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ID: past-exam-of-the-short-options-paper-shared-by-parts-a-and-b-of-the-physics-course-of-the-university-of-oxford/2019/s16/1/c/solution

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Past exam of the short options paper shared by Parts A and B of the physics course of the University of Oxford / 2019 / S16 / 1 / c / Solution by

Codex  0 2026-10-25

At fixed volume, logarithmic differentiation of the Saha ionization equation gives

(\frac{2}{ζ} + \frac{1}{1 - ζ}) ζ^{'} = \frac{1}{T} (\frac{3}{2} + \frac{E _{I}}{k _{B} T}), T ζ^{'} = \frac{ζ ( 1 - ζ )}{2 - ζ} (\frac{3}{2} + \frac{E _{I}}{k _{B} T}) .

Differentiation the internal energy then yields

C_{V} = \frac{3}{2} R (1 + ζ) + R \frac{ζ ( 1 - ζ )}{2 - ζ} (\frac{3}{2} + \frac{E _{I}}{k _{B} T})^{2} .

At low temperature

ζ \to 0

and

C_{V} \to 3 R /2

. At high nonrelativistic temperature the gas is fully ionized,

ζ \to 1

, and

C_{V} \to 3 R

. The large intermediate excess reflects energy spent increasing ionization, as well as heating and creating additional translational particles. “Very high” here remains below regimes where relativistic, radiative or pair-creation contributions invalidate the model.

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