CHAPTER 7 PURE SUBSTANCES IN SINGLE PHASES
7.8 CHEMICALPOTENTIAL ANDFUGACITY 184
By integrating dDVmdpfrom pressurep^00 to pressurep^0 , we obtain
^0 ^00 D
Z 0
^00
dD
Zp 0
p^00
Vmdp (7.8.10)
Equating the two expressions for^0 ^00 and dividing byRTgives
ln
f^0
f^00
D
Zp 0
p^00
Vm
RT
dp (7.8.11)
In principle, we could use the integral on the right side of Eq.7.8.11to evaluatef^0 by
choosing the lower integration limitp^00 to be such a low pressure that the gas behaves as
an ideal gas and replacingf^00 byp^00. However, because the integrandVm=RT becomes
very large at low pressure, the integral is difficult to evaluate. We avoid this difficulty by
subtracting from the preceding equation the identity
ln
p^0
p^00
D
Zp 0
p^00
dp
p
(7.8.12)
which is simply the result of integrating the function1=pfromp^00 top^0. The result is
ln
f^0 p^00
f^00 p^0
D
Zp 0
p^00
Vm
RT