B � 1; fB D .1/.1:321/ D 1:321 atm

about 4% higher.

c) The gibbs energy of saturated liquid is

HLD .3099 J/mol/ � .90 K/.56:02 J/mol K/ D �1942:8 J/mol

In the previous part we found that �L D 0:96, from which we obtain the residual Gibbs energy

GR D RT ln�L D .8:314/.90/ ln.0:96/ D �30:356 J/mol

Therefore,GigD GL �GR D 1912:44 J/mol

If we use �L � 1, thenGig� GL D �1942:8 J/mol




d) State C is compressed liquid. Using the Poynting equation

fC D fB � (Poynting factor)

We calculate the Poynting factor based on the following data:

V D1

34:47l/mol; P sat

D 1:321 atm; P D 50 atm; T D 90 K ) (Poynting factor) D e0:189 D 1:21

For the fugacity we finally obtain:

fC D .1:27/.1:21/ D 1:54 atm

e) The PV is shown on the previous page. You should draw this graph before you eve begin the calculations.





Problem 7.11 Solution Calculations inside boxes represent alternatives ways of answering the problem.These are based on more detailed calculations but in all cases, the final answer does not improve by much.Therefore, the quicker but more approximate solutions are acceptable in this particular problem.

a) At 1 bar, 300 K bromine is compressed liquid (P sat D 0:31 bar).

f D �satP sat.Poynting/

Because of the proximity of the pressure to P sat, we may neglect the Poynting factor. Moreover, �sat � 1,because at 300 K, 0.031 bar the vapor may be assumed ideal (why?). Therefore,

f D 0:31 bar

For a more accurate estimation, use ln�sat � ZV � 1, where ZV is the compressibility factor of the vapor atsaturation, and calculate the Poynting factor. We find:

ZV D 0:993511

�satD 0:993532

Poynting D 1:00143

f D 0:3084 bar

The difference is about 0.5%, therefore, the quicker estimation is justified.

b) For the saturated liquid,f sat;L

D f sat;V and �sat;LD �sat;V

At 1 bar, Pr D 1=103:4 D 0:00967, T r � 0:5 therefore, it is reasonable to assume ideal-gas state. Then,

�sat;V� 1 and f sat;L

� 1 bar

For a more accurate estimation, we may use ln� � Z � 1 where Z is the compressibility factor of saturatedvapor at 1 bar. From the given table we have:

T D 320 K P sat D 0:680 bar vv D 0:2425 m3/kg Z D 0:990511

T D 340 K P sat D 1:330 bar vv D 0:1309 m3/kg Z D 0:98424

By interpolation at P D 1 bar we find,Z D 0:987423

Then,�sat;V

D e0:987423�1 D 0:9875

sot that finally:� D 0:9875 f D 0:9875 bar

This improves the answer by 1.25%.

Note: Here we opted to interpolate the for the value of Z. If instead we interpolate for V at 1 bar and use theinterpolated value (v D 0:187558 m3/kg) to calculate the compressibility factor, we find Z D 1:09298, whichdiffers by more than 10% from the value obtained above. This is because the volume is a non-linear function ofP . As we can see, between 0.680 bar and 1 bar, the volume changes by almost a factor of 2 while Z changes byabout 1%. Therefore, linear interpolation in Z is far more accurate.




c) The fugacity coefficient of saturated liquid bromine is equal to the fugacity of saturated vapor at the sametemperature. At 300 K, 0.031 bar the vapor may be assumed ideal (see above) and thus we have

� � 1

At 150 bar the state is compressed liquid and for the fugacity we have

f D �satP sat exp

"V L.P � P sat/

RT

#

where

�satD 1

P satD 31000 Pa

P D 150 � 105 Pa

V L D 5:17 � 10�5 m3/mol

We findPoynting D 1:364

f D 0:423 bar

� D0:423

150D 2:82 � 10�3

Here again we made the approximation �sat � 1. If we use the more accurate value, �sat D 0:993532, thefugacity and fugacity coefficient are

f D 0:4199 bar; � D 2:799 � 10�3

Again, the difference is less than 1%.

The various states in this problem are shown in the figure below

300 K

329.8 K

0.031 bar

1 bar

150 bar

Pre

ssu

re

Volume

φ≈1

φ=2.82 10-3

φ≈1





Problem 7.12 Solution a) Yes, because Z D 0:9829.

b) f D 8:89554 bar.

c) f D 10:6234 bar, � D 0:177056 ) GR D �2159:09 J/mol.

d) G D �7353:83 J/mol, GR D �2159:09 J/mol ) Gig D �5194:74 J/mol.





Method 1 We may perform the calculation using

lnf

f0DG �G0

RT

where 0 refers to a pressure P0 that is low enough that we may set f0 � P0. At the lowest pressure of thetables we find

P0 D 1:013 bar Z D 0:995845

We consider this close enough to ideal gas state. With P0 D 1:013 bar, we calculate the following valuesfor f , �:

P (bar) G (kJ/kg) f (bar) �

1:013 �2860:4 1:013 1:

5 �2640:5 4:95243 0:990486

10 �2535:6 10:5583 1:05583

20 �2451:5 19:3723 0:968613

40 �2359:1 37:7381 0:943452

60 �2306:3 55:2414 0:920691

80 �2269:3 72:149 0:901862

100 �2241:6 88:1145 0:881145

150 �2191:7 126:312 0:842078

200 �2202:2 117:094 0:58547

300 �2057:8 331:98 1:1066

400 �2067:3 309:982 0:774956

500 �1683:1 4960:13 9:92026

Method 2 Alternatively, we may use the relationship between � and Z:

ln� DZ P

0

Z � 1

PdP .at const. T / Œ2�

Method 3 We recall that an approximation for the above integral is provided by the truncated virial equa-tion:

ln� � Z � 1 Œ3�

The approximation is valid only over the range of pressures that Z is a linear function of P . Although theproblem asks for � and f at 300 K, 100 bar, we will obtain an equation to calculate these properties at anypressure along the 300 K isotherm.

Calculations for Method 2 The numerical procedure is based on the following steps:

1. calculate and plot the compressibility factor at 300 K as a function of pressure

2. fit an appropriate function to Z as a function of P

3. use the fitted equation for Z to calculate the integral in Eq. [2]

For comparison, we will also calculate � from Eq [2].





The calculated values of Z are shown below.

P (bar) v (m3/kg) Z

1:013 1:3622 0:995845

5 0:2747 0:99122

10 0:1365 0:985085

20 0:0674 0:972817

40 0:0329 0:949723

60 0:0215 0:93096

80 0:0158 0:912196

100 0:0124 0:894876

150 0:00805 0:871422

200 0:00605 0:873226

300 0:00433 0:937455

400 0:00361 1:0421

500 0:00322 1:1619

The graph below shows Z as a function of P . Points are calculated values from Perry’s data; the solid lineis a fitted cubic polynomial and is shown below:

Z D �3:738945 � 10�9P 3 C 5:7037123 � 10�6P 2 � 0:00158915P C 1

in which P must be in bar.

0 100 200 300 400 500P !bar"

0.2

0.4

0.6

0.8

1.0

1.2Z

Applying the fitted equation for Z to Eq [2] we obtain the following result for the fugacity coefficient:

ln� D �1:24631 � 10�9P 3 C 2:85186 � 10�6P 2 � 0:00158915P

Finally, the fugacity is obtained as f D �P . The two graphs below show the calculated values of � and fas a function of P .




0 100 200 300 400 500P !bar"

0.2

0.4

0.6

0.8

1.0

1.2!

0 100 200 300 400 500P !bar"

100

200

300

400

500f !bar"

The solid lines are calculated by the procedure outlined above. The dashed line is calculated using Eq. [2].We see that the approximate equation provides a good estimate for � up to about 100 bar.

Answers at 300 K, 100 bar

Method 1

� D 0:8811; f D 88:1 bar

Method 2

� D 0:8767; f D 87:7 bar

Method 3

� D 0:9002; f D 90:0 bar

Comments

1. If you used Method 1 or method 2, you will receive full credit

2. If you used Method 3 and you plotted Z versus P to confirm that the virial approximation is ok, youwill receive full credit

3. If you used Method 3 but did not justify the virial approximation, you will receive half credit.






[Note 1 cu ft/lb = 6:2428 � 10�2 m3/kg]Before we proceed we need the molecular weight of compound X. We can reasonably assume that at 160 ıF,10 psi, the compound is in the ideal gas. At these conditions X is indeed a gas (why?) and Pr < 10=500 D0:02. That is, the reduced pressure is at most 0.02 and, as we can verify from a generalized Z graph, at thisreduced pressure a gas is essentially in the ideal-gas state. Therefore, the molar volume of compound X atthese conditions is

V DRT

P

With T D 160 ıF = 344.26 K, P D 10 psi = 6:8045�104 Pa we find V D 0:04206m3/mol. From the tablewe find that the specific volume V 0 is 11.291 cu ft/lb = 0.7049 m3/kg. The molecular weight is, therefore,

MW DV

V 0D0:04206 m3/mol0:7049 m3/kg

D 0:0597 kg/mol

a) Since we have volumetric data for this substance, we will use the compressibility factor to find the fugacitycoefficient:

ln� DZ P

0

.Z � 1/dP

PD

Z P

0

.Z � 1/d lnP

From the given data we calculate the compressibility factor, Z. We then make a table of Z � 1 as a functionof lnP and integrate from the lowest pressure in the table to the desired pressure (161.1 psi = 1096217 Pa)at constant temperature. The table below summarizes these steps:

P (Pa) V (m3/mol) Z lnP Z � 1 integral68046 4.206E-02 1.0000 11.1279 0.0000 0.0000

100000 2.847E-02 0.9945 11.5129 -0.0055 -0.0011136091 2.078E-02 0.9880 11.8211 -0.0120 -0.0037204137 1.367E-02 0.9748 12.2265 -0.0252 -0.0113272183 1.011E-02 0.9618 12.5142 -0.0382 -0.0204340229 7.972E-03 0.9477 12.7374 -0.0523 -0.0305408274 6.547E-03 0.9339 12.9197 -0.0661 -0.0413544366 4.761E-03 0.9056 13.2074 -0.0944 -0.0644680457 3.684E-03 0.8757 13.4305 -0.1243 -0.0888850572 2.816E-03 0.8367 13.6537 -0.1633 -0.1208

1020686 2.229E-03 0.7948 13.8360 -0.2052 -0.15441096217 2.025E-03 0.7757 13.9074 -0.2243 -0.1698

The last column shows the calculation of the integral using Simpson’s rule, as shown below:

[integral]i D [integral]i�1 C1

2.xi � xi�1/.fi C fi�1/

where [integral]i is the value of the integral of the function f .x/ from 0 to xi . In our case xi D lnPi andfi D Zi � 1. From the above table we find

ln� D �0:1698 ) � D 0:844

The fugacity isf D �P D .0:844/.1096217/ Pa D 9:25 bar D 136 psi




b) At 160 ıF, 200 psi the system is compressed liquid. The fugacity here will be calculated using thePoynting factor. From the table we find V L D 0:0329 cu ft/lb = 2.025E-03 m3/kg. With P D 200 psi =13.61 bar, P sat D 10:96 bar, T D 160 ıF = 344 K, the Poynting factor is

Poynting D exp.P � P sat/V L

RTD 1:0114

As we might have expected, the Poynting factor is nearly 1. The fugacity of the compressed liquid is

f D �satP sat(Poynting) D 9:36 bar D 138 psi

Notice that �sat was calculated in the previous part.

The graphs below show the compressibility factor Z plotted against pressure (on the left), and a graph ofZ � 1 against lnP , showing the area that is calculated by numerical integration (on the right).





Problem 7.15 Solution a) At 0.1 bar, Pr D 0:1=36:48 D 0:0027. This is sufficiently low that we mayassume ideal gas state:

� D 1 f D 0:1 bar

b) We first calculate the fugacity of the saturated vapor using

ln�V D ZV � 1

with

Z DPV V

RTD 0:882375

we find�V D 1 � e0:882375 D 0:889; f V D .0:889/.5 bar/ D 4:445 bar

At saturation, the fugacity and fugacity coefficient of the vapor and the liquid are the same. Therefore:

�L D 0:889 f L D 4:445 bar

c) We need the Poynting factor:

Poynting D exp

"V L.P � P sat/

RT

#D 1:1721

Therefore:f D .4:445 bar/.1:1721/ D 5:21 bar; � D

5:21 bar40 bar

D 0:130

d) Assumptions:

1. Ideal gas state for vapor at 0.5 bar, 310.5 K. This is justified because the pressure is very low (Pr D5 � 10�8).

2. Virial equation valid at 5 bar, 310.5 K. We have no direct way of proving this - we need to check witha Z � P graph.




Problem 7.16

Solution a) We seek by trial-and error a pressure such that the fugacity coefficient of the liquid (calculatedusing ZL) is equal to the fugacity coefficient of the vapor (calculated using ZV ). The iterations are shownbelow:

P (bar) �L �V �V =�L

1 3:68271 0:981010 0:26638

2 1:84782 0:962070 0:52065

3 1:23621 0:943166 0:76295

4 0:93040 0:924284 0:99341

5 0:74693 0:905404 1:21216

6 0:62463 0:886507 1:41925

7 0:53727 0:867564 1:61475

8 0:47176 0:848542 1:79867

9 0:42081 0:829399 1:97096

10 0:38005 0:810075 2:13147

The saturation pressure is between 4 bar and 5 bar because the ration �V =�L goes from being below 1 toabove 1. By further refinement we find

P satD 4:0293 bar

The results at P D 4:0293 bar are summarized below:

RESULTS for ethane: T = 214.65K P=4.0293 bar

a = 6.8614µ10-1

J m3êmol2

b = 4.5079µ10-5

m3êmol

dadt= -1.5379µ10-3

J m3êmol2ês

A = 8.6808µ10-2

-

B = 1.0178µ10-2

-

Real Roots 1.4111µ10-2

6.8233µ10-2

9.1766µ10-1

TWO possible phases - you MUST pick ONE!

ZL= 0.0141107 ZV= 0.917656 -

VL= 0.0000624973 VV= 0.00406435 m3êmol

HRL= -14002.4 HRV= -395.616 Jêmol

SRL= -64.5743 SRV= -1.18348 Jêmol K

phiL= 0.923737 phiV= 0.923731 -

phiVêphiL= 0.999993

Comments: The following graphs show the isotherms. The first graph shows the raw isotherm with valuesof � marked on selected pressures. P sat corresponds to the pressure that gives equal fugacity coefficients onthe liquid and vapor sides.





20

15

10

5

0

Pre

ssure

(bar)

10-4

10-3

10-2

Volume (m3/mol)

f = 0.962

f = 0.887f = 0.625

f = 0.380

f = 0.472

f = 0.380

f = 0.924

f = 1.848

f = 0.849

f = 0.9244.029 bar

50

40

30

20

10

0

Pre

ssure

(bar)

10-4

10-3

10-2

Volume (m3/mol)

Psat

= 4.029 bar

L V

Tc

Once the phase boundary is known, the isotherms are drawn easily: Vary the volume from very small(V � b) to V L and from V V to very large, and plot the corresponding pressure against volume. Noticethat the calculation is not done for V L < V < V V . Instead, the isotherm inside the two-phase region isrepresented by a horizontal straight line. Notice the gap in the two-phase boundary near the critical point.This is because it becomes difficult to obtain a numerical solution too close to Tc . In this example thecalculation was carried out up to Tr D 0:986.





The real question here is to determine the phase of the system based on the given information. Recall thatthe stable phase is the one with the lowest fugacity, or equivalently, with the lowest fugacity coefficient. Wecalculate the fugacity coefficient as

� D exp

"GR

RT

#D exp

"HR � TSR

RT

#

We find:1 2 3

Z 0:0332418 0:342355 0:624404

HR (J/mol) �34046:0 �6020:41 �3384:35

SR (J/mol K) �71:4816 �13:4281 �6:42023

� 0:124201 � 0:748424

We conclude that the phase is liquid, i.e., Z D 0:0332418. The required volume is

V tankD nZ

RT

PD 0:01589 m3






The solution procedure is outlined in the problem statement. The results of this calculation are summarizedin the figure below.

ΦV= Φ

L= 0.7203

Psat= 29.9 bar

1´10-4 5´10-40.001 0.005 0.010 0.050 0.1000

10

20

30

40

50

60

volume Hm3�molL

pre

ssu

reHb

arL




Problem 7.19 Solution a) Saturation properties at 350 K:

T D 350 K P satD 29:878 bar

ZL D 0:141242 ZV D 0:588493

VL D 0:0001376 m3=mol VV D 0:000573148 m3=mol

HLR D �12462:7 J=mol HV

R D �3951:77 J=mol

SLR D �32:8804 J=mol K SRV D �8:56358 J=mol K

�H vapD HV

R �HLR D 8510:88 J=mol

�SvapD SVR � S

LR D 24:3168 J=mol K

Notice that these values satisfy the relationship

�H vapD T�Svap

b) The calculation of the phase boundary is done as described in the problem statement. The calculationsare summarized below:

T .K/ P sat .bar/ ZL ZV HVR .J=mol/ HL

R .J=mol/ SRL .J=mol K/ SRV .J=mol K/170 0:0212918 0:0001089 0:998558 �21408:7 �5:06709 �125:922 �0:017823

190 0:10269 0:000483851 0:99472 �20702:9 �21:1621 �108:919 �0:0675891

210 0:353902 0:00156011 0:985809 �19984:1 �64:2182 �95:0453 �0:188571

220 0:59747 0:00256155 0:978641 �19615:8 �102:377 �88:987 �0:289471

230 0:95841 0:00401053 0:96927 �19239:2 �155:741 �83:3967 �0:425151

240 1:47133 0:00603094 0:957439 �18852: �227:727 �78:2027 �0:601717

250 2:17469 0:00876374 0:942925 �18451:8 �322:016 �73:3447 �0:825564

260 3:11024 0:0123697 0:925525 �18035:9 �442:605 �68:77 �1:10359

270 4:32219 0:0170346 0:905054 �17600:2 �593:891 �64:4297 �1:4435

280 5:8574 0:0229775 0:88132 �17141:3 �780:923 �60:2845 �1:85445

290 7:76422 0:0304652 0:854106 �16653:8 �1009:62 �56:2933 �2:34766

300 10:0925 0:0398346 0:823142 �16131:1 �1287:29 �52:4172 �2:93772

310 12:8934 0:0515319 0:788058 �15564:2 �1623:38 �48:6149 �3:64446

320 16:2191 0:0661823 0:748309 �14940:6 �2030:89 �44:8394 �4:49651

330 20:1223 0:0847279 0:703035 �14241:3 �2529:03 �41:0299 �5:53806

340 24:6571 0:108736 0:65075 �13435:4 �3149:3 �37:0974 �6:84427

350 29:878 0:141242 0:588493 �12462:7 �3951:77 �32:8804 �8:56358

360 35:8408 0:190076 0:508514 �11170:7 �5089:4 �27:9782 �11:0858





Problem 7.20 Solution The table below demonstrates a calculation at T D 350 K, P D 10 bar. Inaddition to the properties of interest (Z, V ,H , S ), the table shows results for various intermediate quantitiessuch as a, b, da=dT , etc. Also shown are the properties of the saturated phases at 350 K. The enthalpy andentropy are calculated as follows

H D �H igCHR

�HR0

S D �S igC SR � SR0

with HR0 D �19203:9 J=mol, SR0 D �82:82 J=mol K and

CigP D 31:984C 0:0426591T C 0:000499755T

2� 6:56224 � 10�7T 3 C 2:55988 � 10�10T 4

with T in K and C igP in J/mol K. The complete phase diagram is obtained by repeating this calculation.

Tc (K) 369:83

Pc (bar) 42:48

! 0:152

T (K) = 231:15 Psat (bar) = 1:01

P (bar) = 10 phase = LIQUID

(sat Liq) (sat Vap)Z 0:0416 0:0042 0:9681 -V 8: 8:021 � 10�5 1:846 � 10�2 m3=molH 36:90 0:00 19041:02 J/molS �0:15 0:00 82:38 J/mol K

HR �19166:97 �19203:87 �162:85 J/molSR �63:89 �82:82 �0:44 J/mol K

a 1:258 Jm3=mol2

da=dT �2:678 � 10�3 Jm3=mol2Kb 6:272 � 10�5 m3=molA 3:406 � 10�1 -B 3:263 � 10�2 -




Problem 7.21 Solution Toluene is liquid because at T D 383:3 K, P D 8 bar the pressure is higher thanthe saturation pressure (1.013 bar). The acceptable value of Z is the smallest root, that is:

Z D 0:0332418 HRD �34046:0 J/mol SR D �71:4816 J/mol K

a) The enthalpy is

H.P; T / D

TZT0

CigP dT CH

RD HR

D �34046:0 J/mol

Notice that since the reference temperature is the same as the final temperature, the temperature integral iszero. For the entropy,

S.P; T / D

TZT0

CigP

TdT �R ln

P

P0C SR D �R ln

P

P0C SR D �58:1007 J/mol K

The Gibbs energy isG D H � TS D �11775:9 J/mol K

b) The fugacity coefficient is� D eG

R=RT

whereGR D HR

� TSR D .�34045:95/ � .383:3/.�71:4816/ D �6647:06 J/mol

It follows that

� D exp�

.�6647:06/

.383:3/.8:314/

�D 0:124203

f D .0:124203/.8/ D 0:993 bar

c) To calculate the fugacity at 40 bar, 383.3 K, we start at 8 bar, 383.3 K and apply the poynting factor:

f2 D f1 exp�V.P2 � P1/

RT

�where P2 D 40 bar, P1 D 8 bar, T D 383:3 K and V is the molar volume at 383.3 K, 8 bar, which iscalculated from the known compressibility factor:

V D ZRT=P1 D 0:000132417 m3/mol

Notice that the algebra can be simplified if we use V D ZRT=P1 to write

exp�V.P2 � P1/

RT

�D exp

�ZP2 � P1

P1

�D 1:142

Then,f2 D .0:993 bar/.1:142/ D 1:135 bar

and�2 D

1:135 bar40 bar

D 0:028







8 Phase Behavior of Mixtures




8. PHASE BEHAVIOR OF MIXTURES

Problem 8.1 Solution a) At 70.82 ıC, the system is below the bubble point, hence the phase is all liquidwith 30% methanol (see point A on T xy graph.)

b) The maximum concentration of methanol corresponds to point V on the graph, namely, it is the com-position of the vapor at the bubble temperature of a 30% solution. The bubble temperature at x1 D 0:3 is73.19ıC. We need to interpolate in the dew line at T D 731:19ıC:

y1 D 0:4 Tdew D 73:49ıC

y1 D 0:5 Tdew D 72:20ıC

At T D 73:19ıC:y1 D

0:4 � 0:5

73:49 � 72:20.73:19 � 72:20/C 0:5 D 0:423

The maximum amount of ethanol corresponds to pointL0 on the attached graph. This corresponds to a liquidwhose bubble point is 74.7ıC. From the data, at x1 D 0:2 the bubble temperature is 74.72ıC. That’s prettyclose to 70.7ıCso no interpolation is required. The corresponding mol fraction ethanol is x2 D 1�0:2 D 0:8.

c) See graph below.

80

79

78

77

76

75

74

73

72

71

70

69

68

67

66

65

64

63

62

61

60

Tem

pera

ture

(C)

1.00.90.80.70.60.50.40.30.20.10.0

x1, y1

Methanol(1)/Ethanol(2) at 1 bar

A

VL

L'

V'




Problem 8.2 Solution a and b) In the i th separator we have

Vi

LiDzi � xi

yi � ziD 0:5 Œ1��

where zi D yi�1 and Fi D Vi�1. First we fit polynomial expressions for the liquid and vapor compositionas a function of temperature:

x1 D 78:0093 � 17:2325T C 4:08657T2� 0:681239T 3

y1 D 78:0215 � 10:5164T � 1:1627T2� 2:14534T 3

Next we solve Eq. [1] for T and move to the next separator. The results are summarized below:

i T z x y F L V

1 75:1921 0:2 0:170651 0:258698 1 0:666667 0:333333

2 74:361 0:258698 0:22355 0:328994 0:333333 0:222222 0:111111

3 73:3663 0:328994 0:288635 0:409713 0:111111 0:0740741 0:037037

4 72:2271 0:409713 0:365696 0:497747 0:037037 0:0246914 0:0123457

5 70:9908 0:497747 0:452549 0:588143 0:0123457 0:00823045 0:00411523

6 69:7309 0:588143 0:544723 0:674983 0:00411523 0:00274348 0:00137174

7 68:5324 0:674983 0:636043 0:752862 0:00137174 0:000914495 0:000457247

8 67:4694 0:752862 0:720168 0:81825 0:000457247 0:000304832 0:000152416

9 66:5866 0:81825 0:792364 0:870023 0:000152416 0:000101611 0:0000508053

10 65:8946 0:870023 0:850502 0:909066 0:0000508053 0:0000338702 0:0000169351

The process requires 10 flash separators and produces a product stream with concentration 0:909066 andmolar flow rate of 0:0000169351 mol on a basis of 1 mol if feed stream. The recovery of methanol is

recovery D.0:909066/.0:0000169351/

.0:2/.1/D 0:0077%

c)

i T x y F L V V=L

1 75:3071 0:2 0:163431 0:248759 1 0:571429 0:428571

2 74:6367 0:248759 0:205854 0:305965 0:428571 0:244898 0:183673

3 73:8459 0:305965 0:257003 0:371247 0:183673 0:104956 0:0787172

4 72:939 0:371247 0:317221 0:443282 0:0787172 0:0449813 0:0337359

5 71:9338 0:443282 0:385991 0:51967 0:0337359 0:0192777 0:0144583

6 70:8645 0:51967 0:461614 0:59708 0:0144583 0:00826186 0:0061964

7 69:7795 0:59708 0:541094 0:671727 0:0061964 0:0035408 0:0026556

8 68:7337 0:671727 0:62045 0:740096 0:0026556 0:00151749 0:00113811

9 67:7778 0:740096 0:695448 0:799626 0:00113811 0:000650351 0:000487763

10 66:9482 0:799626 0:76253 0:849088 0:000487763 0:000278722 0:000209041

11 66:2615 0:849088 0:819505 0:888531 0:000209041 0:000119452 0:0000895891

12 65:7159 0:888531 0:865733 0:918929 0:0000895891 0:0000511938 0:0000383953

This process requires 12 separators and produces a product stream of 0.0000383953 mol on a basis of 1 molin the feed. The recovery of methanol is

recovery D.0:918929/.0:0000383953/

.0:2/.1/D 0:017%





d) With a higher value for L=V we receive more product stream and recover more methanol, but it takesmore separators.

