1

Chapter 7.

Applications of the Second Law

2

Consider entropy changes in various reversible (!!!) processes

We have: )1(dsvdT

P

T

du

T

qvPdduq

(a) Adiabatic process constants0dsso0q Hence a reversible adiabatic process is isentropic.

(b) Isothermal process:T

qs

T

qds

2

1

2

1

(c) Change of phase (isothermal and isobaric)In this case the energy transfer as heat is the latent heat of

transformation, so

Ts

(d) Isochoric process.In general u=u(v,T) and so, for an isochoric process u=u(T)

and dTcdu V

đđ

đ

đ

3

From (1) 2

1 T

dTcs

T

dTc

T

duds VV

If the specific heat is constant over the temperature range of interest

1

2lnT

Tcs V

(e) Isobaric process:In isobaric processes it is often convenient to employ the

enthalpy. PdvdudPvvPddudhvPuh (2)vPddhdu

)3(dhT

1dsvd

T

Pvd

T

Pdh

T

1ds

vPdTdsdu The 1st Law is:

T

vPd

T

duds Using equation (2) to eliminate du

so

(isochoric)

Since we are considering the case P=constant

dTcdTT

hdh P

P

For constant specific heat

1

2lnT

Tcs P

Consider h=h(P,T)

dPP

hdT

T

hdh

TP

T

dTcds PPlacing this in equation (3)

(isobaric)

5

Entropy of an ideal gas

Although it is possible to determine an absolute value for the entropy, we do not do so at the present time.

We have already shown for an ideal gas (students should review):

Q = CPdT - VdP _ _ _ (1)

Q = CVdT + PdV _ _ _(2)

T

dPV

T

dTCdS P Note that from (1) and so

P

dPnR

T

dTCdS P -------(1a)

SimilarlyV

dVnR

T

dTCdS V ------(2a)

đ

đ

6

Introducing molar quantities and integrating equation (1a):

Similarly, using equation (2a):

)3_(____l PPf

Pf

innRcnS T

dT

i

)4_(__ln VVf

Vf

inRcnS T

dT

i

)5_(__lnlni

nRT

TncS

i

f

VV

Vf

From this last equation we deduce the following:(a) V fixed: the greater the temperature rise , the greater

the increase in entropy.(b) T fixed: the greater the volume expansion, the greater

the increase in entropy

7

Although we have been considering an ideal gas, these conclusions hold quite generally.

Consider, for example, an isentropic expansion of a gas.

)0( S The increase in volume will result in an increase in the entropy. This increase must be balanced by an equal decrease and this must come about by a drop in temperature.

8

T S Diagrams

A convenient representation of processes involving reversible cycles for various heat engines is a diagram with T along the y-axis and S along the x-axis.

For an isothermal process the curve (isotherm) will be a straight horizontal line.

For an adiabatic process, since đQ = 0 and so dS = 0, the curve (isentrope) will be a straight vertical line.

9

isochor

isobar

isotherm

isentrope

T

S

From equation (1a)P

dPnR

T

dTCdS P and so, for an isobaric process

PP C

T

S

T

From equation (2a)V

dVnR

T

dTCdS V and so, for an isochoric process

VV C

T

S

T

Since CP>CVPV S

T

S

T

The plots then look like:

(slide 5)

10

The TS diagram for a Carnot cycle (2 isotherms and 2 isentropes) will then be a rectangle.

T2

a b

dT1

T

S1 S2

S

c

11

Area under ab is the heat absorbed (Q2) by system. Area under cd is the heat rejected (Q1) by system. The shaded area is then the work done during the cycle. W = Q1 + Q2 and the efficiency, , can be obtained from the graph.

Remember that is negative.

Remember that, with a PV diagram, the area under the curve is W.

We wish to consider all the entropy changes which take place when there is a reversible exchange of energy between a system and its surroundings dS (system) + dS (surroundings) = dS (universe)

2QW

1Q

12

The temperature of the system and its surrounding differ by only an infinitesimal amount at any time.

