 Solution

ID: past-exam-of-the-thermal-and-statistical-physics-course-of-the-university-of-cambridge/2018/section-a/a1/a/solution

V (T, p, N) = \frac{N k _{B} T}{p} - N A (T) .

The Gibbs free energy has differential

d G = - S d T + V d p + μ d N

. At fixed temperature and particle number, integrate

d G = V d p

between the two pressures:

Δ G = \int_{p_{0}}^{p_{1}} [\frac{N k _{B} T}{p} - N A (T)] d p .

Consequently

Δ G = N k_{B} T ln \frac{p _{1}}{p _{0}} - N A (T) (p_{1} - p_{0}) .

Use

S = - (\partial G / \partial T)_{p, N}

, keeping both endpoint pressures fixed while differentiating the difference:

Δ S = - N k_{B} ln \frac{p _{1}}{p _{0}} + N A^{'} (T) (p_{1} - p_{0}) .

Equivalently, the Maxwell relation

(\partial S / \partial p)_{T, N} = - (\partial V / \partial T)_{p, N}

gives

- N k_{B} / p + N A^{'} (T)

, whose pressure integral is the same result. These are the Gibbs free-energy change for a pressure-linear gas correction identities. The physical path requires

p_{0}, p_{1} > 0

and positive volume throughout. The correction

A (T)

has units of volume per particle, so each term in

Δ G

is an energy and each term in

Δ S

has units of energy per kelvin.

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