Physical Chemistry , 1st ed.

Making the same substitution for dS dq/T, and also substituting for the de-

finition ofpVwork, we get

dU p dVT dSdwnon-pV

If temperature and pressure are constant (the crucial requirements for a use-

ful Gstate function), then we can rewrite the differential as

d(U pVTS) dwnon-pV

U pVis the definition ofH. Substituting:

d(HTS) dwnon-pV

Also,HTSis the definition ofG:

dGdwnon-pV

which we can integrate to get

Gwnon-pV (4.11)

That is, when non-pVwork is performed,Grepresents a limit. Again, since

work performed by a system is negative,Grepresents the maximum amount

of non-pVwork a system can perform on the surroundings. For a reversible

process, the change in the Gibbs free energy is equal to the non-pVwork of the

process. Equation 4.11 will become important to us in Chapter 8, when we

discuss electrochemistry and electrical work.

Example 4.2

Calculate the change in the Helmholtz energy for the reversible isothermal

compression of 1 mole of an ideal gas from 100.0 L to 22.4 L. Assume that

the temperature is 298 K.

Solution

The process described is the third step in a Carnot-type cycle. Since the

process is reversible, the equality relationship Awapplies. Therefore we

need to calculate the work for the process. The work is given by equation 2.7:

wnRTln V

V

f

i



Substituting for the various values:

w(1 mol)8.314 

mo

J

l K

(298 K) ln 

1

2

0

2

0

.4

.0

L

L



w3610 J

Since for this reversible process Aw, we have

A3610 J

Since many processes can be made to occur isothermally (or at least re-

turned to their original temperatures), we can develop the following expres-

sions for Aand G:

AUTS

dAdUT dSS dT

dAdUT dS for an isothermal change

94 CHAPTER 4 Free Energy and Chemical Potential