 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ii/paper-4/35a/i/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

k_{B} = 1

Z = Z_{kin} \frac{1}{N !} \int_{V^{N}} e^{- U (x_{1}, \dots, x_{N}) / T} d^{3 N} x,

whereas the ideal gas has

Z_{ideal} = Z_{kin} V^{N} / N!

. Hence

\frac{Z}{Z _{ideal}} = \frac{1}{V ^{N}} \int_{V^{N}} e^{- U / T} d^{3 N} x = 1 + \frac{1}{V ^{N}} \int_{V^{N}} (e^{- U / T} - 1) d^{3 N} x .

F = - T lo g Z

therefore gives

F = F_{ideal} - T lo g {1 + \frac{1}{V ^{N}} \int_{V^{N}} (e^{- U / T} - 1) d^{3 N} x} .

 New to topics? Read the docs here!