Comment: This design is very impractical because it does not recycle the liquid streams.





a) 0.280 bar

b) vapor

c) PB D 0:22 bar, PD D 0:125 bar (by interpolation).

0.30

0.25

0.20

0.15

0.10

0.05

P (

bar)

1.00.80.60.40.20.0x1 or y1

1,4 dioxane (1)/methanol(2) at 308.5 K





Problem 8.4 a) 42.9 ıC.

b) Maximum boiling azeotrope.

c) y1 D x1 D 0:52

d) Vapor

e) The initial composition is x1 D 0:652, or x2 D 0:348. To produce x02 D 0:5 we must add

ın Dx2 � x

02

1 � x02D 0:304 moles of chloroform





0.50

0.45

0.40

0.35

0.30

0.25

0.20

Pre

ssu

re (

To

rr)

1.00.80.60.40.20.0

mole fraction of methanol

methanol (1) / CCl4 (2)

35 °C

A

B

C D

E

F

We start at point A (pure tetrachloride), and end up at F , almost pure methanol. At each point the molfraction of methanol is

xMeOH Dn

1C n

where n is the moles of methanol that have bee added. Solving for n,

n DxMeOH

1 � xMeOH

We observe the following transitions:

� A: liquid

� At xB � 0:05 (n D 0:0526) the liquid boils

� BC vapor liquid mixture

� At xC � 0:35 (n D 0:538) the vapor condenses and the phase is liquid

� CD liquid

� At xD � 0:75 (n D 3) the liquid begins to boil again

� DE vapor/liquid mixture

� At xE � 0:95 (n D 19) boiling stops and the vapor fully condenses.

� At xF D 0:99 (n D 99) the phase is liquid again.





Problem 8.6 Solution We first fit y and x into polynomial expressions:

x D 14:425 � 0:31757T C 0:0027789T 2 � 0:000011277T 3 C 0:00000001757 � T 4

y D 0:90716C 0:005:5548 � 103T � 9:1222 � 105T 2 C 8:1846 � 107T 3 � 3:7472 � 109T 4

To check how good this fit is, we plot the fitted and given values together. As we see, the fit is very good.

Numerical procedure Let the feed of the i th separator be Fi with heptane mole fraction zi , and with outletstreams Vi , Li , whose composition (mol fraction of heptane) are yi , xi , respectively. By lever rule:

Vi Dzi � xi

yi � xiFi ; Li D

yi � zi

yi � xiFi Œ1�

where the feed is the vapor stream of the previous separator, i.e.

zi D yi�1; Fi D Vi�1

Notice that Fi is not 1 except for the first separator. Dividing Vi by Li we havezi � xi

xi � ziDVi

LiD 3

Since both x and i are functions of temperature,in this equation the only unknown is T : this temperature isfound by trial and error. The solution moves sequentially from on separator to the next until yi � 0:5. Theresults are summarized in the table below:

180

170

160

150

140

130

120

110

100

90

Tem

pera

ture

(C

)

1.00.90.80.70.60.50.40.30.20.10.0

Composition (x1, y1)

nC7(1)/nC10(2) at 1.013 bar

Data

fitted equations

i Ti (ıC) zi xi yi Fi Vi Li Vi=Li

1 153:4 0:400 0:124 0:492 1:000 0:750 0:250 3:00

2 147:2 0:492 0:172 0:599 0:750 0:562 0:188 3:00

3 139:0 0:599 0:247 0:716 0:562 0:422 0:141 3:00

4 128:2 0:716 0:369 0:832 0:422 0:316 0:105 3:00

5 116:0 0:832 0:559 0:923 0:316 0:237 0:079 3:00

6 106:0 0:923 0:774 0:973 0:237 0:178 0:059 3:00




In the first separator, F D 1, z D 0:4. In all other cases, the feed is the same as the vapor stream of the lastseparator. We find that we need 6 stages and that the last vapor stream has a purity f 97.3%.

recovery of heptane Dy6V6

z1F1D.0:973/.0:178/

.0:4/.1/D 43:3%

Less than 50% of the heptane is captured in the purified stream.





Problem 8.7 Solution The process is shown below.

1 2

z1, F1

y1, V1

x1, L1

y2, V2

x2, L2

We count the number of variables. Since z1 is given, we have:

Separator 1: F1; V1; L1; y1; x1; T1

Separator 2: V2; L2; y2; x2; T2

There are 11 unknowns. We now count the equations:

z1F1 C x2L2 D x1L1 C y1V1 Œ1�

F1 C L2 D L1 C V1 Œ2�

V1y1 D x2L2 C y2V2 Œ3�

V1 D L2 C V2 Œ4�

x1 D xŒT1� Œ5�

y1 D yŒT1� Œ6�

x2 D xŒT2� Œ7�

y2 D yŒT2� Œ8�

The first four equations represent material balances in the two separators; the last four express the fact thatx and y in each stage are functions of the temperature in that stage. Since we have 11 unknowns and 8equations, we have 3 degrees of freedom, i.e., we need to specify three independent variables.

Design Specifications First we pick the basis for the calculation to be F1 D 1 mol. Next, we choose tospecify the temperatures in the two separators. We note, though, that these temperatures cannot be arbitrary.The temperature of separator 1 must be between the dew temperature (158ıC) and the bubble temperature(126ıC) of the feed stream. We choose T1 D 150ıC. This immediately fixes the composition of streams V1and L1. We find

x1 D 0:15I y1 D 0:553

The temperature of the second separator must now be between the dew temperature (150ıC) and the bubbletemperature (116.3ıC) of stream V1. We choose T2 D 135ıC. This now fixes the compositions of streamsV2 and L2. We find:

y2 D 0:763621I x2 D 0:28874




The remaining unknowns, V1, L1, V2, L2, are obtained from Eqs [1–4]. The results are summarized below:

T .ıC/ V L y x

Sep 1 150 0:732216 0:592281 0:553168 0:149687

Sep 2 135 0:407719 0:324497 0:763621 0:28874

The properties of the product streams are summarized below:

component molar rate purity recovery (%)heptane 0:4077 0:7636 77:8

decane 0:5923 0:1497 83:9

The states of the various streams are shown in the attached T xy.

180

170

160

150

140

130

120

110

100

90

Tem

pera

ture

(C

)

1.00.90.80.70.60.50.40.30.20.10.0

Composition (x1, y1)

nC7(1)/nC10(2) at 1.013 bar

T1=150 C

T2=135 C

F1

L1

V1

V2

L2

Feed

ProductStream

ProductStream

Question:

A junior member of our design team suggested that it should be possible to run separator 1 attemperatures above the dew point of the feed stream (158 ıC). Do you agree with her opinion?






Note 1: In this problem we use the notation xi , yi , to indicate the liquid and vapor mol fraction ofmethanol in separator i (i D 1; 1). This is different from the usual notation where subscript i refersto component i .

Note 2: We report results on the basis of 1 mol in the feed of separator 1. Results can be easilyconverted into a basis of 10,000 kmol/h but since the problem asks only for fractions, not actualflows, this conversion is not necessary.

Note 3: For a graphical solution, first make a plot of of the T xy diagram based on the equations given.

a) Feed: z1 D 0:55, F1 D 1.

The first flash separator is at 1 atm. and 75ıC. Reading directly from the T xy phase diagram, the composi-tion of the liquid and vapor phase are: x1 � 0:41 and y1 � 0:74.

Since equations are given, the results may be also obtained as follows: For x1 we solve the equationTbubble.x1/ D 75

ıC from which we findx1 D 0:405

For y1 we solve the equation Tdew.y1/ D 75ıC from which we find

y1 D 0:737

Application of the lever rule, the fraction vaporized is

V1

F1Dz1 � x1

y1 � x1D

0:55 � 0:405

0:737 � 0:405D 0:435; L1 D 0:565

where F1 is the feed to separator 1. Therefore, 43.5% of the feed is vaporized in the first flash separator.The answer for the total fraction of the feed vaporized is answered requires completion of part (b).

b) From part (a):x1 D 0:405; y1 D 0:737

T2 and y2 are variables of the second flash separator. In order to determine the composition of the feed intothe second flash separator, we analyze the liquid coming into the second separator which is the liquid stream(same composition) coming out of first separator. Therefore z2 D 0:405 and we know that x2 D 0:15. Wecan directly draw the corresponding tie-line. If we do that we find that the temperature (T2) and y1 in thesecond separator is 84.1ıC and 0.505, respectively.

More accurate values can be determined from the equations given, as explained below. For the temperatureof separator 2 we use T2 D Tbubble.x2/. With x2 D 0:15 we find

T2 D 84:2ıC

The composition y2 is such that the dew temperature is 84.2ıC. We solve the equation Tdew.y2/ D 84:2ıC

and findy2 D 0:514

By level rule in the second separator

V2

F2Dz2 � x2

y2 � x2) V2 D

0:405 � 0:15

0:514 � 0:15.0:565/ D 0:397; L2 D F2 � V2 D 0:168




where F2 D L1 D 0:565 is the feed to separator 2.

Note: V2 and L2 are reported here on the basis of 1 mol in separator 1. To obtain the vapor and liquidfractions in separator 2 on the basis of the feed stream in that separator, we must ratio these numbers toF2 D 0:565:

vapor fraction in separator 2 D0:397

0:565D 0:703

liquid fraction in separator 2 D0:168

0:565D 0:297

The total fraction vaporized is

total fraction vaporized D V1 C V2 D 0:832

The material balances are summarized in the table below.

Separator 1 (75ıC)

Feed Vapor Liquidmol 1 0.435 0.565mol fraction of methanol 0.55 0.737 0.406

Separator 2 (84.2ıC)

Feed Vapor Liquidmol 0.565 0.397 0.168mol fraction of methanol 0.406 0.514 0.15

100

90

80

70

60

Tem

pera

ture

(C)

1.00.80.60.40.20.0

Mol fraction of Methanol

separator 2

separator 1

F1

L1 V1

V2L2





Problem 8.9 Solution REQUIRES MATERIAL FROM CHAPTER 13

y*=0.661CCl4�water at 100 °C

HaL

HbL

HcL

HdL

0.0 0.2 0.4 0.6 0.8 1.00

1

2

3

4

5

mol fraction CCl4

PHb

arL

It will be easier to answer these questions if we construct the Pxy diagram first. The bubble pressure of thetwo-liquid system is

Pbubble D Pw C PCCL4 D 2:95 bar

and the composition of the vapor at this point is

y�CCL4 DPCCL4

PbubbleD 0:661

a) Two liquids, each essentially pure.

b) All vapor.

c) Water.

d) yCCl4 D y�CCl4 D 0:661




Problem 8.10 a) Since the system starts as one liquid and by the addition of water it becomes two phases,we conclude that initially we were somewhere in the 2-butanol-rich phase (point A on the graph ) and afterthe addition of water the system is right at the phase boundary (point B), at which point x01 D 0:48. Thus,the addition of 0:46 mol of water to one mole of solution with x1, changes the mole fraction of 2-butanol to0.48:

x1

1C 0:46D 0:48 ) x1 D 0:7

b) The system is somewhere along the line CD. It cannot be at C because when the butanol rich phase boilsoff, the fraction of the vapor is

V D0:2 � 0:04

0:48 � 0:04D 0:471

which is less than 75%. Therefore, the state is on a tie-line such that V=L D 0:75=0:25 D 3=1. By trial anderror we locate this to be point E at T � 92ıC. At this point we have vapor with y1 � 0:26 and water-richliquid with x1 � 0:01 � 0:02.

c) The solution is given by point F on the graph.

T D 95ıC x1 D 0:90 y1 D 0:76

V D0:9 � 0:82

0:9 � 0:76D 57%

L D 1 � V D 43%

recovery in vap. stream D.0:76/.0:57/

0:82D 53%

recovery in liq. stream D.0:90/.0:43/

0:82D 47%

100

95

90

85

80

75

70

Tem

pera

ture

, C

1.00.90.80.70.60.50.40.30.20.10.0x1, y

1

2-butanol(1)/water(2) at 1 atm

AB

C

D

E

F





Problem 8.11 Solution a) The overall composition of the mixture (1=acrylonitrile):

z1 D5

7C 5D 0:417 D 41:7%

This composition is between 7.3% and 96.8% of the two liquid phases, therefore, the system consists of twophases. The fraction of the acrylonitrile-rich phase is

LA

FD41:7 � 7:3

96:8 � 7:3D 0:384

where F are the total mols in the mixture. With F D 12 mol we have:

phase amount (mol) x1

LA 4:61 96:8

LB 7:39 7:3

total 12:00 41:7

b) We need to add enough water to bring the mol fraction of acrylonitrile below 7.3%. The required molesof water, n0, are

5

n0 C 12� 0:073 ) n0 � 56:5 mol

c) See graph below. In reality, mutual solubility increases with temperature but here we have assumedsolubility to be independent of T , as indicated by the vertical lines

AB

LB

LAV

add water





a) The overall mol fraction of butanol is 2=7 D 0:714. This places the state in the two-liquid region. Thereis no vapor because the temperature is below the boiling point of the two-liquid system.

b) The butanol-rich liquid will boil off first.

c) Given 5 mol of water and 2 mol of butanol, we must either add 18.974 mol of water, or 17.8756 mol ofbutanol in order to create a one-phase system.





Problem 8.13 Solution a) With z1 D 40=.40C 50/ D 0:8, the butyraldehyde will boil off first.

b) Three phases: Butyraldehyde-rich phase = 34.9%, water-rich phase = 15.1%, vapor phase= 50%

c) Three phases: Butyraldehyde-rich phase = 50%, water-rich hase = 16.3%, vapor-phase = 33.7%




Problem 8.14 Solution [DUPLICATE PROBLEM – APPEARS IN CHAPTER 13 AS PROBLEM 13.13]






acetophenone (1)/glycerol(2)

La

LbV

B

C

D

a) See graphb) At z1 D 0:25, 0.2 bar, the system consists of two liquid phases.c) The boils boils at point C. It stay at that point as long as there are two liquids present; when one liquiddisappears, point moves to just below the bubble pressure. In this region, the system consists of vapor andglycerol-rich liquid. Therefore, it is the acetophenone-rich phase that disappears first.

d) At this point we have vapor, with y1 D 0:95, and glycerol-rich liquid (Lb) with xB1 D 0:1. By lever rule,

V D0:25 � 0:1

0:95 � 0:1D 0:177 D 17:7%

e) Point D shows the state where the vapor fraction os 50%. The composition of the two phases are readoff the ends of the tie line. However, since these lines have been drawn qualitatively, these numbers are notaccurate so we do not report them.





0.0 0.2 0.4 0.6 0.8 1.00.0

0.2

0.4

0.6

0.8

1.0

3-Methyl-1-butanol

etha

nol

b) 3-Methyl-1-butanol and water are only partially miscible. All other combinations (water/ethanol, water/3-methyl-1-butanol) are fully miscible.

c) The overall composition of the mixture is

z1 D 0:185494; z2 D 0:0447041; z3 D 0:769802

This places the system in the two phase region and right on the thrird tie-line, whose compositions are

x1 x2 x3

pahse 1 0:534 0:079 0:387

pahse 2 0:006 0:027 0:968

By lever rule on component 1, the amounts in each phase are:

L1 D0:185494 � 0:534

0:534 � 0:534D 34%

You should be able to get the same answer by applying the lever rule to any of the other two components.







9 Properties of Mixtures




9. PROPERTIES OF MIXTURES


We treat the mixture as a pure (ideal) gas whose CP is

CP D .0:25/.40:8/C .0:75/.029:4/ D 32:25 J/mol

Reversible operation:

�S120 D 0 ) T 02 D T1

�P2

P1

�R=CP

D 585:41 K

Wrev D CP .T02 � T1/ D 8930:62 J/mol

Actual operation

Wactual DWrev

�D 9922:91 J/mol

Exit temperature:Wactual D CP .T2 � T1/ ) T2 D 616:187 K

Entropy generation: Since process is adiabatic, entropy generation is equal to entropy change of system

Sgen D �S12 D 1:65197 J/mol K




Problem 9.2 Solution The process is shown in the flow diagram below.

A B C

The parameters in the equation for the ideal-gas heat capacity are:

c1 c2 c3 c4 c5

N2 29110 8610 1701:6 100 909:79

O2 29100 10040 2526:5 9360 1153:8

Te enthalpy and entropy are calculated as follows (“1” = nitrogen, “2” = oxygen):

H D x1H1 C x2H2 Œ1�

S D x1S1 C x2S2 �R.x1 ln x1 C x2 ln x2/ Œ2�

whereHi , Si refers to properties of the pure components. For changes in enthalpy and entropy between twostates,

�Hi D

T2ZT1

CPidT Œ3�

�Si D

T2ZT1

CPidT

T�R ln

P2

P1Œ4�

where we have taken into consideration that the composition of the mixture does not change during thisprocess.

1) Compression: For isentropic compression, �SAB D 0, or

�S D

T2ZT1

�x1CP;1 C x2CP;2

� dTT�R ln

P2

P1D 0

Hint: To obtain an approximate temperature that can serve as the first guess in the trial-and error procedure,assume air to be pure nitrogen with constant CP , which you may take to be the CP of nitrogen at 300 K.

This is solved by trial error. We findTB D 873:616 K

The work isW D HB �HA D 17579 J/mol

2) Heat exchanger: The heat isQ D Hc �HA D �17525 J/mol





Using Reference States: The calculations can be streamlined using reference states. We calculate theenthalpy of the pure component i using

Hi D

T2ZT0

CPidT CHi;0 Œ3b�

Si D

T2ZT0

CPidT

T�R ln

P2

P0C Si;0 Œ4b�

where Hi;0, Si;0 is the enthalpy and entropy of pure i at the reference state T0, P0. These values arethen used in Eqs [1] and [2] for the enthalpy and entropy of the mixture. Fir the reference state we pickarbitrarily the following conditions:

T0 D 25ıC D 298:15 K; P0 D 1 bar;H1;0 D H2;0 D 0; S1;0 D S2;0 D 0

We now summarize the properties in each state in the table below:

T (K) P (bar) H (J/mol) S (J/mol K)

A 298:15 1 0: 4:27304

B 873:616 50 17579 4:27305

C 300 50 53:977 �28:071

The energy balances are now computed easily as follows:

W D HB �HA D .17579 � 0/ D 17579 J/mol

Q D HC �HA D .53:977 � 17579/ D �17525 J/mol




Problem 9.3 Solution The temperature of stream C is obtained by energy balance. Since there is no heator work, �H D 0:

�H D nA1CP1.TC � TA/C nB1CP1.TC � TB/C nB2CP2.TC � TB/ D 0

where nAi refers to molar flow rate of component i in stream A, and similar for streams B and C . Solvingfor the temperature we find:

TC D 365:242 K

We now compute all properties using as reference state the pure components at TC , PC . The results aresummarized blow:

stream A stream B stream C

H (J) �433:94 433:94 0:

S (J/K) �1:26231 5:78567 5:59542

G (J) �26:0245 �1724:98 �2043:68

It follows:�H D 0 J

�S D .5:59542/ � .�1:26231/ � .5:78567/ D 1:07206 J/K

�G D .�2043:68/ � .�26:0245/ � .�1724:98/ D �292:674 J

W idD Tsur�S ��H D .293:15 K/.1:07206 J/K/ � .0 J/ D 314:27 J






We define the following symbols:

nAi moles of component i in compartment AnBi moles of component i in compartment BnCi moles of component i in final mixturen total mol in the boxyBi mole fraction of component i in compartment ByCi mole fraction of component i in final mixtureTA; PA temperature and pressure in compartment ATB ; PB temperature and pressure in compartment BTC ; PC final temperature and pressureVA; VB volume of compartment A, B

with 1 for methane and 2 for ethane.

By mass balance we find that the final mixture consists of 0.6 mol of methane, 0.4 mol of ethane, i.e:

n1 D y1 D 0:6 n2 D y2 D 0:5 n D 1

a) The system is closed, adiabatic, and involves no work:

�U t D QCW D 0 Œ1�

where �U is the total change in internal energy before and after mixing. For the internal energy of an idealgas mixture:

U igmD x1U1 C x2U2

where Ui is the internal energy of pure ideal gas (to show this, start with H D U C PV and use the knownequations for H and V of ideal-gas mixture). Equation [1] becomes:

�U D nA1CV1.TC � TA/C nB1CV1.TC � TB/C nB2CV 2.TC � TB/ D 0

where CV i is the constant V heat capacity of component i . We solve for TC and find

TC DnA1CV1 C nB1CV1TB C nB2CV 2TB

nA1CV1 C nB1CV1 C nB2CV 2D 365:5 K

For the final pressure we write:

PC DnCRTC

VA C VBD

nRTC

nARTA=PA C nBRTB=PBD 1:006 bar

For the following parts it is convenient to adopt a reference state for each component and calculate allproperties with respect to these states. We choose P0 D PC , T0 D TC , for both methane and ethane.Properties are now calculated as follows:

Enthalpy in state x, x D A, B, or C:

Hx D nx Œyx1CP1.Tx � T /C yx2CP2.Tx � T0/�




Sx D nx

hyx1Sx1 C yx2Sx2 �R.yx1 lnyx1 C yx2 lnyx2/

iGx D Hx � TxSx

withSxi D CPi ln

Tx

T0�R ln

Px

P0

The results are summarized blow:

compartment A compartment B after mixing

H (J) �437:093 417:159 0:

S (J/K) �1:26 5:78351 5:59542

G (J) �29:925 �1740:96 �2045:39

It follows:�H D .0/ � .�437:093/ � .417:159/ D 19:9342 J

�S D .5:59542/ � .�1:26/ � .5:78351/ D 1:0719 J/K

�G D .�2045:39/ � .�1740:96/ � .�29:925/ D �274:508 J

W idD Tsur�S ��H D .293:15 K/.1:0719 J/K/ � .19:9342/ D 294:294 J






Calculation of enthalpy, entropy The enthalpy of stream is calculated by

H.P; T; x/ D xAHA.P; T /C xBHB.P; T /

where P , T , is the pressure and temperature of the stream, and xA, xB are the mole fractions of the twocomponents in the stream. HA andHB are the enthalpies of pure components at the pressure and temperatureof the stream and are calculated using a reference state at Pref, Tref as follows:

HA.P; T / D CPA.T � Tref/; HB.P; T / D CPB.T � Tref/

The entropy of stream is calculated as

S.P; T; x/ D xASA.P; T /C xBSB.P; T / �R.xA ln xA C xB ln xB/

with the entropy of pure components calculated with respect to the same a reference state at Pref, Tref:

SA.P; T / D CPA lnT

Tref�R ln

P

Pref; SB.P; T / D CPA ln

T

Tref�R ln

P

Pref

Combining, and after some simplifications, the equations for enthalpy and entropy become:

H D NCP .T � Tref/

S D .xACPA C xBCPB/ lnT

Tref�R ln

P

Pref�R.xA ln xA C xB ln xB/

where NCP is the mean heat capacity of the stream:

NCP D xACPA C xBCPB

Note Reference states are needed for the calculation of the properties of pure components. Accordingly,we need one reference state per component. Although we could pick a different reference for eachcomponent, calculations are simpler if we choose the same reference for all components.

For the calculations shown here we choose:

Pref D P1; Tref D T1

It is left as an exercise to show that the final results will not change if a different state is chosen asreference.

1. Mass balances around mixing point

xA1 Pn1 C xA2 Pn2 D Pn2xA3

Pn1 C n2 D n3

From this we obtain Pn3 and xA3.

We also note that P1 D P2 D P3 D P4 because we are ignoring pressure drops.




2. Energy balances around mixing point Since mixing is adiabatic and no work is involved,

�H D 0 ) � Pn1H1 � Pn2H2 C Pn3H3 D 0

In this equation the only unknown is T3. We find

T3 D 410:1 J/mol K

3. Energy balances in HE

PQ D Pn4H4 � Pn3H3 D 4:42 kW

4. Entropy Analysis

Mixing PointPSmix D �Pn1S1 � Pn2S2 C Pn3S3 D 0:734 W/K

Heat exchanger

PSHE D �Pn3S3 C Pn4S4 �PQ

T 0

where T 0 is an estimate for the temperature of the heat exchanger fluid. Since the heat exchanger isused to raise the temperature of stream 3 to T4 D 873:15 K, the fluid temperature must be at least873.15 K. Using T 0 D 873:15 K we find

PSHE D 2:149W=K

The total entropy generation is

PSgen D PSmix C PSHE D 2:884 W/K

5. Ideal and lost work The ideal work is calculated as

PWideal D P�H � T0 P�S

where P�H and P�S are the total enthalpy and entropy change in the control volume of interest. Thelost work is calculated using

PWlost D T0 PSgen

where T0 is the assumed temperature of the surroundings, and PSgen is the entropy generation in thecontrol volume of interest. Notice that in this problem, PWactual is zero in both the mixing point and theheat exchanger.

Summary The results are summarized below.





Summary of Streams

1 2 3 4

Pn (mol/min) 12 5 17 17

P (bar) 3 3 3 3

T (K) 308:15 623:15 410:1 873:15

xA 0:250 0:670 0:374 0:374

H (J/mol) 0: 11667 3431:46 1901

S (J/mol K) 4:675 31:355 15:114 40:550

Summary of Energy and Entropy Balances

Mixing HE Total

PWactual (W) 0 0 0

PQ (W) 0 0 4416

PSgen (W/K) 0:734346 �5:05741 2:88373

PWideal (W) 220:304 644:815 865:119

PWlost (W) 220:304 644:815 865:119




Problem 9.6 Solution (Outline) The enthalpy and entropy of each stream is calculated as follows (A =oxygen, 2=carbon monoxide):

H D xAHA C xBHB

S D xASA C xBSB �R .xA ln xA C xB ln xB/

The enthalpy and entropy of pure component A is calculated as follows:

HA D CPA.T � Tref/

SA D CPA lnT

Tref�R ln

P

Pref

where P , T , is the pressure and temperature of the stream, and Pref, Tref are the pressure and temperature atthe reference state. The properties of pure B are calculated in the same manner.

The turbine is calculated in the usual manner and leads to the calculation of T2. The composition of stream4 is obtained by mol balance around the mixing point. The mixing point is adiabatic and by applying theenergy balance we obtain the temperature T4. After these calculations, the states of all streams are known.

