Gaseous State(Lecture Notes) - The UraniumGaseous State : Dr. S. S.Tripathy Gaseous State : Dr. S....

Gaseous State : Dr. S. S. Tripathy

Gaseous State : Dr. S. S.Tripathy

Gaseous StateIdeal Gas Laws:1. BOYLE’S LAW

Temperature remaining constant, volume of a given mass of gas is inversely proportional to its pressure.Mathematically,

P

V1

(constant ‘T’ and ‘n’)

kPVP

kV

2211 VPVP

Isotherms: P-V graphs at constant T.

Note that the order of fixed temperatures can be understood by keeping pressure constant and applyingCharles law to the three points. Greater the temperature greater is the volume. Hence the order: T

1<T

2<T

3

Limitations of Boyle’s law: It is an ideal gas law and hence valid at low pressure (< 2 atm.) and high temperature(≥ Boyle’s temp; T

B)

Gas He H2

N2

CH4

NH3

CO2

C4H

10

TB(0C) –249 –156 59 226 587 753 1164

2. Charles Law :

Pressure remaining constant the volume of a given mass of gas increases or decreases by

15.273

1

part of its volume at 00C for every 10C rise or fall of temperature respectively.Historical Significance : 1 L of gas at 00C became 1.366 L at 1000C while pressure remaining constant(1 atm).

For 1000C, increase in volume = 0.366 L; So for 10C, increase in volume = 15.273

1L

Derivation:

Let volume at 00C = V0. Hence, volume at 10C =

15.2730

0

VV

Volume at 20C =15.273

2 00

VV ; Hence volume at t0C =

15.2730

0

tVV

PV

P

T1

T2

T3

T1<T2<T3

V

1/P

T1

T2

T3T1<T2<T3

T1

T2T1<T2

logP

logV

logP=logk–logV

P

V

Isotherms

T1<T2<T3

T1T2 T3



TVtV

tVVt

15.273

15.273

15.2730

[where; T = (273 +t) is temp. in Kelvin or Thermodynamic temperature scale]

2

2

1

1

T

V

T

Vk

T

V (P and ‘mass’ remaining constant)

Alternative Definition of Charles Law:Pressure remaining constant, volume of a given mass of gas is directly proportinal to Kelvin temperature.

Isobars: V-T graphs at constant pressures.

`

The order of fixed pressures can be understood by applying Boyle’s law for the three points taken at fixedtemperature. Greater the pressure less is the volume. Hence the order : P

1 > P

2 > P

3

Importance of 0K: At absolute zero(0K), volume gases become theoretically zero. It has, however, nopractical significance as every gas turns into a solid before –273.150C or 0K. 0K is practially not attainable.

3. Amonton’s Law(Gay Lussac’s law):Volume remaining constant, the pressure of a given mass of gas is directly proportional to Kelvin

temperature. (Similar to Charles law in the original format)

2

2

1

1

T

P

T

Pk

T

PTP ( V and ‘mass’ remaining constant)

Isochores:



N.B Graphs are similar to Charles law graphs. The order in fixed volumes can be understood by applyingeither Charles law at fixed pressure or Boyles law at fixed temperature. Hence the order V

1>V

2>V

3.

Combined Gas Equation:

PV

1 (Constant T, and mass) Boyles law

TV (Constant P and mass) Charles law

When both P and T vary; TP

V

1(mass constant)

2

22

1

11

T

VP

T

VPk

T

PV (Combined Gas equation)

STP(NTP) : According to new IUPAC convention, STP for a gas is 1 bar of pressure (105 Pa) and 00C(273.15K)of temperature.However, the old convention of 1 atm. (1.01325 × 105 Pa = 101.325 kPa) is rather commonly used in manytexts for the standard pressure.GRAM MOLAR VOLUME(GMV):

At standard state according to new convention(1 bar), GMV = 22.71098 L ≈ 22.7 LAccording to old convention (1 atm.), GMV = 22.4139L ≈ 22.4 L

IDEAL GAS EQUATION(PERFECT GAS EQUTION) or Equation of State:

PV

1 (Constant T, and mass) Boyles law

TV (Constant P and mass) Charles law

nV (Constant P and T) Avogadro’s law (n = number of moles)When P, T and ‘n’ vary;

nRTPVnTP

V

1(where R = Universal Gas constant)

11..082.015.2731

4.221

KmolatmLnT

PVR

11117 .3.8.103.815.2731

224009806.1376

KmolJKmolerg

nT

PVR

1135

.3.815.2731

107.2210

KmolJ

nT

PVR

(Note that the value of R remains the same whether calculated on the basis of new or old conventions)SAQ: Calculate the molecular mass of a gas 1.0 g of which occupies 395 mL at 270C and 1520 mm of Hgpressure.Solution:

94.31385.0

7601520

300082.01

PV

wRTMRT

M

wPV

( Pressure is converted to atm. and volume to L.)



Density of gas from Ideal Gas equation:

RT

PM

V

mdRT

M

mnRTPV

''

SAQ: Find the density of CO gas at 270C and 1.5 atm. pressure.

Solution:1707.1

300082.0

285.1

gLRT

PMd (note that we used R=0.082

L.atm.K–1mol–1 and Molecular Mass = 28 g.mole–1 and P = 1.5 atm. Hence the density will be in g/L)

(4) Dalton’s Law of Partial Pressure:Temperature remaining constant, the total pressure exerted by a mixture of non-reacting gases is equal to thesum of the partial pressure of the individual gases. Partial pressure of a gas is the pressure which the gas aloneexerts in the volume occupied by the mixture.(note that since under ideal conditions, the molecules do not have any interaction within themselves, the presenceof one gas does not affect the motion of the other and hence its pressure.)

P = pA + p

B + p

C +...... Where P = total pressure, p

A, p

B, p

C... are the partial pressure

of the individual gases.

Mixing of two gases:

Two vessels containing non-reacting gases A and B separately at P, V and P’, V’ respectively. The two arenow connected with a narrow tube having negligibe volume. Temperature remains constant.Applying Boyle’s law to gas A, we have,

'

'VV

PVpVVpVP AA

(where pA = partial pressure of A)

Applying Boyle’s law to gas B, we have

'

'''''

VV

VPpVVpVP BB

(where pB = partial pressure of B)

According to Dalton’s law; P = pA + p

B; ( where P = total pressure of the mixture)

'

''

'

''

' VV

VPPV

VV

VP

VV

PVP

Partial Pressure from Ideal Gas Equation:Let two non-reacting gases A and B are present in a vessel of volume V.No. of moles of A = n

A;No. of moles of B = n

B



Applying Ideal Gas equation to the individual gas, we shall get its partial pressure.

Gas A : V

RTnpRTnVp A

AAA (1) Gas B: V

RTnpRTnVp B

BBB (2)

Applying Ideal gas to the mixture; V

RTnnPRTnnPV BA

BA

(3)

Diving eqn. (1) by eqn. (3) we have, ABA

AA xnn

n

P

p

(where x

A = mole fraction of A)

PxPxp BAA 1 ( xA + x

B = 1)

Dividing eqn.(2) by eqn. (3) ,we have; BBA

BB xnn

n

P

p

(where x

B = mole fraction of B)

PxPxp ABB 1 OR pB = P – p

A

So, partial pressure of a gas in a mixture is equal to its mole fraction in the mixture multiplied by the totalpressure.

Volume Percent/Volume fraction : Note that in a gaseous mixture volume % actually means mole %. This isaccording to Avogadro’s law at constant P and T.For example, a gaseous mixture contains 80% of N

2 by volume and rest O

2. This means their mole ratio is

80 :20 . Mole fraction 0.8 and 0.2 respectively.

Mole fraction = Volume fraction = Pressure fractionMole % = Volume % = pressure %

SAQ: A mixture of 60% of N2 and 40% of O

2 by volume at certain temperature is at 1 atm. pressure. Calculate

the partial pressure of each gas. Presume no reaction between them.

Solution: 4.0;6.022 ON xx ; So, atmpN 6.0;16.0

2 ; atmpO 4.0;14.0

2

SAQ: A 2 L vessel containing N2 at 500 mm pressure is connected with another 3 L vessel containing O

2 at

700 mm pressure by means of a narrow tube having negligible volume. Calculate partial pressure and volumepercent of each gas. Temperature remains constant.

Solution: Applying Boyle’s law to N2, we have; mmpN 200

32

50022

Applying Boyle’s law to O2, we have, mmpO 420

32

70032

So, P = 200 + 420 = 620mm; ;3226.0620

2002

2

P

px N

N Hence 6774.03226.012

Ox

So volume % of N2 is 32.26% and O

2 of 67.74%.

N.B: Partial pressure of water vapour in gaseous mixture(say air) is called Aqueous Tension.



SAQ: 4 g of H2 and 8g of O

2 were mixed and kept in a 2 L vessel at 270C. Calculate the partial pressure of

each gas, total pressure of mixture and volume % of H2.

Solution: 25.0328;22

422

OH nn

Applying Ideal Gas Equation separtely, we shall get the partial pressures.

atmpH 6.242

300082.022

and atmpO 075.32

300082.025.02

So, P = 24.6 + 3.075 = 27.675 atm.

So volume % of H2 = %89.88100

)25.02(

2100

2

Hx

(Note that the high pressure of 24.6 atm, does not indicate ideal behaviour. Please take the above numericalfor the sake of understanding the calcuation aspects)SAQ: The entire contents of a 2-L vessel containing N

2 gas at 800 mm pressure was transferred to a 4 L vessel

containing O2 gas at 200 mm pressure. Calculate partial pressure of each gas, total pressure and volume

percent of N2.