Suppose the surroundings (reservoir) absorbs an amount of heat đQ, then its entropy increases by đQ/T. In this process an equal amount of heat flows out of the system and its entropy decreases by đQ/T. Hence dS (universe) = 0.

In a reversible process, the change in entropy of the universe is zero.

13

Tds Equations. (Study derivations and examples in textbook.)

1st Tds equation: vdT

dTcvdT

PTdTcTds V

VV

2nd Tds equation: dPvTdTcdPT

vTdTcTds P

PP

3rd Tds equation: vdv

cdP

cdv

v

TcdP

P

TcTds PV

PP

VV

One way to begin the derivations is as follows:First equation: start with s(T,v)Second equation: start with s(T,P)Third equation: start with s(P,v)

Remember that in a reversible process d q=Tds

14

These equations follow from the first and second laws of thermodynamics and thus contain much of the content of thermodynamics. They are often a good starting point for working problems or obtaining other useful relationships. Some examples of the use of these equations are: (a) to find the heat transferred in a reversible process

(Slide 19)(b) the entropy can be obtained by dividing by T and integrating.(c) equations express heat flow or entropy in terms of

measurable properties, such as(d) can be used to determine the difference in the specific heat

capacities at constant P and constant V(Slide 21)

(e) can provide relationships between pairs of coordinates in areversible adiabatic process ( ds=0 )(Slides 18,20)

and,, PV cc

15

Derivation of the 1st Tds equation

Consider )1(vdv

sdT

T

sds)v,T(ss

Tv

sovPdvdv

udT

T

uPdvduq

Tv

)2(vdPv

u

T

1dT

T

u

T

1ds

Tv

Comparing (1) and (2)

Pv

u

T

1

v

s

T

u

T

1

T

s

TTvv

Since s is an exact differential:or

v

s

TT

s

vvTTv

đ

16

vTTv

Pv

u

T

1

TT

u

T

1

v

vvTT2

Tv T

P

T

1

v

u

TT

1P

v

u

T

1

T

u

vT

1

=

vT T

PTP

v

u

Placing in equation (2)

vdT

PTdT

T

uTds

vv

vdT

PTdTcTds

vV

17

Helpful expansion:

1x14

x

3

x

2

xx)x1ln(

432

18

Example: What is the change in temperature when a solid or liquid is compressed adiabatically and reversibly?

Using the 1st TdS equation and remembering that for an adiabatic process, dS=0

0vdT

dTcV

Assuming that the experimental quantities are approximately constant over the range of interest and integrating:

vdcT

dT

V

)vv(cT

Tln if

Vi

f

)vv(cT

TTln if

Vi

i

In such a process the change in temperature is generally small, so

)vv(cT

Tif

Vi

constantS)vv(c

TT if

V

i

If the volume decreases, the temperature must increase!

19

To obtain the change in temperature in terms of pressure, we could start with the 2nd TdS equation and go through a similar analysis.The result is (show):

constantS)PP(c

TvT if

P

i

Example: How much energy is released during an reversible isothermal compression of a solid or liquid?

We use the 2nd TdS equation and for an isothermal process, dT=0

TdPvdTcTds P TdPvTds

We make the approximations that are nearly constant. vand

)PP(Tv)ss(T ifif

constantT)PP(Tvq if

If the pressure increases, the temperature must increase!

20

Example: During a reversible, adiabatic compression, how are the volume and pressure changes related?

Use the 3rd TdS equation and again dS=0

0vdv

cdP

cTds PV

vdv

cdPc P

V

Assuming constant specific heats and experimental quantities:

iP

i

fPifV v

vc

v

vlnc)PP(c

)PP(c

cv)vv( if

P

Viif

(reversible adiabatic process)

Of course, we already know that if the pressure is increased, the volume will decrease. The above expression permits a calculation of the change in the volume for a given change in the pressure.

21

EXAMPLE: Determination of Is this difference constant? If not, what does it depend upon?

