The one-particle partition function is

z_{1} = \frac{A}{h ^{3}} \int_{R^{3}} e^{- β p^{2} / (2 m)} d^{3} p \int_{0}^{\infty} e^{- β m g z} d z = \frac{A}{λ ^{3} β m g},

(1)

where

λ = (2 π ℏ^{2} β / m)^{1/2}

. Thus

Z_{N} = \frac{z _{1}^{N}}{N !}, lo g z_{1} = constant - \frac{5}{2} lo g β .

(2)

Canonical differentiation now gives

⟨ E ⟩ = - \partial_{β} lo g Z_{N} = \frac{5 N}{2 β} = \frac{5}{2} N k_{B} T

(3)

and

(Δ E)^{2} = \partial_{β}^{2} lo g Z_{N} = \frac{5 N}{2 β ^{2}} .

(4)

Therefore

\frac{Δ E}{⟨ E ⟩} = \frac{2}{5 N} .

(5)

The factor

5/2

consists of

3/2

from kinetic energy and

1

from gravitational potential energy.

ii

 0  0

Solution

 0  0

The normalized one-particle height density is exponential:

f_{Z} (z) = β m g e^{- β m g z}, z > 0.

(1)

Consequently

⟨ z ⟩ = \frac{1}{β m g} = \frac{k _{B} T}{m g} .

(2)

The number density and local ideal-gas pressure are

n (z) = \frac{N}{A ⟨ z ⟩} e^{- z / ⟨ z ⟩}, p (z) = n (z) k_{B} T = \frac{N m g}{A} e^{- z / ⟨ z ⟩} .

(3)

This also obeys hydrostatic balance

p^{'} (z) = - m g n (z)

iii

 0  0

Solution

 0  0

Since

⟨ z ⟩ = (β m g)^{- 1}

Z_{N} = \frac{1}{N !} (\frac{A ⟨ z ⟩}{λ ^{3}})^{N} .

(1)

Using

lo g N! ≃ N lo g N - N

and

S = k_{B} (lo g Z_{N} + β ⟨ E ⟩)

gives

S = N k_{B} [lo g (\frac{A ⟨ z ⟩}{N λ ^{3}}) + \frac{7}{2}] .

(2)

This has the Sackur-Tetrode form with the effective volume

A ⟨ z ⟩

, but its additive constant is

7/2

rather than

5/2

. The extra unit arises because the atmosphere's vertical extent grows with temperature and its mean gravitational energy per atom is

k_{B} T

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Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 1 / 36C / b

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