Solution: Note that volume of the mixture is same as the initial volume of O2 gas i.e 4 L. Hence the partial

pressure of O2 will be same as its initial pressure. Note that these are valid only under ideal conditions.

mmpN 4004

80022

mmpO 2004

20042

P = 400 + 200 = 600 mm; Volume % of N2 = %67.66100

600

400100

2

Nx

(Note that mole fraction is same as pressure fraction)

(5) Graham’s Law of Diffusion(Effusion):Temperature and Pressure remaining constant, the rate of effusion of gases is inversely proportional to thesquare root of their densities.(Diffusion of gas through a small orifice in one direction is called effusion)

d

kr

dr

1, Comapring the rates of effusion of two gases A and B

A

B

A

B

A

B

B

A

M

M

D

D

d

d

r

r (where D= V.D of gas, M= Molecular Mass of gas)

Rate of Effusion of gases is usually determined as follows:

t

Vrate (where V= volume of gas diffused, t = time required for it)

Time of effusion(t) remaining constant:;

A

B

B

A

M

M

V

V

Volume of effusion(V) remaining constant;

B

A

B

A

M

M

t

t



Graham’s Law with variable Pressure:

A

B

B

A

B

A

M

M

P

P

r

r

d

Pr

Rate of effusion is directly proportional to gas pressure.SAQ: CO gas effuse through an effusiometer in 200 s, while the same volume of an unknown gas effuses in53.5 s at constant temperature and pressure. Calculate the molecular mass of the unknown gas.

Solution:

004.228

5.53

20028 x

xxt

t

X

CO

SAQ: SO2 and CH

4 separately were allowed to effuse at 8 atm. and 1 atm. respectively at the same temperature.

What the ratio of rate of effusions of SO2 and CH

4.

Solution:

1

4

64

16

1

8

2

4

4

2

4

2 SO

CH

CH

SO

CH

SO

M

M

P

P

r

r

Note that CH4 gas effuses 2 times faster than SO

2 at the same pressure, because the former is a lighter

gas(lower MM) than the latter. But at CH4 and SO

2 pressure ratio 1:8, SO

2 effuses 4 times faster than CH

4.

Hence the effect of pressure on the rate of effusion is quite prominent.

SAQ: What volume of SO2 gas will effuse when 200 mL of O

2 effuses through an effusiometer under same

conditions.Solution: Here time of effusion remains constant.

mLxxM

M

V

V

SO

O

O

SO 4.14164

32200

2

2

2

2

SAQ: Certain vaolume of a mixture of ozone and O2 with volume % of 60% O2 was effused in 360 s. Same

volume of an unknown gas effused under same conditions in 307 s. Calculate the molecular mass of theunknown gas.Solution:

4.38100

48403260

M

94.274.38

307

3602323

xxx

M

t

tOO

X

OO



KINETIC THEORY OF GASES:Postulates:1. A gas is made up of tiny particles called molecules which are in a state of constant rapid motion in allpossible directions.2. Gas molecules collide with one another and also on the walls of the container. The latter collision givesrise to gas pressure.3. All the collisions are pefectly elastic i.e there always a conservation of momentum and kinetic energyduring collisions.4. The actual volume of gas molecules which are believed to be non-compressible is negligible comparedto the volume of gas i.e volume of the container.5. There is no intermolecular force of attraction or repulsion.6. The average kinetic energy of gas molecules is directly proportional to absolute temperature which isindependent of the nature of gases.

kTETE KK Kinetic Gas Equation:

A B

CD

E F

GH

u

uxuy

uz

u

uxx

z

y

uy

uz

u2 = ux2 + uy2 + uz2

Let a molecule is moving with a velocity of u in a cubic container having dimension of ‘l’. This velocity vector

can be resolved into three mutually perpendicular components ux, u

y and u

z. Geometrically, we can prove that

u2= ux2 + u

y2 + u

z2.

Analysis of ux Component:

ux component oscillates between BCFG and ADEH faces.

Momentum before colliding with DCFG face = mux

(where m = mass of one molecule)4Momentum after collision on the same face = – mu

x.

So change in momentum in one collision = 2 mux.

The distance for 1 complete oscillation = 2l; Hence time period(T) for one oscillation = xu

l2

Hence frequency of oscillation = l

u

Tx

2

1

Hence change in momentum in one second at one face = 2mux

l

u x

2 =

l

mux2



Hence change in momentum per second in two opposite faces = l

mux22

Hence force exerted by the Ux component on the two opposite faces of ‘x’ axis = l

mux22

uy and u

z components:

Like Ux component, the force exerted by uy and uz components in the respective opposite faces are

l

muy22

and l

muz22

.

Hence the total force exerted by 1 molecule in the six faces of the cube is as follows:

l

muuuu

l

mF zyx

2222 22

Let there are ‘n’ molecules having velocities u1, u

2, u

3,........ u

n. Hence the total force exerted by ‘n’ molecules

is as follows.

n

uuuu

l

mnuuuu

l

mF n

n

223

22

2122

322

21

...........2...........

2

⇒ u2F = 2mn

lwhere u2 =

n

uuuu n22

322

21 .......

= Mean Square Velocity

Pressure(P):u22mn

P =FA

=l 6l2

⇒ u2P =

13

mnV

u213

mnPV = Kinetic Gas Equation

Note that here n = number of molecules (not number of moles. Usually in ideal gas equation ‘n’ is taken asnumber of moles, but in Kinetic Gas equation, n is the number of molecules)

Root Mean Square Velocity (RMS velocity):

U2 = n

uuuu n22

322

21 ..........

From kinetic Gas eqation :

u213

mnPV = ⇒ U2 = '

33

m

PV

mn

PV

(where m’ =mn = mass of gas sample )For one mole of an ideal gas PV =RT; then n = N

A, where NA = Avogadro’s number)

U2 = M

RT

mN

RT

A

33 (1) (M = molecular mass of gas)



U2 = d

P

Vm

P

m

PV 3'

3

'

3 (where d = density of the gas)

SAQ: Calculate the RMS velocity of 0.75 g of a gas at 1.5 atm pressure occupying a volume of 2 L.Solution :

U2 =15101.1

75.0

20009806.13765.13

'

3

cmsm

PV

(note that the pressure is determined from the formula : P = hdg, and in CGS scale, the height of Hg column of1 atm. is 76 cm. g = acceleration due to gravity; d = density of Hg)SAQ: Calculate the RMS velocity of CO

2 gas at 270C at 2 atm. pressure.

Solution: Note that the velocity is only dependent on temperature and is independent of pressure. So we shalluse the relation which contains temperature only.

U2 = 147

1012.444

300103.833

cmsM

RT

The second way to express the velocity of gas moelcules is Average Velocity.

Average Velocity(u ):n

uuuuu n

.......321

M

RTu

8

(2) (See later for derivation)

Most Probable Velocity(uM) : It the velocity with which or near which maximum number molecules move.

M

RTuM

2 (3) (See later for derivation)

U2 : u : uM

= 1 : 0.92 : 0.82

SAQ: Calculate the RMS, average and most probable velocity of a gas having density 0.00075 g/cm3 at apressure of 1 atm.Solution:

U2 = 1410635.6

00075.0

6.1398076133

cmsd

P

So 144 1085.510365.692.0 cmsuu

M = 0.82 × 6.365 × 104 = 5.219 × 104 cms–1.

(Note that RMS velocity is the largest and the most probable velocity is the smallest among the three)Average Kinetic Energy:

From kinetic Gas equation:

u213

mnPV = ⇒ U2PV = 2

3n 1

2 m = n 3

2 EK



kTTN

R

n

PVE

AK 2

3

2

3

2

3

(Average KE per molecule)

Where k = Boltzmann Constant =(R/NA) = 1.37 × 10–16 ergK–1 = 1.37 × 10–23 JK–1

Average KE per mole : RTNTN

RmoleE A

AK 2

3

2

3)(

Average KE per ‘n’ moles of gas : nRTmolenEK 2

3)'(' (here n = number of moles)

SAQ: Calculate the average KE per molecule and per mole of CO2 gas and H

2 gas at 270C.

Solution: (i) CO2

ergkTmoleculeEK1416 102.63001037.1

2

3

2

3)(/

ergRTmoleEK107 10735.3300103.8

2

3

2

3)(/

Althernatively

ergergNmoleculeEmoleE AKK10102314 10735.310735.310022.6102.6)(/)(/

(ii) H2 : Same as CO

2

as KE is only dependent on temperature.

SAQ: Calculate the total KE of 0.5 mol of an ideal gas at 273KSolution: Total KE = (3/2) nRT = (3/2) × 0.5 × 8.3 × 273 = 1702 J (here n = number of moles)

Validity of Ideal Gas Laws in the light of Kinetic Theory of Gases:Boyle’s Law:

PV = 13 m n u2 =

32

21n m u2 = n2

3k'T (I)

⇒ PV = Const.(when T is constant) Boyle’s law(Since from the postulate of Kinetic Theory of Gases ; average KE ∝ T)

When volume of a given mass of gas is decreased at constant T, the total force exerted per unit area increasesas the density of gas increases; so more number of molecules hit on the unit surface area of the wall of thecontainer. Hence pressure increases. Opposite happens when the volume of a gas is increased.Charles Law:

From eqn. (I), we get;n2

3V = P

T = k'' T (constant P)

ConstT

V (At constant P) (Charles Law)

When temperature of a given mass of gas is increased, the average KE increases according Kinetic theory.Hence the force per unit area i.e pressure is expected to increase. But when pressure is kept constant by usinga movable piston on the gas vessel, the volume of the gas is bound to rise(piston is raised up) till the initialconstant pressure is established. Since the volume increases, the number of molecules hitting on unit surfacearea of the wall decreases, although individual molecule hits with greater force. Thus F/A remains constant. In



the initial state, more number of molecules with lesser force were hitting on the unit area and in the final statelesser number of molecules with greater force are hitting on the unit area. Conversely speaking, volume willincrease on increasing temperature till the intial pressure is reestablished, i.e pressure remains constant.Graham’s Law of Diffusion(Effusion):

u2 = 3RTM (From Kinetic Gas equation for one mole)

⇒r1r2

=u2

1

u22

=M2M1

(Graham’s Law of effusion)

The rate of diffusion of gas molecules is inversely proportional to its molecular mass, while T remainingconstant. Note that actually this is valid only when both T and P remain constant.(note that similarly Avogadr’s law can be derived from Kinetic Gas equation)

SAQ: Calculate the pressure exerted by 1023 gas particles each of mass 10–22 g in a container of volume 1dm3. The rms velocity is 105 cm/s. What is the total kinetic energy of these particles ? What will be thetemperature of the gas ?

