 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2005/ii/paper-2/34d/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

For a reversible change of a simple compressible system with fixed composition, the first law is

d E = T d S - P d V .

An adiabatic change exchanges no heat. A reversible adiabatic change is isentropic,

d S = 0

; adiabatic alone does not imply constant entropy for an irreversible process.

F = E - TS

. Then

d F = - S d T - P d V

(\frac{\partial S}{\partial V})_{T} = (\frac{\partial P}{\partial T})_{V} .

Holding

T

fixed in the first law and using this relation gives

(\frac{\partial E}{\partial V})_{T} = T (\frac{\partial P}{\partial T})_{V} - P .

For equilibrium radiation,

E = V e (T)

and

P = e (T) /3

, so this identity reads

e = T e^{'} /3 - e /3

, or

T e^{'} = 4 e

. Integrating gives

e = a T^{4}, E = aV T^{4} .

Finally in a reversible adiabatic expansion,

d E = - P d V

becomes

4 aV T^{3} d T + a T^{4} d V = - a T^{4} d V /3

. Dividing gives

3 d T / T = - d V / V

, hence $V T^{3}$ is constant. This assumes equilibrium radiation throughout the expansion.

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