Work in the turbine:

PW D Pn1.H2 �H1/

Heat exchanger duty:

PQ D Pn4.H5 �H4/

Ideal Work:

PWideal D � PH � T0� PS

where

� PH D �Pn1H1 � Pn3H3 C Pn5H5

� PS D �Pn1S1 � Pn3S3 C Pn5S5

Entropy generation

PSgen D � PS �PQ

T5





1 2 3 4 5

Pn (mol/min) 100 100 70 170 170

P (bar) 5: 1 1 1 1

T (K) 500 390:825 500 437:448 300

xA (–) 0:8 0:8 1: 0:882353 0:882353

xB (–) 0:2 0:2 0 0:117647 0:117647

H (J/mol) 0 �4583:18 0 �2695:99 �8620

S (J/mol K) 4:16035 7:19948 13:3809 10:632 �5:6243

PWturbine D �7:64 kW PQturbine D �16:8 kW

PWideal D �12:9 kW PWlost D 5:2 kW

PSgen D 17:46 W/K





Note: these results are obtained using CP ’s from Perry’s Handbook (Table 2-198, 7th edition)

CP D

�c4

�c5

T cosh .c5=T /

�2C c1 C

�c3

T sinh .c3=T /

�2c2

�with T in K, CP in J/mol, and the parameters c1 through c5 given by

c1 c2 c3 c4 c5

CH4 33300: 79930: 2086:9 41600: 991:96

C2H6 40330: 134220: 1655:5 73220: 752:87

CO2 29370: 34540: 1428: 26400: 588

Since we are treating the system as an ideal-gas mixture, the equations for the enthalpy and entropy are:

H D x1H1 C x2H2 C x3H3

S D x1S1 C x2S2 C x3S3 �R.x1 ln x1 C x2 ln x2 C x3 ln x3/

where 1, 2, and 3 refer to methane, ethane and CO2, respectively. In this problem the mixture undergoeschanges from initial state P1, T1 to some final state P2, T2, but we notice that the composition remains thesame during such change. Then, the changes in H and S are

�H D x1�H1 C x2�H1 C x2�H2

�S D x1�S1 C x2�S1 C x2�S2

where �Hi and �Si refer to the corresponding change of the pure component i , namely

�Hi D

T2ZT1

CPidT

�Si D

T2ZT1

CPidT

T�R ln

P2

P1

with CPi referring to the ideal-gas CP of component i , obtained from Perry’s.

a) For reversible isentropic process, �S D 0, or

�S D

T2ZT1

�x1CP;1 C x2CP;2 C x3CP;3

� dTT�R ln

P2

P1D 0

This equation is solved for T2 by trial and error. As a first approximation, assume the system to be puremethane with CP D 38:2291 J/mol K, (corresponding to T1 D 180ıF). For isentropic expansion then wehave,

T2 D T1

�P2

P1

�R=CP

D .355:372 K/�14:5 psi5000 psi

�8:314=38:2291D 99:72 K





Therefore, we seek the solution in the neighborhood of 99.72 K. By tabulation we find:

T2 (K) �S (J/mol K)

85 �2:15782

86 �1:7583

87 �1:3634

88 �0:973016

89 �0:587039

90 �0:205375

91 0:172073

92 0:545397

93 0:914686

94 1:28003

95 1:6415

The solution is between 90 and 91 K. By further refinement we find

T2 D 90:5 K

b) In throttling, �H D 0 and since in the ideal-gas state H depends only on T , �H D 0 implies �T D 0,i.e. T2 D T1 D 355:372 K.

Comments: The true process is somewhere between the isentropic and isenthalpic cases, meaning that theexit temperature is somewhere between 90.5 K (�188ıC, �182ıF) and 355.372 K (82ıC, 180ıF). That’sa huge range. The isentropic calculation predicts a very low temperature because the pressure difference isvery large from 5000 psi to 14.5 psi. If the isentropic assumption is correct, then the stream must emerge atvery high velocity, having converted its energy into kinetic energy. We must point out, however, that evenif the isentropic assumption is correct, the ideal-gas calculation should be seriously questioned at such lowtemperature.

The point of this exercise is to demonstrate the use of the ideal-gas equations but also to highlight the factthat the engineer is ultimately responsible for accepting or rejecting the result of a calculation.




Problem 9.8 Solution a) Summary of results

A B C D

Molar flow rate of methane (mol/min) 10: 70: 80: 0:

Molar flow rate of ethane (mol/min) 55: 20: 0: 75:

Mole fraction of methane 0:153846 0:777778 1: 0:

Mole fraction of ethane 0:846154 0:222222 0: 1:

P (bar) 5 10 1 1

T (K) 250 600 300 300

H (kJ/min) �2511:54 11700: 0 0

S (kJ/K/min) �18:9696 12:293 0 0

n mol/min 65: 90: 80: 75:

H (kJ/min) �163:25 1053: 0 0

S (kJ/K min) �1:23303 1:10637 0 0

Note: The enthalpy of a stream, e.g. stream A, is calculated as

HA D nAhA

where nA is the molar flow in the stream and hA is the molar enthalpy of the stream (J/mol). HA is in J/min(or kJ/min).

b) Ideal Work:�H D �HA �HB CHC CHD D �889750 J/mol

�S D �SA � SB C SC C SD D �126:654 J/min

W idD �H � T0�S D .�889750/ � .300/.�126:654/ D �927:746 kJ/mol

The work is negative, therefore, the system can produce work!

Why is the ideal work negative?

To answer this condition we must understand the meaning of the ideal work. Ideal work looksat the inlet and outlet streams (not the processes inside the system). If the result is negative,it means that it is possible to design a process that takes the feed streams at the specified con-ditions, delivers the outlet streams at the specified conditions, and the process produces work.How can we accomplish that? The ideal-work calculation doesn’t give that answer, but we canuse engineering intuition to make suggestions. For example, the inlet streams are at higherpressure than the outlet streams; we could use turbines to extract work from this process. Oneinlet stream is much hotter than the rest; we could use it to run a power cycle between 400 Kand the surroundings and produce some additional work.

The actual process may or may not produce work. If it does not produce any work, we shouldlook for ways to get some work out of the process. If it does produce some fraction of the idealwork, we should look for ways to increase production. Whether the improvements are worththe costs is a separate question. The point here is that the ideal-work calculation allows us toevaluate a process with respect to how it utilizes the energy resources.

But didn’t we say that to separate we must do work?





Yes – if the pressure and temperature of the mixture and purified streams are the same, i.e. ifthe only thing that changes between inlet and outlet streams is the composition. The processunder consideration is not a pure separation process because in addition to separation we havechanges in temperature and pressure. Such process may result in either positive or negativeideal work.





a) Enthalpy

H D x1Hig1 C x2H

ig2 CH

R

H1 D CigP 1.T � Tref/ D 0

H2 D CigP 2.T � Tref/ D 0

) H D �7211 J/mol

b) Entropy

S D x1Sig1 C x2S

ig2 �R.x1 ln x1 C x2 ln x2/C SR

S1 D CigP 1

lnT

Tref�R ln

P

PrefD �32:5 J/mol K

S2 D CigP 2

lnT

Tref�R ln

P


R.x1 ln x1 C x2 ln x2/ D �4:675 J/mol K

) S D �63:8493 J/mol K

therefore, S D �63:8493 J/mol K.

c) We need to calculate �H and �S between inlet and outlet of the separation process (see schematic).

The molar properties of stream A (inlet) have been calculated. We need those of streams B and C. Theseinvolve pure gas in the ideal-gas state.

HB D CigP 1.TB � Tref/ D 4306:65 J/mol

SB D CigP 1

logTB

Tref�R ln

PB

PrefD 19:9608 J/mol K

HC D CigP 2.TC � Tref/ D 6152:36 J/mol

SC D CigP 2

logTC

Tref�R ln

PC

PrefD 28:5154 J/mol K

We now have�H D 0:75HB C 0:25HC �HA D 11979:1 J/mol

and�S D 0:75SB C 0:25SC � SA D 85:9488 J/mol K

For the ideal work finally:

Wideal D DeltaH � T0�S D .298:15/.85:9488/ � .11979:1/ D 13646:5 J/mol

The calculations are summarized in the following table:





A B Cmoles 1 0.75 0.25T (K) 150.15 298.15 298.15P (bar) 50 1 1x1 0.75 0.75 0H (J/mol) -7211 4306.65 6152.36S (J/mol K) -63.8493 19.9608 28.5154H igm(J/mol) 0 4306.65 6152.36HR(J/mol) -7211 0 0S igm(J/mol) 27.8493 19.9608 28.5154SR(J/mol K) -36 0 0

Note: You should get in the habit of summarizing your results in a table like this even though the problemdoes not ask for it.





A (inlet) B (outlet) C (outlet)molar flow (mol/s) 100 50 50mol fraction of CO 0.5 0.1 0.9P (bar) 100 50 1T (ıC) 25 0 25

Sig1 (J/mol K) -5.763 -2.267 32.525S

ig2 (J/mol K) -5.763 -5.255 32.525SR (J/mol K) �3:1 �30 0S (J/mol K) -3.1 -32.289 35.226






The number of moles in 100 kg of this mixture is

n Dm

x1Mw1 C x2Mw2D 5874:72 mol

where we used Mw1 D 16:04 g=mol and Mw2 D 30:07 g=mol. The required volume is

V tankD n

ZRT

PD 0:31 m3

where Z D 0:00384449 and is calculated from the SRK with k12 D 0, as summarized in the table below.

Pc (bar) Tc (K) !

methane (1): 45:99 190:56 0:011

ethane (2): 48:72 305:3 0:1

T= 165 P= 1:

x1= 0:93 phase= L

b a dadTmethane 2:985 � 10�5 0:249712 �6:769 � 10�4

ethane 4:514 � 10�5 0:771659 �1:871 � 10�3

mixture 3:092 � 10�5 0:276911 �7:414 � 10�4

Z 0:00384449 Z1= 0:00384449

HR �7324:14 Z2= 0:00873609

SR �64:6349 Z3= 0:987419





b) First we collect the critical parameters and acentric factors and calculate ai and bi of pure components.

nitrogen methane

Tci 126:2 190:6 K

Pci 33:9 46 bar

! 0:04 0:008 �

ai 0:142594 0:283255 Jm3=mol2

bi 2:682 � 10�5 2:985 � 10�5 m3=mol

dai=dT �6:204 � 10�4 �8:374 � 10�4 Jm3=mol2K

k12 D 0

Using the mixing rules, we calculate the a and b parameters of the mixture.

Mixture

amix 0:206949 Jm3=mol2

bmix 2:83311 � 10�5 m3=mol

damix=dT �7:3160510�4 Jm3=mol2K

From these we calculate A and B:

A Damix

.RT /2D 0:207913 B D

bmixP

RTD 0:028397

The cubic equation is�0:00590408C 0:178709Z �Z2 CZ3 D 0

Solving for the roots we find three real roots: 0.042889, 0.176305, 0.780806. For each root we obtain theresiduals:

Residuals1 2 3

Z 0:042889 0:176305 0:780806

HR (J/mol) �6240:7 �2375:4 �590:3

SR (J/mol K) �48:32 �19:75 �3:29

In this problem we are told that the system is in theliquid phase, therefore, we select the smallest root andthe corresponding residuals:

Z D 0:042889; HRD �6240:73 J/mol; SR D �48:3233 J/mol K

b) The enthalpy of the mixture isH D x1H

ig1 C x2H

ig2 CH

R





with

Hig1 D

Z T

Tref

CP1dT D �5162:28 J/mol

Hig2 D

Z T

Tref

CPwdT D �3998:78 J/mol

The enthalpy of the mixture isH D �10821:3 J/mol

The entropy is calculated using

S D x1Sig1 C x2S

ig2 �R .x1 log x1 C x2 log x2/C SR

with

Sig1 D

Z T

Tref

CP1dT

T�R ln

P


Sig2 D

Z T

Tref

CP2dT

T�R ln

P


The entropy of the mixture isS D �92:8161 J/mol K





The problem requires trial and error. The general procedure is as follows:

1. Set the inlet of stage 1 to P1 D 1 bar, T1 D 25 ıC and the outlet temperature to T2 D 200 ıC.

2. Set th

3. Guess an outlet pressure P2

4. Calculate the efficiency of the compression for the chosen value of P2; if higher than 80%, decreaseP2; if lower than 80%, increase P2.

5. Repeat the steps until the efficiency is sufficiently close to the given value.

Once stage 1 is solved, repeat the calculation of stage 2 where now the inlet pressure is P2 determinedabove. The table below shows the summary of the calculations using the SRK equation with k12 D 0. Eachrow shows the actual exit state of each stage as well as the reversible exit (the inlet state is at the pressure ofthe previous stage and at temperature 25 C). The calculated efficiency is shown in the last column. Due tothe trial and error method, this efficiency is not exactly 80%.

T P H S Trev hrev Srev �

.K/ .bar/ .J=mol/ .J=mol K/ .K/ .J=mol/ .J:mol K/(inlet) 298:15 1 �60:6105 �0:201058

1 473:15 4: 5046:36 1:71429 443:15 4168:93 �0:197579 0:83

2 473:15 12: 5037:43 �7:46626 442:15 4126:05 �9:45433 0:82

3 473:15 28: 5020:95 �14:6018 438:15 3981:5 �16:8795 0:80

4 473:15 56: 4996:44 �20:5168 438:15 3939:8 �22:8324 0:80

5 473:15 150: 4950:93 �29:1587 428:15 3526:48 �32:3166 0:74





Problem 9.14 Solution a) The results of the SRK calculation for a vapor mixture at 1.2 bar, 130ıC aresummarized below:

nC6 nC8

xi 0.25 0.75 -

Tci 507.6 568.7 K

Pci 30.25 24.9 bar

wi 0.301 0.4 -

ai 3.05667 5.26199 SI Units

bi 1.209µ10-4

1.645µ10-4

SI Units

dadTi -5.75µ10-3

-1.015µ10-2

SI Units

MIXTURE PROPERTIES

a= 4.65485 SI units

b= 0.000153606 SI units

dadT= -0.00893366 SI units

A= 0.0497203 -

B= 0.00549937 -

Z= 0.953978 -

HZ Roots: 0.00742621 0.0385959 0.953978 L

HR= -463.223 Jêmol

SR= -0.774086 Jêmol K

The enthalpy and entropy are calculated as follows:

H D x1CP1.T � T0/C x2CP2.T � T0/CHR

S D x1

�CP1 ln

T

T0�R ln

P

P0

�C x2

�CP2 ln

T

T0�R ln

P

P0

��R.x1 ln x1 C x � 2 ln x2/C SR

where 1, 2, stand for n-hexane and n-octane respectively, CPi stands for the ideal-gas heat capacity ofcomponent i , and P0, T0, are the conditions at the reference state. We find:

H D 23183:9 J/mol S D 15:6101 J/mol K

b) For a liquid mixture at 90ıC, 1.2 bar:

T=363.15 K; P=1.2bar Hphase = L

PURE COMPONENTS

nC6 nC8

xi 0.25 0.75 -

Tci 507.6 568.7 K

Pci 30.25 24.9 bar

wi 0.301 0.4 -

ai 3.29724 5.68683 SI Units

bi 1.209µ10-4

1.645µ10-4

SI Units

dadTi -6.292µ10-3

-1.112µ10-2

SI Units

MIXTURE PROPERTIES

a= 5.02875 SI units

b= 0.000153606 SI units

dadT= -0.00978355 SI units

A= 0.0661987 -

B= 0.00610511 -

Z= 0.00771228 -

HZ Roots: 0.00771228 0.0559675 0.93632 L

HR= -35573.3 Jêmol

SR= -90.6263 Jêmol K




For the enthalpy and entropy we obtain:

H D �21096:2 J/mol S D �78:9181 J/mol K

c) Since we have already calculated H at the initial and final state, the easiest way to calculate the heat is

Q D H2 �H1 D .�21096:2/ � .23183:9/ D �44280:1 J/mol

Alternatively, we could do calculate �H between the two states as follows:

Q D �H D x1CP1.T2 � T1/C x2CP2.T2 � T1/CHR2 �H

R1 D �44280:1 J/mol

——————————-







10 VLE of Mixture




10. VLE OF MIXTURE

Problem 10.1 Solution Outline:

a) zero

b) 0.611 kPa

c) Gice D GL D 0 (using steam tables)

d) Must use the Poynting factor. Using V D 1 g=cm3, we find:

f D ftriple exp�.P � Ptriple/VL

RT

�D .0:611 kPa/.1:268/ D 0:775 kPa




Problem 10.2 Solution a) From the definition of fugacity,

d lnfi Dd�i

RTŒ1�

If the system were treated as an ideal gas under these conditions,

d lnyiP Dd�

igi

RTŒ2�

Taking the difference we find

d ln�i Dd.�i � �

igi /

RTDd.�i � �

igi /

RTDd�RiRT

Dd NGRiRT

Œ3�

where NGRi is the residual partial molar Gibbs energy, namely

NGRi D

@nGR

@ni

!P;T;nj

Œ4�

Integration of Eq. [3] from the ideal-gas state (�i D 1, NGRi D 0) to the actual state results in

ln�i DNGRiRT

Œ5�

If the fugacity coefficient of the entire mixture is defined to be

ln� DGR

RTŒ6�

by combining Eqs. [4], [5], [6] we find

.RT / ln�i D NGRi D

@nGR

@ni

!P;T;nj

D .RT /

�@n ln�@ni

�P;T;nj

Œ7�

or

ln�i D�@n ln�@ni

�P;T;nj

Œ8�

b) The procedure for obtaining the fugacity of species from an equation of state such as the SRK is asfollows:

1. Obtain an expression for GR D HR � TSR using the expressions derived for HR and SR.

2. Obtain an expression for ln� using Eq. [6]. This expression contains a, b, andZ, all of which dependon composition (recall that a and b of the mixture are obtained from those of the pure componentsusing an appropriate mixing rule).

3. Multiply ln� with n D n1Cn2C� � �niC� � � ; replace all the mole fractions in the resulting expressionwith ni=.n1 C n2 C � � � /.

4. Treat the resulting expression as a function of ni . Differentiate with respect to ni (keeping everythingelse constant). The result should be Eq. (5.54) in the Notes. Analogous is the derivation of Eq. 5.57which is based on the Peng-Robinson equation.




10. VLE OF MIXTURE

Problem 10.3 Solution (Outline)

a) For any component i :�Vi � 1

f Li D fVi � yiP

�Li Df LixiP

�yi

xi

b) For the chemical potential:

�i D Gigi CRT ln

f LiP

The fugacity f Li is calculated in part (a). We need Gigi at T D 50 ıC, 1.2 bar, using as reference state the

ideal-gas state at T0 D 200 K, P0 D 35 bar. At the reference state, H0 D 0, S0 D 0. Therefore:

Gig.P; T / D H0 C CigP .T � T0/ � T

hS0 C C

igP ln

T

T0�R ln

P

P0

i





�L1 Dy1�

V1

x1D.0:9876/.0:9479/

0:2D 4:68

�V2 Dx2�

L2

y2D.0:8/.0:0126/

0:0124D 0:813

b) From table at x2 D 1 (x1 D 0) P sat2 D 0:074 bar

c)f2 D �

sat2 P

sat2 Poynting D .0:9953/.0:074/.1/ D 0:07365 bar

d) State is compressed liquid - Poynting may be significant and must be included:

f2 D fsat2 Poynting

Poynting D exp�V L2

P � P sat2

RT

�D exp

�Zsat2

�P

P sat � 1

��D exp

�0:0004

�300

0:074� 1

��D 5:059

Thereforef2 D .0:07365/.5:059/ D 0:3726

e) P � Pbubble D 6:148 bar

f)

V D NZLRT

PD .106/.0:031/

.8:314/.250/

6:148 � 105D 105

m3

MM moles




10. VLE OF MIXTURE

Problem 10.5 Solution In all cases the work to unmix the components is given by

W D �H � T0�S

where T0 D 298:15 K.

a) Treating the Ne/H2 system as an ideal-gas mixture with

�H D 0

�S D RX

yi lny1 D �4:675 J/mole K

Here,�H ,�S refer to unmixing, i.e., going from a mixture to the pure components. The work for unmixingis

W D �.298:15/.�4:675/ D 1393:9 J/mole

b) Treating the system methane/ethane as an ideal gas, the result is the same as before:

W D 1393:9 J/mole

c) For the system ethanol/water we must calculate �H and �S from the data given.

Enthalpy

after Hafter D x1H1 C x2H2 D 0 J/mole

before Hbefore D x1 NH1 C x2 NH2 D �16:82 J/mole

change �H D Hafter �Hbefore D C16:82 J/mole

Entropy

after Safter D x1S1 C x2S2 D 0 J/mole

before Sbefore D x1 NS1 C x2 NS2 D 2:3245 J/mole K

change �Smix D Safter � Sbefore D �2:3245 J/mole K

Therefore, the work for unmixing is

W D .C16:82/ � .298:15/.�2:3245/ D 710 J/mole

d) Cases (a) and (b) give identical results because the systems are treated as ideal gases. In this case theresults depend only the composition of the mixture, but not on the identity of the components or the pressureand temperature of the system. Case (3) involves a liquid whose enthalpy changes when the componentsmix. This is indication of interactions between the two components. The resulting work is less than that (a)and (b), implying that the water and ethanol molecules do not like each other thus making the separationeasier.





a) We assume that the components form ideal solution and that the vapor phase is ideal. The fugacity ofcomponent i in the vapor is calculated as

fi D yiP

The fugacity coefficients in the vapor are 1 since we have assumed ideality.

The fugacity coefficients in the liquid are calculated as follows:

fi D xi�Li P ) �Vi D

fi

xiP

The results are summarized below:

1 2 3fi (bar) 1:044 0:12 0:036

�Vi 1 1 1

�Li 1:45 0:5 0:15

b) The chemical potential of n-heptane (component 3) is

�3 D GV3 CRT lny3 D GL3 CRT ln x3

The first equality in the right-hand side is the chemical potential in the vapor, and the second one is thechemical potential in the liquid. Here, GV3 is the Gibbs energy of pure heptane vapor at 50ıC, 1.2 bar, andGL3 is the chemical potential of pure liquid heptane at 50ıC, 1.2 bar. The two are related as follows:

GL3 D GV3 CRT ln

P sat3

PŒ1�

(This result is explained at the end of the solution.) Therefore, the chemical potential can be calculated aseither

�3 D GV3 CRT lny3 Œ2�

or

�3 D GVCRT ln

P sat3

PCRT ln x3 Œ3�

The calculation is done most easily using the vapor phase (Eq. [2]). With

G3 D H3 � TS3

H3 D CP .T � T0/

S3 D CP ln.T=T0/ �R ln.P=P0/

we find:

H3 D 24630 J/mol

S3 D 124:003 J/mol K

G3 D �15441:6 J/mol

�3 D �24862:6 J/mol

Alternatively, we may use Eq. [3]. Noting that P sat3 D y3P=x3 D 0:18 bar, the result turns out exactly the

same.




10. VLE OF MIXTURE

Justification of Eq. [1]: Recall that GV is the Gibbs energy of pure heptane vapor at 50ıC, 1.2 bar, andGL is the Gibbs energy of pure heptane liquid at 50ıC, 1.2 bar. Since the saturation pressure of heptaneat 50ıC is 0.18 bar, pure heptane at 1.2 bar, 50ıC exists as liquid, not as vapor. This means that GV

refers to a hypothetical liquid heptane at 1.2 bar, 50 ıC. These states are illustrated on the qualitativePV chart below.

Psat

=0.18 bar

P=1.2 bar GL

GL,sat GV,sat

GV

50 C

P (bar)

V

The Gibbs energyGV is calculated by extrapolating the vapor isotherm past the saturation point (dashedline). The calculation is now performed as follows. Treating the vapor phase as an ideal gas,

GV D GV;satCRT ln.P=P sat/

Noting that GV;sat D GL;sat (due to VLE) and GV;sat � GL (since the Poynting factor between P andP sat is small), we finally have

GV D GL CRT ln.P=P sat/

Solving for GL we obtain Eq. [1].

Notice that this detour is necessary only if we want to use Eq. 3 for the calculation. It is much simplerto use Eq. [2] instead.




Problem 10.7 Solution The calculations are done using the following equations:

fi D xiP

�i D Gig0 CRT ln xi D RT ln xi

The results are summarized below

A B after

f N2 (bar) 0:666 0 0:1665

f O2 (bar) 0:334 1: 0:8335

G N2 (J/mol) �1; 013:81 �1 �4; 471:5

G O2 (J/mol) �2; 735:18 0: �454:248




10. VLE OF MIXTURE


a) 84.562 kPA

b) Vapor liquid mixture (why?)

c) At z1 D 0:2 the system is in the two-phase region. The composition of the liquid and the vapor can beobtained either by graphing the data in Pxy form, or by interpolation in the given table. We find x1 D 0:09,y1 D 0:3. The chemical potential is

�i D Gi CRT lnyi

withHi D CPi .T � T0/; Si D CPi ln

T

T0�R ln

P

P0

We find:Hi (J/mol) Si (J/mol K) Gi (J/mol) �i (J/mol)

MeOH 1624:35 15:2283 �3448:95 �6783:7water 1120:47 13:6351 �3422:08 �4410:0




Problem 10.9 Notes on the calculation of T xy graphs

The calculation of T xy starts with a known P and T and solves for x1 and y1. The full T xy graph is obtained byrepeating the calculation at various temperatures between the boiling points of the two components at the pressure ofinterest. This calculation requires an initial guess that is guaranteed to converge. You may obtain such a guess byrunning the a Pxy calculation first. Supposing we want the T xy graph at 1 bar, we do the following:

1. Run the Pxy calculation at some some chosen temperature over a range of mole fractions. If the calculatedrange pressures does not include the desired pressure of the T xy graph (in our case 1 bar), repeat the Pxycalculation at a different temperature.

2. Locate in the output generated in the previous step the pressure that is closest to the desired pressure of theT xy graph. Copy the values of T , x1 and y1 that correspond to this pressure. These will serve as the initialguess of the T xy calculation. Enter these values as well as the value of y2 (which is calculated as 1 � y1) inthe corresponding cells of the input spreadsheet.

3. Run a T xy calculation from temperature T (of the initial guess) to some higher temperature. If at the end of thiscalculation the last mole fractions are still less than one, repeat the calculation to an even higher temperature; ifthe mole fractions exceed one, discard these values from the output.

4. Save the output generated by the previous T xy calculation.

5. Starting with the same initial guess as in step 2, initiate a new T xy calculation, this time to a final temperaturethat is below the initial guess.

6. By combining the calculation at temperatures above the initial guess with those below, you will obtain the fullT xy graph.