Solution: P = 13

m n u2V

Pa

m

msKg 733

2132325

103

1

10

101010

3

1

12

Total KE = n m u2 JmsKg 500010102

110

2132523

KJK

TkTn 24141037.1103

10000

2

35000

12323

SAQ: A bulb of capacity 1 dm3 contains 1.03 × 1023 molecules of H2 and the pressure exerted by these

molecules is 101.325 kPa. Calculate the mean square velocity of gas molecules and the temperature of gas.

Solution: 171.010023.6

1003.1)(

23

23

molesn

KJ

m

nR

PVT 27.71

3.8171.0

1010325.101 333

u2 2151088.8002.0

27.713.833

msM

RT

Maxwell-Boltzmann Distribution of Molecular Velocities:Velcocity of gas molecules constantly change due to collisions. Thus there is a distribution of molecular velocitiesamong molecules which take values from very small to very large with maximum number of molecules movingat and near the Most Probable Velocity(u

M) represented by the maximum point in the graph of velocity and

fraction of molecules. The fraction of molecules having velocities between any definite range is constant at agiven temperature and pressure although the individual molecules constantly change their velocities.The fraction of molecules having velocity within u, and (u+du) is given by the following Maxwell’s law.

duueRT

M

n

dn RT

Mu

u 222

3 2

24



Maxwell put forth the distribution of molecular velocities based on theory of probability as follows.

222

3 2

24

1ue

RT

M

n

dn

duRT

Mu

u

( M = molecular mass of gas)

Or22

23 2

24

1ue

kT

m

n

dn

dukT

mu

u

( m = mass of one molecule; k = Boltzmann constant)

udn is the number of molecules having velocities betwen u and (u+du) and n is the total number of molecules.

Hence the ratio n

dnu represents the fraction of molecules having velocities u and (u+du). At at given T, this

fraction is constant. Consequently, the average kinetic energy of a gas is also constant at a given temperature.

Maxwell distribution of velocities is customarily plotted with

n

dn

duu1

as ordinate and u(absolute velocities)

as the abscissa. The ordinate term gives the fraction of molecules having velocity between u and (u+du) per unitinterval of velocity. In fact, this term gives the probability of finding a molecule with velocity between u and(u+1).

The graph rises parabolically(sharply) as the u2 term is dominant and the exponential term is approximatelyequal to unity. However, at higher velocities, the exponential term gradually comes to prominence where thegraph attains its maximum point(u

M) and therafter the graph falls exponentially(gradually) due to dominance of

the exponetial term. Thus the fractions of molecules having either very small velocities and very high velocitiesare small. Majority of molecules move with velocites equal to or nearer to Most Probable Velocity(u

M) which

is intermediate in the range.At higher temperature, the graph becomes broader, shorter in height and elongated to the right. However, thearea under the graph which gives the total number of molecules remains constant with change in temperature.Although at higher temperature, u

M is shifted to right, the fraction of molecules having velocity equal to u

M has

reduced(shorter height). Since the curve has been broader at uM

, more number of molecules have velocitiescloser to u

M at higher temperature. Fraction of molecules having velocities greater than u

M will increase at the

expense of the decrease in the fraction of molecules having velocities lesser than uM

. This can be known fromthe areas shown under the two graphs at two temperatures above their u

M values.



Narrowness/Broadness of Distribution Graphs:

The distribution is dependent on the ratio

T

M. Greater the ratio, narrower is the distribution(graph) and

smaller the ratio, broader is the distribution. For a fixed gas(M fixed), greater the T, smaller is the ratio andhence broader is the distribution. For a fixed T, heavier the gas, greater is the ratio and hence narrower is thedistribution. At fixed temperature, SO

2 gas has a narrower distribution than O

2 gas. But SO

2 gas will have the

same ditribution at temperature 2T as O2 gas at temperature T.

At maximum point, 0

1

dududn

nd u

RT

Mu

RT

Mu

eRT

Muueu

RT

M

du

ueRT

Md

dudu

dn

Nd

RT

Mu

u

2

2

2

2

2

2

22

24

241

22

3

222

3

When the 0

1

du

du

dn

Nd u

, u in RHS is replaced by uM

.

02

22

24 222

23 22

RT

MuM

MRT

Mu

M

MM

eRT

Muueu

RT

M

022

42

22

3 2

RT

Mueu

RT

M MRT

Mu

M

M

Since all other tems except the last term inside square bracket cannot be zero, we have

022

RT

MuM

m

kT

M

RTuM

22

Where m= mass of one molecule and k = Boltzmann constantSAQ: Calculate the fraction of nitrogen molecules at 101.325 kPa and 300K whose speeds are in the range of(u

M – 0.005u

M) to (u

M + 0.005u

M)

Solution:

smM

RTuM /422

028.0

3003.822

; smuM /11.2422005.0005.0

smdu /22.411.22

duueRT

M

N

dn RT

Mu

u 222

3 2

24

92

3

1039.23003.814.32

028.0

2

RT

M

;



3679.0718.2

113003.82

422028.02

2

eeeRT

Mu

329 103.822.43679.04221039.214.34 N

dnu

Derivation of Average Velocity:

i

in u

nn

uuuuu

1........321

This can be expressed in definite integral form as follows.

00

1

n

dnuudn

nu u

u

(where dn= number of molecules having velocity between u and (u+du))

Substituting the value of n

dnu , we get

dueuRT

Mudn

nu

RT

Mu

u

232

3

00

2

24

1

0

232

3 2

24 dueu

RT

Mu

RT

Mu

On integration, we get;22

3

22

4

M

RT

RT

Mu

;

m

kT

M

RTu

88

(where m = mass of one molecule, k=Boltzmann const.)

Kinetic Energy Distribution :

212

1

212

1

2

2

12

2

1

E

mdE

duE

mumuE

⇒ dEEm

du 212

1

2

1

We alraedy have,

duueRT

Mndn

RT

Mu

u22

23 2

24

Substituting the value of u2 and du in the Maxwell’s law for velocity distribution we have,



dEEmM

Ee

RT

Mndn RT

E

E2

121

23

2

12

24

dEeEkT

ndn kT

E

E

2

123

12

dnE is the number molecules having energy between E and (E+dE). Hence the fraction of molecules having

energy in the above range is given by

dEeEkTn

dn kT

E

E

212

31

2

Plot of

dE

dN

nE1

vs. E gives the distribution of KE among molecules. This graph, though similar to velocity

distribution graph, rises more sharply and falls more gradually compared to the latter.

Fraction of molecules having KE greater than some specified energy :

We can derive the experssion of the fraction of molecules having energy greater than a specific value, forexample, threshold energy in studying chemical kinetics etc.The fraction of molecules having energies greater than E' is given by

'

212

3

'

' 12

E

kT

E

E

EE dEeEkTn

dn

n

n

This, on integration yields

kT

EE e

kT

E

n

n '21

' '2

(where E’ > kT)

SAQ: Calculate the fraction of molecules of nitrogen gas at 101.325 kPa and 300K whose kinetic energy are

in the range of EE 005.0 and EE 005.0 .

Solution:

JkTE 2123 10213.63001037.12

3

2

3

JE 2121 10031.010213.6005.0005.0

JJdE 2321 10213.610031.02

292

3

10742.61

kT 223.0

kT

E

e

dEeEkTn

dn kT

E

E

212

31

2

32321

2129 10624.410213.6223.010213.610742.614.32 n

dnE



SAQ: Calculate the number of molecules in one mole of an ideal gas that have energies greater than four timesthe average thermal energySolution:

kTE2

3 . So the required energy, kTkTE 6

2

34

The fraction of molecules having energy greater than E is given by

62

1621

21

62

622

ee

kT

kTe

kT

E

N

n kT

kT

kT

E

A

E

Upon solving by log and antilog method, we get

31086.6 A

E

N

n213 10131.41086.6

AE Nn

Note that in this case, the fraction A

E

N

n is independent of temperature, as the T term is cancelled out. Hence the

number of molecules above a multiple of average energy will remain same at temperature 250C and 500C. Theareas bound by the graphs above the respective energies at two temperatures are the same.

Variation of Gas Pressure with Height: (Barometric Distribution Law):As a result of the influence of gravitational field, there is a decrease in pressure and density with increase inheight relative to the sea level(ground). This is ignored for a small gas vessel particularly for low molecular massgas. However if there is long gas column or say the atmosphere above us, or a macromolecular gas sample(havinghigh molecular mass), the pressure of gas decreases with increase in height. This is given as follows(derivationis not given).

RT

Mgh

p

p

0

ln RT

Mgh

epp

0 RT

Mghpp 0lnln

Where M = molecular mass or average molecular mass of the gas/gas mixture.h = height above the sea level at which pressure is determined.p

0 = pressure at the sea level (h =0). T = Mean tempeature at height =0 and height = h

Similar equations are valid for densities.

RT

Mgh

0

ln

RT

Mgh

e

0

Where ρ= density at a height of ‘h’. and ρ0 = density at sea level



Effect of Temperature on Barometric Distribution:Note that for general purpose, and for height not very high, we use the average temperature at h=0 and h=h .However, in reality, the temperature is decreased as we go higher and hence the effect of temperature on thepressure can be accurately determined as per the following law(derivation not shown). Decrease of temperaturefurther decreases the pressure.

ahTRa

MgT

Ra

Mg

p

p 00

0

lnlnln

Where T0 = Temperature at sea-level, and temperature at any height is expressed as linear funtion of h

i.e T = (T0 - ah), where a = constant (not V.W constant)

SAQ: Determine the molecular mass of a gas if its pressure is to fall to one half of its value in a vertical distanceof one metre at 298K.Solution: p = p

0/2; h = 1 m, g = 9.8 m/s2.

gKgMM

p

p175133133.175

2983.8

18.9

2ln

0

0

SAQ: Calculate the pressure of a barometer on an aeroplance which is at an altitudeof 10 Km. Assume thepressure to be 101.325 kPa at sea level and the mean temperature 243 K. Use the average molar mass ofair(80% N

2 and 20% O

2).

