Physical Chemistry Third Edition

4.4 Gibbs Energy Calculations 175

4.4 Gibbs Energy Calculations

For a closed simple system at constant temperature, Eq. (4.2-19) is

dGVdP (simple system,Tandnconstant) (4.4-1)

Integration of this formula at constantTandngives

G(T,P 2 ,n)−G(T,P 1 ,n)

∫P 2

P 1

VdP (4.4-2)

The Gibbs Energy of an Ideal Gas

For an ideal gas of one substance Eq. (4.4-2) becomes

G(T,P 2 ,n)G(T,P 1 ,n)+nRT

∫P 2

P 1

1

P

dP

G(T,P 2 ,n)G(T,P 1 ,n)+nRTln

(

P 2

P 1

)

(ideal gas) (4.4-3)

The molar Gibbs energy,Gm, is equal toG/n,

Gm(T,P 2 )Gm(T,P 1 )+RTln

(

P 2

P 1

)

(ideal gas) (4.4-4)

Thestandard statefor the Gibbs energy of an ideal gas is the same as for the

entropy: a fixed pressure ofP◦, thestandard pressure, defined to be exactly equal to

1 bar100000 Pa. Specifying the standard state does not specify a particular temper-

ature. There is a different standard state for each temperature. At one time a value

of 1 atm was used forP◦. Use of this choice forP◦makes no difference to the

formulas that we write and makes only a small difference in numerical values. For

highly accurate work, one must determine whether the 1-atm standard state or the

1-bar standard state has been used in a given table of numerical values. If state 1

is chosen to be the standard state and if the subscript is dropped onP 2 , Eq. (4.4-4)

becomes

Gm(T,P)G◦m(T)+RTln

(

P

P◦

)

(ideal gas) (4.4-5)

whereG◦m(T) is the molar Gibbs energy of the gas in the standard state at tempera-

tureT.

EXAMPLE4.13

Obtain a formula to change from the 1-atm standard state to the 1-bar standard state for an

ideal gas.