 Solution

For a high-temperature linear rigid rotor, $q_{\mathrm{rot}}=T/{\theta }_{\mathrm{rot}}$ up to temperature-independent symmetry/spin factors. Therefore $q^{\prime }/q=1/T$ and $q^{\prime \prime }=0$, givingThis agrees with the equipartition theorem for two quadratic rotational coordinates. It is the rotational contribution; adding free three-dimensional translation gives $C_{V,m}=5R/2$ when vibrations remain unexcited. A constant rotational symmetry number does not affect this derivative.