Physical Chemistry Third Edition

178 4 The Thermodynamics of Real Systems

b.For carbon dioxide at 323.15 K,B 2 −103 cm^3 mol−^1. Find the value of the fugacity of

carbon dioxide gas at 5.000 atm and 323.15 K.

c.Find the value of the fugacity of carbon dioxide gas at 15.000 atm and 323.15 K.

The Gibbs Energy of Solids and Liquids

The value of the isothermal compressibility of a typical solid or liquid is roughly

equal to 10−^9 Pa−^1 so that a change in pressure of 10 atm (roughly 10^6 Pa) changes

the volume by only a tenth of a percent. We assume the volume to be approximately

constant in the integrand of Eq. (4.4-2), giving

G(T,P 2 ,n)−G(T,P 1 ,n)≈V(P 2 −P 1 ) (4.4-13)

Thestandard stateof a substance in a condensed phase (liquid or solid) is the actual

pure substance at the standard pressureP◦and whatever temperature is of interest.

Equation (4.4-13) implies that

Gm(T,P)G◦m(T)+Vm(P−P◦) (4.4-14)

EXAMPLE4.15

CalculateGm−G◦mfor liquid water at 298.15 K and 10.00 bar.

Solution

Gm−G◦m(18. 0 × 10 −^6 m^3 mol−^1 )(9.00 bar)

(

105 Nm−^2

1 bar

)

 16 .2 J mol−^1  0 .0162 kJ mol−^1

This difference can be neglected for many purposes.

Exercise 4.12

a.FindGm−G◦mfor solid copper at 293.15 K and 1.100 bar.

b.FindGm−G◦mfor solid copper at 293.15 K and 10.00 bar.

The Temperature Dependence of the Gibbs Energy

From Eq. (4.2-19), if the pressure is constant and the system is closed,

dG−SdT (closed system,Pconstant) (4.4-15)

Integration of this equation at constant pressure would give a relation for∆Gfor a

constant-pressure temperature change.

G(T 2 ,P)−G(T 1 ,P)−

∫T 2

T 1

S(T,P)dT (equationnotusable) (4.4-16)