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ID: past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2017/paper-2/section-c/c11/c/solution

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Past exam of the physics B course of the University of Cambridge / 2017 / Paper 2 / Section C / C11 / c / Solution by

Codex  0 2026-10-18

For a reversible adiabatic change,

d S = 0

, equivalently

C_{V} d T = - p d V

. Substitute

p = RT / (V - b)

:

C_{V} \frac{d T}{T} + R \frac{d V}{V - b} = 0.

Integration, with fixed positive reference values

T_{*}, W_{*}

, gives

C_{V} ln \frac{T}{T _{*}} + R ln \frac{V - b}{W _{*}} = constant, T (V - b)^{R / C_{V}} = constant .

Since

T = p (V - b) / R

and

γ = 1 + R / C_{V}

,

p (V - b)^{γ} = constant .

The relation applies to the quasistatic reversible adiabat. Adiabatic heat isolation alone does not enforce it for an irreversible process which generates entropy. The shifted volume

V - b

, rather than

V

itself, is the appropriate coordinate in this model.

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