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

Table of contents
- Solution i

Solution

A triangular lattice has area per particle

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.

 Ancestors (14)

 View article source

 Discussion (0)

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version