Pxy at 350 K T xy at 1 barP(bar) x1 y1 T(K) x1 y10.206 0 0.000 395 0.028 0.1070.311 0.1 0.398 390 0.077 0.2330.416 0.2 0.598 385 0.130 0.3750.523 0.3 0.717 380 0.188 0.4980.630 0.4 0.797 375 0.254 0.6050.737 0.5 0.855 370 0.327 0.6960.846 0.6 0.898 365 0.410 0.7740.955 0.7 0.932 360 0.505 0.8401.066 0.8 0.959 355 0.614 0.8951.177 0.9 0.981 350 0.741 0.9291.289 1 1.000 350 0.741 0.940

349 0.769 0.941348 0.797 0.949347 0.827 0.957346 0.857 0.964345 0.889 0.971344 0.922 0.978343 0.956 0.985342 0.990 0.991




10. VLE OF MIXTURE

Problem 10.10 Solution a)f L1 D f

V1 D 7:150 bar

x1 Df L1

�L1 PD

7:150 bar.21:7 bar/.0:531/

D 0:62

x2 D 1 � x1 D 0:38

f L2 D fV2 D 9:686 bar

y2 Df V2

�V2 PD

9:686 bar.0:844/.21:7 bar/

D 0:5282

y1 D 1 � y2 D 0:4718

�V1 Df V1y1P

D7:150

.0:4718/.21:7/D 0:697

�L2 Df L2x2P

D9:686

.0:38/.21:7/D 1:173

b)S D y1S

ig1 C y2S

ig2 �R.y1 lny1 C y2 lny2/C SR

Sig1 D S

ig2 D 0 (why?)

S D 0C 0 � .8:314/�.0:3/ ln.0:3/C .0:7/ ln.0:7/

�� 1:8 D 3:283J/mol K




Problem 10.11 Solution a & c) The Pxy and T xy graphs are shown below.

500

450

400

350

300

250T

em

pera

ture

(K

)

1.00.80.60.40.20.0

mole fraction of CO2

CO2 – nC5 40 bar; k12 = 0.12

40 bar & 290 K

pure CO2

critical point

B

C

60

50

40

30

20

10

0

Pre

ssure

(bar)

1.00.80.60.40.20.0

mole fraction of CO2

CO2 – nC5 290 K; k12 = 0.12

40 bar & 290 KA

D

0.73239 0.9823

P-xy T-xy

b) The state is shown by point A and is found to lie in the two phase region. From the graph (or, moreaccurately, from the table produced by the spreadsheet) we read x1 D 0:73239, y1 D 0:9823. By lever rulewe find

V D 0:271; L D 0:729

d) The bubble temperature of the solution is 287 K and the dew temperature is 360K. The two temperaturesdefine the initial and final states, shown on the T xy graph as points B and C , respectively. For constant-pressure heating, Q D �H D HC �HB . With reference state 298 K, 1 bar, for both components, the SRKenthalpies are:

HB D �14; 247:04 J/mol

HC D 2; 031:97 J/mol

Therefore,Q D 2; 031:97 � .�14; 247:04/ D 16; 279:01 J/mol K

NOTE: It does not matter what reference state is used in this calculation because we are calculating a changein enthalpy.

e) The initial state is shown as point D on the Pxy graph and we see that it is vapor. The equations to besolved are:

�S D 0

W D �H

We first run an SRK calculation at the inlet conditions (state D); secondly, we run an SRK calculationin which we set the pressure to 40 bar and iterate the temperature until the exit entropy is the same as at




10. VLE OF MIXTURE

the inlet. The work is then obtained by taking the difference in the enthalpy. The results, using the samereference states as above, are summarized below:

Before AfterT (K) 290 486.3H (J/mol) -497.7 11,996.8S (J/mol K) 2.5414 2.5391

from which it followsW D �497:7 � 11; 996:8 D 12; 494:5 J/mol

The positive sign in our convention indicates that this amount of work must be added to the system.




Problem 10.12 Solution a) The Pxy module of VLE produces the results shown on the graph below:

0.0 0.2 0.4 0.6 0.8 1.00

5

10

15

20

x1,y1PHb

arL

b) The state is shown as A in the above graph. The system is in the liquid phase.

c) The state is shown as B . The pressure and compositions are read off the graph (or by interpolation in thetable generated by the spreadsheet). We find

P D 0:72 bar

x1 D 0:02

y1 D 0:9




10. VLE OF MIXTURE

Problem 10.13 Solution a) The T xy calculation requires a good initial guess. This guess is obtainedfrom the Pxy. To do this we run a Pxy calculation until we get a pressure close to the desired 25 bar.Notice that the Pxy graph of the previous part goes only up to about 18 bar. This means that we must runthe Pxy at higher temperature. The procedure is outlined below:

1. Start with T D 250 K, x1 D 0:5 and run a Pxy calculation. The resulting pressure is 16.46 bar < 25bar, thus, a higher temperature is required.

2. Increase the temperature to 270 and repeat the Pxy calculation keeping x1 D 0:5. The new pressureis 20.6 bar.

3. Continue until the pressure is “close” to 25 bar. In fact, at T D 280 K we find P D 25:13 bar andy1 D 0:98378. Thus the first guess is

T D 280 K; P D 25 K; x1 D 0:5; y1 D 0:98378

4. From here on we run the T xy module. Using the above initial guess we find

T D 280 K; P D 25 K; x1 D 0:4966; y1 D 0:9837

5. Changing the temperature by small increments (i.e.. ˙10 K) we can obtain the full T xy graph. Theresults are shown in graphical and tabular form.

500

450

400

350

300

250

T (

K)

1.00.80.60.40.20.0

mol fraction of CO2

CO2 - nC5 at 25 bar

(SRK with k12=0.12)

T (K) x1 y1 Lz1D0:5 T (K) x1 y1 Lz1D0:5

b) The state is defined by the conditions P D 25 bar, z1 D 0:5 and L D 0:5. To find the exact conditions weuse the lever rule to calculate L for all the entries in the above table and locate the entry for which L D 0:5.




By inspection of the table we find

P D 25 bar; T D 350 K; x1 D 0:1818; y1 D 0:8239; L D 0:50

(Notice that L < 0 or L > 1 indicates one phase system).

c) State is indicated by B on the T xy graph.

d) The heat is equal to the change in enthalpy:

Q D �HAB D HB �HA

We calculate AA, HB , using SRK and an arbitrary reference state. Foe simplicity I chose ideal-gas state at0 ıC, 1 bar. We find the following results:

State B: P D 25 bar, T D 400 K, x1 D 0:5, phase = vapor

HB D 8; 758:44 J/mole

State A

Liquid phase: P D 25 bar, T D 350 K, x1 D 0:1818, phase = liquid

HLA D �12; 537:20 J/mole

Vapor phase: P D 25 bar, T D 350 K, x1 D 0:0:8239, phase = vapor

HVA D 2; 817:64 J/mole

Overall system:HA D LH

LA C VH

VA D �4; 859:78 J/mole

Enthalpy change:�HAB D HB �HA D 13; 618:22 J/mole




10. VLE OF MIXTURE


[THIS PROBELM REQUIRES MATERIAL FROM CHAPTER 12]

a) The temperature is above the critical temperature of hydrogen sulfide. We expect critical behavior atelevated pressures.

b) The fugacity of benzene at x2 D 0:7 is

f2 D x2�L2 P D .1 � 0:3/.0:1558/.36:99 bar/ D 4:034 bar

The activity coefficient of benzene is

1 Df1

x1Psat1

D4:034 bar

.0:7/.5:76 bar/D 1:00053

The activity coefficient of benzene t xbenzene D 0:7 (xH2S=0.3) is essentially 1, therefore the solution may betreated as ideal.

c) At 5.76 bar, benzene is pure (x1 D 0, x1 D 1) but at at 36.99 bar is in mixture.




Problem 10.15 Solution THIS PROBLEM REQUIRES MATERIAL FROM CHAPTER 13

Throughout this problem we use the equilibrium conditions

x1k1 D y1P (1)

.1 � x1/Psat2 D .1 � y1/P (2)

a) P sat2 D 0:0403 bar

b) First we calculate Henry’s law constant for CO2:

f1 D x1k1 D x1�11 P ) k1 D �

11 P D .374:04/.0:0403/ bar D 15:0738 bar

The solubility is obtained by solving Eqs 1 and 2 for x1

x1 DP � P sat

2

k1 � Psat1

D 0:0663 bar

c) The pressure is obtained by solving Eqs 1 and 2 for P with y1 D 0:8:

P D1

y1=k1 C .1 � y1/=Psat2

D 0:199368 bar

d) The mol fraction of CO2 in the liquid is

x1 DP � P sat

2

k1 � Psat2

D 0:0105809

By level rule, the fraction of the gas phase is

V Dz1 � x1

y1 � x1D 0:3666

The fraction of CO2 that is recovered in the gas is

recovery Dy1V

z1D 97:8%







11 Ideal Solution




11. IDEAL SOLUTION

Problem 11.1 Solution a) The answer is the bubble point of a solution with x1 D 0:7 (point B in graph):

P D x1Psat1 C .1 � x1/P

sat2 D .0:7/.463/C .0:3/.132:4/ D 363:8 kPa

b) The answer is the mole fraction at point C of the Pxy graph:

y1 Dx1P

sat1

PD.0:7/.463/

.363:8/D 0:891

c)f L2 D x2P

sat2 D .0:3/.132:4/ D 39:7 kPa

This is also equal to f V2 D y2P D .1 � 0:891/.363:8/ D 39:7 kPa.





a) The bubble pressure is

Pbubble D x1Psat1 C x2P

sat2 C x3P

sat3 D 0:926 bar

with x1 D x2 D x3 D 0:33.

The dew pressure is

Pdew D1

y1

P sat1

Cy2

P sat2

Cy3

P sat3

D 0:468 bar

with y1 D y2 D y3 D 0:333.

b) At 1.7 bar we are above the bubble pressure, therefore, the system is liquid.

c) At 0.5 bar the system is between the bubble and dew point, therefore it is a two-phase (V/L) system. Toobtain the composition and amounts of the two phases we solve the flash problem. As sown in the notes, theequations to be solved can be reduced to:

yi D Kixi

xi Dzi

1 � V C VKi

1 DX zi

1 � V C VKi

where Ki D P sati =P . The last equation is solved first (requires trial and error) and the rest by backward

substitution. The final results are:

L D 0:110 V D 0:890

y1 D 0:3627 x1 D 0:0953

y2 D 0:3417 x2 D 0:2653

y3 D 0:2955 x3 D 0:6396

d) This is still a flash calculation and the above equations apply. In this case V D 0:75 and the flash equationis to be solved for the unknown pressure P . The results now are

P D 0:548 bar V D 0:75

y1 D 0:4055 x1 D 0:1168

y2 D 0:3462 x2 D 0:2947

y3 D 0:2482 x3 D 0:5888




11. IDEAL SOLUTION

Problem 11.3 Solution a) Using the ideal-solution equations:

H D x1H1 C x2H2 D �39675:4 J/mol

S D x1S1 C x2S2 �R.x1 ln x1 C x2 ln x2/ D �103:443 J/mol K

V D x1=�1 C x2=�2 D 0:177691 l/mol

b) The heat isQ D �H D x1�H1 C x2�H2

where

�H1 D CP1.TB � TA/ D 3367:5J/mol

�H2 D CP2.TB � TA/ D 4679:4 J/mol

Therefore,Q D 4187:44 J/mol K

c) For the entropy, we have�S D x1�S1 C x2�S2

Notice that the term �R.x1 ln x1C x2 ln x2/ drops out because it is not a function of temperature. �S1 and�S2, is the change in entropy of the pure component for a change of state from .P; TA/ to .P; TB/. Forconstant-pressure process,

�S1 D

TBZTA

CP1dT

TD CP1 ln

TB

TAD 11:0197 J/mol K

�S2 D

TBZTA

CP2dT

TD CP2 ln

TB

TAD 15:3127 J/mol K

Finally,�S D 13:7029 J/mol K

Q: Why is �S positive?




Problem 11.4 Solution a) We take the reference point to be the pure liquid at 40 ıC, 1 bar:

�1 D RT ln x1 D 2078:94 J=mol

b)f1 D x1P

sat1 D 0:055621 bar

c)Pbubble D x1P

sat1 C x2P

sat2 D 0:0784253 bar

d)

Pdew D1

y1=Psat1 C y2=P

sat2

D 0:0591513 bar

e) The bubble temperature is obtained by solving the equation below for T :

P D x1Psat1 .T /C x2P

sat2 .T / ) Tbubble D 115:59

ıC

f) The dew temperature is obtained b solving the following equation for T :

P D1

y1=Psat1 .T /C y2=P

sat2 .T /

) Pdew D 115:59ıC:

g) We solve the flash problem:

V D 0:51956

L D 0:48044

x1 D 0:587255

y1 D 0:323079




11. IDEAL SOLUTION

Problem 11.5 Solution a) At the dew temperature:

y1P D x1Psat1

y2P D x2Psat2

with P D 1 bar, P sat1 D 1:16 bar, P sat

2 D 0:516 bar. Adding the two equations and solving for x1 we find


sat2 ) x1 D

P � P sat2

P sat1 � P

sat2

D 0:756

b) At the dew temperature,

P D1

y1=Psat1 C y2=P

sat2

Here, P D 1 bar, y1 D 0:756, y2 D 0:244. The only unknown is T . We solve for T by trial and error: weguess a value, calculate the right-hand side of the above equation and compare the result to P : if it higherthan P we reduce the temperature and try again; if it is lower than P , we increase the temperature andrepeat. We find

Tdew D 107ıC

c) The flash equation isz1

1C .K1 � 1/VC

z2

1C .K2 � 1/VD 1

With T D 120ıC, P D 1 bar:

x1 D 0:657624

y1 D 0:802952

L D 0:322735

V D 0:677265

In flash calculations with deal solutions, it is not necessary to use the flash equation. Instead, theproblem can be solved more easily as follows. Using the bubble equation,


sat2

If T and P are known, as in part (c), this is immediately solved for x1. The other unknowns followeasily from Raoult’s law.

recovery Dy1V

z1D 72%

d) The equation to be solved in

z1

1C .K1 � 1/VC

z2

1C .K2 � 1/VD 1

and the unknown is T , which is needed in the calculation of the Ki . When all known quantities are substi-tuted, the flash equation becomes

0:75605

0:85C 1043:89e�2911:32

tC216:64

C0:24395

0:85C 1118:35e�3120:29

tC209:52

D 1




This can be solved numerically, or by trial-and-error. Most programmable calculators can solve equationslike this. A simple procedure is to tabulate the values of the left-hand side of the flash equation at varioustemperatures, as done below.

T .ıC/ flash equation50 1:15486

60 1:14454

70 1:13074

80 1:11285

90 1:09033

100 1:06281

110 1:03011

120 0:99233

130 0:94982

140 0:90322

150 0:85339

Clearly, the solution lies between 110ıC and 120ıC. By further refinement we find:

T D 118:072 ıC

x1 D 0:738152

y1 D 0:85747

In Mathematica, this equation can be solved as follows:

flash =0.24395

0.85 + 1118.35 ‰-

3120.29

209.52+t

+0.75605

0.85 + 1043.895 ‰-

2911.32

216.64+t

;

FindRoot@flash ã 1, 8t, 100<D

8t Ø 118.072<

This command instructs Mathematica to solve the equation by trial-and-error for t , starting with the guess,t D 100.




11. IDEAL SOLUTION


a) With z7 D 0:4, zt D 0:6:Pbubble D z7P7 C ztPt D 1:32 bar

Pdew D1

z7=P7 C zt=PtD 1:22 bar

b) Since we need to increase the bubble pressure from 1.32 bar to 4 bar, we must add a component withhigher saturation pressure: normal pentane. Let x be the moles of C5 added per mol of mixture. The newmol fractions are:

z5 D D x=.1C x/

z7 D D 0:4=.1C x/

zt D D 0:6=.1C x/

The bubble pressure of this mixture is

x

1C x.9:2/C

0:4

1C x.1:8/C

0:6

1C x.1:1/ D 4

Solving for x:x D 0:516

Therefore, we need 0.516 mol of normal pentane per mol of original mixture.

c) The mol fractions in the feed are calculated easily. For heptane in the liquid and vapor streams,

Py7 D P7x7 ) P DP7x7

y7D.0:3728/.1:8/

.0:2527/D 1:22 bar

With P known, the remaining mol fractions in the liquid and vapor are calculated using

yiP D xiPi

The vapor mol fraction isV D

zi � xi

yi � xiF

Using any of the three components (they all give the same answer) and with F D 100 mol/min we find

V D 0:671F D 67:1 mol/min

Feed Vapor Liquidmol fraction of n-heptane 0:333 0:3728 0:2527

mol fraction of n-octane 0:167 0:1609 0:1784

mol fraction of toluene 0:5 0:4663 0:5689

mol/min in stream 100 67:1 32:9

T D 120ıC; P D 1.22 bar





a) The problem is asking for the bubble pressure:

P D x1Psat1 C x2P

sat2 D .0:62/.3 bar/C .0:38/.1:81 bar/ D 2:55 bar

b) We are given the bubble pressure and want to fond the dew pressure. We need the composition of themixture first. This is obtained from the known bubble pressure:


sat2 ) x1 D

P � P sat2

P sat1 � P

sat2

D 0:1597

The dew pressure is

Pdew D1

y1=Psat1 C .1 � y1/=P

sat2

With y1 D 0:1597 we findPdew D 1:932 bar

c) We must fist determine whether the phase of the system is liquid, vapor, or vapor/liquid. We calculate thebubble and dew pressures at z1 D 0:61:

Pbubble D .0:61/.3 bar/C .0:39/.1:181 bar/ D 2:536 bar

Pdew D1

.0:61/=.3 bar/C .0:39/.1:18 bar/D 2:39 bar

At P D 2:45 bar the system is in two-phases. The fugacity of component 1 is

f1 D y1P D x1Psat1

To proceed we need to calculate the composition in each phase. We perform a flash calculation at P D2:45 bar, T D 120ıC, z1 D 0:61:

y1P D x1Psat1

y2P D x2Psat2

We add:


sat1 ) x1 D

P � P sat2

P sat1 � P

sat2

D 0:5378

Though not necessary, we also calculate y1:

y1 Dx1P

sat1

PD.0:5378/.3 bar/

2:45 barD 0:6585

Now the fugacity can be calculated either as

f1 D x1Psat1 D .0:5378/.3 bar/ D 1:613 bar

orf1 D y1P D .0:6585/.2:15 bar/ D 1:613 bar




11. IDEAL SOLUTION

Problem 11.8 Solution We need to calculate the bubble and dew pressure at 25 ıC: if P > Pbubble, thephase is liquid; if P < Pdew, the phase is vapor; if Pdew < P < Pbubble the system is a Vapor/Liquidmixture.

Pbubble D .0:833/.2:34/C .0:167/.0:0175/ D 1:9521 bar

Pdew D1

.0:833/=.2:34/C .0:167/=.0:0175/D 0:101 bar

The tank contains a vapor/liquid mixture - you must sample each separately.




Problem 11.9 Solution For all parts of this problem we use the equilibrium conditions:

y1P D x1Psat1 .T / Œ1�

y2P D x2Psat2 .T / Œ2�

along withx1 C x2 D 1 Œ3�

y1 C y2 D 1 Œ4�

a) We know x1 and T and seek P and y1 (“bubble P ” problem). Add Eqs [1] and [2] to eliminate theunknown yi :

P D x1Psat1 .T /C x2P

sat2 .T / D 2:517 bar

y1 D x1=Psat1 .T /=P D 0:281; y2 D 1 � y1 D 0:7185

For the dew pressure, we know y1 D 0:2, T D 100ıC and seek xi , P : Eliminate the unknown xi by solving[1] and [1] for xi and adding them up:

y1 Dx1P

sat1 .T /

Py2 D

x2Psat2 .T /

P

1 Dx1P

sat1 .T /

PCx2P

sat2 .T /

P)

P D1

y1=Psat1 .T /C y2=P

sat2 .T /

D 2:437 bar

x1 D y1P=Psat1 .T / D 0:137

b) To construct the Pxy graph we set T D 100ıC, fix the value of x1, and solve Eq. [1] for P and then Eq.[2] for y1. We repeat this calculation with values of x1 in the interval 0 to 1.c) For the T xy graph we set P D 1:2 bar, fix the value of x1, and solve Eq. [1] for T . This requires trialand error. Once T is known, we use Eq. [2] to obtain y1. We repeat this calculation with values of x1 in theinterval 0 to 1.

T xy at 1.2 bar Pxy at 100 ıC

x1 y1 T ıC x1 y1 P (bar)

0 0 82:6703 0 0 2:26009

0:1 0:156348 81:0019 0:1 0:148332 2:38836

0:2 0:295443 79:4087 0:2 0:281545 2:51662

0:3 0:419544 77:8861 0:3 0:401838 2:64488

0:4 0:530578 76:4298 0:4 0:511003 2:77314

0:5 0:630191 75:0357 0:5 0:610517 2:9014

0:6 0:719796 73:6999 0:6 0:701604 3:02966

0:7 0:800605 72:4188 0:7 0:785293 3:15792

0:8 0:873662 71:1891 0:8 0:862449 3:28618

0:9 0:939867 70:0077 0:9 0:933808 3:41444

1 1 68:8718 1 1 3:5427

d) In addition to [1]–[4] we also haveV D

z1 � x1

y1 � x1D 0:75 Œ5�




11. IDEAL SOLUTION

Here we know z1 D 0:32, T D 100ıC and want to calculate P , x1, x2, y1, y2. We have 5 equations for the5 unknowns, hence problem is solved. Eliminate x2, y2:

P D 2:26009C 1:28261x1

y1 D .3:5427x1/=P

0:75 D .0:32 � x1/=.y1 � x1/

Eliminate P between the first two of these:

1:64461 x1y1 � 4:54259x1 D �2:89798y1

Solve for y1 and substitute into the last equation:

1:56842x21 C 13:7526x1 � 3:53757 D 0

we findx1 D 0:25 y1 D 0:343 P D 2:581 bar





a) At 85 ıC the saturation pressures are P1 D 2:15805 bar, P2 D 1:31188 bar. The pressure is

P D Pbubble D 1:52342 bar

y1 D x1P1=P D 0:354144

b)P D Pdew D 1:45446 bar

x1 D y1P=P1 D 0:1685




11. IDEAL SOLUTION

Problem 11.11 Solution a) We calculate the bubble and dew pressure of the mixture at 85 ıC. At thistemperature the saturation pressures are P1 D 1:147 bar, P2 D 0:599 bar.

Pbubble D .0:82/.1:147/C .0:18/.0:599/ D 1:048 bar

Pdew D1

.0:82/=.1:147/C .0:18/=.0:599/D 0:98 bar

At 1 bar, the system is in the vapor-liquid region. To determine the composition and amounts of the phaseswe begin with the equilibrium conditions:

x1P1 D y1P

x2P2 D y2P

from whichx1P1 C x2P2 D P ) x1 D

P � P2

P1 � P2D 0:7322

andy1 D x1P1=P D 0:8395

Finally by lever rule:V D 0:819; L D 0:181

b) The calculations are done the same way as in part (a) and the results are:

P D 2:7 barT D 120 C

Pbubble D 2:79608 barPdew D 2:67675 barz1 D 0:82

x1 D 0:744162

y1 D 0:833492

L D 0:15104

V D 0:84896





a) At 120ıC the saturation pressures are:

P1 D 6:43239 bar; P2 D 4:34246 bar; P3 D 3:02411 bar; P4 D 1:75725 bar

The bubble pressure is

Pbubble D z1P1 C z2P2 C z3P3 C z4P4 D 3:89 bar

The dew pressure is

Pdew D1

z1=P1 C z2=P2 C z3=P3 C z4=P4D 3:11 bar

b) Vapor.




11. IDEAL SOLUTION


a) Relative humidity

RH DywPtot

P satw .T /

D.0:1/.2/

1:013D 19:75%

b) Dew Temperature:

P satw .Tdew/ D ywPtot

P satw .Tdew/ D exp

hA � B

Tdew

i) Tdew D

B

A � ln.ywPtot/D 334:5 K D 61:3 ıC

c) To avoid condensation at 25 ıC, the new mole fraction of water after dilution must be at most such thatthe dew temperature is 25 ıC:

y0w DP satw .25

ıC/Ptot

D0:0297

2D 0:0149

If we add n moles of dry air per mol of wet gas we then have

y0w ;Dyw

1C n) n D

yw

y0w� 1 D

0:1

0:0149� 1 D 5:7

Therefore we need 5.7 moles of dry air per mole of wet gas.




Problem 11.14 Solution a) At the dew temperature, 25ıC, the air is saturated. Therefore, the molefraction of water in air at the unknown temperature T is equal to the mole fraction at saturation at 25 ıC.

ywP D Psatw ) yw D

P satw .25

ıC/P

At 25 ıC, P satw D 3:166 kPa therefore

yw D3:166

100D 0:03166

b) At the unknown temperature T the saturation mole fraction of water in air is

ysatw .T / D

yw

RHD0:03166

0:75D 0:0422

From the equilibriums condition

ysatw .T /P D P

satw .T / ) P sat

w .T / D 0:0422kPa

From the steam tables we find that temperature where the saturation pressure of water is 0.0422 kPa isapproximately 29.5 ıC.

c) By equilibrium at T 0 D 12 ıC:

ysatw .T

0/P D P satw .T

0/ ) ysatw .T

0/ DP satw .12

ıC/P

D1:401kPa100kPa

D 0:01401

This is the remaining vapor at the exit of the dehumidifier. The amount removed is

% removed D0:03166 � 0:01401

0:03166D 56%

d) The air that exits the dehumidifier is saturated. Its dew temperature is the temperature of the dehumidifier,12 ıC.

e)We have assumed that the amount of air that dissolves in water is negligible. We have also assumed airand water vapor at 1 bar to form an ideal-gas mixture at the temperatures of this problem.




11. IDEAL SOLUTION


a) At the bubble point,P D x1P1 C x2P2

with P1 D 3:024 bar, P2 D 1:757 bar, P D 1 bar we solve for x1 to find

x1 D 0:732

b) At the dew point:

Pdew D1

y1=P1 C y2=P2

With P1 D 1:4799 bar, P2 D 0:7963 bar, Pdew D 1 bar we solve for y1 and find

y1 D 0:4449

Therefore, a mixture that contains 44.49% nitromethane at 111 ıC, 2 bar, is liquid (why?)