)( VP cc

We equate the first two Tds equations. Brings in the two specific heats.

vdT

PTdTcdP

T

vTdTc

vV

PP

Volume expansivityPT

v

v

1

Isothermal compressibilityTP

v

v

1

vdT

PTdTcdPvTdTc

VVP

vTPV T

P1

P

v

v

T

T

P

vdT

dTcdPvTdTc VP

22

Solving for dT dP)cc(

vTvd

)cc(

TdT

VPVP

But we can write dPP

Tvd

v

TdT

VP

Comparing gives )cc(

vT

P

T

)cc(

T

v

T

VPVVPP

vTcc)v(

T

T

vT

vTT

cc2

VPP

P

VP

This shows that always. is easy to determine experimentally, but is difficultIf is measured, then can be obtained if the quantities on the right hand side have been measured. An example is given on the next slide.

VP cc Pc Vc

Pc Vc

23

EXAMPLE: Calculation of the specific heat at constant volume forCu at 1000K and 1 atm. For Cu:

PaKKkmole

JcP

1253 105.9105.6

1029

atomic weight=63.6 331096.8

m

kg

kmole

m1010.7v

mkg

1096.8kmole1

kg6.63

n

m

n

Vv

33

33

vT

cc2

VP

112

2153

33

3

105.9

)105.6(1010.7)10(

1029

Pa

Kkmole

mK

Kkmole

JcV

)R3c :note(Kkmole

J1026c V

3

V

24

EXAMPLE: Consider a reversible adiabatic process, then ds=0

From the 3rd Tds equation: giving:vdv

cdP

c0 PV

)constants(vddPvvdc

dPvc PV

SS P

v

v

1v

P

v

We define an adiabatic ( or isentropic) compressibility

SS P

v

v

1

SS 1

We can understand how this comes about.

25

An increase in pressure results in a decrease in the volume.

Consider a small increase in pressure ΔP.For an isothermal process (heat will leave the system) there will be a certain decrease in volume (ΔV)T

Now consider an isentropic process (no heat leaves the system) with the same increase in pressure. The increased pressure leads to an increase in temperature which tends to increase the volume. Hence, in this case, (ΔV)S will be less than (ΔV)T

TS )V()V(

S

But and S

S P

v

v

1

TP

v

v

1

Hence

The isentropic compressibility is smaller than the isothermal compressibility

26

Example: How much work is performed during a reversible, adiabatic compression of a solid or liquid?

Starting with the definition of the isentropic compressibility:

)constantS(dPvvdP

v

v

1S

SS

2

PPvdPPvdPvPvdPW

2i

2f

SSS

2f

2i

S PP2

vW

(reversible, adiabatic)

Work is done on the system!

27

Irreversible processesWe defined entropy in terms of reversible processes. What about irreversible processes?

Entropy is a state variable and change of entropy between any two equilibrium states is the same regardless of the process. Hence we can calculate S for any irreversible process between two equilibrium states by choosing any convenient path (reversible) between the same two states!

Of course, regardless of the process, the change in entropy of the system must be the same, but the change in entropy of the universe will depend on the process.

28

As an example consider the following situation in which we have a large reservoir at temperature T2 and a small body (system) at a lower temperature T1. We let the system come into thermal equilibrium by means of some isobaric process, which is necessarily irreversible.

body T1

reservoir T2

Qadiabatic walls

29

Now we imagine a reversible process, by which we bring about the change in T of the body through a whole series of reservoirs between T1 and T2, and keeping constant the pressure. In this manner the body passes through a large number of equilibrium states. For the reversible process, and using the second Tds equation and since dP = 0

dS = CP

Upon integration S (body) = CP ln

TdT

1

2

TT

The heat absorbed is )( 12

2

1

TTCdTCQ P

T

T

P

30

The temperature of the reservoir remains constant and the heat flow out of the reservoir is -CP(T2 -T1). Hence, for the reservoir,

This will always be greater than zero, regardless of whether T2 > T1 or T1 > T2.