Solution:

0288.0032.02.0028.08.0 M

kPapp

027.252433.8

100008.90288.0325.101ln



Real GasesGases which really exist are called real gases. This means that all gases that we know are real. Only underrestricted conditions of high temperature(≥ T

B) and low pressure( ≤2 atm), real gases behave ideally i.e obey

all the gas laws such as Boyle’s law, Charles law, Avogadro’s law, Dalton’s law etc. Under such conditions onecan brand them as ideal gases on a temporary basis. Strictly speaking, no gas can be designated as ideal, instead, we can say a real gas shows ideal behavior at lower pressure and higher temperature conditions. Underopposite conditions they behave non-ideally i.e deviate from ideal behavior. The product of P and V, no longerremains constant at constant temperature. PV - P graph no longer remians a straight line parallel to P-axis.Nature of Deviation:Compressibility Factor(Z):

For non-ideal behavior : PV ≠ k; PV ≠ nRT (at constant T)

Compressibility Factor(Z) = nRT

PV

(When Z = 1, the gas behaves ideally, Z ≠ 1, the gas behaves non-ideally)

)(

)(

idealV

realVZ

m

m

Where Vm(real) is the actual molar volume at a particular P and V

m(ideal) is the ideal volume at that P.

AMAGAT’S CURVES for Real Gases:

Real gases show two types of deviations.

(1) Positive Deviation(Z>1):If the gas is kept at or above its Boyle’s temperature((T

B), the PV-P graph increases linearly with increase in

pressure in stead of remaining constant as expected from ideal behavior. The graph remains above the ideanRT line.This is positive deviation. At this condition, Z > 1. This is due to intermolecular repulsion for which thegas is less compressible than idea value. V

m(real) > V

m(ideal).

However, under low pressure region(< 2 atm), PV remains nearly constant and shows ideal behaviour. Thegraphs shown above have been obtained for high pressure ranging upto 1000 atm. So the linearity at very lowpressure region is not observed.Examples: H

2 and He at 00C.

Note that since these gases have subzero TB values(Refer the table given before), at all temperatures above it,

including room temperature which is above the TB, these gases show positve deviation only.



(2) Negative Deviation (Z<1):If ths gas is kept below its T

B, PV-P graph initially decreases with increase in pressure to a minimum and

thereafter increases. The PV-P graph remains below the ideal nRT line as long as the gas is showing negativedeviation and Z <1 for it. Beyond a certain high pressure, the graph crosses the ideal line and then it startsshowing +ve deviation(Z>1). The neative deviation is due to intermolecular attraction for which the gas is morecompressible than its ideal value. V

m (real) < V

m(ideal)

Example: N2 at 250C and CO

2 at 250C.

Note that gases which have higher TB values show such deviations. More the difference between the working

temperature and TB for a gas, more is the extent of deviation. CO

2 has a T

B =7530C, while for N

2, T

B=+590C.

Therefore, the dip in the PV-P curve for CO2 is much greater than that of N

2, both of which show negative

devaitions first when the fixed working temperature is below their respective TBs.

Conclusion: Every gas is real and can show negative deviation first followed by positive deviation if the workingtemperature remains below its T

B. For H

2 and He, such a situation will arise if the working temperatue goes

below –156 0C and –249 0C respectively. For higher temperature, these gases never show negative deviation.Gases like N

2 and CO

2 also will not show negative deviations if their working temperature goes above 59 0C

and 753 0C respectively.(3) Ideal Behaviour :

As the temperature increases, the extent of negative deviation i.e the dip in the PV-P curve decreases and theminimum point gradually shifts to y-axis. At T

B, there is no minimum point in the graph, as the dip or the negative

deviation vanishes at TB and the gas shows positive deviation only. For example, in case of N

2 gas, the dip

vanishes at 590C and in case of CO2 at 7530C.

At this condition, when working temperatue becomes equal to TB or above, the PV-P graph becomes parallel

to P axis only under lower pressure conditions( i.e upto 2 atm.). As the pressure increases beyond 2 atm. moreand more, the extent of positive deviation increases. more and more.Just like an aeroplane goes straight in the ground in the runway of the airport upto some distance and then takesoff above the ground, the PV remains constant upto nearly 2 atm(1720 mm of Hg) pressure and thereafterincreases with further increase in temperature. Note that this happens, only when the working temperature is ator above T

B. Below T

B, there is negative deviation even at low pressure(<2 atm) conditions. The first graph

shown above are taken under high pressure conditons(upto say 1000 atm) at differnt fixed temperature conditionswhile the graph above under low pressure(upto 2 atm) conditions at or above T

B.



nT

PV vs. P graph:

The above graph is a straight line parallel to P-axis when the gas behaves ideally and has a value of R(8.3JK–1mol–1). In otherwords, for all real gases the the above plots when extrapolated to P=0, meet at the samepoint i.e a value of R.

Cause of Deviation:

Because there is intermolecular attraction and repulsion at high pressure and low temperature and also at thoseconditions, the volume of gas molecules can be negligible, we observe deviation from ideal behaviour. Thesewere the two fundamental defects present in the postulates of kinetic theory of gases. However, under hightemperature and low pressure conditions, ideal behaviour is observed, as has been explained before.

Van der Waals Equation for Real Gases:

Van der Waals modified the ideal gas equation; PV = nRT, by introducing (a) pressure correction and(b) volume correction terms in it. He used ideal pressure and ideal volume for the real gas by introducing thesecorrections in the observed pressure and volume.Pressure Correction :A molecule moving inside the container is attracted by other molecules from all direction at a given point of timeand the resultant force acting on it is zero. However, the molecule which is about to strike on the wall isattracted by molecules on one side(as the other side is wall). Hence it is acted upon by a net resultant inwardforce(R) for which the observed pressure is lowered from the ideal pressure(in the absence of attractiveforces).

R



Let p' = reduction in pressure due to the net resultant inward attractive force(pull)= pressure correction term = cohesive pressure.P = Observed pressure

So ideal pressre(PI) = P + p'

Measurement of p' :Van der Waals proposed that the p’ is dependent on two attraction terms.

1 2

p' ∝ attractive force that molecule 1 is exerting on molecule 2 which is striking the wall. ∝ attractive force that molecule 2 is exerting on molecule 1 which is inside the container.

Each attractive force is directly proportional to number density(n/V)

Number density V

n (where n = number of moles of gas, V= volume)

V

n

V

np '

2

2'

V

anp

Where a = Van der Waals constantUnit of ‘a ‘ = atm.L2mol–2.

Ideal Pressure 2

2

V

anPPI

Volume Correction:The volume of gas(volume of container) is made up of two parts

(a) total compressible volume/ideal Volume (VI) :

(b) total non-compressible volume(excluded volume):While compressible volume consists of empty space present in the container, non-compressible volume consitsof the actual volume of gaseous molecules and some associated empty space which also cannot be compressed.

Let total non-compressible volume = v' = volume correction term = covolumeVolume of the gas = volume of the container = VIdeal volume of the real gas = V

I = V – v'

Ideal gas is thought to be completely compressible as the volume possessed by the gas molecules wereassumed to be negligible, which is actually not. Gas molecules are actually believed to be rigid, non-comressibleand nearly spherical particles. Hence some more empty space is carried away with the molecules as part oftotal non-compressible volume. Imagine when large number of sphercal glass beads are packed in a container,there are large empty space embedded within the packed spheres which cannot be further compressed alongwiththe glass beads.

Determination of v' :Since gas molecules execute all types of motions such as translational, vibrational, rotational etc. Van

der Waals proposed the imaginary sphere(bigger sphere) containing two spherical molecules(smaller spheres)in contact with each other, is the the total non-compressible volume for two molecules.



R

r

If R= radius of the imaginary bigger sphere and r = radius of the molecule, then R = 2r

Volume of bigger sphere =

33

3

48

3

4rR = total non-compressible volume for 2 molecules

So non-compressible volume for one molecule =

3

3

44 r = 4 × volume of one molecule

So non-compressible volme for one mole(NA molecules) = 4 × N

A × volume of one molecule = b

Where ‘b’ = Van der Waals constant(note that ‘a’ was one Van der Waals constant and ‘b’ is the other)

So for ‘n’ moles of a gas total non-compressible volume = nb

Unit of ‘b’ : nb = L, so b = Lmol–1.

Ideal Volume(VI) = V – v' = V – nb

Van der Waals Equation:Ideal pressure (P

I) × Ideal volume (V

I) = nRT

nRTnbVV

anP

2

2

For one mole Van der Waals equation becomes

RTbVV

aP

2 (for one mole)

Significance of ‘a’ and ‘b’:‘a’:

‘a’ is a measure of intermolecular force of attraction. Greater the value of ‘a’ greater is this force. ‘a’value increases with increase in polarity and size of the molecules. The value of ‘a’ for Xenon is 4.19atm.L2mol–2 due to its larger size(polarizability) and that for NH

3 it is almost the same (4.17 atm.L2mol–2) and

is due its polarity. The value of ‘ a’ are lowest for H2, He and Ne. Greater the ‘a’ value greater is the

TB(Boyle’s temperature) and hence greater is the deviation from ideal behaviour at room temperature.

Greater the value of ‘a’’ for a gas, greater is its boiling point. Greater heat is required to break the greaterintermolecular attraction.



‘b’:‘b’ is a measure of total excluded volume of the gas, which is dependent on(a) size (b) shape (c) polarity (d) bond angle.

‘a’ and ‘b’ values of some common gases:Gas a (atm.L2.mol–2) b (L.mol–1)He 0.0341 0.0237Ne 0.211 0.0171H

20.244 0.0266

Ar 1.34 0.0322O

21.36 0.0318

N2

1.39 0.0391CO 1.49 0.0399CH

42.25 0.0428

Kr 2.32 0.0398CO

23.59 0.0428

HCl 3.67 0.0408NH

34.17 0.0371

Xe 4.19 0.051H

2O 5.46 0.0305

Cl2

6.49 0.0562CCl

420.4 0.1383

NB: Note that the ‘a’ and ‘b’ values of He, Ne and H2 are the lowest.