Problem 11.16 Solution The calculations are based on the following equations:

f Vi D yi�Vi P

f Li D xi�Li P

f Vi D fLi

Vapor Under the conditions the vapor phase is ideal (why?). Therefore

�V1 � 1; �V2 � 1

The fugacity of each species is

f V1 D y1�V1 P D .0:37/.1/.1:32/ D 0:488 bar

f V2 D y2�V2 P D .1 � 0:37/.1/.1:32/ D 0:832 bar

Liquid Since the system is at equilibrium, f Li D fVi . Therefore,



The fugacity coefficients in the liquid are

�L1 Df L1x1P

D 1:479

�L1 Df L2x2P

D 0:840

Calculation of chemical potentials The chemical potential �i is related to fugacity, fi :

�i D Gigi CRT ln.fi=P / D Gi �GR CRT ln.fi=P /

We need a means of calculating the residual Gibbs energy of pure component i , and a reference state tocalculate the Gibbs energy of pure component i . For the reference state we pick the ideal-gas state at T0,P0. By this convention, the enthalpy, entropy and Gibbs energy at P , T , are:

H.P; T / D

TZT0

CigP dT CH

R.P; T /

S.P; T / D

TZT0

CigP

TdT �R ln

P

P0C SR.P; T /

G.P; T / D H.P; T / � TS.P; T /

(with the subscript i dropped for clarity). If we pick T0 D T D 80ıC, P D 1:32 bar, the Gibbs energy issimply

Gi D Gi .P; T / D HRi .P; T / � TS

Ri .P; T / D G

Ri .P; T /

and the chemical potential is simply:�i D RT ln.fi=P /




11. IDEAL SOLUTION

In general, it is much more work to calculate chemical potential than fugacity because the chemicalpotential requires a reference state while fugacity does not.




12 Non-Ideal Solution




12. NON-IDEAL SOLUTION

Problem 12.1 Solution By writing this as

GE

RTD Ax1x2

we recognize this as the simplified Margules equation with corresponding activity coefficients,

ln 1 D Ax2; ln 2 D Ax21

The value of A is calculated from the data at x1 D 0:3:

1x1Psat1 C 2.1 � x1/P

sat2 D P1 C P2 ) A D 0:881

0.0 0.2 0.4 0.6 0.8 1.0100

150

200

250

300

350

400

x1

PHb

arL





a) The excess enthalpy isHED H � x1H1 � .1 � x1/H2

withH1 D �50 J/mol; H2 D 635 J/mol

b) The partial molar excess enthalpies are obtained from the intercepts of the tangent line onHE at x1 D 0:2:

NH1 D 2249:32 J/mol; NH2 D 112:616 J/mol

c) For isothermal, constant-pressure mixing:

Q D �H D H � x1H1 � x2H2 D HE

At x1 D 0:2:Q D �HE

D 540 J/mol

The heat is added to the solution from the heat bath that maintains the temperature constant.

d) For adiabatic mixing without any work:

Q D �H D 0

Here, �H is the enthalpy if solution at the unknown final temperature T 0 minus the enthalpy of the purecomponents at T D 20ıC:

�H D H.T 0/ � x1H1.T / � x2H2.T / D 0

The solution enthalpy at T 0 is

H.T 0/ D x1H1.T0/Cx2H2.T

0/CCHE.T 0/ D x1

hH1.T /CCP1.T

0�T /

iCx2

hH2.T /CCP2.T

0�T /

iCHE.T 0/

Combining the last two results:�x1CP1 C x2CP2

�.T 0 � T /CHE.T 0/ D 0

To complete the calculation we need the excess enthalpy at the unknown temperature T 0 but we only havedata at T D 20ıC. In the absence of other information we must make an assumption for this unknownenthalpy: we will assume that the excess enthalpy is not significantly different at T 0 and T , i.e.,

HE.T 0/ � HE.T /

This assumption allows us to solve for the unknown final temperature:

Tf � T D�HE

x1CP1 C x2CP2D �2:6ıC

Since the temperature change is small, the assumption HE.T 0/ � HE.T / is acceptable.





This system is endothermic, i.e. it absorbs heat when the components are mixed isothermally. If themixing is done adiabatically without access to a heat bath, the system draws that energy from itself, thusdropping its temperature.

2500

2000

1500

1000

500

0

Excess E

nth

alp

y (

J/m

ol)

1.00.80.60.40.20.0

mol fraction of Benzene

Benzene(1)/Heptane(2) at 20 C, 1 bar

partial molar excessenthalpy of benzene

partial molar excessenthalpy of heptane





a) The partial molar enthalpy, NHi , and the excess partial molar enthalpy NHEi are related as follows:

NHEi D

NHi �Hi

To prove this start with the definitionHED H �H id

and take the partial molar of both sides. To obtain the final result we note that NH idi D Hi (why?)

In this problem, H1 D H2 D 0, therefore, NHEi D

NHi . The calculation of NHi is shown graphically in thegraph below. The solid line is a quadratic fit to four points in the vicinity of x1 D 0:5. The dotted line is thetangent line calculated based on that fit. Because the polynomial fit is pretty much linear, the fitted line andits tangent don’t look much different in this case. From the fitted line at x1 D 0:5 we find

HED �167:3 J/mol; dHE=dx1 D 1698:83 J/mol

The partial molar properties are now obtained as follows:

NH1 D

�HEC .1 � x1/

dHE

dx1

�x1D0:5

D �167:3C .1 � 0:5/.1698:83/ D 682:1 J/mol

NH2 D

�HE� x1

dHE

dx1

�x1D0:5

D �167:3 � .0:5/.1698:83/ D �1016:7 J/mol

with the derivatives calculated numerically from these data. [NEED FIGURE]

b) At 2 bar, 25ıC, the enthalpy of the pure liquids is essentially the same as at 1 bar, 25ıC. This is because theenthalpy, as well other –but not all – properties of liquids are essentially independent of pressure. Therefore,the results with the new reference state are identical to those in the previous part.

c) We mix nA moles of solution A (composition xA1 ) with nB moles of (composition xB1 ) solution B to createa final solution F whose composition is xF1 ). For this isothermal process,

Q D .nA C nB/�H D .nA C nB/HsolutionF � nAHsolution A � nBHsolution B

where�H is the enthalpy change expressed per mole of the final solution. Each of the enthalpies on the lefthand side can be written in terms of the excess enthalpy and the enthalpy of the pure components as follows:

H D x1H1 C x2H2 CHE

where x1, x2 is the composition of that solution and HE is its excess enthalpy (corresponding to the com-position of the solution). Putting everything together we find

�H D HEF �

nA

nA C nBHEA �

nB

nA C nBHEB

With nA D nB , xA1 D 0:5, xB1 D 0:8, the final composition of the solution is xF1 D 0:65. By interpolationin the data we obtain the following values of the excess enthalpy:

x1 = 0.65 HE D 100:7 J/molx1 = 0.5 HE D �167:57 J/molx1 = 0.8 HE D 256:2 J/mol

For the amount of heat we find

Q D �H D 100:7 � .0:5/.�167:57/ � .0:5/.256:2/ D 56:4 J/mol






0 10 20 30 40 6050 70 80 90 100

weight percent hydrazine

1000

500

-500

0En

thal

py

(k

J/k

g)

80 °C100 °C

105 °C

110 °C

115 °C

115 °C

118 °C

118 °C

119 °C

119 °C120.2 °C

60 °C

40 °C

20 °C

-20 °C

-40 °C

-60 °C

0 °C

SOLID/

LIQUID

SOLID/

LIQUID80 120

410

A

B

2

1

a) Exothermic, because the excess enthalpy is negative.

b) The excess enthalpy (or enthalpy of mixing) is

HED HA � x1H1 C x2H2

Using data from the figure above, we find:

H1 D 410 kJ=kJ

H2 D 80 kJ=kJ

HA D 120 kJ=kJ

HED �158 kJ=kg

Notice the the excess enthalpy is the distance between points A and B , and that B lies on the straight linethat connects the pure components at 20 ıC.





0 10 20 30 40 6050 70 80 90 100


1000

500

-500

0Enth

alpy (

kJ/

kg)

80 °C100 °C

105 °C

110 °C

115 °C

115 °C

118 °C

118 °C

119 °C

119 °C120.2 °C

60 °C

40 °C

20 °C

-20 °C

-40 °C

-60 °C

0 °C

SOLID/

LIQUID

SOLID/

LIQUID100

450

A

B

C

B

a) The heat is equal to the distance between points A and B:

Q D HB �HA D 350 kJ=kg

c) The state (point C ) is vapor.






0 10 20 30 40 6050 70 80 90 100


1000

500

-500

0En

thal

py

(k

J/k

g)

80 °C100 °C

105 °C

110 °C

115 °C

115 °C

118 °C

118 °C

119 °C

119 °C120.2 °C

60 °C

40 °C

20 °C

-20 °C

-40 °C

-60 °C

0 °C

SOLID/

LIQUID

SOLID/

LIQUID

A

B

80

410

a) 60 ıC (point A)

b) With H1 D 410 kJ=kg, H2 D 80 kJ=kj, HB D 80 kJ=kg, we find

Q D HB � x1H1 � x2H2 D �132 kJ=kg





a) With x1 D 0 we find 11 D e

AD 2:72

and, similarly, with x2 D 0, 12 D e

AD 2:72

b) At the azeotropic composition (yi D xi ) the equilibrium conditions for the two components give:

xi iPsati D yiP ) iP

sati D P

It follows that

1Psat1 D 2P

sat2 ) eAx

22P sat

1 D eAx2

1P sat2 ) eA.x

22�x

21/ D

P sat2

P sat1

Using the identity x22 � x21 D .x2 C x1/.x2 � x1/ and x1 C x2 D 1 we solve for x1 to find:

x1 D1

2

�1 �

1

AlnP sat2

P sat1

�If this equation has an acceptable solution, i.e.. 0 < x1 < 1, then the system forms an azeotrope at thatcomposition. In our case we find

x1 D 0:535

and you can verify that this leads to a maximum pressure azeotrope with P D 277:6 Torr.

c) The bubble pressure isP D x1 1P

sat1 C x2 2P

sat2

At x1 D 0:3 we find 1 D 1:632, 2 D 1:094, and P D 269:2 Torr.

For the dew pressure we havey1P D x1 1P

sat1 D x1e

x22P sat

1 ŒE1�

y2P D x2 1Psat2 D x2e

x21P sat

2 ŒE2�

with y1 D 0:3, y2 D 0:7. These are to be solved for x1, x2 and P . Solving by trial and error we find

x1 D 0:171; P D 254 bar

(Note: By dividing E1 by E2 and using x2 D 1 � x1 we obtain an equation for x1. Once x1 is known, P iscalculated from either equation E1 or E2.





Problem 12.8 Solution a) At the azeotropic point, yi D xi , therefore, the equilibrium criterion takes thesimplified form

P D iPsati or i D

P

P sati

where i D e

Ax2j

The ratio 1= 2 then becomes

1

2DP2

P1D eA.x

22�x

21/ D eA.x2�x12/

which solved for A gives

A D

lnP sat2

P1satx2 � x1

With x1 D 0:416, P sat1 D 0:81 bar, P sat

2 D 0:41 bar we find A D �4. We may now calculate the pressure atthe azeotropic point as follows:

P D x1 1Psat1 C x2 2P

sat2 D 0:2 bar

b) At 50 ıC, 0.5 bar, pure acetone is vapor and pure chloroform is liquid. To determine the phase of themixture we need to place the pressure of the system in relation to the bubble and dew pressures. For thebubble pressure we have

PB D x1 1Psat1 C x2 2P

sat2

With x1 D x2 D 0:5 and 1 D 2 D 0:363 we find PB D 0:22 bar. Since the system pressure is higherthan the bubble pressure, the mixture is in the liquid phase.

c) The activity coefficients of this system are less than one, indicating preferential attraction between unlikemolecules. We expect full miscibility.





38

37

36

35

34

33

32

31

30

29

28

27

26Pr

essu

re (

kPa)

1.00.90.80.70.60.50.40.30.20.10.0mol fraction of benzene, x 1, y1

Liquid

Vapor

F

E(after mixingthe liquids)

a) P sat1 D 29:8 kPa, P sat

2 D 27:8 kPa

b) Compressed liquid

c) f L1 � 29:8 kPa, f L2 D 27:8 kPa

d) x1 D 2=5 D 0:4: vapor

e) See graph

f) From graph at G: P � 36:8 kPa

g) f L1 D fV1 � y1P . From graph, y1 � 0:47, P D 36:8 kPa, therefore f L1 D 17:3 kPa.

h) We need a model for the activity coefficient. We choose the simple model ln i D Ax2j . At the conditionof the azeotrope:

1 DP

P1D 1:248 ) A D 1:004

2 DP

P2D 1:338 ) A D 1:036

The two values of A calculated above are very close, therefore the model is acceptable. Using an averagevalue A � 1:02,

ln 11 D A D 1:02 ) 11 D 2:77:





Problem 12.10 Solution a) From the given information, A D ln 2:89 D 1:0613. Activity coefficients arelarger than 1, therefore, deviations from ideality are positive.

b) With x1 D 0:634, the activity coefficients are

1 D e.1:0613/.0:634/2

D 1:153; 2 D e.1:0613/.0:366/2

D 1:532

The bubble pressure is

Pbub D .1:153/.0:634/.1:126 bar/C .0:366/.1:532/.0:847 bar/ D 1:298 bar





a) Positive, because 1i > 1.

Note: Activity coefficients are functions of composition. Though not very common, it is possible that the behaviorof a system can turn from positive deviations to negative. For most systems, however, activity coefficients varymonotonically from i D 1 at x1 D 1 to 1i at xi D 0. In the absence of additional information it is reasonableto guess that this system exhibit positive deviations at all compositions.

b)A12 D ln 11 I A21 D ln 12 I ) A12 D A21 D 1:386 D A

c) For an azeotrope, xi D yi , i.e.:

y1P D 1x1Psat1

y2P D 2x2Psat1

�)

1

2DP sat2

P sat1

) A.x22 � x21/ D ln

P sat2

P sat1

) x1 D1

2�

lnP sat2 =P

sat1

2A

If this equation has an acceptable solution (0 � x1 � 1), the system indeed forms an azeotrope. We find

x1 D 0:371

which is acceptable. The pressure is


sat2 D 1:21 bar

d)

0.371 0.950.7920.45

1 bar

0.7 bar

liq.

vap.

0.84 bar

e) First, we need to determine whether the process is feasible in a single flash. With x1 D 0:95 we findP D 0:84 bar and y1 D 0:792. It is impossible to flash a mixture with z1 D 0:45 so that x1 D 0:95. Thisflash can only work if the inlet is 0:792 � z1 � 0:95.






a) To answer this question we need the activity coefficients, so we must first solve for the unknown parameterA. At the azeotropic composition (xi D yi ): At the azeotropic composition (yi D xi ) the equilibriumconditions for the two components give:

xi iPsati D yiP ) iP

sati D P

It follows that

1Psat1 D 2P

sat2 ) eAx

22P sat

1 D eAx2

1P sat2 ) eA.x

22�x

21/ D

P sat2

P sat1

Using the identity x22 � x21 D .x2 C x1/.x2 � x1/ and x1 C x2 D 1 we solve for x1 to find:

A D1

x2 � x1lnP sat2

P sat1

D �4:05284

The activity coefficients at the azeotropic point are 0.256 and 0.501, respectively. The bubble pressure is

P D x C 1 1Psat1 C x2 2P

sat2 D 0:203 bar

b) Acetone is vapor (because P < P sat1 ) and chloroform is liquid (because P > P sat

2 ). To find the phaseof the mixture we must calculate the bubble and dew pressure. With x1 D 0:5 we find 1 D 0:368 and 2 D 0:368. The bubble pressure is

Pbub D x1 1Psat1 C x2 2P

sat2 D 0:221 bar

Since P D 0:5 bar > Pbub, we do not need to check the dew pressure: the system is all liquid.




Problem 12.13 Solution a) We begin with

GE

RTD x1 ln 1 C x2 ln 2

From the van Laar equation with simple manipulation we obtain

ln 1 Dx22B12B

221

.B12x1 C B21x2/2)

x1 ln 1x1x2

D B12B21x2B21

.B12x1 C B21x2/2

A similar expression is obtained for x2 ln 2=x1x2. Thus, for the excess Gibbs energy we find find

GE

x1x2RTD

B12B21

B12x1 C B21x2)

x1x2RT

GE D

�1

B21�

1

B12

�x1 C

1

B12

A graph of x1x2RT=GE versus x1 must be a straight line with slope and intercept

slope D1

B21�

1

B12; intercept D

1

B12

The parameters are then obtained as

B12 D1

intercept; B21 D

1

slopeC intercept

Alternatively, 1=B12 and 1=B12 can be easily obtained from the fitted line at x1 D 0 and x1 D 1, respec-tively (see graph below). The calculations are summarized in the table below:

x1 y1 P (bar) 1 2 ln 1 ln 2 GE=x1x2RT x1x2RT=GE

0.0000 0.0000 0.1230 1.0000 0.00000.0895 0.2716 0.1551 1.3042 1.0088 0.2656 0.0087 0.3893 2.568430.1981 0.4565 0.1861 1.1883 1.0255 0.1725 0.0251 0.3420 2.923570.3193 0.5934 0.2163 1.1138 1.0504 0.1078 0.0492 0.3124 3.200750.4232 0.6815 0.2401 1.0713 1.0779 0.0689 0.0750 0.2967 3.370650.5119 0.7440 0.2592 1.0438 1.1053 0.0429 0.1001 0.2834 3.528500.6096 0.8050 0.2796 1.0231 1.1354 0.0228 0.1270 0.2667 3.749020.7135 0.8639 0.3012 1.0105 1.1633 0.0104 0.1512 0.2484 4.025120.7934 0.9048 0.3175 1.0033 1.1894 0.0033 0.1735 0.2345 4.265110.9102 0.9590 0.3415 0.9970 1.2676 -0.0030 0.2372 0.2268 4.408301.0000 1.0000 0.3609 1.0000 0.0000

When x1x2RT=GE is plotted against 1=x1 we obtain a fairly linear plot. The fitted parameters are:

intercept D 2:4334; slope D 2:2191

from which we obtainB12 D 0:4506; B21 D 0:2149

Using these parameters we now calculate the activity coefficients from the van Laar equation. For example,at x1 D 0:5 we find

1 D 1:0481; 2 D 1:1035





0.0 0.2 0.4 0.6 0.8 1.02.0

2.5

3.0

3.5

4.0

4.5

5.0

x1

x1x

2R

T�

GE

b) The Pxy graph can be easily constructed by solving the bubble P problem. The relevant equations are:

yi Dxi iP

sati

P


sat2

The procedure is as follows:

1. set x1

2. calculate 1, 2

3. calculate the bubble pressure

4. calculate the vapor-phase compositions

5. repeat with different value of x1.

From the gpxy graph we see that the agreement is very good we may conclude that the van Laar equationis good for this particular system. Recall that the Margules equation was also found very good as well. Inthis particular case there is no compelling reason to choose one equation of the other.

0.0 0.2 0.4 0.6 0.8 1.0

0.15

0.20

0.25

0.30

0.35

x1, y1

PHb

arL




Problem 12.14 a) We calculate the activity coefficients:

1 Dy1P

x1Psat1

D.0:053/.48955:97/

.0:2/.6759:44/D 1:919

2 D.1 � y1/P

.1 � x1/Psat2

D.1 � 0:053/.48955:97/

.1 � 0:2/.54382:19/D 1:066

Although 2 is very close to 1, 1 is not. The solution is non ideal.

b)fw D wxwP

satw D .0:2/.1:919/.6759:44/ Pa D 2594:67 Pa

�w Dfw

xwPD .2594:67/=.0:2/.48955:97/ D 0:265






1 2 3 4 5

P bar 1 5:07 5:07 5:07 5:07

T (ıC) 200 247 152 152 152

n (mol/s) 100 100 100 40 60

mol. fraction of heptane 0:6 0:6 0:6 0:525 0:65

H (J/mol) 0 8299 �25813 �8204 �37500

S (J/mol K) 5:56 8:82 �51:3 �26 �68

Reference state (both components): ideal-gas at 200 ıC, 1 bar

W D 829943 J/s Q D 3411200 J/s Sgen D 2338:25 J/K s

(Reference state (both components): ideal-gas at 200 ıC, 1 bar/




Problem 12.16 Solution a) For a system that obeys the Margules equation, the quantity g

g Dx1 ln 1 C x2 ln 2

x1x2

plotted against x1 is a straight line; if the system is described by the van Laar equation, then a graph of 1=gagainst x1 is a straight line. Below we calculate the values of 1, x2, g and 1=g from the data:

x1 y1 P (Torr) 1 2 g 1=g

0:1 0:7385 415:3 2:4535 1:0056 1:053 0:950

0:2 0:8485 649:3 2:2037 1:0250 1:111 0:900

0:3 0:8921 828:3 1:9703 1:0643 1:176 0:850

0:4 0:9149 959:1 1:7551 1:1332 1:250 0:800

0:5 0:9286 1049:7 1:5596 1:2489 1:333 0:750

0:6 0:9375 1108:9 1:3862 1:4439 1:429 0:700

0:7 0:9440 1147:0 1:2374 1:7858 1:538 0:650

0:8 0:9504 1175:9 1:1175 2:4324 1:667 0:600

0:9 0:9621 1208:6 1:0336 3:8146 1:818 0:550

We see that the plot of 1=g against x1 is substantially more linear than that of g versus x1, thus we concludethat the van Laar equation is a better fit. Fitting a straight line we find

slope D �0:5; intercept D 1

From this we find (how?)B12 D 1:0; B21 D 2:0

or,

ln 1 D1�

1C 0:5x1

x2

�2 ; ln 2 D2�

1C 2x2

x1

�2 ;1.0

0.8

0.6

0.4

0.2

0.0

1/(x

1 ln

γ 1 +

x2 ln

γ 2)

1.00.80.60.40.20.0x1

van Laar2.0

1.5

1.0

0.5

0.0

x1 ln

γ 1 +

x2 ln

γ 2

1.00.80.60.40.20.0x1

Margules

b) Positive






a) The Pxy graph is constructed based on the equations below:

ln 1 D0:83�

0:257764x1

x2C 1

�2

ln 2 D3:22�

3:87952x2

x1C 1

�2


sat2

y1 D 1x1P

sat1

P

0.0 0.2 0.4 0.6 0.8 1.00.0

0.2

0.4

0.6

0.8

1.0

x1, y1

PHb

arL

Note: The wrinkle in the graph in the region x1 � 0:6 � 1 indicates that this system separates into twophases when the mol fraction of water is close to 1. The calculation of the Pxy graph of two partiallymiscible liquids is discussed in chapter 13. Here we ignore this separation since the next question refers tothe region of full miscibility.

b) For the flash problem we solve the equations


sat2

y1 D 1x1P

sat1

P

L Dy1 � z1

y1 � x1

with P D 0:5 bar, z1 D 0: We find

x D 0:218764

y D 0:912643

L D 0:378514





The equilibrium conditions are:y1P D x1 1P

sat1 Œ1�

y2P D x2 2Psat2 Œ2�

The activity coefficients are given by:

1 D e�0:327x3

1C1:0221x21�1:0632x1C0:3681

2 D e0:5316x2

1�0:327x31

These are obtained from the standard form of the Margules equation using x2 D 1 � x1 and the numericalvalues of A12, A21.

a) Adding [1] and [2] we obtainP D x1 1P

sat1 C x2 2P

sat2 Œ3�

With x1 D 0:3, we find 1 D 1:05248, 2 D 1:09639 and it follows that

P D 0:257 bar

The mole fraction in the vapor is calculated from [1] to be

y1 Dx1 1P

sat1

PD 0:7380

b) The dew pressure calculation requires trial and error: we guess x1, calculate 1, 2, then from Eq. [3] wecalculate P and y1 (as in part a of the problem), and finally from [1] a new value for x1:

xnew1 D

y1P

1Psat1

The new value of x1 becomes the new guess and we repeat until convergence:





x1 1 2 P (bar) y1 xnew1

0:5000 1:052 1:096 0:2573 0:7380 0:2033

0:2033 1:211 1:019 0:1887 0:4707 0:1295

0:1295 1:280 1:008 0:1678 0:3566 0:109

0:1090 1:302 1:006 0:1615 0:3172 0:1031

0:1031 1:309 1:005 0:1596 0:3050 0:1014

0:1014 1:311 1:005 0:1590 0:3015 0:1009

0:1009 1:311 1:005 0:1589 0:3004 0:1007

After seven iterations, x1 converges to within 3 decimal places. The answer is

x1 D 0:100 P D 0:1589 bar y1 D 0:3

c) The calculation is outlined: given x1 D 1 � 0:88 D 0:12, calculate P and y1 as in part a. We find

P D 0:164879 bar y1 D 0:338847

The liquid fraction is calculated by the lever rule:

L D 0:1775

The recovery of toluene is.0:88/.0:1775/

.0:7/D 22:8%





ln 11 D A ) A D 1:09861; ln 12 D B ) B D 1:38629

b) At x1 D 0:3, the activity coefficients are D1:86432, 2 D 1:09255. The bubble pressure is


sat2 D 0:08 bar < 0:7 bar

Since the actual pressure is higher than the bubble pressure, the system is liquid. (Note: If P were less thanthe bubble pressure, we would have to determine the dew pressure before we could draw a conclusion aboutthe phase.)

c)f1 D x1 1P

sat1 D 0:0678 bar

d)

�1 Df1

x1PD 0:323

e) With x1 D 0:6 the activity coefficients are 1 D 1:25988, 2 D 1:51621. The pressure is equal to thebubble pressure of the liquid:


sat2 D 0:1012 bar

f)

y1 Dx1 1P

sat1

PD 0:906

g) With z1 D 0:8, x1 D 0:6, y1 D 0:906:

V Dz1 � x1

y1 � x1D 0:654 L D 0:346






We write the equilibrium conditions for the two components:

1Ax1APsat1 D 1W x1W P

sat1 Œ1�

2Ax2APsat1 D 2W x2W P

sat1 Œ2�

To solve these equation we write them as

ln 1A D ln 1W C ln.x1W =x1A/ Œ3�

ln 2A D ln 2W C ln.x2W =x2A/ Œ4�

with:

ln 1A D x22A.A12 C 2.A21 � A12/x1A/

ln 1W D x22W .A12 C 2.A21 � A12/x1W /

ln 2A D x21A.A21 C 2.A12 � A21/x2A/

ln 2W D x21W .A21 C 2.A12 � A21/x2W /

After numerical substitution and some algebra, Eqs. [3], [4], become

0:00198246A21 � 0:000958464A12 D 0:733867A12 C 0:125462A21 � 2:58477 Œ5�

0:0599695A12 C 0:877054A21 D 0:00987997A12 � 0:00455097A21 C 3:36622 Œ6�

These are easily solved for A12 and A21:

A12 D 2:90363; A21 D 3:65331

Therefore, the Margules equations for this system are:

1 D e.1:49935x1C2:90363/x22

2 D e.3:65331�1:49935x2/x21

Note: Since x1A and x2W are both close to unity, the corresponding activity coefficients are nearly 1. Wecould have used this fact to simplify the algebra in the calculation of the unknown Margules coefficients byrewriting Eqs [1] and [2] as

x1A D 1W x1W 2Ax2A D x2W

) 1W D x1A=x1W D 13:2603

2A D x2W =x2A D 28:9687)

which are now easier to solve for the Margules parameters. We find

A012 D 2:89986; A021 D 3:63981

which are slightly different from the accurate set obtained above.

a) We obtain the following values for the activity coefficients:

1A D 1:00447 1W D 29:3203

2A D 13:3195 2W D 1:01213




b) The bubble pressure is

P D 1Ax1APsat1 C 2A C x2AP

sat2 D 1:06161 bar

Notice that the same result is obtained if we use:

P D 1W x1W Psat1 C 2W C x2W P

sat2 D 1:06161 bar

The gas-phase composition is

y1 D 1Ax1AP

sat1

PD 0:724

c) The Pxy is shown below (see table and graph). Notice that calculations should only be done in the range0 � x1 � 0:073 and 0:968 � x1 � 1.





x1 y1 1 2 P (bar)

0: 1: 18:2403 1: 0:31177

0:0073 0:9927 17:6755 1:00012 0:41159

0:0146 0:9854 17:1281 1:00046 0:50516

0:0219 0:9781 16:5976 1:00105 0:59277

0:0292 0:9708 16:0836 1:00188 0:67471

0:0365 0:9635 15:5855 1:00295 0:75124

0:0438 0:9562 15:1030 1:00427 0:82263

0:0511 0:9489 14:6355 1:00584 0:88912

0:0584 0:9416 14:1827 1:00767 0:95096

0:0657 0:9343 13:7442 1:00977 1:00839

0:073 0:9270 13:3195 1:01213 1:06161

two-phase region

0:968 0:0320 1:00447 29:3203 1:06161

0:9712 0:0288 1:00362 30:1196 1:04143

0:9744 0:0256 1:00286 30:9449 1:01993

0:9776 0:0224 1:00219 31:7971 0:99702

0:9808 0:0192 1:00161 32:6771 0:97265

0:984 0:0160 1:00112 33:5860 0:94674

0:9872 0:0128 1:00072 34:5248 0:91919

0:9904 0:0096 1:00040 35:4947 0:88994

0:9936 0:0064 1:00018 36:4967 0:85888

0:9968 0:0032 1:00005 37:5321 0:82593

1: 0: 1: 38:6021 0:79098

1.4

1.2

1.0

0.8

0.6

0.4

0.2

0.0

Pre

ssure

(bar)

1.00.80.60.40.20.0

Mole fractionof Acrylonitrile

Acrylonitrile/Waterat 70 C




Problem 12.21 Solution are shown below, along with the results for the pure components (1=nitrogen,2=oxygen).

