Physical Chemistry Third Edition

122 3 The Second and Third Laws of Thermodynamics: Entropy

For isothermal reversible volume changes in a system consisting of an ideal gas,qrev

is given by Eq. (2.4-10), so that

∆SnRln

(

V 2

V 1

)

(ideal gas, reversible isothermal process) (3.3-3)

EXAMPLE 3.2

Find∆S,∆Ssurr,q,w, and∆Ufor the reversible isothermal expansion of 3.000 mol of argon

(assumed ideal) from a volume of 100.0 L to a volume of 500.0 L at 298.15 K.

Solution

∆S(3.000 mol)(8.3145 J K−^1 mol−^1 )ln

(

500 .0L

100 .0L

)

 40 .14JK−^1

Since the process is reversible,

∆Suniverse∆S+∆Ssurr 0

∆Ssurr−∆S− 40 .14JK−^1

Since the system is an ideal gas,∆U 0.

q(3.000 mol) (8.3145 J K−^1 mol−^1 ) (298.15 K) ln

(

500 .0L

100 .0L

)

11,970 J

w−q−11,970 J

Exercise 3.5

Find∆S,∆Ssurr,q,w, and∆Uif 3.000 mol of argon (assumed to be ideal) expands reversibly

and isothermally from a volume of 50.0 L to a volume of 250.0 L at 298.15 K. Compare your

answers with those of the previous example and explain any difference.

For a nonideal gas, the entropy change of a reversible isothermal volume change

can be calculated from Eq. (3.3-2) if an expression forqrevis obtained.

EXAMPLE 3.3

Find∆Sand∆Ssurrfor the reversible expansion of 1.000 mol of argon from 2.000 L to

20.00 L at a constant temperature of 298.15 K as in Example 2.15. Argon is represented by

the truncated virial equation of state as in that example.

Solution

Using the result of Example 2.15a forqrev

∆S

qrev

T



5757 J

298 .15 K

 19 .31JK−^1