The pressure calculated by using Van der Waals equation under non-ideal conditions closely matched with theobserved pressure. Thus the Van der Waals equation was highly useful for real gases and serves as convincingmodification of ideal gas equation.

Note that the constants ‘a’ and ‘b’ are independent of temperature and only are characteristic constants ofeach gas.



SAQ: Calculate the total non-compressible volume in 1 mole of water vapour. Refer ‘b’ value of H2O from

table. Approximately what volume of it constitutes the empty space embedded in it.Solution: 1 mole of liquid water = 18 g = 18 mLIf we presume that there is negligible empty space in liquid state and the the actual volume of H

2O molecules in

one mole = 18 mL(which is not true as some empty space is still there in liquid state to allow molecular motion)b = 0.0305 L.mol–1 = 30.05 mL mol–1 = total excluded volume for 1 mole of water vapour. In

other words, we cannot keep 1 mole of water vapour in a volume lower than 30.05 mL. Note that watervapour exists at a higher temperature. But since on compression, it can be liquefied(see later), practically wecannot reduce the volume of one mole of water vapour to 30.05 mL by applying pressure above its criticaltemperature of 647.4K. But one can verify this above its Critical temperature(Please see later). Above thistemperature, we cannot compress 1 mole of water vapour to a volume less than 30.05 mLSo in the vapour state, approximate volume of empty space embedded in the exluded volume =

30.05 – 18 = 12.05 mLSAQ: Calculate the radius of a O

2 molecule. Given that ‘b’ for O

2 = 0.0318 L mol–1 = 31.8 mL mol–1.

Solution:

323

3

410023.64 rb

323 14.3

3

410023.648.31 r

466.110466.114.3410023.64

38.31 83

23

cmr Å

SAQ: Calculate the pressure exerted by 10 moles of CO2 gas in a 2L vessel at 470C using Van der Waals

equation. . (a = 3.59 atm.L2.mol–2 ; b = 0.0428 L.mol–1. Also calculate the ideal pressure of the gas. If theobserved pressure is 82 atm., comment on the valididy of Van der Waals modification.Solution: Applying Van der Waals equation ,we have to solve this.

atmV

an

nbV

nRTP 065.77

2

1059.3

0428.0102

320082.0102

2

2

2

Ideal Pressure: atmV

nRTP 2.131

2

320082.010

The ideal pressure(in the absence of intermolecular attraction), the pressure would have been 131.2 atm, butin reality there is intermolecular attraction at temperatue below T

B(Boyle’s temp.), hence the actual pressure is

less( 82 atm.). Van der Waals equation gives the pressure close to the observed value. Hence it is approximatelyvalid for the real gas at the given conditions, though not fully valid. No doubt, it is a significant advancementover the ideal gas equation.

SAQ: Two real gases have the same value of ‘b’ but different ‘a’ values. which of these could occupy greatervolume under identical conditions ? If the gases would have same ‘a’ value but different ‘b’ values, then whichwould be more compressible ?Solution: (a) The gas having greater ‘a’ value will occupy lower volume, as the intermolecular forces will bemore in that gas. (b) If the ‘b’ values are different, then gas having lower ‘b’ value is more compressible as thetotal non-compressible volume is less for that gas.



Validity of Van der Waals Equation at different conditions:Case-I: Under Low Pressure:(for all gases except He, Ne and H

2 whose ‘a’ values are low and hence T

B subzero)

Note that for all other gases, TB are greater than room temperature. So below T

B, i.e at room temperature, first

negative deviation is observed at low pressure conditions.At low pressure condition, volume is large, the term ‘nb’ term is neglected from the V.W equation.

nRTVV

anP

2

2

nRTV

anPV

2

12

VnRT

an

nRT

PVVRT

anZ 1

Since Z < 1, it shows negative deviation under low pressure conditions at room temperature.Case-II: At high pressure :This is valid for all gases except He, Ne and H

2 at room temperature.

Since pressure is high, the term 2

2

V

an is neglected from V.W equation.

nRTnbVP nRTPnbPV

1nRT

Pnb

nRT

PV

RT

PbZ 1

Since Z > 1, the gas shows positive deviation at high pressure conditions.Case-III: At high temperature and low pressure :

Under the above conditions both the 2

2

V

an and ‘nb’ are neglected from V.W equation, which is reduced to

ideal gas equation; PV = nRT or Z = 1The gas shows ideal behaviour.Case-IV: For He, Ne and H

2:

Since the value of ‘a’ for these three gases are very low(see table), the 2

2

V

anterm is neglected from

V.W equation (analogous to case-II) under all conditions of pressure at room temperature(which is muchabove their T

B).Hence they show +ve deviation at all pressure conditions(low and high).

RT

PbZ 1



Experimental Validity of Van der Waals Equation: (Critical Pheonomena andLiquefaction of gases)

(I) Andrews Experimental Isotherms: P – V curves for CO2 at different temperature.

P - V graphs below a certain temperature called Critical Temperature(TC) look different compared to usual P-

V graph(smooth curve) which appear above TC. For CO

2 gas, T

C = 30.980C. Below this temperature, the P -

V graphs show three distinct linear regions, AB, BC and CD as shown for temperature 21.50C.(Note that in many literature, T

C of CO

2 is recorded as 31.050C)

AB: This portion represents gaseous phase. In fact below TC, the gas is called VAPOUR.

BC: This represents the equilbrium between vapour phase and liquid phase. Liquefaction starts at point B andends at point C, with gas pressure remaining constant. The pressure of the gaseous phase is called the VapourPressure at that temperature. There is a drastic volume reduction(horizontal portion) during liquefaction. Thishorizontal portion diminishes as the fixed temperature is increased. At T

C, the horizontal portion is reduced to

single point, called the Critical Point(X). The pressure at this point is called Critical Pressure(PC) at which gas

is liquefied without change in volume. There is no phase separtion while liquefaction occurs.Critical Pressure(P

C) :Hence P

C is defined as the pressure at the critical temperature at which liquefaction of

a real gas occurs without change in volume. At point X, the the liquid and gas phases are indistinguishable. Thedensities liquid and vaopur become identical and there is no boundary surface of separation between the twophases.Critical Volume(V

C): The volume of one mole of real gas at T

C and P

C is called Critical Volume(V

C).

Critical Temperature(TC) : It is temperature which is fixed for each gas, at or below which a gas can be

liquefied by applying pressure.Gases like O

2, N

2, He, He have subzero T

C values and hence can only be liquefied by first cooling the gas to

or below their respective TC values and then applying pressure, liquefaction occurs. For example, Oxygen gas

has a TC of 155K, the gas is cooled by repeated Joule Thomson Effect just below 155K. Then the gas is

compressed isothermally and at certain optimum pressure liquefaction occurr. If it is exactly TC, then liquefaction

occurs at PC occuping a volume of V

C per mole.

CD: This represents liquid phase, which is much less compressible than the gas phase, and hence the portion issteep.* At any constant temperature higher than T

C, the gas cannot be liquefied by applying any amount of

pressure. P-V graph becomes smooth curves, apparently looking like that for ideal behaviour. However PV-



P graph does give –ve deviation. Because TB is much higher than T

C. Only above T

B, ideal behavious is

observed under low pressure conditions.

Surface of Discontinuity:The line joining the two ends of the horizontal poritions of P-V graphs below critical point produces a hillarea(shaded differently) which is called surface of discontinuilty.. To the right of this area, there is only gasphase and to the left(also left of line EH), only liquid phase exists. But within this area, there exist both liquidand gas phase in equilibrium. Above this area and to the right of the line EH, there exist only the gas phase.Note that one can directly convert a gas to liquid without discontinuity(equilbrium state) by travelling above thisarea of discontinuity. For example, in case of CO

2 gas at point A at 21.50C having volume V

4, can be heated

above 21.50C in a sealed vessel(volume constant) so as to increase its pressure linearly(Amonton’s law) andthe graph is a vertical line AF as shown in the P - V graph. Point F is at temperature of 31.20C. Here if againtemperature is kept constant and pressure is increased using a vessel having movable piston to apply pressure,then it gives continuous curve FG where the gas can never be liquefied. However, in the constant volumevessel, if temperature is further increased above T

C say to 800C and the pressure attained is 84.5 atm(which

is above PC of CO

2=72.85 atm), and then cooled at constant pressure. A fixed load(weight) is now kept on the

movable pistion so as to keep pressure constant, the volume will decrease with decrease in T (Charles law) ina horizontal line(not shown) in the P-V graph. At certain temperature the line goes to the left of the line EH andturns abruptly to liquid state without any discontinuity.

Supercritical fluid: When the T > TC, and P > P

C i,e above the critical point the fluid in the close vicinity of the

line EH, a real gas is called as a supercritical fluid . At this state, there is no distinct gas or liquid phase as thereis no phase boundary. The surface tension of such fluid is almost zero. Such fluid has flow and diffusionproperty like a gas and solvent property like liquid. Such fluid can effuse through solids like a gas and dissovematerials like a liquid.. Close to critical point, its density can be tailor-made by small change in T and P andfine-tune its properties for specific application.. Supercritical fluids have great application as solvent fornanoparticles and other materials Supercritical CO

2, H

2O etc are used as substitutes for damaging organic

solvens in “green chemistry”, the latter for power generation. The sky blue region close to the vertical line(EHin the previous diagram) is P, T region for the supercritical fluids.

Critical Temperature(TC) of some commom gases:

Gas He H2

N2

O2

CH4

CO2

NH3

C4H

10

TC(0C) –267 –239 –147 –118 –83 31 132 153



Critical Constants of some common gases.