P (bar) T (K) x1 y1 �L1 �V1 �L2 �V2 ZV ZL

10.30 120 0.00 0.000 2.1486 0.8834 0.8608 0.8608 0.8384 0.035120.87 120 0.70 0.789 0.8612 0.7638 0.5025 0.7153 0.6634 0.101622.17 120 0.79 0.852 0.8074 0.7488 0.4916 0.6964 0.6372 0.114525.37 120 1.00 1.000 0.7108 0.7108 0.4787 0.6468 0.5636 0.1556

The activity of oxygen in the gas at the dew pressure is

a2 D y2�V2

Pdew

P ıD 3:15

The calculations for the activity of oxygen in the liquid at the bubble pressure are summarized below. Allthe required information is obtained from the table above

Pbubble D 22:17 bar

T D 120 K

�L2 D 0:4916

�sat2 D 0:8608

P sat2 D 10:30 bar

V L2 D ZL2

RT

P sat2

D 3:4 � 10�5 m3/mol

exp�P � P sat

2

RTV L2

�D 1:04

2 D 1:18

x2 D 0:21

exp�P � P ı

RTV L2

�D 1:07

a2 D 0:27






a) The activity coefficient is calculated as follows:

2x2�sat2 P

sat exp

".P � P sat

2 /VL2

RT

#D x2�2P ) 2 D

�2P

�sat2 P

sat2 exp

�Zsat2 .P=P

sat2 � 1/

�where Zsat

2 D Psat2 V

L2 =RT . With

x2 D 0:3

�L2 D 0:520

�sat2 D 0:765

P sat2 D 11:02 bar

Zsat2 D 0:093

we find: 2 D 1:00802

b) First we calculate H1 from the data at x1 D 0:

x1H1 D x1�L1 P ) H1 D �

L1 P D .1:912/.11:02 bar/ D 21:1 bar

Using this value, the fugacity at x1 D 0:3 is

f1 D .21:1 bar/.0:3/ D 6:32 bar

From the data in the table,f1 D .0:3/.1:237/.17:22 bar/ D 6:39 bar

Henry’s law agrees to within 1% of the value in the table.

c) At 570 K, the system is above the critical temperature of heptane. Raoult’s’s law cannot be applied.

d)component 1 = n-heptane; component 2 = n-decane

P (bar) T (K) x1 y1 �L1 �V1 �L2 �V2 ZV ZL

11:02 570 0:00 0:000 1:912 0:938 0:765 0:765 0:685 0:093

12:99 570 0:10 0:179 1:627 0:911 0:659 0:723 0:664 0:109

13:50 570 0:125 0:217 1:572 0:905 0:638 0:713 0:659 0:114

17:22 570 0:30 0:431 1:237 0:861 0:520 0:640 0:611 0:148

e, f) Left as an exercise.





a) The pure components correspond to xi D 1:

H1 D 100 kJ=mol; H2 D 80 kJ=mol

b) Infinite dilution corresponds to xi D 0:

H11 D 108 kJ=mol; H12 D 88 kJ=mol

c) The excess partial molar enthalpy is NHEi D

NHi �Hi :

NHE1 D �8x

22

NHE2 D �8x

21

At infinite dilution,NHE11 D �8 kJ=mol NHE1

2 D �8 kJ=mol

d) The enthalpy of mixture isH D x1 NH1 C x2 NH2 D 92 kJ=mol

e) The enthalpy of mixing is

�HmixD x1 NH1 C x2 NH2 � x1H1 � x2H2 D 2 kJ=mol

f) Exothermic





Problem 12.24 Solution The enthalpy of mixing can be expressed in terms of the excess partial molarenthalpies of components

�Hmixx1 NHE1 C x2

NHE2

and the partial molar enthalpies can be expressed in terms of activity coefficients:

NHEi D �RT

2

�@ ln i@T

�P;x

To calculate�Hmix from UNIFAC we need to calculate the above derivatives: we will calculate the activitycoefficients at two temperatures and will calculate the derivative numerically. We use T1 D 10 ıC, T2 D30 ıC to obtain the following values:

T .K/ ln AA ln W283:15 0:216079 0:184396

303:15 0:224609 0:184044

We findNHE1 D �304:73 J=mol NHE

2 D 12:57 J=mol

The enthalpy of mixing is�Hmix

D �104:83 J=mol

This of course is an estimate that is based on the regressed parameters in the UNIFAC database.





a) From the Margules equation with x1 D 0 we find

ln 11 D A12 ) A12 D 1:70475

and similarly (with x2 D 0)ln 12 D A21 ) A21 D 0:94391

b) The bubble pressure is calculated from


sat2

At 100 ıC the Antoine equation gives

P sat1 D 1:013 bar; P sat

2 D 0:570 bar

With x1 D 0:63 the Margules equation gives

1 D 1:1075; 2 D 1:8187

For the bubble pressure we finally findP D 1:090 bar

c) The equations for the dew problem are

x1 Dy1P

iPsat1

P D1

y1

1Psat1

Cy2

2Psat2

The two unknowns are P and x1. We cannot simply solve for P and then back-substitute for x1 becausethe calculation of P requires the activity coefficients which depend on the unknown liquid mole fractions.A trial and error method will be needed. We can easily do this as follows. Rewrite the equation for x1 as

x1 1Psat1

y1PD 1

Iterate now as follows: guess x1, calculate the activity coefficients, calculate the dew pressure, and finallycalculate the ratio x1 iP sat

i =yiP . Continue changing x1 until this ratio becomes 1. The results of thiscalculation are summarized below:

x1 D 0:5917 1 D 1:143

x2 D 0:4083 2 D 1:230

andPdew D 1:088 bar





Problem 12.26 Solution a) The working equation is yi�iP D xi�Li P :

P (bar) x1 y1 �L1 �V1 �L2 �V2 ZV ZL

0:5 0 0:000 128:211 1:009 0:979 0:979 0:979 0:003

30:1 0:5 0:978 1:645 0:841 0:021 0:481 0:799 0:116

34:8 0:6 0:980 1:334 0:817 0:021 0:421 0:761 0:125

42:5 0:8 0:984 0:955 0:777 0:026 0:329 0:692 0:134

53:5 1 1:000 0:716 0:716 0:052 0:221 0:578 0:147

b) This requires the molar volume of the liquid:

VL DZLRT

PD 92:9 cm3=mol

The required volume is 92:9 � 103 cm3.

c) f1 D x1�1P D 24:75 bar.

d) The Poynting factor is

(PF) D exp�P � P sat

1

RT

�with VL obtained from the last row (pure carbon dioxide):

V L1 DZL;sat1 RT

P sat1

We find(PF) D 0:938

e)

i Dx0�

Li P

x1�sat1 P

sat1 (PF)

D 1:378

f) Raoult’s law applies to ideal solutions at low pressure. Here pressure is high (�Vi ¤ 1) and the liquidphase is not ideal ( i D 1:378).





We collect the following from the UNIFAC tables:

�1 �2 GROUP GROUP NO SUBGROUP k R Q MW1 0 CH3OH 6 CH3OH 15 1:4311 1:432 32:042

0 1 CH2 1 CH3 1 0:9011 0:848 15:034

0 1 COOH 20 COOH 42 1:3013 1:224 45:018

A sample calculation of a tie line is shown below:

x1 y1 P .mmHg/ 1 2 GE=RT

0:8 0:965251 1:48741 0:991522 0:935322 �0:0201839

a) See graph below.

0.0 0.2 0.4 0.6 0.8 1.00.0

0.5

1.0

1.5

2.0

methanol

PHb

arL

b) The excess Gibbs energy is positive at all composition (it becomes very slightly negative near x1 D 1).We surmise positive volume of mixing due to positive deviations from ideality.

c) At x1 D 0:8, P D 0:4 bar, T D 80ıC, the system is vapor:

f1 D y2P D .0:8/.0:4 bar/ D 0:32 bar

d) At the bubble point the phase is liquid.

f1 D 1x1Psat1 D .0:8/.0:991522/.1:81/ D 1:436 bar

e) The chemical potential is

�1 D G1 CRT ln 1x1 D 0C .8:314 J=mol K/.353:15 K/ ln.0:991522 � 0:8/ D �680:2 J=mol

where G1 is made zero by the choice of the reference state.

f) The new reference state is at the same temperature but at P D 1:81 bar instead of 0.4 bar. Since pressurehas negligible effect, the answer is essentially the same. For a more accurate answer we would need tocalculate the Poynting factor but this is left as an exercise.






DUPLICATE PROBLEM 12.27





�1 �2 GROUP GROUP NO SUBGROUP k R Q MW1 0 CH2 1 CH3 1 0:9011 0:848 15:034

1 0 COOH 20 COOH 42 1:3013 1:224 45:018

0 1 H2O 7 H2O 16 0:92 1:4 18:016

0.0 0.2 0.4 0.6 0.8 1.00.5

0.6

0.7

0.8

0.9

1.0

1.1

acetic acid

PHb

arL

UNIFAC predicts an azeotrope at 100 ıC, however, no azeotrope is reported for this system.







13 Miscibility, Solubility and other PhaseEquilibria




13. MISCIBILITY, SOLUBILITY AND OTHER PHASE EQUILIBRIA


We need a model for the activity coefficients. Here we will assume that the Margules equation is acceptable.Other equations may be used, e.g. Van Laar, UNIFAC, etc., but the results may vary. Recall that theMargules constants are related to the infinite dilution activity coefficients as follows:

A12 D ln 11 ; A21 D ln 12

a) Hexane in methanol - For this system we have

A12 D e18:97

D 2:943; A21 D e21D 3:045

We construct a graph of �G=RT versus x1, where

�G

RTD x1 ln 1x1 C x2 ln 2x2

This graph is shown below.

-1.0

-0.8

-0.6

-0.4

-0.2

0.0

∆G/RT

1.00.90.80.70.60.50.40.30.20.10.0

Mol fraction of hexane

hexane-decane

-0.10

-0.05

0.00

0.05

0.10

∆G/RT

1.00.90.80.70.60.50.40.30.20.10.0

Mol fraction of hexane

AB

hexane/methanol

We observe the presence of concave part, therefore we conclude that this is system exhibits partial miscibil-ity. To obtain the composition of the two phases we draw a line such that it is tangent to the curve at bothcontact points. The two contact points represent the composition of the two phases. Reading from the graphwe find:

� Methanol-rich phase: xH � 0:08

� Hexane-rich phase: xH � 0:94

b) The infinite dilution activity coefficients are close to and somewhat less than 1. This indicates nearlyideal behavior, with somewhat preferential attractive interactions. Recall that at infinite dilution we normallyobserve the largest deviation from ideality, so at intermediate concentrations we expect behavior even closerto ideal. Based on these observations we expect complete miscibility of the two components. To verify thisconclusion one may draw the plot of �G=RT versus x1 based on these activity coefficients. This plot isindeed convex at all concentrations.




Problem 13.2 Solution a) By equilibrium we have:

ywP D xw wPsatw � P

satw

yCP D xCkH

where we set xw � 1 from which it follows w � 1. Adding the two equations we find

P D P satw C xCkH

From the steam tables we find P satw D 0:3116 bar and from the above graph kH D 4000 bar. Therefore,

P D 0:3116C 5 � 10�4.4000/ D 2:31

The vapor composition is

yw DP satw

PD0:3116

2:31D 0:134

yC D 1 � yw0:865

b) We need a temperature such that

P D P satw .T /C x

0CkH .T /

where x0C D 25:0 � 10�4 is the new mole fraction of carbon dioxide. In this equation, the saturationpressure of water may be dropped because it makes a small contribution to the total pressure.1 With thisapproximation,

kH DP

x0CD

2:31

25:0 � 10�4D 924 bar

That is, at the desired temperature, kH D 924 bar. From the graph we find T D 5 ıC. (From the steamtables at 5 ıC we find P sat

w D 0:00872 bar. Therefore, the assumption that P satw is negligible was justified in

this case.)

c) The new vapor composition is

yw DP satw

PD0:00872

2:31D 3:8 � 10�3

yC D 1 � yw D 0:996

d) We have assumed that the activity coefficient of water in the liquid is approximately 1. This is justifiedbecause of mole fraction of water in the liquid is nearly 1. We have also assumed that all the fugacitycoefficients and the Poynting factor can be neglected.

1This approximation is not always correct. If the temperature is sufficiently high, the saturation pressure of water may be asignificant fraction of the total pressure.





Problem 13.3 Solution a) The fugacity of CO2 can be calculated either using Henry’s modified law,fi D

Hi Hixi , or using the general equation fi D xi�Li P . Setting the two expressions equal to each other

we find

Hi Hixi D xi�Li P ) Hi D

�Li P

Hi

At xi D 0 we have by definition Hi D 1, therefore,

Hi D �L;1i P

where �L;1i is the fugacity of species i in the liquid at infinite dilution (xi D 0). We now run a Pxycalculation for this system at T D 344:15 K, x1 D 0 (1=CO2, 2=n-pentane) and find

�L;1CO2D 39:2; P D 2:93 bar

from which it followsHCO2 D .39:2/.2:93/ D 114:9 bar

b) The fugacity plots are shown below. In the SRK calculation the fugacity is computed as

f Li D xi�Li P

with �Li obtained from the SRK. The fugacity from Henry’s law is computed as

f Li D xiHi

We note that Henry’s law is good at low concentrations (up to about x1 � 0:2tp0:3) but it deviates at higherconcentrations. We also notice that the Henry line is tangent to the actual fugacity.

c) The dependence of of Hi on pressure at constant temperature is given by

Hi .P2/ D Hi .P1/ exp��P2 � P1

RTNV1i

�We now HCO2 at P D 2:36 bar but we want to know how much it changes at higher pressures. To usethe above equation we need NV1CO2

, the partial molar volume of CO2 at infinite dilution. We leave it as anexercise (these derivations are shown in the textbook) to show that

NV1 D V C .1 � x1/dV

dx1

(1=CO2). At x1 D 0 we obtain

NV11 D Vx1D0 CdV

dx1

ˇx1D0

We can obtain NV11 as follows: calculate the molar volume of solutions as a function of x1, fit a line throughthe points, use the fitted line to calculate the derivative dV=dx1 at x1 D 0, and plug into the above equationto calculation NV11 .




The molar volume V of the solution can be easily obtained from the SRK equation. If we know the com-pressibility factor of the liquid, ZL, the molar volume is

V L D ZLRT

P

(Recall that ZL is computed as part of the calculation of fugacity). The graph of V versus x1 is shownabove. If we fit a polynomial

V D K0 CK1x1 CK2x21 C � � �

we see thatVx1D0 D K0

anddV

dx1

ˇx1D0

D K1

from which it follows thatNV11 D K0 CK1

That is, NV11 is obtained from the first two coefficients of the fitted polynomial. Here we have used a 10th

degree polynomial to fit all the points of the graph (a lower-order would be sufficient if we only fitted theregion close to x1 D 0) and we find

K0 D 1:39 � 10�4; K1 D �7:54 � 10

�5) NV11 D 6:40 � 10

�5 m3/mole

We now compile the following results for H1:

P (bar) Poynting Factor HCO2 (bar)2.39 1.00 115.020 1.04 119.440 1.09 124.980 1.19 136.5

That is, at the highest pressure (80 bar) Henry’s law constant increases by about 20%. The change is notnegligible but notice that it took a pressure change of 3,000% to produce a 20% change in HCO2 .

� NOTE: Our calculation is not exact. We have used NV11 corresponding to P D 2:39 bar to docalculations at higher pressures while we should be calculating a new NV11 at each pressure. That is,we have assumed that NV1i is not a strong function of pressure. How good is this assumption?





Problem 13.4 Solution The material balance in the scrubber is

A.yin�yout/ D Wxout Œ1�

where A and W are the molar flow rates of the air and water streams, respectively, and the subscripts “in”and “out” refer to the inlet and outlet streams respectively. Since the mole fraction of DMA is so much lessthan unity, we take the molar flow rates of air and water to be approximately constant between inlet andoutlet.

Assuming that the DMA in the gas and the liquid streams exiting the scrubber are at equilibrium, we alsohave:

youtP D xoutH Œ2�

These are the two equations that solve this problem. We will need Henry’s law constant for DMA in waterat 25ıC. This is H D 1:84 bar.

a) In this case we know yin D 0:01, yout D 100 � 10�6 D 0:0001, P D 1 bar. Equations [1] and [2] aresolved for W=A and xout. We find

W=A D 181:689 mol of water/mol air D 113 kg water/kg air

xout D 54 ppm

b) In this part of the problem we know W=A but not P . With

W=A D 50 kg water/kg air D 80:55 mol water/mol air

Eqs [1] and [2] are solved for P and xout:

P D 2:26 bar

xout D 123 ppm

c) The two streams meet in counter-current configuration. This means that the the purified air stream and thewater/DMA streams are not in direct contact, therefore, they are not necessarily equilibrated. The assump-tion would have been acceptable if the contents of the tank were vigorously mixed, allowing equilibration.Our calculation is approximate. A more accurate calculation would have to take into consideration howquickly DMA is transferred from one phase into the other. This is covered in courses on mass transfer.




Problem 13.5 Solution a)P D xH C .1 � x/P sat

w Œ1�

at 70ıC, H D 3900 bar, P satw D 0:3166 bar and we find

P D 2:26 bar

yCO2 DHxCO2

P

yw DxwP

satw

P

from which we findyw D 0:1396; yCO2 D 0:8604

b and c) The working equation is still given by Eq. [1] except that now we know P and need to calculate T .This would require trial and error but his can be avoided if we notice that in order to increase the amount ofCO2 in the liquid the temperature must be decreased. Since P sat

w decreases very rapidly with temperature,we assume that P sat

w �D 0 at the temperature of part (b). This assumption will need to be checked at theend of the calculation. With this, Eq. [1] becomes

P � xH ) x DP

HD2:26 bar0:00025

D 904 bar

From the given chart we findT D 6ıC

At 6ıC, the saturation pressure of water is 0.00935 bar. This is less than 0.5% of the total pressure in thechamber. Therefore, the assumption that P sat

w can be neglected is valid. To confirm, calculate new estimateof gas phase fractions based on the tabulated value of P sat

w at 6 ıC:

xwPsatw D ywP ) yw D 0:0041; yCO21 � yw D 0:9959

The mol fraction of CO2, is nearly 1, as in the initial guess.

d) Assumptions:

1. equilibrium

2. H law applicable

3. vapor phase ideal

4. Poynting factor negligible





Problem 13.6 Solution a) We begin with the equilibrium conditions in the tank:

y1P D k1x1ywP D xwPw

) x1 DP � Pw

k � PwŒ1�

with k1 D 879:758 bar and Pw D 0:1244 bar. Thus we obtain x1 and by back substitution all the othermole fractions:

x1 D 0:000995 x2 D 0:999 y1 D 0:8757 y2 D 0:1243

b) The conditions of stream G are: P D 5 bar, T D 25 ıC, z1 D 0:8757. To determine whether this is asingle or two-phase system we calculate the dew pressure:

Pdew D1

y1=k1 C y2=Pw

With k1 D 539:7 bar, Pw D 0:0314 bar we find Pdew D 0:252 bar. Since PG > Pdew we conclude that thesystem is two-phase system, i.e. that some vapor condenses. To calculate the amounts of the two phases, wefirst calculate the composition of each individual phase. These are calculated from Eq. [1] with k1 D 539:7bar, Pw D 0:0314. We find

x1 D 0:0092 y1 D 0:9938

By lever rule:

L Dy1 � z1

y1 � x1D0:9938 � 0:8757

0:9938 � 0:0092D 0:12





a) We write the dissolution of ammonia as a chemical reaction:

NH3.g/! NH3.aq/

with

K25 D exp��

�10050

.8:314/.298:15/

�D 57:6477

The activity of ammonia in the gas is ag D PA=P0 where PA D 2 bar, and in the liquid it is aaq D c=c0.Therefore:

c=c0

PA=P0D K ) c D

Kc0PA

PAD 115:295 mol/kg

This corresponds to a mol fraction of ammonia,

x DcMW

1C cMW

where MW D 18 � 10�3 kg/mol is the molecular weight of water. Therefore,

xA D 0:6748

b) The Gibbs energy of formation can be computed for various standard states. If s1 and s2 are two suchstandard states, then the corresponding Gibbs energies are related as follows

�Gı.s1/ ��Gı.s2/ D RT ln

f ıi .s2/

f ıi .s1/(2)

where f ıi .s1/ and f ıi .s2/ is the fugacity of the species in standard state 1 and standard state 2, respectively.

Proof The chemical potential based on standard state s1 is

� D �Gıs1CRT ln

f

f ıs1

where fs1 is the fugacity of component in the standard state. Using this formula to calculate the Gibbs energy atstandard state s2 we obtain

�Gıs1D �Gıs1

CRT lnf ıs2

f ıs1

With s1 D .g/, s2 D .aq/, the corresponding fugacities are:

fgi D P0

(since the “g” standard state prescribes “pure gas in the ideal gas state at P0, T ”) and

faqi D c0H

0D x0H

(since the “aq” standard state prescribes a“solute with molality c0 D 1 mol/kg obeying Henry’s law at P0,T ”). Here, H 0 is Henry’s law expressed in bar kg/mol (works with the concentration of solute expressed as





molality), H is the regular Henry’s law constant (works with the solute expressed in mol fraction) and x0 isthe mol fraction of solute that corresponds to molality 1 mol/kg:

x0 Dc0MW

1C c0MW

(MW is the molecular weight of solvent, in kg/mol). Putting everything together,

exp

"��Gıaq ��G

ıg

RT

#Dx0H

P0D

c0MWH

P0.1C c0MW/

and solving for H :

H DP0.1C c0MW/

c0MWexp


ıg

RT

#The numerical substitutions are left as an exercise.





[NOT AVAILABLE]






We collect the Antoine parameters for the two components:

Pw=bar D 105:20389�1733:93

T=ıCC233:665

P7=bar D 104:02832�1268:64

T=ıCC216:951

a) The bubble temperature is obtained by solving the equation

P D Pw.T /C P7.T / ) T D 78:8ıC

b) The dew temperature is obtained by solving the equation

P DPW .T /

yW) T D 81:3ıC

c) We first calculate the triple point at 50ıC:

P � D PW .T /C P7.W / D 0:312 bar

The corresponding mole fractions are

y�W DPW

PbubbleD 0:395

In a mixture with yW D 0:5, water will condense first (why?). The dew point is defined by the condition

Pdew DPW

P �D 0:247 bar





First, we calculate the temperature and vapor composition at the bubble point of the two-liquid system atP D 2 bar:

P D PW .T�/C PO.T

�/ ) T � D 108:9 C

y�W DPW .T

�/

PD 0:6898

For y < y�W , the dew line is give by the equation

.1 � yW /P D PO.T /

For y > y�W , it given by:yW P D PW .T /

The solution is much easier to do if we sketch the T xy graph:

160

140

120

100

80

60

40

20

Tem

pera

ture

1.00.80.60.40.20.0

mol fraction of water

(a)

(b)

(c)

y*=0.6898

(1-yW) P = PO(T)

yW P = PW(T)

a) Since yW < y�W , the calculations are based on octane, which condenses first:

.1 � yW /P D PO.T / D e9:3222�3120:29=.TC209:52/

Solving for T :T D 113ıC

The first liquid drop is octane (no water).

b) At 115 ıC we are above the bubble point; we need to determine the dew point of the mixture. Since themol fraction of water is above y�W we must work with the equilibrium equations for water:

yW P D PW .T / D e11:6832�3816:44=.TC227:02/

Solving for T we findT D 111:4ıC





c) In a vapor liquid mixture with zw D 0:75 > y�W , water is always present. Since the problem says that wealso have octane, we must be on the triple line. LetW , O and V be the moles of the water phase, the octanephase, and the vapor phase. By mol balance:

W CO C V D 100

1C y�W V D zW

With O D 10, zW D 0:75, y�W D 0:6898, we find

W D 41:6; V D 48:4




Problem 13.11 Solution First we calculate P � and y�w . At 50 ıC, the saturation pressures are

P satw D 0:1232 bar P sat

o D 0:0671 bar

Then we findP � D P sat

w C Psato D 0:1232C 0:0671 D 0:1904 bar

and

y�W DP satW

P �D0:01232

0:1904D 0:647

y�O D 1 � y�W D 0:353

The questions can be answered more easily if we construct a qualitative Pxy graph based on the informationwe have for the system.

