 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ii/paper-1/36c/c/solution

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

δ Q = 0

. For a reversible process it is also isentropic. The first law then gives

C_{V} d T = d E = - p d V .

Using

p = N k_{B} T / V

\frac{d T}{T} = - \frac{N k _{B}}{C _{V}} \frac{d V}{V} = - (γ - 1) \frac{d V}{V} .

Integration yields

T V^{γ - 1} = constant .

Substituting

T = p V / (N k_{B})

p V^{γ} = constant .

 New to topics? Read the docs here!