[Physics] Joule-Thomson effect of Van der Waals gas

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I'm supposed to calculate the inversion pressure $p_i$ of a Van der Waals gas. The state equation of the Van der Waals gas is: $$(p + \frac{a}{V^2})(V-b) = RT.$$ To get a hold of the inversion temperature, I differentiate this equation with respect to $T$, while keeping $p$ constant. This gives me:
$$\frac{(\frac{\partial V}{\partial T})_p \cdot (2ab – aV + pV^3)}{V^3} = R,$$ which with the condition for the inversion curve $T(\frac{\partial V}{\partial T})_p = V$ I get:
$$\frac{1}{T_i}(\frac{2ab}{V^2} – \frac{a}{V} + pV) = R,$$ which gives me an easy to solve equation for $T_i$. But to get $p_i$, I'd have to solve the equation for $V$ and put it back into the state equation. But solving this equation for $V$ gets really messy. That's why I'm thinking I went wrong somewhere.

Can you guys help me out here?

Best Answer

Solve Van der Waal's equation and approximate the $PV$ term by $RT$, and you will have your required equation. Although we are introducing an error, it will hardly affect the final result since there will be both $a$ and $b$.

For Van der Waal gas: $$\left(p+\frac{n^2q}{v^2}\right)\left(v-nb\right) = nRT$$ $$\Rightarrow pv+\frac{n^2q}{v}-pnb-\frac{n^3ab}{v^2}=nRT$$ $$\Rightarrow pv = nRT - \frac{n^2q}{v}+pnb+\frac{n^3ab}{v^L}$$ $$\Rightarrow V = \frac{nRT}{p} - \frac{n^2q}{pV}+nb+\frac{n^3ab\cdot P}{pV^2\cdot P}$$ $$\Rightarrow V = \frac{nRT}{P}-\frac{nq}{RT}+nb+\frac{nabP}{R^2T^2}$$ Note: Replacing $Pv$ by $nRT$ in the terms, the error becomes negligibly small and it hardly affects the final result. $$\because \left(\frac{\partial V}{\partial T}\right)_p = \frac{nR}{P}+\frac{nq}{RT^2}-\frac{2nabP}{R^2T^3}$$ We know, then, that $$M_{JT}=\left(\frac{\partial T}{\partial P}\right)_H$$ $$\Rightarrow M_{JT}=-\frac{1}{cp}\left(\frac{\partial H}{\partial P}\right)_T = -\frac{1}{cp}\left[V-\left(\frac{\partial V}{\partial T}\right)_P T\right]$$


Note from editor: some of the handwriting was a bit unclear, so here is the original picture posted for reference (I did, however, rotate the image so it is easier to read):

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