0.4

0.3

0.2

0.1

0.0

Pre

ssure

(bar)

1.00.80.60.40.20.0

mol fraction of water

water/octane at 50 C

a) The liquid will begin to boil at P D P � D 0:1904 bar.

b) The composition of the first bubble is y�o D 0:352.

c) Water, because the overall composition of the solution, zo D 0:68, is above y�o D 0:352.

d) As long as there are two liquid phases present the pressure remains the same and equal to P � D 0:1904

bar. This is the pressure at the moment that the last drop of the water phase disappears.

e) The second liquid (octane) phase boils off when the system reaches the dew point. At this point thecomposition of the vapor is equal to the initial mol fraction of water, yo D 0:68 and the remaining liquid isoctane. The equilibrium criterion for octane gives

yoP D oxoPsato � P

sato ) P D

P sato

yoD0:0671

0:68D 0:099

f) yo D 0:68 (see above).





Though the problem does not ask for it, we outline the procedure for generating the T xy graph shownhere.

For yO � y�O , the equation for the bubble line is

ywP D xwPw ) 1 � yO DP satw .T /

P

For yo � y�0 , the equation for the bubble line is

yoP D xoPsato ) y0 D

P sato .T /

P

These equations establish the relationship between y0 and T . To construct the graph, set the value ofy0 and then solve for P .




Problem 13.12 Solution In this problem the pressure is fixed, therefore, we should be looking at T xygraph. First we calculate T � and y�w .At the triple point (two liquids plus vapor)

P D P satw .T

�/C P sato .T

�/

where P � D 1 bar. Solving for T � by trial and error we find

T � D 89:1 ıC; y�o DPo.T

�/

PD0:324

1D 0:324

140

120

100

80

60

40

Tem

pera

ture

(C

)

0.80.60.40.20.0

Mol fraction of octane

yo = 0.2

a) At T D T � D 89:1 ıC.

b) yo D y�o D 0:324.

c) Octane, because once the system moves above T � it is found to contain vapor and liquid water only.

d) T D T � D 89:1 ıC, because temperature remains constant as long as there are three phases present.

e) When the water phase boils off too, the vapor composition is yo D 0:2. The equilibrium criterion forwater gives

ywP D wxwPsatw � P

satw .T / ) P sat

w .T / D .0:8/.1/ D 0:8 bar

Solving for T we find T D 93:5 ıC.

f) yo D 0:2 (see above).

Though the problem does not ask for it, we outline the procedure for generating the T xy graph shownhere.


ywP D xwPw ) 1 � y0 DP satw .T /

P






yoP D xoPsato ) y0 D

P sato .T /

P

These equations establish the relationship between y0 and T . To construct the graph, either set the valueof y0 and then solve for T by trial and error, or, set the value of T and solve for yo. The latter methodis easier.




Problem 13.13 Solution Before we answer the specific questions it is useful to calculate some basicresults. First, we calculate the activity coefficients of the minority component in each phase:

nw D1 � 0:147

0:083D 10:27 wn D

1 � 0:083

0:147D 6:238

where the subscript indicates the component and the superscript the phase. The bubble pressure of the twophase system is

Pbubble D nwx

nwPw C

wn x

wn Pn D 687 Torr

and the composition of the vapor at the triple point is

y�w D nwx

nwPw

PbubbleD 0:971

With this information we sketch the Pxy graph below:

water

pre

ssure

(T

orr

)

21

760

667

0.083 0.853 0.97

water-richnitrobenzene-rich

vapor

a) Assuming the liquid phases to be dilute enough in the minority component, the activity coefficients atinfinite dilution are those calculated above:

nw D1 � 0:147

0:083D 10:27 wn D

1 � 0:083

0:147D 6:238

b) Gibbs’s phase rule with two components and three phases (two liquids and a vapor) gives

F D N C 2 � �4 � 3 D 1

With one degree of freedom, if pressure is fixed, so is temperature. Therefore, boiling at constant pressuremeans that temperature is also constant.

c) At 50-50 the state is in the two-phase region. The pressure is that of the triple point, 667 Torr.

d) The vapor contains 85.3% by mol water.

e) A tie-line just below the triple line connects the vapor to the nitrobenzene-rich phase. There is no waterrich phase present, i.e., the water phase has disappeared at this point.

f) See above.





Problem 13.14 Solution This problem is analogous to example 8.2 in the notes. We notice that A D2:8 > 2, therefore we predict partial miscibility. To obtain the composition of the two phases we shouldgraph �G=RT versus x1, but in this case the solution is simpler because of the simplicity of the equationthat gives the activity coefficients. From example 8.2 we see that the graph of�G=RT is symmetric, that is,the double tangent line contacts the Gibbs energy at the points where �G=RT has a minimum. This pointis defined by the condition that

d

dx1

��G

RT

�D 0

and using Eq. 8.22 in the notes the derivative is

ln x1 � ln.1 � x1/C A.1 � 2x1/ D 0

with A D 2:8. Solving by trial and error we find (1=hexane, 2=ethanol)

x1 D 0:093 and x1 D 0:907

(Notice that x1 D 0:5 is also also a solution but it does not correspond to either minimum – this is themaximum of the curve at the middle.)

a) The Pxy is calculated as usual using the following equations:


sat2

y1 D Dx1 1P

sat1

P

For a calculation, we fix x1, calculate the two activity coefficients using ln i D 2:8x2j , and then obtain thebubble pressure and vapor composition from the above equations. The calculation is done from x1 D 0 tox1 D 0:093 and from x1 D 0:907 to x1 D 1. Results are summarized in the table below:

x1 1 2 y1 P (bar)0.00 16.44 1.00 0.00 1.310.02 14.72 1.00 0.27 1.770.04 13.20 1.00 0.41 2.130.06 11.87 1.01 0.48 2.410.08 10.70 1.02 0.53 2.630.093 10.02 1.02 0.56 2.740.907 1.02 10.02 0.56 2.740.92 1.02 10.70 0.58 2.660.94 1.01 11.87 0.63 2.490.96 1.00 13.20 0.70 2.270.98 1.00 14.72 0.81 1.991.00 1.00 16.44 1.00 1.64

Notice that the bubble pressure at x1 D 0:093 and x2 D 0:907 is the same, as it should.

b) The specified state is shown as point A on the Pxy graph. The given composition is between the that ofthe two liquids so the system consists of two liquid phases. The bubble point is at 2.74 bar, which the bubblepoint of the two-liquid phase system.




Problem 13.15 Solution The solution is much easier to understand if we first sketch the T xy graph forthis system.

100

80

60

40

Tem

pera

ture

(°C

)

1.00.80.60.40.20.0

Mole fraction of benzene

V+W B+V

Vapor

B+W

benzene/water at 1 bar

a

c

b

y* = 0.7066

To proceed we must also know the composition y�B corresponding to the bubble point of the two-liquidsystem. This point is defined by the conditions:

P D P satB C P

satW

y�B DPB

P

We solve the first equation by trial and error for T � (the boiling temperature of the two-phase system) andsubstitute to the second equation to obtain y�B . We find

T � D 68:6 ıC

y�B D 0:7066

a) T D T � D 68:6 ıC.b) yB D y�B D 0:7066.c) Benzene.d) T D T � D 68:6 ıC.e) This is point c on the T xy graph. The equilibrium condition for water is

yW P D PW

with yW D 0:75. We solve for T by trial and error and find

T D 91:8 ıC

f) The composition of the vapor is yB D 0:25.






The equilibrium conditions are

ywP D wxwPsatw

ybP D bxbPsatb

Since xw D 0:96 we assume that w � 1, therefore

yw D wxwP

satw

PD.1/.0:96/.0:7849/

1:013D 0:744

where P satw D 0:7849 bar.

When 3 phases are present the equilibrium equations are:

ywP D IwxIwP

satw D

IIw xIIw P sat

w

ybP D Ib xIbP

satb D

IIb xIIb P sat

b

where the superscripts I and II indicate the water-rich and butanol-rich phases, respectively. For the activitycoefficients we now have

IIw DywP

xIwPsatw

D.0:744/.1:013/

.0:6/.0:7849/D 1:60

Similarly for butanol

IIb DybP

xIbP satb

D.0:256/.1:013/

.0:4/.0:387/D 1:67

Similarly one can calculate the activity coefficient of butanol in the water-rich phase:

IIb DybP

xIbP satb

D.0:256/.1:013/

.0:04/.0:387/D 16:7

As expected, the activity coefficients are larger than one (positive deviations from ideality).

P = 1.013 bar

T=93 C

0 0.04 0.4 1.0

100 C





semipermeable

membrane

membrane

sea

water

clean

water

brine1

3

2

Stream 3 is at osmotic equilibrium at P D 42 bar. The mol fraction of the salt in this stream is

x2 D1

2

�1 � exp

��…Vw

RT

��D 0:01494

with the factor of 2 accounting for the fact that both NaC and Cl� contribute to the osmotic pressure (i.e.,the concentration of the NaCl is half of that of the ions). The corresponding wight fraction is

w3 D 0:0468929

By mass balance on salt (no salt in stream 2), the mass flow in stream 3 is

m3 Dm1w1

w3D 0:746 kg

where we used a basis of m1 D 1 kg. The mass flow of stream 2 is

m2 D m1 �m3 D 0:254 kg

Therefore, the process recovers 25.4% of the water in the feed in the form of purified water.







14 Reactions




14. REACTIONS


a) The stoichiometric coefficients are shown below:

Na2S2O3 C 2H2O2 ! 0:5Na2S3O6 C 0:5Na2SO4 C 2H2O

b) We setup a stoichiometric table as follows (assuming 1 mole of solution initially):

species: 1 2 3 4 5 (sum)�i �1 �2 0.5 0.5 2ni0 0.1 0.1 0 0 0.8ni 0:1 � � 0:1 � 2� 0:5� 0:5� 0:8C 2� (1)xi 0:1 � � 0:1 � 2� 0:5� 0:5� 0:8C 2� (1)

At completion we have � D 0:05 (why?), therefore, the mole fractions are as follows:

x1 D 0:05

x2 D 0

x3 D 0:025

x4 D 0:025

x5 D 0:9




Problem 14.2 Solution The stoichiometric table is shown below:

H2 N2 NH3 He

�i �3 �1 2 0

n0i 1 1 1 1

ni 1 � 3� 1 � � 1C 2� 1

yi1 � 3�

4 � 2�

1 � �

4 � 2�

1C 2�

4 � 2�

1

4 � 2�

a) All moles ni must be positive:�min D �1=2 �max D 1=3

b) With yHe D 0:3 we find:

� D 0:33; yH2 D 0: yN2 D 0:2 yNH3 D 0:5 yH2 D 0:3

The conversion of hydrogen is

conversion H2 D3�

1D 100%

c) With yHe D 0:215 we find:

� D �0:325581 yH2 D 0:425 yN2 D 0:285 yNH3 D 0:075 yH2 D 0:215

The conversion of hydrogen is

conversion H2 D3�

1D �97:6%

It makes more sense here to report the conversion of ammonia:

conversion NH3 D�2�

1D 65:1%




14. REACTIONS


The reaction isCH4 .g/C 2O2 .g/! CO2 .g/C 2H2O .g/

We construct the stoichiometric table:

CH4 O2 N2 CO2 H2O

� �1 �2 0 1 2 .�/

�H ıf

�74520 0 0 �393509 �241818 .J=mol/�Gı

f�50460 0 0 �394359 �228572 .J=mol/

CP .25ıC/ 51:0308 31:5446 30:1466 46:4956 34:9572 .J=mol=K/

n0 1 2:4 9:029 0 0 .�/

n 1 � � 2:4 � 2� 9:029 � 2� .�/

The calculation path is:

1. Un-mix the inlet stream into pure components at constant T D Tinlet, P D Pinet. The enthalpychange for this step is to a very good approximation zero because the components form essentially anideal-gas mixture, and they are already at the required pressure and temperature:

�H1 D 0

2. Bring each pure component to P0 D 1 bar, T0 D 25ıC. The enthalpy change for this step is zerobecause as it happens, the inlet temperature and pressure are exactly the same as T0, P0.

�H2 D 0

3. Run the reaction at 25ıC, 1 bar, to form all products at their respective standard states. The enthalpychange is equal to the standard heat of reaction at 25 ıC:

�H3 D ��Hırxn D �802625� J/mol

Notice that if the reaction does not proceed to completion, this standard heat must be multiplied bythe extent of reaction. In this problem � D 1 but we will leave � in the formula to show the moregeneral case.

4. Bring all pure species from their standard state at P0, T0 to their standard state at P , T , at the reactorexit. Since we’re dealing with ideal-gas states, the enthalpy change is

�H4 DXi

ni

TZT0

CigPidT

with ni given in the stoichiometric table. This expression includes all products, as well as unreactedcomponents and inert species.

5. Mix the pure components at constant T , P , to form the effluent stream. Since all components areessentially in the ideal gas state, the enthalpy change for this step is approximately zero:

�H5 D 0




If we add the intermediate steps, we obtain the enthalpy change for the inlet-outlet process. Since thisprocess is adiabatic, we have:

0C 0C ��H ırxn CXi

ni

TZT0

CigPidT D 0 Œ1�

The only unknown is temperature.

Numerical calculations

1. Assuming constant Cp: Equation [1] in this case becomes

��H ırxn C .T � T0/Xi

niCigPi D 0 ) T D T0 C

��H ırxnPi niC

igPi

With � D 1, �H ırxn D �802625 J/mol and the C igP ’s in the stoichiometric table, we find:

T D 2298:6 K D 2025ıC D 3678ıF:

2. Using temperature-dependent heat capacities:

Xi

"ni

Z Tout

Tin

CP;idT

#D .T � T0/

�Xniai

�CT 2 � T 20

2

�Xnibi

�where X

niai D 385:403 J/mol KXnibi D 0:053055 J/mol K2

The equation for T now is

�919891:C 385:403T C 0:0265275T 2 D 0

and we findT D 2087 K

Notice that the calculation using temperature-dependent CP ’s gives a final temperature that isabout 10% lower.




14. REACTIONS

Problem 14.4 Solution The reaction is

CH4 C 2O2 ! CO2 C 2H2O

with�H ı298 D �802; 625 J/mol

The standard heat at T refers to the heat of reaction when the products and reactants are in the standard stateat temperature T . From tables we have the heat of reaction when products and reactants are at their standardstate at T0 D 298:15ıC. We devise the following path:

1. Bring reactants from their standard state T to their standard state at T0. The enthalpy involved is

�H1 D

T0ZT

CigP;CH4

dT C 2

T0ZT

CigP;O2

dT D �

TZT0

CigP;CH4

dT C�2

TZT0

CigP;O2

dT

On the far right, we have inverted the integration limits and took a minus sign. This facilitates thealgebra below.

2. Run the reaction at T0 with all species at their standard state. The enthalpy change is

�H2 D �Hı298

3. Bring the products from their standard state at T0 to their standard state at T :

�H3 D

TZT0

CigP;CO2

dT C 2

TZT0

CigP;H2OdT

Adding the intermediate values we obtain the standard heat of reaction at T :

�H ı.T / D �H ı298 C

TZT0

h�C

igP;CH4

� 2CigP;O2C C

igP;CO2

C 2CigP;H2O

i„ ƒ‚ …

�CP

dT

Notice that quantity in parenthesis is the sum of the heat capacities multiplied by their stoichiometric coef-ficients:

�Cp DX

�iCP;i Œ1�

This represents the change in heat capacities from the reactants to the products and allows us to write thefinal result in the compact form:

�H ı.T / D �H ı298 C

TZT0

�CPdT Œ2�

Note: Equations [1] and [2] are general and apply to any reaction. The CP ’s in this equation must matchthe standard state of species: for the standard state (g), use ideal-gas heat capacities; for (l) one mustuse the heat capacity of the liquid; for (s), the heat capacity of the solid.

Since �CP is a difference, it could be positive or negative!




In this problem, since all CP ’s are given by the same mathematical formula, �CP can be calculated asfollows:

�CP D �aC�bT

where

�a DX

�iai D 4:0102 J/mol/K

�b DX

�ibi D �0:00577 J/mol/K2

For later use we construct the stoichiometric table:

CH4 O2 N2 CO2 H2O

� �1 �2 0 1 2 .�/

�H ıf

�74520 0 0 �393509 �241818 .J=mol/�Gı

f�50460 0 0 �394359 �228572 .J=mol/

CP .25ıC/ 51:0308 31:5446 30:1466 46:4956 34:9572 .J=mol=K/

n0 1 2:4 9:029 0 0 .�/

n 1 � � 2:4 � 2� 9:029 � 2� .�/

a) We find:1073:15Z298:15

�CPdT D �a.T � T0/C�b

2.T 2 � T 20 / D 41:85 J/mol

and

�H ı1073 D �802; 625C 41:85 D �802; 583 J/mol

Note: In this case, �CP is small and makes little difference in the heat of reaction. This is not alwaysso.

b) Since inlet and outlet are at the same temperature, the heat is simply equal to the standard heat of reactionat that temperature:

Q D �H D ��H ı.T /

With � D 1 we find

Q D �802; 583 J/mol

The basis for this result is the same basis used in the stoichiometric table. In our case, this basis is 1 mol ofmethane: this follows from the fact that complete combustion corresponds to � D 1.

b) In this case we must also include the enthalpy needed to bring the reactants to the temperature of thefurnace:

�H DX TZ

Tin

ni0CP;idT C ��Hı.T / Œ1�




14. REACTIONS

with the summation going over all species present in the inlet stream.

ni0CPi

Z T

Tin

ni0CPidT

CH4 .1/.44:2539C 0:02273T / 45; 607:

O2 .2:4/.30:5041C 0:00349T / 60; 051:9

N2 .9:029/.29:2313C 0:00307T / 215; 189:

CO2 0 0

H2O 0 0

320; 847 .J=mol/

Finally, for the heat we find

Q D �H D 320; 847 � 802; 583 D �481; 736 J/mol

� Note that the heat is less by about 40% because the cold inlet must be heated and that energycomes from the reaction. Notice that most of that energy goes to heat up nitrogen, which is aninert specie in this reaction.

� Pressure makes no difference in this problem. This is because all states are in the ideal-gas andas we know, enthalpy in the ideal-gas state is independent of pressure.

� Exercise: Repeat the calculations assuming the Cp’s to be constant and equal to their value at25ıC.

� Prove that Eq. [1] is equivalent to Eq. (9.6) in the book. The two equations look different becausethe formula in the book uses the heat of reaction at T0 while Eq. [1] in this solution uses the heatof reaction at T . Using the relationship between the two you should be able to show that the twoequations are identical.





From tables we find

H ı.g/ D 82; 930 J/mole

H ı.l/ D 49; 080 J/mole

ThereforeHvap D 82; 930 � 49; 080 D 33; 850 J/mole

b) The absorption of ammonia in water can be represented by the following reaction:

NH3.g/C H2O! NH3.aq/C H2O

with equilibrium constant

K D exp��Gı.aq/ �Gı.g/

RT

�The gibbs energy of water does not show in this equation because it is essentially the same in pure water(lhs) as in water containing 1 molality ammonia. Also,

K DaNH3.aq/

aNH3.g/

Dc=cı

y�P=P ıD

cP ı

cı�yP

where y and � are the mole fraction and fugacity coefficient of ammonia in the vapor. By Henry’s law wealso have

�yP D Hx � HMwc

where Mw is the molecular weight of water. Combining the results

exp��Gı.aq/ �Gı.g/

RT

�D K D

cP ı

cıHMwc

from which

H DP ı

Mwcıexp

��Gı.aq/ �Gı.g/

RT

�with

Gı.g/ D �16; 450 J/mole

Gı.aq/ D �26; 500 J/mole

cı D 1 mole/kg

P ı D 1 bar

Mw D 0:018 kg/mole

the final result isH D 0:964 bar

c) Solving the equation derived above for Gı.aq/ we find

Gı.aq/ D Gı.g/CRT ln�C ı

P ıMwH

�379




14. REACTIONS

The numerical substitution is left as an exercise.

d) The equilibrium conditions are (g stands for the gas, N2O)

ygP D xgHg

ywP D xwPsatw

from which we obtain

xg DP � P sat

w

Hg � P satw

At 10 ıC, Henry’s law constant for N2O is 1433.5 bar. The saturation pressure of water at this temperaturecan be neglected when compared to P and H .The mole fraction of N2O in the liquid is

xg DP

HgD

12

1433:5D 0:000837

andxw D 1 � xg � 1

The concentration of N2) in units of molality is

cg �xg

.1 � xg/MW =1000D 0:465 mol=kg

where Mw D 18 is the molecular weight of the solvent and the factor of 1000 is used to convert the mass ofthe solvent into kg.

cg �xg

Mw

For the activity of the two components we now have:

ag (aq) D 0:465

aw (aq) � xw D 1

You should be able to verify the Poynting correction for water at 10 ıC, 12 bar, is negligible.




Problem 14.6 Solution The gas (g) standard state refers to the reaction:

H2 .g/C1

2O2 .g/! H2O .g/ Œ1�

This reaction can also be accomplished in the following sequence:

1. Run the reaction at 1 25 ıC, 1 bar using the (l) standard state. The enthalpy and entropy change forthis step are found from the tabulated values in the standard state (l):

�H1 D �285; 830 J/molI �G1 D �228; 572 J/mol

2. Change the pressure to P D P sat.25ıC/ D 0:03166 bar. The change in the enthalpy and Gibbsenergy of the liquid is essentially zero because pressure has negligible effect on liquid properties:

�H2 D 0 J/molI �G2 D 0 J/mol

3. Evaporate the liquid at 25ıC, 0.03166 bar. The changes are found from the stream tables:

�H3 D �HvapD 43; 965 J/molI �G3 D �H

vapD 0 J/mol

Notice that the Gibbs energy does not change upon evaporation.

4. Remove the residual enthalpy and residual Gibbs energy because we are seeking properties in thehypothetical ideal state:

�H4 D �HR� 0 J/molI �G4 D �G

R� 0 J/mol

The residuals are nearly zero because at the low pressure of 0.03166 bar the vapor is essentially in theideal gas state.

5. Change the pressure to P0 D 1 bar, treating the vapor as ideal. That’s because after the previous step,the vapor is in the hypothetical ideal-gas state:

�H5 D 0 J/mol

�G5 D �RT lnP0

P sat D .8:314:25/ ln1

0:003166D 8; 558:62 J/mol J/mol

We now add all of the intermediate results:

H D .�285; 830/C .43; 965/ D �241; 865 J/mol

G D .�228; 572/C .8; 558:62/ D �228; 570 J/mol

The calculated values compare very well with the tabulated values. Some conclusions worth pointingout:

1. If we know the enthalpy or Gibbs energy of formation at one standard state, we can calculate it atany other standard state.

2. Standard states should be treated as reference state with precise instructions as to how to calculateproperties.

3. When we refer to the “ideal-gas standard state”, we must always understand it as hypotheticalideal-gas standard state. For example, pure water at 1 bar, 25ıC is liquid. Yet, this does notprevent us from calculating properties at the hypothetical ideal-gas state at 25ıC, 1 bar.




14. REACTIONS


�Gı

RTD � lnK D � ln a �

b

T) �Gı D �Rb �RT ln a

At 25 ıCwe find�Gı D 18547:3 J/mol

For the heat of reaction recall thatd

dT

��Gı

RT

�D �

�H ı

RT 2

With �Gı D �Rb �RT ln a we find

�H ı D �Rb D 9976:8 J/mol

The entropy of the reaction is calculated as follows:

�Gı D �H ı � T�Sı ) �Sı D�H ı ��Gı

TD �95:72 J/mol

b) The stoichiometric table for this reaction is as follows:

A B C�i �2 �3 2

ni0 1 1 1ni 1 � 2� 1 � 3� 1C 2�

yi1 � 2�

3.1 � �/

1 � 3�

3.1 � �/

1C 2�

3.1 � �/

Assuming the gases to be in the ideal-gas state and since the reaction is isothermal,

Pf

PiD

ni

ni0D3.1 � �/

3D 1 � �

Therefore, the value of � at equilibrium is

� D 1 �Pf

PiD �0:25

The corresponding mole fractions are:

yA D 0:4; yB D 0:4667; yC D 0:133

The equilibrium constant that produces this distribution is

K Dy2C

y2Ay3B

�Pf

P ı

�2�2�3D 3:224 � 10�4

The temperature that corresponds to this equilibrium constant is

T Db

lnK

a

D 345 K




Problem 14.8 Solution a) The reaction is

3H2 C N2 D 2NH3

Using tables we find at 25 ıC:

�Gırxn D 2�GıNH3D �32; 900 J/mole

since �GıH2D �GıN2

D 0. Accordingly, the equilibrium constant at 25 ıCis

K D exp��

�32; 900

.8:314/.298:15/

�D 5:81 � 105

b) The heat of reaction at 25 ıCis

�H ırxn D 2�HıNH3D �92; 220 J/mole

Increase P ; decrease T .




14. REACTIONS

Problem 14.9 Solution a) From tables we find

�H ı298 D �92; 220 J/mole; �Gı298 D �32; 900 J/mole; K298 D 5:81 � 105

We also have

�Cp D �28:34 J/mol;Z T

T0

�CpdT D �28:34T (J/mol)

with T in K. The heat of reaction at T is

�H ı.T / D �H ı298 C

Z T

T0

�CpdT D �83770:4 � 28:34T

The equilibrium constant at T is

lnK.T /

K.T0/D

Z T

T0

�H ı.T /

RT 2dT D

Z T

T0

��10075:8=T 2 � 3:40871=T

�dT

D �14:373C10075:8

T� 3:40871 lnT

At 1000ıC we findK D 2:37 � 10�8

b)H2 N2 NH3

� �3 �1 2

n0 1 1 1

n 1 � 3� 1 � � 1C 2�

y1 � 3�

3 � 2�

1 � �

3 � 2�

1C 2�

3 � 2�

Assuming all species to be in the ideal gas phase, ai D yiP=P0, and the equilibrium constant is

.3 � 2�/2.2� C 1/2

P 2.1 � 3�/3.1 � �/

P 20P 2D K1000 D 2:37 � 10

�8

Solving for � we find:

� D �0:499; y1 D 0:6246; y2 D 0:3749; y3 D 0:0004657

c) The amount of heat is

Q D ��H ı D .�0:499/.�92; 220/ J/mole D 59814:2 J/mol

This heat must be added.

