VIII. Entropy

T

dQdS for a reversible process at constant T

dQ is path dependentdS is path independent S is function of state S is additive function

T

dQdS for any process (including irreversible)

For closed, isolated system (dQ = 0): 0dS

2.The second law of thermodynamics

any reversible cycle: ΔS=0any irreversible process in closed isolated system: ΔS>0

1. Macroscopic definition of entropy

Example 1: This P-V diagram represents a system consisting of a fixed amount of ideal gas that undergoes three different processes in going from state A to state B. Rank the change in entropy of the system for each process.

V

State A

I

P

State B

23

ΔS1 = ΔS2 = ΔS3 = SB - SA

The same as: ΔT1 = ΔT2 = ΔT3 = TB - TA

ΔU1 = ΔU2 = ΔU3 = UB - UA

Example 2: Which of the following statements is false?

A. The change of entropy in a cyclic process is zeroB. The change of entropy for any adiabatic process is zeroC. The change of entropy for any isothermal process is zeroD. Entropy for a closed, isolated system is constantE. Entropy of a system can decrease

Example 3: 50.0 kg of water is converted to ice at 0.0ºC. What is the change in entropy of water?

m = 85.0 kg

T = 0.0ºC

L = 334*103 J/kg

ΔS - ?

KkJK

kgJkgS

T

mL

T

QS

/61273

/103340.50 3

Example 4: The isolated system is 50.0 kg of ice at 0 ˚C plus the temperature reservoir at slightly above 0 ˚C that is used to melt the ice. What is the change in entropy of the system when the ice is melted ?

Solution:

•The system is isolated: Qice = -Qreservoir

•The ice and the reservoir are at almost the same temperature: Tic = Treservoir=T•The system consists of both the ice and the temperature reservoir:

0

T

Q

T

Q

T

Q

T

QSSS iceicereservoirice

reservoiricesystem

ΔS = 0. Therefore the process is reversible!

3. Entropy of ideal gas

VnRTdVPdVdW

dTnCdUV

dVnR

T

dTnC

T

VnRTdVdTnC

T

dWdU

T

dQdS

V

VV

1

2

1

212 lnln

V

VnR

T

TnCSSS V

3a. Free expansion of ideal gas

1

2

1

2 lnlnV

VnR

T

TnCS V

•A given amount of an ideal gas undergo free expansion from volumeV1 to V2

•Gas forms a closed and thermally isolated system. Because of that: W=0, Q=0 ΔU=0 ΔT=0 T1 = T2

•General equation for the entropy change of any ideal gas:

0ln1

2

V

VnRS

Closed and thermally isolated system with ΔS>0: the process is irreversible!

Example: Two moles of an ideal gas undergo an adiabatic free expansion from V1 = 1.00 L to V2 = e1.00 L = 2.72 L. (The gas is an isolated system). The change in the entropy of the gas is __ J/K.

0/6.1672.2ln/31.800.2 KJKmolJmolS

The process is irreversible!

3b. Reversible isothermal expansion of ideal gas

Ideal Gas :

Sgas nCV lnT2

T1

nR ln

V2

V1

Reversible Isothermal Expansion (V2 V1)

T2 T1 T , so ln(T2 / T1) 0 and

Sgas nR lnV2

V1

But Sgas 0, so doesn't this mean that

the process is not reversible?

Answer: No. One must consider only

isolated systems to answer that.

Isolated system =

gas + T reservoir (heat source)

The heat absorbed by the gas is

Qgas Wgas nRT lnV2

V1

.

The change in entropy of the gas is

Sgas Qgas

TnR ln

V2

V1

, which agrees

with our previous result. The heat

absorbed by the reservoir is Qreservoir Qgas .

The change in entropy of the system is

Ssystem Sgas Sreservoir Qgas

T

Qgas

T0.

Thus the process is reversible.

3c. Reversible Adiabatic Expansion of Ideal Gas

Ideal Gas :

Sgas nCV lnT2

T1

nR ln

V2

V1

(1)

Reversible Adiabatic Expansion (V2 V1)

Here the gas is an isolated system (W Q 0).

Therefore we expect Sgas 0. Adiabatic:

TV 1 constant = T1V1 1 T2V2

1, so

T2

T1

V1

1

V2 1

V1

V2

1

V2

V1

1

1

and

lnT2

T1

( 1)ln

V2

V1

1

( 1)lnV2

V1

.

Substituting this expression into Eq. (1) gives

Sgas n CV( 1) R ln V2

V1

. (2)

For an ideal gas,

Cp

CV

CV R

CV

. Therefore,

1 CV R

CV

1 R

CV

and

CV( 1) R. (3)

Substituting Eq. (3) into (2) gives

Sgas 0 .

Thus the process is indeed reversible.

4. Microscopic interpretation of entropy

lnBkS

number of possible microscopic states1

212 lnlnln

BBB kkkS

Example 6: A thermally insulated box is divided by a partition into to compartments, each having volume V. Initially one compartment contains n moles of an ideal gas at temperature T, and other compartment is evacuated. We then break the partition, and gas expands to fill both compartments. What is the entropy change in this free-expansion process?

?

2

12

12

S

nnn

VV

RNk

nNN

AB

A

12

12

2

2

N

VV

1

1

V

2ln2ln2lnln1

2 nRNkkkS BN

BB