Physical Chemistry Third Edition

27.5 Thermodynamic Functions in the Classical Canonical Ensemble 1143

this theorem as follows: Leturepresent some variable such as a coordinate or a

momentum component, and assume that the contribution of this variable to the energy

of one molecule is

εau^2 (27.5-10)

whereais a constant and whereucan take on any real values. The average energy

contribution is

U〈E〉N〈ε〉N

∫∞

−∞

au^2 e−au

(^2) /kBT
du
∫∞
−∞
e−au^2 /kBTdu

(27.5-11)

When the integrals are evaluated from a table such as that of Appendix C, we obtain

U

Naπ^1 /^2 (kBT/a)^3 /^2 / 2

(πkBT/a)^1 /^2



NkBT

2

(27.5-12)

This result is independent of the constantaand corresponds to equipartition of energy.

Exercise 27.6

Look up the integrals and verify Eq. (27.5-12).

The energy of a dilute diatomic or polyatomic gas is a sum of contributions. If we

ignore the vibrational and electronic contributions,

UNkBT^2

(

∂ln(zcl)

∂T

)

V

NkBT^2

(

∂ln(zcl,tr)

∂T

)

V

+NkBT^2

(

∂ln(zcl,rot)

∂T

)

V

Utr+Urot (27.5-13)

The translational energy is the same as that of a monatomic gas:

Utr

3

2

NkBT (27.5-14)

The rotational energy is independent of the moment of inertia:

UrotNkBT (diatomic or linear polyatomic substance) (27.5-15)

Urot

3

2

NkBT (nonlinear polyatomic substance) (27.5-16)

These contributions conform to the equipartition of energy and also agree with the

quantum statistical mechanics results.

According to equipartition of energy the vibration of a diatomic molecule should

make a contribution ofNkBT to the energy, since there is one coordinate and one

momentum component and both enter quadratically in the energy expression. However,

this is a very poor approximation.