Note: The ammonia synthesis reaction is exothermic; however, under the conditions of the problemthe reaction proceeds in the reverse direction, therefore, heat must be added in order to maintain thetemperature constant.




Problem 14.10 Solution The reaction is

C6H6.g/C 3H2.g/! C6H12.g/

All the activities are calculated asai D

yi�iP

P ı�yiP

P ı

where we assumed �1 � 1. The equilibrium constant is

K DyC6C12

yC6H6y3H2

�P

P ı

�1�1�3)

yC6C12

yC6H6y3H2

D K

�P

P ı

�3� Effect of pressure: Increasing P increases the mole fraction of the product - run at high pressure.

� Effect of temperature: We calculate �H ı for this reaction:

�H ırxn D �HıC6H12

��H ıC6H6� 3�H ıH2

D �123; 140 � 82; 930 � 0 D �206:070 J/mole < 0

The reaction is exothermic, therefore, decrease T to increase conversion to products.




14. REACTIONS

Problem 14.11 Solution We construct the stoichiometric table below:

C6H12 C6H6 H2 N2

� �1 1 3 0

n0 1 0 0 1

n 1 � � � 3� 1

y1 � �

2C 3�

�

2C 3�

3�

2C 3�

1

2C 3�

�H ıf 298

�123140 82930 0 0 .J=mol/�Gı

f 29831920 129665 0 0 .J=mol/

a) �H ı298 D 206; 070 J/mol, �Gı298 D 97; 745 J/mol

K298 D 7:49659 � 10�18

b) Assuming �H ı is independent of temperature:

Q D �H ı D.0:85/.206; 070/ J.mol

2 molD 175; 160 J/mol

This heat must be added to the reactor.

c) With � D 0:85, the stoichiometric table gives:

yC6H12 D 0.032967

yC6H6 D 0.186813

yH2 D 0.56044

yN2 D 0.21978

c) The equilibrium constant is

K D27P 3�4

.1 � �/.3� C 2/3P 30

With � D 0:85, P D 5 bar, P0 D 1 bar we find

K D 124:688 Œ1�

On the other hand, the equilibrium constant as a function of temperature is

K D K298 exp��H ı

298

�1

T�1

T0

��D exp

�43:7002 �

24785:9

T

�Œ2�

Solving for T between [1] and [2]:T D 637 K D 364ıC




d) SinceP�i > 0, to shift the equilibrium further to the right and thus increase conversion, pressure must

be decreased. Operate at 1 bar.

e) Assumptions:

1. All species are in the ideal-gas state. This allows us to:

– take the activity of all species to be ai D yiP=P0 (i.e. �i D 1)

– ignore mixing and residual terms in enthalpy so that the heat can be assumed to be equal to thestandard enthalpy of reaction

2. Ignore �Cp. This allows us to assume �H ı is independent of temperature.




14. REACTIONS


a) The heat of reaction is

�H ı D RT 2d lnKdT

D RT 2d.a � b=T /

dTD �bR

With b D 6673 we find�H ı D �55479:3 J=mol

The heat of reaction is negative, therefore the reaction is exothermic.

b) We construct the stoichiometric table:

IPA PA H2 N2� �1 1 1 0

n0 0:7 0 0 0:3

n 0:7 � � � � 0:3

y 0:7��C1:

��C1:

��C1:

0:3�C1:

The equilibrium constant at 500 K, 3 bar is

K D 9:64733 � 10�8

The equilibrium condition is3�2

.� C 1:/.0:7 � �/D 9:6473 � 10�8

Solving for � we obtain� D �0:00015004; � D 0:00015003

The acceptable solution is the positive root (why?). Substitution into the stoichiometric table gives

IPA PA H2 N2� �1 1 1 0

n0 0:7 0 0 0:3

n 0:69985 0:00015003 0:00015003 0:3

y 0:699745 0:000150007 0:000150007 0:299955

The conversion of IPA isconversion D 1 �

n1

n0;1D 0:021% (14.1)

The equilibrium constant is too low to produce significant conversion.




Problem 14.13 Solution We construct the stoichiometric table:

EtOH AcAH H2� �1 1 1

G �168490 �128860 0

H �235100 �166190 0

cp 118 90 30

n0 1 0 0

n 1 � � � �

y 1��C1

��C1

��C1

a) The standard Gibbs energy of formation at 298 K is

�Gı298 DX

�iGıi D 39630: J=mol

The equilibrium constant at 298 K is

K298 D exp��Gı298R.298 K/

�D 1:13043 � 10�7

b) The heat of reaction is�H ı298 D

X�iH

ıi D 68910 J=mol

The reaction is endothermic.

c) Assuming the enthalpy of reaction to be constant, the equilibrium constant at 550 ıC (823 K) is

K823 D K298 exp��H ı298R

�1

T�1

T0

��D 7:53312

d) We calculate �Cp:�CP D

X�iC

igP iD 2 J=mol K

The value of �CP is fairly small; for every 500 K above 298 K the enthalpy changes by 1000 J/mol, or1.5% of its value at 298 K. we may conclude that �H ı is fairly constant with temperature. If we calculatethe equilibrium constant including the effect of temperature on the heat of reaction, the result is

lnK

K298D�H ı298 C�CP .T � T0/

RT 2) K D 6:2953

which represents a difference of about 20%. Even though the effect of temperature on �H ı is mall, theeffect on the equilibrium constant is higher due to the exponential relationship between K and �H ı. Forthe rest of the calculations we will use

K823 D 7:53312

e) The equilibrium condition is�2P

.1 � �/.� C 1/P0D K

With P D 2 bar, P0 D 1 bar, K D 7:53312 we find � D 0:888935. The corresponding conversion ofethanol is

(conversion) D 0:888935 D 88:9%




14. REACTIONS

Substitution into the stoichiometric table gives

EtOH AcAH H2� �1 1 1

Gı �168490 �128860 0 J=molH ı �235100 �166190 0 J=molC

igP 118 90 30 J=mol Kn0 1 0 0 moln 0:111065 0:888935 0:888935 moly 0:0587978 0:470601 0:470601

f) If the conversion is 98%, then � D 0:98. We now solve the equilibrium condition for P :

�2P

.1 � �/.� C 1/P0D K ) P D 0:041 bar





We construct the stoichiometric table below:

C3H8 C3H6 H2 N2�:i -1 1 1 0

Hi J/mol -104,680 19,710 0 0Gi J/mol -24,290 62,205 0 0ni0 1 0 0 x

ni 1 � � � � x

yi1 � �

1C � C x

�

1C � C x

�

1C � C x

x

1C � C x

In anticipation of part (d) we have included a row for inert nitrogen.

a) From the above table, �Gı298 D 62; 205 � .�24; 290/ D 86; 495 J/mol. Therefore,

K298 D exp��

86; 495

.8:314/.298:15/

�D 7:01275 � 10�16

b) For the heat of reaction we find

�H ı298 D 19; 710 � .�104; 680/ D 124; 390 J/mol

The heat of reaction is positive therefore the reaction is endothermic.

c) Neglecting fugacity coefficients, the equilibrium constant for this reaction is

K D�2

.1C �/.1C x � �/

�P

P ı

�1C1�1ŒE1�

For 95% conversion, � D 0:95. With x D 0, P D 5 bar, P ı D 1 bar we find K D 46:28. We also have

lnK

K298D �

�H ı

R

�1

T�

1

T ı

�Solving for T we find:

1

TD

1

T ı�

R

�H ıln

K

K298) T D 1306 K ŒE2�

d) In the presence of nitrogen, x D 1 in the above stoichiometric table. The equilibrium constant is cal-culated from Eq. (E1) and now we find K D 30:59. The temperature is calculated from Eq. (E2) and wefind

T D 1261 K

e) The procedure is the same as in the previous step except that P D 1 bar. We find

K D 6:118

T D 1110 K

f) We assumed all the species to behave as ideal-gases in the conditions of the experiment; we have alsoassumed the heat of reaction is independent of temperature and that the reaction reaches equilibrium.




14. REACTIONS


a) We start with �i D �ıi CRT ln ai which we write as

�i D �GıCRT ln

fi

f ıi(1)

This equation gives the chemical potential of component at any pressure, temperature and composition. Weuse it to calculate the chemical potential at some different standard state, at temperature T , pressure P0, andcomposition as specified by that standard state:

.�Gı/0 D �Gı CRT0 ln.f ıi /

0

f ıi(2)

With primed quantities referring to s1 and unprimed quantities to s1, this proves part (a).

Alternative solution: Consider the reaction for a component A,

A.s1/• A.s2/

where the left-hand side is referenced to standard state Œs1� and the right-hand side to standard state Œs2�.The equilibrium constant is

exp��Gı.s2/ ��G.s1/

RT0

�Da.s2/

a.s1/(3)

For the activities we have

a.s1/ DflhsA =fA.s1/ (4)

a.s2/ DfrhsA =fA.s2/ (5)

Here, fA.s1/ is the fugacity of A in standard state s1, fA.s2/ is the fugacity of A in standard state s2, f lhsA is

the fugacity of A on the left-hand side of the reaction, and f rhsA the fugacity of A on the right-hand side of

the reaction. Next, we note thatf lhsA D f rhs

A

That’s because at equilibrium, the fugacity of A is the same throughout the system. Combining (3–5) weobtain

�Gı.s2/ ��G.s1/ D RT0 lnfA.s2/

fA.s1/(2b)

which the same as Eq. (2).

b) To calculate the Henry’s law constant, we set s1 D (g/ and s2 D aq. For the fugacity we have:

fgi D P0

(since the “g” standard state prescribes “pure gas in the ideal gas state at P0, T ”) and

faqi D c0H

0D x0H

(since the “aq” standard state prescribes a“solute with molality c0 D 1 mol/kg obeying Henry’s law at P0,T ”). Here, H 0 is Henry’s law expressed in bar kg/mol (works with the concentration of solute expressed as




molality), H is the regular Henry’s law constant (works with the solute expressed in mol fraction) and x0 isthe mol fraction of solute that corresponds to molality 1 mol/kg:

x0 Dc0MW

1C c0MW

(MW is the molecular weight of solvent, in kg/mol). Putting everything together,

exp


ıg

RT

#Dx0H

P0D

c0MWH

P0.1C c0MW/

and solving for H :

H DP0.1C c0MW/

c0MWexp


ıg

RT

#The numerical substitutions are left as an exercise.

c) This part is left as an exercise.




14. REACTIONS

Problem 14.16 Solution a) �H ı298 D 180500 J/mol, �Gı298 D 173100 J/mol, from which we find

K298 D exp��Gı298RT0

�D 4:7044 � 10�31

The reaction is endothermic.

The equilibrium constant at temperature T is1

K D K298 exp��H ı298R

�1

T�

1

298:15

��D exp

�2:98529 �

21710:4

T

�Œ1�

b) Pressure has no effect; increasing temperature shifts the right (increases production of NO).

c) We construct the stoichiometric table using two independent reactions: combustion of methane (�1, �1),and NO formation (�2, �2):

CH4 O2 N2 CO2 H2O NO

�1 �1 �2 0 1 2 0

�2 0 �1 �1 0 0 2

n0 1 2:4 9:02857 0 0 0

n1 1 � �1 2:4 � 2�1 � �2 9:02857 � �2 �1 2�1 2�2

Assuming complete combustion of methane (�1 D 1), the mol fractions of N2, O2, NO, are

yN2 D 0:0804598.9:02857 � �2/; yO2 D 0:0804598.0:4 � �2/ yNO D 0:16092�2�2

If the required mol fraction of NO is 500 ppm, then �2 D 0:00310714 and

yN2 D 0:726187; yO2 D 0:0319339 yNO D 0:0005

These values correspond to an equilibrium constant,

K D.0:0005/2

.0:725937/.0:0316839/D 1:07805 � 10�5 Œ2�

The temperature is calculated by equating Eqs [1] and [2]:

T D 1506 K

We conclude that the furnace must be operated below 1506 K.

1Note: The equilibrium constant can be written as

K D K298 exp

"�

�H ıT0

R

�1

T�

1

298:15

�#D exp

�lnK298 C

�H ı298RT0

��H ı298RT

�D exp

�2:98529 �

21710:4

T

�





SO2 C 0:5O2 D SO3

b) The standard enthalpy and entropy of formation of the three species at 298 K are:

H ıCH4 D �296830; H ıO2 D 0; H ıSO3 D �395720 (J/mol)

GıCH4 D �300194; GıO2 D 0; GıSO3 D �371060 (J/mol)

From these values we find:

�H ırxn D �98890 (J/mol) �Gırxn D �70866 (J/mol)

c) we construct the stoichiometric table:

CH4 O2 N2 SO3

� �1 �0:5 0 1

n0 1 0:6 2:25714 0

n 1 � � 0:6 � 0:5� 2:25714 �

y1 � �

3:85714 � 0:5�

0:6 � 0:5�

3:85714 � 0:5�

2:25714

3:85714 � 0:5�

�

3:85714 � 0:5�

Adiabatic temperature We have two unknowns, � and T but we also have tow equations: One is theadiabatic energy balance and the second is given by the equilibrium constant, which relates the mol fractionsof the stoichiometric table to T . We will also need the heat capacities of the four species. These heatcapacities will be calculated from the equation

CP =R D AC BT CD=T2

where R is the ideal-gas constant, T is in K and the parameters A, B , D of the four species are:

CH4 O2 N2 SO3

A 5:699 3:639 3:28 8:06

B 0:000801 0:000506 0:000593 0:001056

D �110150: �22700: 4000: �202800:

(Depending on the source, other equations for the heat capacity may be used). We now construct the equa-tions for the energy balance and the equilibrium constant.

� Energy Balance Since the initial temperature is the same as the temperature T0 of the tables, the heatthat is exchanged at T is

�H.T / D ��H ı.T0/C

Z T

T0

XniCPidT

For adiabatic conditions,�H.T / D 0. Using ni from the stoichiometric table and the CP parametersabove, the energy balance becomes

0 D 8:314 � 10�6�T 2 C 0:0101559T 2

C 4:50203�T C 127:086T � 102500� � 41993:2C675928�

TC953960

T(1)




14. REACTIONS

� Equilibrium According to the van’t Hoff equation, the equilibrium constant at T is

lnK.T /

K.T0/D

Z T

T0

�H ı.T /

RT 2dT (0)

where �H ı.T / is the standard heat of reaction at T and K.T0/ is the equilibrium constant at T0 D298:15 K. To simplify the solution, we will assume that �H ı of the reaction is approximately con-stant in the temperature range T0 to T . We will examine the effect of this assumption at the end. Withthis, the equilibrium constant becomes,

K.T / D K.T0/ exp��H ı

R

�1

T�1

T0

��and with numerical substitutions,

K.T / D e11894:4=T�11:3054 (1)

From the stoichiometric table with P D P0 D 1 bar, we obtain the equilibrium constant in terms ofcomposition:

K D�

.1 � �/

�3:85714 � 0:5�

0:6 � 0:5�

�0:5(3)

Equating [2] and [3] we obtain the second equation:

e11894:4=T�11:3054 D�

.1 � �/

�3:85714 � 0:5�

0:6 � 0:5�

�0:5(4)

� Solution Solving [1] and [4] simultaneously we get:

T D 855:6 K � D 0:769495

and the mole fractions at the exit of the reactor are:

CH4 O2 N2 SO3

y 0:0663821 0:0619896 0:650025 0:221604

� Is it OK to assume �H ı is constant? To answer this question we repeat the calculations withoutmaking this assumption. The complete expression for the heat of reaction is

�H ı.T / D �H ı.T0/C

Z T

T0

�CPdT

Numerical substitution gives

�H ı.T / D �102500C 675928:=T C 4:50203T C 8:314 � 10�6T 2

We insert this into the van’t Hoff equation and perform the integration:

K.T / D e10�7T�15:39C12328:6=T�40650=T 2C0:5415 lnT (2b)

This equation is to be used in place of Eq. [2]. That is, Eq. [4] now becomes

e10�7T�15:39C12328:6=T�40650=T 2C0:5415 lnT

D�

.1 � �/

�3:85714 � 0:5�

0:6 � 0:5�

�0:5(4b)




Solving [4b] simultaneously with [1] we get

T D 856:6 K; � D 0:770883

and the mole fractions are

CH4 O2 N2 SO3

y 0:0659955 0:0618021 0:650155 0:222048

Therefore, the assumption that �H ı is constant is not very bad in this case.




14. REACTIONS


CH3COOH.l/C CH3CH2OH.l/! CH3COOCH2CH3.l/C H2O.l/

We form the stoichiometric table as follows:

AcH EtOH AcEt H2Oi 1 2 3 4�i -1 -1 +1 +1ni0 0.8 0.2 0 0ni 0:8 � � 0:2 � � � �

Pni D 1

xi 0:8 � � 0:2 � � � �Pxi D 1

�Gı -389,900 -174,780 -332,200 -237,129 �Gırxn D �4; 649 J/mole

The equilibrium constant at 25 ıC is

K D exp��Gı

RT

�D 6:524

b) The equilibrium constant is also given by

K Da3a4

a1a2

The standard state for all species is (l). Ignoring the Poynting factor (we assume that the reaction is run atnot very high) the activity is

ai D ixi

thus

K D

� 3 4

1 2

��x3x4

x1x2

�D

� 3 4

1 2

��2

.0:8 � �/.0; 2 � �/

� Ideal solution In this case i D 1 and the above equation becomes quadratic in �. We find

� D 0:1908

x1 D 0:6092

x2 D 0:0092

x3 D 0:1908

x4 D 0:1908

� Non-ideal solution The activity coefficients must be included but now the solution requires trial anderror. we do this as follows:

1. guess �.

2. calculate xi

3. with the calculated xi use UNIFAC to calculate i

4. calculate the quantity

K D

� 3 4

1 2

��x3x4

x1x2

�D

� 3 4

1 2

��2

.0:8 � �/.0; 2 � �/




5. Is it equal to K? if not, repeat.

By this method we find � D 0:1919. The UNIFAC table with the final answers is shown below:

Component Acetic Acid ethanol ester waterx(i) 0.620542559 0.020542559 0.179457441 0.179457441

subgroup n1(k) n2(k) n2(k) n2(k)CH3 1 1 1 0CH2 0 1 1 0OH 0 1 0 0H2O 0 0 0 1COOH 1 0 0 0CH3COO 0 0 1 0

i 1 2 3 4

ln C 0.00593961 -0.005778986 -0.065843047 0.313932042

ln R -0.048853331 0.146293074 0.452032453 0.346373904 0.957994041 1.150865293 1.471363329 1.935384367




14. REACTIONS


a) Exothermic (K decreases with increasing T )

b) KP D K D 0:899784 (dimensionless).

c) We construct the stoichiometric table:

CO H2O CO2 H2�1 �1 1 1

n0 1 5 0 0

n 1 � � 5 � � � �

y .1 � �/=6 .5 � �/=6 �=6 �=6

With the activity of all species expressed as yiP=P0, the equilibrium constant is

�2

.1 � �/.5 � �/D 0:899784

Solving this quadratic equation for � we find � D 0:820826 and with this, the mole fractions at the exit are:

CO H2O CO2 H2y W 0:0298623 0:696529 0:136804 0:136804

d) Pressure has no effect (why?). We will go with the suggestion of the intern, most likely a PSU ChEundergrad. The mol fraction of CO must satisfy 1 � �/=6 � 0:005, from which � D 0:97. With this value,the mol fractions are

CO H2O CO2 H2y W 0:005 0:671667 0:161667 0:161667

and the required equilibrium constant must be

K D.0:161667/2

.0:005/.0:671667/D 7:78246

That is,e�5:057C4951:4=T D 7:78246

from whichT D 696 K





CO.g/ H2O.g/ CO2.g/ H2.g/

� �1 �1 1 1

n0 1 1 1 1

n 1 � � 1 � � 1C � 1C �

y 0:25.1 � �/ 0:25.1 � �/ 0:25.1C �/ 0:25.1C �/

H ı295 �110525 �241818 �393509 0

Gı295 137169 �228572 �394359 0

We find�H ı298 D �4:117 � 10

4 J/molI �Gı298 D �2:862 � 104 J/mol

a) �1 � � � 1

b) The equilibrium constant isK298 D 1:033 � 10

5

At T D 900 K (assuming �H ı � constant):

K D K298 exp��H ı298R

�1

T�1

T0

��D 1:552

Therefore:

K D.1C �/2

.1 � �/2D 1:552 ) � D 0:1094

The percentage of CO that has reacted is 10.94%.

c) Nitrogen has no effect in this case because the number of moles does not change during reaction (do thealgebra to verify this fact). In reactions where the total number of moles changes, the addition of an inertgas changes the mol fractions at equilibrium.

d)Q D ��H D �4505:89 J/mol

Must be removed!

e) The equilibrium constant at temperature T is

K D exp��5:06209C

4951:41

T

�Œ1�

With � D 0:5, the value of K is

K D.1C �/2

.1 � �/2D 9 Œ2�

Solving for T between Eqs [1] and [2]:T D 682 K

e)

1. All species are in the ideal gas state at the conditions of this reaction

2. Heat of reaction is independent of temperature




14. REACTIONS


Na2SO4H2O and Na2SO4H2O � 10H2O are solid, H2O is vapor. The equilibrium constant for this reactionis

y�10w D K298

From tables we collect the information shown below:

Na2SO4 H2O NaSO4 � 10H2O� �1 �10 1

Gı .kJ=mol/ �1265:2 �228:6 �3642:3

from which�Gı298 D �91100 J=mol

The equilibrium constant at 25 ıC is

K298 D exp��

�91100 J=mol.8:314 J=mol K/.298:15 K/

�D 9:1394 � 1015

The equilibrium mol fraction of water is

yequw D .K298/

�1=10D 0:02535:

At T D 25 ıC the saturation pressure of water is 0.0317 bar, and since pressure is 1 bar and relative humidityis 0.85, the actual mol fraction of water is

yactualw D .0:85/

.0:0317 bar/.1 bar/

D 0:026945

This value is higher than the equilibrium value, therefore, the reaction will take place. The minimumhumidity for the reaction to go forward corresponds to the equiibrium mol fraction of water:

Minimum Relative Humidity D.0:02535/.1 bar/.0:0317 bar/

D 0:8

The minimum humidity is 80%.

Here is how to understand this solution: If the mol fraction of water is less than the equilibrium value,reaction is not possible, not only because the equilibrium condition is not satisfied, but also because suchreaction would further decrease the mol fraction of water vapor, i.e., it would move the system further awayfrom equilibrium.

If the mol fraction of water is above the equilibrium value, then reaction will occur, because this causes themol fraction of water to decrease and approach equilibrium. Whether the water mol fraction will eventuallyreach its equilibrium value or not will depend on the amount of Na2SO4 relative to the total amount of watervapor: If Na2SO4 is in excess, then the reaction will stop when water reaches its equilibrium value and theequilibrium state will contain a mixture of Na2SO4 and Na2SO4 � 10H2. If water is in excess, the entireamount of Na2SO4 will be converted. At the final state, the mol fraction of water will be less than initially,but still above its equilibrium value.





1 2 3 5

4

reactor

With reference to the above graph, let ai , bi , be the molar flows of A and B in stream i . By mol balancebetween stream 1 and 5,

a5 D a1 � �

b5 D b1 C �

The compositions stream 5 satisfies the equilibrium condition

yA

yBD K )

b1 C �

a1 � �

We conclude that the composition of stream 5 depends only on:

1. The mol fractions in stream 1

2. The temperature (and possibly pressure) of stream 5

The recycle stream makes no difference in the balance. This is because we are recycling a stream that hasalready reached chemical equilibrium. This portion of the feed will not be changed any further by goingthrough the reactor. To improve conversion we must separate the components and recycle only reactants, ormostly reactants with as little product as possible.




14. REACTIONS


1 2 3

56

7

reactor

sep

arat

or

r

r + 1

1

heat

4

1. We collect the thermodynamic properties of the reaction:

SO2 O2 N2 SO3� �1 �0:5 0 1

H ı �296830 0 0 �395720 J=molGı �300194 0 0 �371060 J=mol

From these data we calculate the equilibrium constant at 850 K:

�H ı D �98890 J=mol

�Gı D �70866 J=mol

K298 D 2:60541 � 1012

K850 D 14:7026

2. Construct the table of molar flows in each stream on a basis of 100 mol in stream 1:

1 2 3 4 5 6 7

SO2 14:8936 a2 a3 a4 a5 a6 0

O2 17:8723 b2 b3 b4 b5 b6 0

N2 67:234 c2 c3 c4 c5 c6 0

SO3 0 d2 d3 0 0 d6 0

3. Perform an overall balance using the known overall conversion:

a5 D 14:8936 � �

b5 D 17:8723 � 0:5�

c5 D 67:234

d7 D �

�

14:8936D 0:95

Solve these equations for a5, b5, c5, d7 and �.

4. Set all molar rates in stream 6 to be r times of those in 5.




5. Perform a balance on the mixing point to obtain the molar rates in stream 2. These rates are expressedin terms of the unknown factor r .

6. Perform a balance on the reactor: for each species i , the moles at the exit are related to the moles atthe inlet according to stoichiometry:

ni;3 D ni;2 C �i�

where � is known from the solution of the overall balance.

7. Obtain the mol fractions in stream 3:yi;3 D

ni;3Po ni;3

These are all expressed in terms of r .

8. Apply the equilibrium condition:y4

y1y1=22

D K

The only unknown in this equation is r . Solve for r and obtain all streams that are expressed in termsof r .

9. Calculate the remaining unknown stream 4 by balance around the separator.

The results are summarized below. All entries are in mol.

1 2 3 4 5 6 7

SO2 14:8936 16:8127 2:66378 2:66378 0:744681 1:9191 0:

O2 17:8723 45:6993 38:6248 38:6248 10:7979 27:8269 0:

N2 67:234 240:501 240:501 240:501 67:234 173:267 0:

SO3 0: 0: 14:1489 0: 0: 0: 14:1489

total 100: 303:013 295:939 281:79 78:7766 203:013 14:1489




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