Gas PC(atm) V

C(cm3.mol–1) T

C(K) Z

CT

B(K)

Ar 48.00 75.25 150.72 0.292 411.5Br

2102 135 584 0.287

C2H

450.5 124 283.1 0.270

C2H

648.20 148 305.4 0.285

C6H

648.6 260 562.7 0.274

CH4

45.6 98.7 190.6 0.288 510.0Cl

276.1 124 417.2 0.276

CO2

72.85 94.0 304.2 0.274 714.8F

255 144

H2

12.8 64.99 33.23 0.305 110.0H

2O 218.3 55.3 647.4 0.227

HBr 84.0 363.0HCl 81.5 81.0 324.7 0.248He 2.26 57.76 5.21 0.305 22.64HI 80.8 423.2Kr 54.27 92.24 209.39 0.291 575.0N

233.54 90.10 126.3 0.292 327.2

Ne 26.86 41.74 44.44 0.307 122.1NH

3111.3 72.5 405.5 0.242

O2

50.14 78.0 154.8 0.308 405.9Xe 58.0 118.8 289.75 0.290 768.0CO 34.99 93.1 132.91 0.295

(II) Theoretical P-V Isotherms from Van der Waals equation:

From V.W equation, we get the following equation for 1 mole of a real gas.

RTbVV

aP

2 (For one mole: hence V = Vm)



023

P

abV

P

aV

P

RTbV (Subscrit m is dropped)

The P-V graphs below TC shows one maximum and one minimum point in the liquefaction zone(unlike the

experimental horizonal line of Andrews isotherms). The above cubic equation of volume has three roots for agiven set of values of ‘a’, ‘b’, P and T. All the roots can be real as shown in the diagram. At a fixed pressure,the there will be three different volumes corresponding to root 1, root 2, and root 3. RHS ascending(from Root3) portion(upto maximum)of the gaseous phase can be realised without liquefaction by very slow change involume/pressure which is called supercooled or supersaturated vapour. Similarly LHS descending(from Root1) portion (upto minimum)can be realised without vapour formation by slow change in volume/pressure, whichis called superheated liquid. Both these states are, however, are metastable states and can quickly revert to aequilibrium mixture of liquid and vapour on slight disturbance. But the portion from minimum point to maximumpoint having +ve slope cannot be realised. Hence, this is a major flaw in Van der Waals equation that there isa serious mismatch between the theoretical graphs from his equation and the exprimental Andrews graphsbelow T

C.

However the maximum and minimum points coalesce to a single point called the Critical point at TC.

0dV

dPand 0

2

2

dV

Pd

From V.W equation, we have

2V

a

bV

RTP

(for one mole) Note that we have written V for V

M.

0

232

CC

C

V

a

bV

RT

dV

dP

32

2

CC

C

V

a

bV

RT

(1)

0

62432

2

CC

C

V

a

bV

RT

dV

Pd

43

62

CC

C

V

a

bV

RT

(2)

Dividing eqn. (1) by (2) we get, VC = 3b (3)

Substitution eqn (3) in eqn. (1) we getRb

aTC 27

8 (4)

Substituing the values of TC and V

C in V.W equation, we get P

C.

227b

aPC

From the above three expressions for TC, V

C and P

C, we can get the Van der Waals’ constants ‘a’ and ‘b’ in

terms of critical constants.

C

CC

P

RTVb

83 (i)

C

CCC P

TRVPa

222

64

273 (ii)

C

CC

T

VPR

3

8 (iii)



Critical Coefficient(ZC):

The ratio C

CC

RT

VP is called critical coefficent(Z

C) and it is constant quantity.

Critical Coefficient (ZC)= 375.0

8

3

C

CC

RT

VP

Failure of Van der Waals equation:Although, V.W equation is a significant advancement over ideal gas equation for a real gas, it has some flaws.(i) Observed Critical coefficients of real gases have values less than 0.30(see previous table) as against of0.375 predicted by V.W equation.

(ii) b

VC for all real gases should be 3, as per V.W equation. However, the observed values are less than 3. It is

nearly 2.

Determination of Boyle’s Temperature(TB) from PV-P curves (Amagat’s Curve):

Note that a gas below its TB, shows negative deviation at low pressure region. When temperature is increased

the extent of deviation(the dip) decreases and the minimum point shifts towards y-axis(PV axis). At TB, the

minimum point falls on the y-axis, at P=0, the slope is the x-axis(P- axis). Mathematically for minimum point,the differential coefficient is to vanish at P=0.

0

TP

PV(at P =0 at T

B)

V

a

bV

RTVPV

TTT P

V

V

PV

P

PV

TP

V

dV

Vda

bV

V

dV

dRT

1

TT P

V

V

a

bV

bRT

P

V

V

a

bV

RTV

bV

RT

2222

At TB,

022

BT

B

P

V

V

a

bV

bRTHence

022

V

a

bV

bRTB

(Since BTP

V

cannot be zero)

22 V

a

bV

bRTB

Since ‘b’ can be neglected in the denominatior as it is too small in the face of ‘V’ at TB. So we have

22 V

a

V

bRTB From this TB is determined as



Rb

aTB

Comparision between TB and T

C :

Rb

aTC 27

8

Rb

aTB

375.38

27

C

B

T

T

TB is expected to be 3.375 times greater than T

C. But the observed ratio has been found to be less than this

with an average value of 3.0. Only exception is Helium having this ratio of 3.65. This is another defect in theV.W equation.SAQ: Indicate whether the gas will show ideal behaviour at the following conditions.Justify your answer.(a) 0.25 mole of CO

2 at 1200K exerting a pressure of 24.63 atm occupying 1 L volume.

(b) 1 mole of CO2 at 300K occupying a volume of 22.42 L at 1 atm. pressure.

Solution:

(a) 0.1001.1082.0120025.0

163.24

nRT

PVZ

The gas behaves ideally at these conditions as Z =1. As the TB of CO

2 is 714.8K, it will behave ideally at

1200K even upto 24.63 atm pressure. Note that at TB, the ideal behaviour is shown upto 2 atm. But as the

temperature increases, the maximum pressure upto which ideal behaviour is shown also increases.

(b) 91.0082.03001

42.221

nRT

PVZ

The gas behaves nonideally with negative deviation. Since the working temperature is much lower than its TB,

even at 1 atm. pressure, it shows negative deviation. Note that when T < TB, negative deviation is shown even

at low pressures.SAQ: The Critical constants for water are 647K, 22.09 MPa and 0.0566 dm3.mol–1. Calculate ‘a’, ‘b’ and R.Explain the departure from their actual values.

Solution: 13133

9.183

100566.0

3

molcmmolcmV

b C

Note that the actual value of ‘b’ for water is 30.05 cm3.mol–1. This is the drawback of V.W equation.

232133362 2123.01.00566.01009.2233 molJmmolmJmVPa CC

The exprimental value of ‘a’ is 5.46 atm.L2.mol–2 = 0.551 Jm3mol–2

We find here also that ‘a’ value obtained from the calculation does not match with the experimental data of ‘a’as shown above. This departure again is the failure of V.W equation.

11153.53

8 molJKT

VPR

C

CC

Again the value of R does not match with its actual value(8.3). This is again because ZC obtained by taking

R=8.3 is only 0.227 as against the expected value of 0.375(3/8). Hence the R value obtained by using aboverelation is less than 8.3. Note that at critical state, simple Van der Waals attraction does not exist as is thecase at temperature above T

C uptill T

B.So the values of ‘a’, ‘b’ and ‘R’ obtained by using critical constants as

per V.W equations will not match with their observed values near critical tempeatrue. Thsu V.W equationseriously fails at the vicinity of T

C and below it, although it does have appreciable matching at higher

temperatures(>TC)



Reduced Van der Waals Equation of State .

RTbVV

aP

2 (for one mole)

CC

C

C TRbVV

aP

22

Where CP

P(reduced pressure)

CV

V(reduced volume)

CT

T(reduced temperature)

Substituting the values of critical constants in terms of ‘a’, ‘b’ and R, we get

8133

2

Van der Waals Reduced equation of State

This equation does not contain the terms ‘a’, ‘b’ and R which is valid for all gases. Any two gases having sameπ and θ, must be at the same φ. This is also called law of corresponding state. Hence all gases whenmeasured at same π and θ, must have the same φ. It was found that this law also is approximately valid as theφ values are found to be slightly different from each other. Note that this not a different gas equation. It is thesame V.W equation expressed differently. So failure of this law also is bound to occur particularly in the vicinitycritical point as has been found for the original VW equation.

Modified Real Gas equations:(A) Dieterici Equation:

Dietereci and Jeans modified the Real Gas equation to a better matching with experimentalresults. It is given below.

RTV

a

ebV

RTP

(for one mole)

By taking the first derivative and second derivative of P w.r.t V at constant T in the P-V graph to find the criticalconstants, the following results were obtained(derivation not shown).

VC = 2b (1)

Rb

aTC 4

(2)

224 eb

aPC (3)

Critical Coefficient: 2706.0C

CCC RT

VPZ (4)

Rb

aTB (5)

Excepting TB, all were different than those obtained from VW equation. This gave close matching of Critical



Coefficient 2706.0C

CCC RT

VPZ and 2

b

VC between experimental result snd those derived from Dietereci

equation. However there is a mismatch in c

B

T

T (=4) ratio which was much greater found in this equation while

the experinmental value is nearly 3. Dietereci equaion, though was an advancement over V.W eqution, it hadstill drawbacks, particularly in the T

B value.

At low pressure, Dietereci equation is reduced to V.W equation.

(B) Berthelot Equation:Berthelot further modified the equation for Real gases as follows.

RTbVTV

aP

2 (for one mole)

The above equation was put forth first, which could not match with experimental data. Then it was furthermodifed as follows.

2

261

128

91

T

T

TP

PTRTPV C

C

C(for one mole) Modified Berthelot equn.

(for n moles, ‘n’ is multiplied in RHS)Reduced Berthelot Equation(derived from modified Berthelot equation)

2

61

4

1

9

32

This equation gave different expressions of VC, T

C and P

C in terms of ‘a’ and ‘b’(not given here) It matched

more closely with the experimental results, particularly the ratio c

B

T

Twhich was found to be less than 3.0(found

to be 2.45).

(C) The Virial Equation (Karmmerlingh-Onnes Equation):

Kammerlingh Onnes suggested an equation of state for Z as the power series in terms of mV

1 for real gases.

