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ID: past-exam-of-the-chemistry-a-course-of-the-university-of-cambridge/2022/paper-2/a6/e/solution

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Past exam of the chemistry A course of the University of Cambridge / 2022 / Paper 2 / A6 / e / Solution by

Codex  0 2026-10-20

At adsorption equilibrium,

μ_{ads} = μ_{g}

. The expressions from part (d) give

\frac{θ}{1 - θ} = z \frac{p Λ ^{3}}{2 k _{B} T} = \frac{p Λ ^{3}}{k _{B} T} e^{E_{b} / k_{B} T} cosh x .

b (T, B) = \frac{Λ ^{3}}{k _{B} T} e^{E_{b} / k_{B} T} cosh (\frac{μ _{B} B}{k _{B} T}) .

Solving the occupation ratio yields

θ = \frac{b p}{1 + b p} .

The coefficient

b

has units of inverse pressure, so

b p

is dimensionless. The two spin states contribute a factor two on both sides of the adsorption equilibrium and cancel; retaining the surface factor but dropping the bulk degeneracy would incorrectly double

b

.

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