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ID: past-exam-of-the-chemistry-course-of-the-year-3-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-3/48/b/solution

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Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge / 2022 / Paper 3 / 48 / b / Solution by

Codex  0 2026-10-20

At sufficiently low packing fraction, neglect overlaps of the excluded regions around different disc centres. Sequential insertion gives the configurational integral

Z_{N} ≃ \prod_{j = 0}^{N - 1} (a - 4 j a_{d}) .

(1)

Taking logarithms and expanding to first order in density,

ln Z_{N} ≃ N ln a - \frac{4 a _{d}}{a} \sum_{j = 0}^{N - 1} j = N ln a - \frac{2 a _{d} N ( N - 1 )}{a} .

(2)

For large

N

, replace

N - 1

N

. The expression

N ln (a - N b)

has the same leading-density expansion when $b = 2 a_{d}$ . Thus the hard-disc excluded-area approximation yields

Q_{ld} ≃ \frac{( a - 2 N a _{d} ) ^{N}}{Λ ^{2 N} N !} .

(3)

The factor

N!

accounts for indistinguishable particles and

Λ^{2 N}

for two-dimensional momentum integration. The factor one-half between pair-excluded area and

b

avoids counting every pair twice. Equality of leading expansions does not make this free-volume resummation exact at high density.

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