 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ii/paper-3/35a/b/iii/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

E = \int_{0}^{E_{F}} ε g (ε) d ε = \frac{V E _{F}^{4}}{4 π ^{2} ℏ ^{3} c ^{3}} = \frac{3}{4} N E_{F} .

Using part (i),

p = \frac{N E _{F}}{4 V} = \frac{D}{4} N^{4/3} V^{- 4/3} .

p V^{a} = K, a = \frac{4}{3}, K = \frac{ℏ c ( 3 π ^{2} ) ^{1/3}}{4} N^{4/3} .

The pressure is degeneracy pressure: it remains nonzero at zero temperature because of the Pauli exclusion principle. By comparison, a classical ultrarelativistic ideal gas at fixed temperature obeys

p V = N k_{B} T

, so its corresponding volume exponent is one and its pressure vanishes as

T \to 0

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