Physical Chemistry Third Edition

26.2 Working Equations for the Thermodynamic Functions of a Dilute Gas 1095

CV,elN

dεel

dT

−

N

z^2 el

dzel

dT

g 1 ε 1 e−ε^1 /kBT+

N

zel

g 1

ε^21

kBT^2

e−ε^1 /kBT

−

N

zel^2

1

kBT^2

(

g 1 ε 1 e−ε^1 /kBT

) 2

+

N

zel

g 1

ε^21

kBT^2

e−ε^1 /kBT



N

zel

e−ε^1 /kBT

(

g 1 ε^21

kBT^2

)(

1 −

g 1 e−ε^1 /kBT

zel

)

Exercise 26.8

Evaluate the electronic contribution to the molar heat capacity of NO at 298.15 K. The ground

electronic level is a^2 Π 1 / 2 term. The first excited level is a^2 Π 3 / 2 term with an energy 2.380×

10 −^21 J above the ground level. Both levels have a degeneracy equal to 2.

The Entropy of a Dilute Gas

The entropy is also a sum of contributions.

SStr+Srot+Svib+Sel (26.2-11)

From Eq. (26.1-5)

Str

Utr

T

+NkBln

(

Ztr

N

)

+NkB (26.2-12a)

Srot

Urot

T

+NkBln(Zrot) (26.2-12b)

Svib

Uvib

T

+NkBln(Zvib) (26.2-12c)

Sel

Uel

T

+NkBln(Zel) (26.2-12d)

There is only one additive term,NkB, and there is only one divisor,N, in Eq. (26.1-5),

and these are placed with the translational contribution to the entropy.

The translational molar energy is 3R/2 and the additive term is equal toR, so that

the translational contribution to the molar entropy of a dilute gas is

Sm,tr

5 R

2

+Rln

(

ztr

NAv

)

(26.2-13)

whereNAvis Avogadro’s constant. Using Eq. (25.3-21) and the ideal gas equation of

state, we write

ztr

NAv



(

2 πmkBT

h^3

) 3 / 2

V

NAv



(

2 πmkBT

h^3

) 3 / 2

kBT

P



(2πm)^3 /^2 (kBT)^5 /^2

h^3 P



(2πM/NAv)^3 /^2 (kBT)^5 /^2

h^3 P◦

(

P◦

P

)

(26.2-14)