Physical Chemistry Third Edition

3.3 The Calculation of Entropy Changes 125

EXAMPLE 3.7

For a gas whose molar constant-pressure heat capacity is represented by

CPa+bT+cT−^2

derive a formula for∆Sif the temperature is changed reversibly fromT 1 toT 2 at constant

pressure.

Solution

∆Sn

∫T 2

T 1

CP, m

T

dTn

∫T 2

T 1

a+bT+cT−^2

T

dT

naln

(

T 2

T 1

)

+nb(T 2 −T 1 )−

nc

2

(

1

T 2

−

1

T 1

)

Exercise 3.7

Calculate the entropy change for heating 2.000 mol of CO 2 gas from− 20. 00 ◦Cto100. 00 ◦C

at a constant pressure of 1.000 atm using the formula from the previous example and data from

Table A.6 of Appendix A.

For temperature changes at constant volume, Eq. (2.4-3) gives

dqdUCVdT (closed system, constant volume) (3.3-7)

so that Eq. (3.3-2) becomes

∆S

∫T 2

T 1

CV

T

dT (closed system, constant volume) (3.3-8a)

If the heat capacity is constant

∆SCVln

(

T 2

T 1

)

(3.3-8b)

EXAMPLE 3.8

Calculateq,w,∆U, and∆Sif 1.000 mol of helium gas is heated reversibly from 25.0◦Cto

50.0◦C at a constant volume. Assume thatCV, mis equal to 3R/2 and is constant.

Solution

w 0

qCV∆T(1.000 mol)

(

3

2

)

(8.3145 J K−^1 mol−^1 )(25.0K) 311 .8J

∆Uq 311 .8J

∆S(1.000 mol)

(

3

2

)

(8.3145 J K−^1 mol−^1 )ln

(

323 .15 K

298 .15 K

)

 1 .004JK−^1

Notice that the volume does not matter so long as it is constant.