 Solution

ID: past-exam-of-the-chemistry-a-course-of-the-university-of-cambridge/2023/paper-2/a7/c/solution

α = \frac{d}{d T} \frac{V ( T )}{V _{0}} = \frac{d P _{1}}{d T} .

Let

y = 3 e^{- ε / (k_{B} T)}

. Since

\frac{d y}{d T} = \frac{ε}{k _{B} T ^{2}} y, \frac{d}{d y} \frac{y}{1 + y} = \frac{1}{( 1 + y ) ^{2}},

we obtain

α (T) = \frac{ε}{k _{B} T ^{2}} \frac{3 e ^{- ε / (k_{B} T)}}{( 1 + 3 e ^{- ε / (k_{B} T)} ) ^{2}} .

It is positive and has units of inverse temperature. Equivalently,

α = [ε / (k_{B} T^{2})] P_{1} (1 - P_{1})

: the response reflects changes in the thermally occupied excited population. The usual instantaneous fractional thermal expansion coefficient

V^{- 1} d V / d T

would instead equal

α V_{0} / V

; substituting that different definition would change the answer.

 New to topics? Read the docs here!