Physical Chemistry Third Edition

4.4 Gibbs Energy Calculations 177

The formula in Eq. (4.4-7) is equivalent to integrating from the standard-state pressure

P◦down to zero pressure with the ideal gas, and then integrating back up to pressureP′

with the real gas. This procedure gives the difference between the real gas at pressure

P′and the ideal gas at pressureP◦.

The integral in Eq. (4.4-9) can be broken into two parts, as follows (we have also

exchanged the limits, which changes the sign):

(last two terms)−

∫P′

0

RT

P

dP−

∫P◦

P′

RT

P

dP

−

∫P′

0

RT

P

dP−RTln

(

P◦/P′

)

(4.4-10)

The left-hand side of Eq. (4.4-7) is equal toRTln(f/P◦), so that iff′denotes the

fugacity at pressureP′, we can combine the two integrals to write

Gm(T,P′)−G◦m(T)RTln

(

f′

P◦

)

RTln

(

P′

P◦

)

+

∫P′

0

(

Vm,real−

RT

P

)

dP (4.4-11)

which is the same as

RTln

(

f′

P′

)



∫P′

0

(

Vm,real−

RT

P

)

dP (4.4-12)

The integrand of this integral is small if the deviation from ideality is small and vanishes

if the gas is ideal.

EXAMPLE4.14

Find an expression for the fugacity of a gas that obeys the truncated pressure virial equation

of state

PVmRT+A 2 P

where the second pressure virial coefficientA 2 is a function of temperature. It was shown in

Example 1.9 that the second pressure virial coefficient,A 2 , is equal toB 2 , the second virial

coefficient.

Solution

RTln

(

f′/P′

)



∫P′

0

(

RT

P

+A 2 −

RT

P

)

dPA 2 P′

or

f′P′eA^2 P

′/RT

Exercise 4.11

a.For argon at 273.15 K,B 2 − 21 .5cm^3 mol−^1. Find the value of the fugacity of argon gas

at 5.000 atm and 273.15 K.