Physical Chemistry Third Edition

4.2 Fundamental Relations for Closed Simple Systems 163

We can now derive the formula for∆Sfor an isothermal volume change in an ideal

gas in a different way from that used in Chapter 4:

∆S

∫V 2

V 1

(

∂S

∂V

)

T,n

dV

∫V 2

V 1

(

∂P

∂T

)

V,n

dV



∫V 2

V 1

(

nR

V

)

dVnRln

(

V 2

V 1

)

(4.2-23)

We can also obtain an expression for∆Sfor nonideal gases, liquids, and solids.

EXAMPLE 4.3

a.Find an expression for (∂S/∂V)T,nfornmoles of a gas obeying the truncated virial

equation of state

PVm

RT

 1 +

B 2

Vm

whereB 2 is a function ofTand whereVmis the molar volume.

b.Evaluate the expression for (∂S/∂V)T,n for 1.000 mol of argon in 25.00 L

at 298.15 K. At this temperature,B 2 − 15 .8cm^3 mol−^1 and dB 2 /dT 0. 25 ×

10 −^6 m^3 mol−^1 K−^1.

c.Find an expression for∆Sfor an isothermal volume change fornmoles of a gas obeying

the truncated virial equation of state in part a. Compare your result with the corresponding

equation for an ideal gas.

d.Find the value of∆Sfor expanding 1.000 mol of argon isothermally at 298.15 K from

25.00 L to 50.00 L. Compare the result with the result assuming argon to be an ideal gas.

Solution

a.KeepingVandnconstant is the same as keepingVmconstant.

(

∂S

∂V

)

T,n



(

∂P

∂T

)

V,n



⎛

⎝

∂

(

RT /Vm+RT B 2 /Vm^2

)

∂T

⎞

⎠

Vm



R

Vm

+

R

Vm^2

[

B 2 +T

dB 2

dT

]

b.

(

∂S

∂V

)

T,n



8 .3145 J K−^1 mol−^1

0 .025 m^3 mol−^1

+

8 .3145 J−^1 mol−^1

(

0 .025 m^3 mol−^1

) 2

×

[(

− 15. 8 × 10 −^6 m^3 mol−^1

)

+(298.15 K)

(

0. 25 × 10 −^6 m^3 mol−^1 K−^1

)]

 332 .6Nm−^2 K−^1 + 0 .58Nm−^2 K−^1

 333 .2Nm−^2 K−^1

 333 .2JK−^1 m−^3

The correction for gas nonideality, 0.58 J K−^1 m−^3 , is numerically almost insignificant

in this case, but for smaller molar volumes it would be more important.