.........132

mmm

m

V

D

V

C

V

B

RT

PVZ

Where Z= compressibility factor; Vm = molar volume

Dropping the subscrit ‘m’, we have

.......1

32mmm V

D

V

C

V

BRTPV (1)

Where B, C, D..... are called 2nd, 3rd, 4th Virial Coefficients and so on. The 1st Virial coefficient A = 1.B, C, D..... are highly dependent on temperature. Usually the value rises from negative values(at lowertemperature) to zero and then to positve values at higher temperature. with increase in temperature. Howeverthese are not dependent on pressure.



This can also be expressed as the power series of Pressure(P) also

.....1 31

211 PDPCPBRTPV

Where B1 , C

1, D

1 ....different set of virial coefficent for power series in P.

SAQ: Calculate the molar volume of N2(g) at 600 K and 600 atm according to (a) ideal gas law and (b) the

virial equation; (B = second virial coefficient = 0.0217 L mol–1.)

Solution: (a) 1.082.0

600

600082.0

molLP

RTV (So under ideal conditions one mole will occupy 82mL)

(b) Neglecting the higher terms involving C, D..., the virial equation becomes

V

BRTPV 1 = RT+BP (2)

BPRTPV

1.1037.066

6000217.0600082.0

molLP

BPRTV

According virial equation state one mole will occupy 103.7 mL. This is acceptable result aas Z>1 in this case.

Van der Waals Equation expressed similar to Virial Equation:

22 V

a

bV

V

V

RT

V

a

bV

RTP

2

1

1V

a

V

b

V

RT

V

a

V

bRTPV

1

1

Expanding in RHS as a power series we have

V

a

V

b

V

bRTPV

....1

2

2

Neglecting higher terms, we get

V

a

V

bRTPV

1

VRT

abRTRTPV

1

VRT

abRTPV

11 (3)

Comparing eqn. (2) with (3), we get

RT

abB

Using T = TB, we get B=0. In other words B = 0 at Boyle’s temperature.



Values of 2nd Virial Coefficients(B) of some Common Gases at 298KGas B (L.mol–1)H

20.0141

He 0.0118N

2–0.0045

O2

–0.0161CO –0.0086CO

2–0.123

(Note that B data can also be obtained from Van der Waals constants as shown in previous expression, butthere will ba mismatch as the data given before are experimental).SAQ: Calculate Van der Waals constants for ethylene, if its T

C = 282.8K and P

C= 50 atm.

Solution:

2222

.47.464

27 molLatmP

TRa

C

C 1.057.08

molLP

RTb

C

C

SAQ: Using modified Berthelot equation, calculate the mass of hydrogen in a vessel of 500 mL capacity whenthe gas is forced to a pressure of 100 atm at 1270C. (T

C = 33.2K, P

C = 12.8 atm)

Solution:

2

261

128

91

T

T

TP

PTnRTPV C

C

C

2

2

400

2.3361

8.12400

2.33100

128

91400082.05.0100 n

7.0 n So mass = 0.7 × 2= 1.4 gSAQ: Use the virial equation to determine the pressure in atm of 1 mole of carbon dioxide gas contained in avolume of 5.0 L at 300 K. Compare your result to the pressure that would have been obtained from theideal gas equation. [B(300K) = –0.126 L.mol–1)Solution: Virial Equation:

atmV

B

V

RTP

mm

80.45

126.01

5

300082.01

Ideal gas equaion: atmV

RTP

m

92.4

Note that low pressure, not exceeding 5 atm. the deviation is appreciable from ideal behavior. The observedpressure(data not given) must be close to that obtained from viirial equation.

SAQ: At 273.15K and under a pressure of 10.1325 MPa, the compressibility factor of O2 is 0.927. Calculate

the mass of O2 necessary to fill a gas cylinder of 100 dm3 capacity under the given conditions.

Solution:

ZRT

PVm

3386

77.207100777.2101325.10

15.273314.8927.0cmmVm



So 207.77 mL of O2 under the given conditions weight 32 g

So 100,000 mL of O2 under the same conditions will weigh 15,401.646 g = 15.401646 Kg

SAQ: Given that Z = 1.00054 at 273.15K and at 101.325 kPa pressure and Boyle’s temperature(TB) of the

gas is 107K, estimate the values of ‘a’ and ‘b’.Solution:

We have to put forth two equations to solve for getting ‘a’ and b’.

Rb

aTB (1)

RT

ab

RTZ

P

RTZ

BP

PRTZ

B

V

BZ 1111 (2) Virial Eqn.

From eqn. (1), a = RbTB , Substituing the value of ‘a’ in eqn. (2) we get,

T

TT

RTZ

Pb

T

bTb

RTZ

P

RT

RbTb

RTZ

PZ BBB

111

Lm

TTP

ZRTZb

B

0199.01099.110715.27310325.101

00054.115.273314.8100054.11 353

22

So, b = 0.0199 L.mol–1.Substuting the value of ‘b’ in eqn. (1) we get,

22.174.00199.0082.0107 molLatmaSAQ: The compressibility factor for N

2 at 223K and 81.06 MPa is 1.95 and at 373K and 20.265 MPa, it is

1.10. A certain mass of N2 occupies a volume of 1.0 dm3 at 223 K and 81.06 MPa. Calculate the volume

occupied by the same quantity of N2 at 373 K and 20.265 MPa.

Solution:

ZnRT

PV 42.22

95.1223314.8

101006.81 336

mPa

RTZ

PVn

Substituing this value of ‘n’ for other conditions, we get the volume.

LmPaP

nRTZV 773.3107739.3

10265.20

10.1373314.842.22 336

SAQ: The molar volume of helium at 10.132 MPa and 273K is 0.011075 of its molar volume at 101.325 kPaat 273K. Calculate the radius of Helium atom. The value of ‘a’ may be neglected.Solution:

RTbVP m Let us write two equations for solving simultaneously for V

m and b.

273314.8011075.010132.10 6 bVPa m (1)

273314.810325.101 3 bVPa m (2)

Solving eqn. (1) and (2), we get, b = 2.386 × 10–5 m3= 23.86 cm3 = 0.02386 L mol–1.Then from ‘b’ , the radius of the atom is calculated as shown before;

3

410023.6486.23

323 r

34.11034.1 8 cmr Å



Molecular Collision and Mean Free Path:For molecular collision, molecules are assumed to be rigid and spherical and move with the same averagevelocity u .Collision Diameter(σ) : This is same as molecular diameter. Two molecules collide when their centres areseparated by σ.If a molecule is moving with an average speed of u , then in unit time, it will sweep a volume of πσ2 u .

LetV

NN

= number of molecules per unit volume.

Hence the number of molecules and hence collisions a single molecule will have with others per unit time

= πσ2 u N .Since all the molecules are moving, we take the resultant of two velocity vectors in perpendicular directions

which is u2 . Average velocity term is to be replaced by u2 in the above expression.(presumption)

COLLISION NUMBER(Z1): It is the number of collisions which a single molecule makes with others per

unit time.

NuZ 21 2 (where

V

NN

)

COLLISION FREQUENCY/BIMOLECULAR COLLISION(Z11

):Total number of collisions(bimolecular collisions) per unit time (Z

11) happening per unit volume:

22111

2

1)(

2

1 NuNZZ

If the molecules are different(two different gases), then

2121212

8NN

kTZ

Where 2

2112

(average molecular diameter); μ = reduced mass of two molecules.

21

21

21

111

mm

mm

mm

N

1 and N

2 are the number of molecules per unit volume of the two gases respectively.

(N.B: In kinetic theory, ‘n’ was used to represent number of molecules. Here ‘N’ is used for this. The readeris not to be confused)Mean Free Path(λλλλλ):

It is the average distance travelled by a molecule between two successive collisions.

n

dddd n..........321

If a molecule(black faced) collides ‘n’ times with distances travelled d1, d

2, d

3 ........, d

n successively, the above



ratio gives the average distance travelled by the molecule in two successive collisions.In this case, we assume that only one molecule(test molecule which is black faced) is moving and all others arestatic.

1Z

u

NNu

u22 2

1

2

Effect of Temperature and Pressure:

kT

P

RT

PN

V

NN A

APN

RT

N 22 22

1

Hence; T ;P

1 and 2

1

. Hence for a given gas, mean free path is directly proportional

to temperature and inversely proportional to pressure. At constant P and T, mean free path is inversely proportionalto the square of the diamter(size). A bulkier gas has a lower mean free path than a thin gas. For example CO

2

gas has a lower mean free path than H2 gas at same T and P.

Note that at constant volume, since the ratio (P/T) is constant, mean free path is independent of P or T.

SAQ: Calculate values of collision diameter, mean free path, collision number(Z1) and collision frequency(Z

11)

for O2 at 298.15 K at the pressure of 101.325 kPa, given Van der Waals const b = 3.183 × 10–3 dm3 mol–1.

Solution:

32323

10147.24615.2981037.1

101325

mkT

PN

dmdm

N

brrNb

AA

93

1

23

3231

3 10467.110023.614.316

10183.33

16

344

mdmr 109 10934.210934.22 ;

125.444032.014.3

15.2983.888

msM

RTu

mN

7

2321021006.1

10147.24610934.214.3414.1

1

2

1

1921 1018.42 sNuZ

1334111 10144.5

2

1 smNZZ

_____________________________________________________________________________________Number of Collisions on Unit Surface Area of the wall per unit time:

= uN

4

1

(Derivation not given)_____________________________________________________________________________________



Viscosity during Gas Flow:Viscosity in case of a gas arises due to transfer of momentum across the layers of the gas. The viscosity of gas isdetermined from the following (derivation not shown).

uumN2

1

2

1

(where ρ = density of gas)

Using the values of u and λ, we get

22

3

21

mkT (where η = coefficient of viscosity)

Note that viscosity of gases is independent of pressure and density. It is directly proportional to square root of T. Alsonote that for liquid –viscosity decreases with increase of temperature contrary to the case of gas.Unit: SI: Nm–2s. CGS : dyne.cm–2.s = poise 1 Nm–2s = 10 poise

SAQ: The Van der Waals const b for n-heptane is 0.2654 dm3 mol–1. calculate coefficient of viscosity of this gas at298K.Solution:

mdmrmoldmrNb A98133 102974.0102974.02654.0

3

44

Molar mass of n-heptane=100g = 0.1 Kg

poisesNmmkT 525

2923

21

2323

223 10328.110328.1

105948.014.3

2981037.110023.6

1.0

SAQ: Calculate the number of collisions per m2 per second of O2 molecules with a wall at a pressure of 101.325 kPa

and temperature 298KSolution:

325104627.2 mkT

PN 11.444 msu

Number of collisions per unit area per sec on the wall 122710734.2

4

1 smuN



Enrichment of a single gaseous component by repeated Effusion steps:

When a mixture of two gases is subjected to repeated effusion, the lighter gas among the two gets progressivelyenriched during the successive effusion steps. One can calculate the number of successive effusion stepsneeded to achieve a desired level of enerichment as follows.

