= Solution
The classical <equipartition theorem> assigns mean <energy> $k_BT/2$ to each independent quadratic term in the Hamiltonian, provided the canonical distribution and its relevant boundary integrals exist. For a quadratic term $aq^2$, the <Boltzmann distribution> is proportional to $e^{-\beta aq^2}$, with $\beta=1/(k_BT)$. Integrating a total derivative gives
$$
0=\int_{-\infty}^{\infty}\frac{d}{dq}\left(qe^{-\beta aq^2}\right)dq=\int e^{-\beta aq^2}dq-2\beta a\int q^2e^{-\beta aq^2}dq.
$$
After dividing by the normalization,
$$
\boxed{\langle aq^2\rangle=\frac1{2\beta}=\frac12k_BT.}
$$
The same argument applies to quadratic momentum terms. Other independent canonical coordinates factor out of the ratio. It is a classical statement: widely spaced quantum energy levels can freeze out a contribution at low <temperature>.
The confining quartic <potential energy> $U(x)=Qx^4$, with $Q>0$, is not quadratic, so the $k_BT/2$ rule cannot be assigned to it. This does not invalidate the canonical <Boltzmann distribution> or its generalized integration-by-parts identity. Indeed,
$$
0=\int_{-\infty}^{\infty}\frac{d}{dx}\left(xe^{-\beta Qx^4}\right)dx=Z_x-4\beta Q\int_{-\infty}^{\infty}x^4e^{-\beta Qx^4}dx.
$$
Hence <equipartition for a homogeneous potential> gives
$$
\boxed{\langle U\rangle=Q\langle x^4\rangle=\frac14k_BT,\qquad C_{\rm potential}=\frac{d\langle U\rangle}{dT}=\frac14k_B.}
$$
Equivalently, $Z_x=\Gamma(1/4)(\beta Q)^{-1/4}/2$ and $-\partial\ln Z_x/\partial\beta=1/(4\beta)$. This is the potential contribution only; an additional standard kinetic term $p^2/(2m)$ would contribute a further $k_BT/2$ to <energy> and $k_B/2$ to <heat capacity>.
For an intact diatomic molecule, the classical degrees of freedom comprise three translational quadratic terms, two rotational quadratic terms, and two terms for its one bond-stretching vibration: one kinetic and one potential. Thus complete classical equipartition predicts
$$
\boxed{C_{V,\rm molar}=\frac72R.}
$$
The <heat capacity of a diatomic gas> has lower-temperature regimes because rotational and vibrational levels are quantized. In a dilute molecular gas, the expected progression is
$$
\frac{C_V}{R}\simeq\begin{cases}
3/2,&\text{rotations and vibrations frozen},\\
5/2,&\text{rotations active, vibrations frozen},\\
7/2,&\text{rotations and vibrations active}.
\end{cases}
$$
The rotational energy scale is normally much smaller than the vibrational scale, so these steps are separated. The high-temperature value treats the vibration as harmonic and excludes dissociation and electronic excitation; the lowest-temperature gas regime is meaningful only while the sample remains dilute and gaseous.<Statistical Physics by David Tong>[§2.4, molecular rotation and vibration, and §3.4, quantum heat capacity of a diatomic gas]{ref}
\Image[/past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2016/paper-2-diatomic-heat-capacity.png]
{title=Original ideal diatomic heat-capacity model showing rotational and vibrational activation, without dissociation}
The plotted example uses a quantum rigid rotor and harmonic vibration, with an illustrative ratio of characteristic temperatures. It is a model calculation, not measured data for a particular molecular species. For the vibrational part,
$$
\frac{C_{V,\rm vib}}R=\left(\frac{\Theta_v}{T}\right)^2\frac{e^{\Theta_v/T}}{(e^{\Theta_v/T}-1)^2},
$$
which tends to zero on freezing out and to one when many vibrational levels are accessible. The classical <equipartition theorem> is recovered when the relevant level spacings are small compared with $k_BT$.
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