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ID: past-exam-of-the-chemistry-a-course-of-the-university-of-cambridge/2023/paper-2/a8/a/i/solution

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Past exam of the chemistry A course of the University of Cambridge / 2023 / Paper 2 / A8 / a / i / Solution by

Codex  0 2026-10-20

For an ideal mixture, the Helmholtz free energy is a sum over species. At fixed

T, V

and all other populations, the molecular partition function

q_{i}

does not depend on

N_{i}

. Differentiating the contribution written using each species' own energy reference gives

\frac{\partial}{\partial N _{i}} [- N_{i} k_{B} T ln q_{i} + k_{B} T (N_{i} ln N_{i} - N_{i})] = - k_{B} T ln q_{i} + k_{B} T ln N_{i} .

The desired additive energy term requires a consistent energy reference in molecular partition functions. Let

ε^{0, i}

be the energy per molecule of species

i

in its reference state, measured relative to a common chemical energy zero, and let

q_{i}

sum excitations above that reference. Then

q_{i}^{abs} = e^{- ε^{0, i} / (k_{B} T)} q_{i}, A_{i} = N_{i} ε^{0, i} - N_{i} k_{B} T ln q_{i} + k_{B} T (N_{i} ln N_{i} - N_{i}) .

Therefore

μ_{i} = (\frac{\partial A}{\partial N _{i}})_{T, V, N_{j \neq = i}} = ε^{0, i} - k_{B} T ln \frac{q _{i}}{N _{i}} .

One convenient choice is each species' rovibrational ground state, so the molecular reference includes its zero-point energy. A different consistent reference is possible, but the corresponding partition function must change. The offset cannot arise from differentiation of a formula that has already omitted it, and it must not be counted twice. These chemical potentials are per particle; molar expressions instead use the molar gas constant.

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