 Solution

ID: past-exam-of-the-chemistry-a-course-of-the-university-of-cambridge/2025/paper-2/a6/a/solution

U = \sum_{i} n_{i} ε_{i} = \frac{N}{q} \sum_{i} ε_{i} e^{- ε_{i} / k T} .

At fixed volume, assume the energy levels themselves are temperature-independent. Differentiating the molecular partition function gives

(\frac{\partial q}{\partial T})_{V} = \frac{1}{k T ^{2}} \sum_{i} ε_{i} e^{- ε_{i} / k T}, U = N k T^{2} \frac{\partial lo g q}{\partial T} .

Choosing the zero of energy changes

U

by the corresponding constant per molecule, so the ground-level convention must be consistent throughout.

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