Physical Chemistry Third Edition

26.2 Working Equations for the Thermodynamic Functions of a Dilute Gas 1091

Solution

zel≈ 2 + 2 e−ε^1 /kBT 2 +2 exp

(

− 2. 380 × 10 −^21 J

(1. 3807 × 10 −^23 JK−^1 )(298.15 K)

)

≈ 2 + 2 e−^0.^57815  3. 1219

Um,el≈

NAv

zel

(

0 +g 1 ε 1 e−ε^1 /kBT

)

≈

6. 02214 × 1023 mol−^1

3. 1219

(

0 +2(2. 38 × 10 −^21 J)e−^0.^57815

)

≈ 515 .1 J mol−^1

The Rotational Energy of a Dilute Gas

For a diatomic substance or a polyatomic substance with linear molecules the rotational

energy is given by

UrotNkBT^2

(

dln(zrot)

dT

)

V

NkBT^2

(

dln(8π^2 IkBT/σh^2

dT

)

NkBT^2

dln(constant)

dT

+NkBT^2

dln(T)

dT

NkBT^2

1

T

UrotNkBTnRT (diatomic or linear molecules) (26.2-6)

For a polyatomic substance with nonlinear molecules,zrotis proportional toT^3 /^2 so

that

Urot

3

2

NkBT

3

2

nRT (nonlinear molecules) (26.2-7)

Exercise 26.5

Show by differentiation that Eq. (26.2-7) is correct.

The Vibrational Energy of a Dilute Gas

If the zero-point vibrational energy is excluded from the vibrational energy, the vibra-

tional energy of a dilute diatomic gas is given by

UvibNkBT^2

(

dln(zvib)

dT

)

V

NkBT^2

d

dT

[

ln

(

1

1 −e−hν/kBT

)]

Uvib

Nhc ̃νe

ehc ̃νe/kBT− 1

(diatomic substance) (26.2-8)