 Solution

ID: past-exam-of-the-chemistry-course-of-the-year-4-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-2/section-e/26/d/i/solution

3 ℓ_{hex}^{2} /2

, so

ℓ_{hex} = 2 A / (3 N)

. Inserting this into the supplied accessible area yields

\frac{a _{c}^{hex}}{Λ ^{2}} = [\frac{2}{Λ} \frac{A - σ 3 N /2}{N}]^{2} .

Therefore

F_{hex} = - 2 N k_{B} T ln [\frac{2}{Λ} \frac{A - B _{hex}}{N}], B_{hex} = \frac{3}{2} N σ^{2} .

The prefactor

C

remains

2/Λ

. The lattice is triangular in terms of particle centres; “hexagonal” here describes its sixfold coordination and cell symmetry.

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