Joule-Thomson inversion curve of a Dieterici gas I'm trying to show that the equation of the Joule-Thomson inversion curve for 1 mole of a gas obeying the reduced Dieterici equation of state is
$$\tilde{P}=(8-\tilde{T})e^{\frac{5}{2}-\frac{4}{\tilde{T}}}$$
Where the tilde indicates a property divided by its critical point value (eg. $\tilde{P}=P/P_{c}$).
From Dieterici's equation of state, $$P(V-b)=RTe^{\frac{-a}{RTV}}$$
I have already found the critical point values to be $P_{c}=\frac{a}{4b^2}e^{-2}$ , $V_{c}=2b$ , and $T_{c}=\frac{a}{4Rb}$ . These reduce the equation of state to
$$\tilde{P}(2\tilde{V}-1)=\tilde{T}e^{2(1-\frac{1}{\tilde{T}\tilde{V}})}$$
I know the equation of the inversion curve is given when 
$$\bigg(\frac{\partial{V}}{\partial{T}}\bigg)_{P}=\frac{V}{T}$$
from solving the Joule-Thomson coefficient equation when the coefficient is equal to $0$. However I'm struggling to find an equivalent differential equation using the reduced properties, which  I can then use to show the equation in the question.
I hope this makes sense. Please could you help with this?
 A: Notice that once you fix $P$, $\tilde{P}$ is automatically fixed as well. Aditionally, since the critical values are constants it follows from chain's rule that
\begin{equation}
\left(\frac{\partial V}{\partial T}\right)_{P} = \frac{V_{c}}{T_{c}}\left(\frac{\partial\tilde{V}}{\partial\tilde{T}}\right)_{\tilde{P}}\ .
\end{equation}
By inserting this expression into the condition for the critical curve, it follows that
\begin{equation}
\left(\frac{\partial\tilde{V}}{\partial\tilde{T}}\right)_{\tilde{P}} = \frac{\tilde{V}}{\tilde{T}}\ .
\end{equation}
In order to show that this expression does lead to the correct answer, let me give additional details towards the solution. For simplicity, let me denote the argument in your exponential by $A$. In the Equation of State (EoS), fix $\tilde{P}$ and carry out the derivations with respect to $\tilde{T}$. The result should be
\begin{equation}
2\tilde{P}\left(\frac{\partial\tilde{V}}{\partial\tilde{T}}\right)_{\tilde{P}} = e^{A}+2e^{A}\left[\frac{1}{\tilde{V}^{2}}\left(\frac{\partial\tilde{V}}{\partial\tilde{T}}\right)_{\tilde{P}}+\frac{1}{\tilde{V}\tilde{T}}\right]\ .
\end{equation}
Rearranging terms,
\begin{equation}
\left(\frac{\partial\tilde{V}}{\partial\tilde{T}}\right)_{\tilde{P}}\left(2\tilde{P}\tilde{V}^{2}-2e^{A}\right) = e^{A}\frac{\tilde{V}}{\tilde{T}}\left(\tilde{V}\tilde{T}+2\right)\ .
\end{equation}
Since we are interested in the critical curve, the partial derivative on the left hand side is equal to $\tilde{V}\tilde{T}^{-1}$, so
\begin{equation}
\left(2\tilde{P}\tilde{V}^{2}-2e^{A}\right) = e^{A}\left(\tilde{V}\tilde{T}+2\right)\ .
\end{equation}
Here we can rearrange terms again and then insert the EoS in order to deduce an explicit relation between $\tilde{V}$ and $\tilde{T}$:
\begin{equation}
2\tilde{P}\tilde{V}^{2} = e^{A}(4+\tilde{V}\tilde{T})\ ,
\end{equation}
\begin{equation}
2\tilde{V}^{2} = (2\tilde{V}-1)\left(\tilde{V}+\frac{4}{\tilde{T}}\right)\ ,
\end{equation}
\begin{equation}
\boxed{\tilde{V} = \frac{4}{8-\tilde{T}}}\ .
\end{equation}
By employing this relation and with some amount of algebra, it can be shown that
\begin{align}
\frac{\tilde{T}}{2\tilde{V}-1}& = 8-\tilde{T}\\
2\left(1-\frac{1}{\tilde{T}\tilde{V}}\right)& = \frac{5}{2}-\frac{4}{\tilde{T}}\ .
\end{align}
Inserting these results into the EoS must lead you to the $\tilde{P}-\tilde{T}$ relation that you wrote and which describes the inversion curve.