Let the intial number of moles of two components A and B = nA and n

B respectively

The finally desired number of moles of the two compnents A and B after ‘x’ number of effusion steps= n

Ax and n

Bx

Let us presume ‘A’ is the lighter gas and ‘B’is the heavier gas.

Enrichment factor in a single effusion step = A

B

B

A

M

M

r

rf '

Overall Enrichment factor:

B

A

Bx

Ax

nn

nn

f

The number of effusion steps ‘x’ needed for the purpose can be found out from the following expression.

xff 'Derivation of the expression:

mole ratio after 1st effusion = A

B

B

A

B

A

M

M

n

n

n

n

1

1

(nA : n

B = P

A : P

B and n

A1 : n

B1 = r

1 : r

2 )

Enrichment factor in one step(1st step) =

B

A

B

A

nn

nn

f 1

1

'

A

B

M

M

Mole ratio after 2nd effusion = A

B

B

A

B

A

M

M

n

n

n

n

1

1

2

2

2

2

2

A

B

B

A

B

A

M

M

n

n

n

n

Similarly mole ratio after 3rd effusion =

3

3

3

A

B

B

A

B

A

M

M

n

n

n

n

In this way, we can continue with 4th..... upto xth effusion step.



So mole ratio after ‘x’ effusion step(last step) =

x

A

B

B

A

Bx

Ax

M

M

n

n

n

n

x

A

B

B

A

Bx

Ax

M

M

nn

nn

xff '

SAQ: Naturally occuring Uranium ore contains 0.7% U235 and the rest U238 isotope. Suppor for a nuclear

power plant we want to enrich the U235 isotope to 6%. This is achieved by preparing hexafluorides( 6235UF

and 6238UF ) which are gases. This mixture is effused for large number of times successively till the desired level

of enrichment is achieved. Calculate the number of effusion steps required for this.Solution:

004288.1349

352' A

B

M

Mf

05.9

3.997.0

946

f

'' loglog fxfffx

88.514004288.1log

05.9log x

Hence nearly 515 successive effusion steps are required to get the desired level of enrichment.

Manometer:A simple manometer was being used to measure the pressure of a gas enclosed in a container. It consists of aU-tube with one open end. The closed end has a side tube for connecting with the enclosed gas container. Thetube contains mercury. For the differential height of the mercury levels on the two arms and barometricpressure(atmospheric pressure), we can find out the pressure of the enclosed gas. Note that barometer recordsatmospheric pressure which is fluctuating every day depending on clamatic and weather conditions of theplace. Standard Pressure/Normal pressure of 760 mm of Hg(1 atm) or 1 bar is an arbitrary selection ofaverage pressure of the atmosphere used for standardisation purpose. But actual atmospheric pressure variesfrom day to day, some day higher than 760(high pressure) and some day lower than 760(low pression ordepression).



Case-I: Gas pressure greater than Atmospheric pressure:The pressure exerting on the surface shown by the line ‘xy’ on both the arms must be

same. On LHS, is the gas pressure and on RHS it is the sum of atmospheric pressure and pressure due to themercury column of the differential height ‘h’.

Gas Pressure = Atmospheric pressure + hCase-II: Gas pressure equals with Atmospheric pressure:

There is no differential height between the two columns in this case. So the gas pressureis equals to the atmospheric pressure.Case-III: Gas pressure lees than Atmospheric pressure:

In this case, the pressure exerted on the surface shown by the imaginary line ‘xy’ is the same.Gas pressure + h = Atmospheric pressureSo Gas pressure = Atmospheric pressure – h

N.B: Nowadays, compact and well calibrated pressure gauges are used to measure pressure of gas frow verylow range from 0.5 m Bar to 7000 Bars. They make use of special materials including copper and steel alloysfor detecting mitute variation of pressure. Vacuum gauges are also available to measure pressure created invacuum tubes



Joule Thomson Effect(Application of 1st Law)When a real gas expands adiabatically from a higher pressure to a lower pressure through a porous

plug or throttle, change of temperature is observed. When the working temperature is below the inversiontemperature of the gas(T

i), then there is a cooling effect. This is a

case of irreversible free adiabatic expansion.Since, q=0, ΔE = W

For cooling effect, T2 < T

1, P

1 >>> P

2

To push one mole of a gas through the porous plug, work amounting +P1V

1 has to be done on the gas by the

piston on the left. Work amounting to –P2V

2 has to be done by the gas on the piston on right. Hence net work

done isW

T = W

1 + W

2 = P

1V

1 – P

2V

2

[ For pushing : W1 = -P

1(0-V

1) = P

1V

1;For expansion on RHS: W

2 = –P

2(V

2-0) ]

E2 –E

1 = P

1V

1 – P

2V

2H

2 = H

1Isoenthalpic process

Joule Thomson Coefficient(μ):HP

T

For Cooling effect : μ = +ve (as both are negative), For heating effect : μ = –ve,When throttled expansion is carried out below INVERSION TEMPERATURE(T

i), there is cooling effect

and above Ti, heating effect is observed. (

Rb

aTi

2 ). So easily liquefiable gases having greater T

C values have

also larger Ti values and are used for producing cooling effect.

H2 and He have subzero T

i, hence show heating effect at room termperature. Most other gases show cooling

effect as their Ti are higher than room temperature.

(Here molar values of E and H are taken). Since enthalpy remains the same, it is called an isenthalpic process.)Derivation of Inversion Temperature: H = f(P,T), The exact differential fulfills the condition;

dTT

HdP

P

HdH

PT

since, in this case, dH = 0,

TPJT

TPH

TP

P

H

C

P

H

CP

T

dPP

HdT

T

H

1

1

Ideal gas vs. Real Gas:

TPTPTPHJT P

PV

CP

E

CP

H

CP

T

)(111 (1) (Since H = E + PV)

Ideal Gas: Under ideal conditions, both TP

E

and TP

PV

)(

terms vanish, so μJT

= 0

P1V1

P2V2

T1 T2



Real Gas: On rearranging eqn. 1 , we get

TPTTP P

PV

CP

V

V

E

C

)(11

For real gas, TV

E

is usually +ve, while TP

V

is always –ve, so the first term becomes +ve. In the second

term TP

PV

)(

is –ve for all gases except H2 and He at room temperature and at lower pressure

conditions(negative deviation to ideal behavior). Hence the second term also is +ve for most gases. In that caseμ is +ve and the JT effect shows cooling effect. Only at higher temperture(above T

B) for most gases and for

H2 and He even at room temperature,

TP

PV

)(

is +ve (positive deviation). So μ =–ve only when the

second term numerically exceeds first term and show heating effect. This means that when μ =0, neithercooling or heating effect is observed. In thas case, +ve and –ve terms numerically become equal. Thecorresponding temperature for a gas is called T

i.

Determination of Inversion Temperature:Since; S = f(P, T)

dPP

SdT

T

SdS

TP

⇒ dPT

VdT

T

CdP

P

SdT

TT

dHdS

P

P

T

1(i)

[Since; CP = dH/dT and from Maxwell relation

TP P

S

T

V

]

We have,dH = TdS + VdP (ii)

Substiting the value of dS from eqn. (i) in eqn (ii), we have

VdPdPT

VdT

T

CTdH

P

P

Since, here dH = 0, we have

dPVT

VTVdPdP

T

VTdTC

PPP

;

⇒

VT

VT

CP

T

PPH

1(iii)

The term inside square bracket in RHS will vanish for an ideal gas.For real gas,

RTbVV

aP

2



232

)(2Va

P

R

PV

aVbVa

R

T

V

P

(neglecting the cubic term)

222

22V

abV

RTR

Va

Va

P

R

T

V

P

⇒V

RTV

bVabVV

RTVbVa

bVV

Va

bVRT

RTV

T

VT

P

1

2

22

)(21)(

)(21

)(2

⇒ b

RT

ab

RTV

bVaV

RTV

bVabVV

T

VT

P

22)(2

12

2

2 (iv)

(Neglecting the higher terms in the power seris and neglecting ‘b’ in parenthesis)Substituting the value in eqn. (iii), we have

bRT

a

CV

T

VT

CP

T

PPPH

211

At low temp., (2a/RT) > b, so μ =+ve (cooling effct) and at high temperature it is the opposite. The temperatureat which the two terms are equal in magnitude, μ=0, and that temperature is called T

i.

Rb

aTb

RT

ai

i

220

Example : Linde Refrigerator : used for liquefaction of gases. Suppose you want to liquefy N2 gas, you have tofirst reduce the temperature

Why Cooling Effect ? At lower temperature(than Ti), the predominant force is intermolecular attraction.

When the gas is expanded, the intemolecular distance increases and hence attraction decreases, energy has tobe absorbed from from the system(adiabatic process). In other words, the work done by the gas brings aboutdecrease in E, hence cooling.Why Heating Effect ? At higher temperature(than T

i), the predominant force is intermolecular repulsion.

When the gas is expanded the the molecules go further apart and hence there will be decrease in repulsion andhence heat is evolved. In otherwords, in this case there is increase in E due to this effect. Thus there is a heatingeffect.At T

i, the gas behaves ideally and hence no effect.

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