Physical Chemistry Third Edition

25.3 The Probability Distribution and the Molecular Partition Function 1063

Exercise 25.16

Verify Eq. (25.3-26) by evaluatingdβ/dT, which is equal to the reciprocal ofdT/dβ.

Consider the effect of changing the zero of energy by adding a constantV′to the

value of the potential energy of a molecule. (See Problem 15.6 for the effect on an

energy eigenvalue.) Letε′jrepresent the new value of the energy eigenvalue of level

numberj:

ε′jεj+V′ (25.3-27)

Denote the new value of the partition function byz′:

z′

∑

j

gje−ε

′

j/kBT

∑

j

gje−(εj+V

′)/kBT

e−V

′/kBT∑

j

gje−εj/kBTe−V

′/kBT

z (25.3-28)

The probability distribution is

p′

1

z′′

gje−(εj+V

′)/kBT



1

ze−V′/kBT

gje−εj/kBTe−V

′/kBT



1

z

gje−εj/kBT (25.3-29)

The probability distribution is independent of the choice of the zero of energy, as we

would expect, but the value of the partition function changes by the factore−V

′/kBT

.

A change in the zero of energy adds a constant to the thermodynamic energy.

Exercise 25.17

Using Eq. (25.3-26), show that the thermodynamic energy using the new zero of energy is

given by

U′U+NV′ (25.3-30)

PROBLEMS

Section 25.3: The Probability Distribution and the
Molecular Partition Function

25.14Write the translational partition function for a monatomic
dilute gas at 298.15 K and a volume of 0.0244 m^3 as a
constant timesM^3 /^2 , whereMis the molar mass in
kg mol−^1. Evaluate the partition function for He, Ne, Ar,
Kr, and Xe. Explain the dependence on molar mass.

25.15 a.Find the value of the molecular partition function of a

neon atom confined in a cubical volume of 1.000 m^3

at 298.15 K.

b.Find the probability of the translational state of a neon

atom corresponding tonxnynz 3. 5 × 1010

(the values in Example 25.4) at this temperature and

volume.