= Solution
At low <temperature>, $T\ll\theta_r$, retain the ground level and the first excited level. Their <energies> are $0$ and $\hbar^2/I$, with degeneracies $1$ and $3$. Therefore
$$
Z_{\rm rot}=1+3e^{-\hbar^2/(Ik_BT)}+O\left(e^{-3\hbar^2/(Ik_BT)}\right),
$$
so in the notation of the question,
$$
\boxed{A=3,\qquad B=\frac{\hbar^2}{Ik_B}=2\theta_r.}
$$
$A$ is dimensionless and $B$ has units of kelvins. The first omitted level has three times the first excitation <energy>, explaining the exponent in the error term.
At high <temperature>, $T\gg\theta_r$, many rotational levels are occupied and the sum can be replaced to leading order by an <integral>:
$$
Z_{\rm rot}\simeq\int_0^\infty(2J+1)e^{-(\theta_r/T)J(J+1)}\,dJ.
$$
Set $u=J(J+1)$, so that $du=(2J+1)dJ$. Then
$$
Z_{\rm rot}\simeq\int_0^\infty e^{-(\theta_r/T)u}\,du=\frac T{\theta_r}
=\frac{2Ik_B}{\hbar^2}T.
$$
Thus
$$
\boxed{C=\frac{2Ik_B}{\hbar^2}=\frac1{\theta_r},\qquad [C]=\mathrm{K^{-1}}.}
$$
The high-temperature approximation here concerns the rotational spectrum. For a real <molecule> it should still be used below <temperatures> where neglected vibrational excitation or dissociation invalidates the rigid-rotor model.
Back to article page
