 Solution

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

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

δ Q = d E = C_{V} d T .

At constant pressure,

δ Q = C_{p} d T = d E + p d V .

Differentiating

p V = N k_{B} T

at fixed

p

gives

p d V = N k_{B} d T .

Consequently,

C_{p} d T = (C_{V} + N k_{B}) d T,

C_{p} - C_{V} = N k_{B} .

γ = \frac{C _{p}}{C _{V}} = 1 + \frac{N k _{B}}{C _{V}} .

 New to topics? Read the docs here!