 Solution

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

Differentiating each Boltzmann factor yields

\frac{\partial q _{3}}{\partial T} = \frac{θ _{1} e ^{- θ_{1} / T} + θ _{2} e ^{- θ_{2} / T}}{T ^{2}} .

The derivative is positive: warming increases occupation of both excited energy levels. At

100 K

\frac{\partial q _{3}}{\partial T}_{100 K} = 1.56692 \times 1 0^{- 3} K^{- 1} .

 New to topics? Read the docs here!