Physical Chemistry , 1st ed.

Example 4.7

What is ( S/ V)Tfor a gas that follows a van der Waals equation of state?

Solution

The Maxwell relationship in equation 4.36 shows that ( S/ V)Tis equal to

( p/ T)V. Using the van der Waals equation,

p

V

n



RT

nb



a

V

n

2

2



Taking the derivative ofpwith respect to Tat constant volume gives



T

p



V



V

nR

nb



Therefore, by Maxwell’s relationships,



V

S



T



V

nR

nb



We do not need to measure the entropy changes experimentally. We can get

the isothermal change in entropy versus volume from the van der Waals pa-

rameters.

Example 4.8

In Chapter 1, we showed that



T

p



V









where is the expansion coefficient and is the isothermal compressibility.

For mercury,1.82 10 ^4 /K and 3.87  10 ^5 /atm at 20°C. Determine

how entropy changes with volume under isothermal conditions at this tem-

perature.

Solution

The derivative of interest is ( S/ V)T, which by equation 4.36 is equal to

( p/ T)V. Using the expansion coefficient and the isothermal compress-

ibility:



T

p



V



3

1

.8

.8

7

2





1

1

0

0





(^5) /
4
a

/

t

K

m

4.70 

at

K

m



These do not seem to be appropriate units for entropy and volume. However,

if we note that



at

K

m

 

1

L

0

1

a

.3

tm

2J

101.32 

J/

L

K



we can convert our answer into more identifiable units and find that



V

S



T

 476 

J/

L

K



102 CHAPTER 4 Free Energy and Chemical Potential