Physical Chemistry Third Edition

1128 27 Equilibrium Statistical Mechanics. III. Ensembles

Solution

ln(Z)Nln(z)−ln(N!)

z

(

2 π(6. 65 × 10 −^27 kg)(1. 3807 × 10 −^23 JK−^1 )

(6. 6261 × 10 −^34 Js)^2

) 3 / 2

(0.0250 m^3 )

 1. 938 × 1029

We use Stirling’s approximation for ln(N!):

ln(Z)Nln(z)−ln(N!)≈Nln(z)−Nln(N)+N

≈−(6. 022 × 1023 )ln(1. 938 × 1029 )

−(6. 022 × 1023 )ln(6. 022 × 1023 )+ 6. 023 × 1023

≈− 4. 06 × 1025 − 3. 297 × 1025 + 6. 023 × 1023 ≈ 8. 24 × 1024

This is a large number, anderaised to this power is unimaginably large.

PROBLEMS

Section 27.1: The Canonical Ensemble

27.1 From the value of the molecular partition function in
Example 25.1, calculate the value of the canonical
partition function for 1.000 mol of argon confined in a
volume of 25.0 L (0.0250 m^3 ) at 298.15 K.

27.2 Calculate the natural logarithm of the canonical partition
function of 1.000 mol of O 2 gas at 298.15 K and a pressure
of 1.000 atm.

27.3 Derive a formula for the canonical partition function when
a different zero of energy is used.

27.4 Calculate the natural logarithm of the canonical partition

function of 1.000 mol of Cl 2 gas at 298.15 K and a pressure

of 1.000 bar. Use the value of the molecular partition

function from Chapter 25, and use Stirling’s approximation

for ln(NAv!).

27.5 a.Estimate the value of the canonical partition function

for 1.000 mol of neon gas in a volume of 25.0 L at

298.15 K. Use Stirling’s approximation forN!.

b.Estimate the probability of a single microstate of this

system with an energy of 3RT/ 2 3718 J at a

temperature of 298.15 K.

27.2 Thermodynamic Functions in the Canonical

Ensemble

We can now write the formula for the energy in Eq. (27.1-20) in terms of the

temperature:

〈E〉−

(

∂ln(Z)

∂β

)

V,N

−

(

∂ln(Z)

∂T

)

V,N

(

dT

dβ

)

−

(

∂ln(Z)

∂T

)

V,N

(−kBT^2 )

〈E〉UkBT^2

(

∂ln(Z)

∂T

)

V,N

(general system) (27.2-1)