(Students: Let x = T2/T1 and examine the function).

Hence S (universe) > 0.

2

12

TTT

PCreservoirS

2

12

1

2ln TTT

TT

PCuniverseS

31

In this example the decrease in entropy of the reservoir is lessthan the increase in entropy of the body

In all real processes the entropy increases or, to put it in another way, entropy is created.

The entropy of the universe is constantly increasing.

It should be kept in mind that, for an irreversible process,

T

QdS đ

and hence this formula cannot be used to calculate any entropy change that occurs.

32

Free expansion of an ideal gas

For a reversible isothermal process we would have P0V0 = P1V1. However, this does not describe free expansion. Initially P1 = 0. This is an irreversible process!.

Regardless of the process there is a fixed difference between the final and initial entropies of the gas.

V0 T0

P0

V1 - V0

vacuum

V1 T0 P1

adiabatic container

33

We can calculate S by assuming a reversible isothermal expansion between the two equilibrium states. Because of the adiabatic walls,

S (surroundings) = 0.

Even in this adiabatic process the entropy of the universe increases. Later we will provide some insight as to why this is so.

0

1

0

1

VV

VV

VV

V

ln

ln

nRuniverseS

nRsystemS

nRCdS dTdT

(1st Tds equation for an ideal gas)

The two processes, the free expansion and the reversible expansion, are completely different. What they have in common is that the initial and final states of the gas are the same.

34

In a reversible isothermal expansion, the work done by the ideal gas is W = nRT0ln

dU = 0 Q = W S = S = nRln

In a free expansion no work is done, but the change in entropy is as if work were done in a reversible, isothermal process between the same endpoints.

To Summarize: reversible process S (universe) =

0 irreversible process S (universe) >

0

0

1

VV

0TQ

0

1

VV

If the entropy of the universe increases in some process, we knowthat it is irreversible.

35

Entropy change for a liquid or a solid

An approximate equation of state for a liquid or solid is

)]PP()TT(1[vv 000

2nd Tds equation: dPT

vTdTcTds

PP

Since and since the volume does not change drastically for a liquid or solid, we write:

PT

v

v

1

P0 T

v

v

1

Substituting into the Tds equation gives: dPvT

dTcds 0P

Integration gives )PP(vT

Tlncss 00

0P0

If T increases, s increases.If P increases, s decreases.

36

Comments:

2nd Law: energy tends to disperse. Natural processes accompany the dispersal of energy

High quality energy is undispersed energy (energy in a lump of coal)or stored in coherent motion (flow of water)

A local reduction of entropy is possible (such as cooling something below surroundings).

37

Example: Express in standard form. Evaluate this partial for an ideal gas. (Not obvious how to start.)

TP

u

v1

PdduT

ds

dPP

dTTT

PdP

P

udT

T

u

Tds

TPTP

vv1

dPPT

P

P

u

TdT

TT

P

T

u

Tds

TTPP

v1v1

But writing s(T,P) dPP

sdT

T

sds

TP

TTTPPP P

v

T

P

P

u

TP

s ,

T

v

T

P

T

u

TT

s 1

1

and comparing

(start with 1st law)

38

Since ds is an exact differential:

PTTP P

s

TT

s

P

PTTTPP PT

P

P

u

TTTT

P

T

u

TP

v1v1

PTT2

PTT2

TPPTP

P

v

TT

P

P

v

T

P

P

u

TT

1

P

u

T

1T

v

PT

P

T

v

T

1

T

u

PT

1

TTP PT

P

P

u

TTT

v1v1

22

TPT PP

TT

P

u

vv

39

PTP

u

T

vv

For an ideal gas P

RTvRTv P

TP

R

T P

1

v

1v

v

1

PP

RT

P T

1

v

1v

v

12

Using these expression in the expression at the top:

gas) ideal(00v1

v1

TT P

uP

PT

TP

u

( measurable quantities)