The Joule-Thomson coefficient of a gas is given by

\[
\left(\frac{\partial T}{\partial P}ight)_{H}=-\frac{(\partial H / \partial P)_{T}}{(\partial H / \partial T)_{P}}=\frac{1}{C_{P}}\left[T\left(\frac{\partial V}{\partial T}ight)_{P}-Vight]=\frac{N}{C_{P}}\left[\left(\frac{\partial \mathrm{v}}{\partial T}ight)_{P}-\mathrm{v}ight] .
\]

By eqn. (10.2.1),

\[
\begin{aligned}
\frac{P \mathrm{v}}{k T} & =1+\frac{a_{2} \lambda^{3}}{\mathrm{v}}+\ldots, \text { so that } \\
\mathrm{v} & =\frac{k T}{P}\left(1+\frac{a_{2} \lambda^{3} P}{k T}+\ldotsight)=\frac{k T}{P}+a_{2} \lambda^{3}+\ldots
\end{aligned}
\]

It follows that

\[
T\left(\frac{\partial \mathrm{v}}{\partial T}ight)_{P}-\mathrm{v}=\left[T \frac{\partial\left(a_{2} \lambda^{3}ight)}{\partial T}-a_{2} \lambda^{3}ight]+\ldots
\]

and hence the quoted result for \((\partial T / \partial P)_{H}\).

With the given interparticle interaction, eqn. (10.2.3) gives

\[
\begin{aligned}
a_{2} \lambda^{3} & =-2 \pi\left[\int_{0}^{D}-1 \cdot r^{2} d r+\int_{D}^{r_{1}}\left(e^{u_{0} / k T}-1ight) r^{2} d right] \\
& =\frac{2 \pi}{3}\left[D^{3}-\left(r_{1}^{3}-D^{3}ight) e^{u_{0} / k T}ight],
\end{aligned}
\]

whence

\[
T \frac{\partial\left(a_{2} \lambda^{3}ight)}{\partial T}-a_{2} \lambda^{3}=\frac{2 \pi}{3}\left[\left(r_{1}^{3}-D^{3}ight)\left(1+\frac{u_{0}}{k T}ight) e^{u_{0} / k T}-r_{1}^{3}ight] .
\]

The desired result for \((\partial T / \partial P)_{H}\) now follows readily.

We note that the Joule-Thomson coefficient obtained here vanishes at a temperature \(T_{0}\), known as the temperature of inversion, given by the implicit relationship

\[
\left(1+\frac{u_{0}}{k T_{0}}ight) e^{u_{0} / k T_{0}}=\frac{r_{1}^{3}}{r_{1}^{3}-D^{3}}
\]

For \(T0\), which means that the Joule-Thomson expansion causes a cooling of the gas. For \(T>T_{0},(\partial T / \partial P)_{H}<0\); the expansion now causes a heating instead.