How to redefine internal pressure in terms of κ and α?

The formula for internal pressure of a system is:
$$\pi\_T=T\left(\frac{\partial p}{\partial T}\right)\_{\!V}-p$$


My professor said we can redefine this equation in terms of either the


1. thermal compressibility, $\kappa$
2. expansion coefficient, $\alpha$


How do we do this?


I was able to derive for a van der Waals gas:


1. $\Large\kappa = \frac{V-nb}{pV-\frac{an^2}{v}+\frac{2abn^3}{v^2}}$
2. $\Large \alpha= \frac{nR}{PV-\frac{an^2}{v}+\frac{2abn^3}{v^2}}$




$\def\d{\mathrm{d}}$I thought I might include the slightly longer approach because the non-standard treatment of partial differentials can be confusing.




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Begin with the differential of volume assuming a constant number of particles.


$$\d V= \left(\frac{\partial V}{\partial T}\right)\_P\d T + \left(\frac{\partial V}{\partial P}\right)\_T\d P \tag1$$


Similarly, assuming that $P = P(T,V)$ gives us


$$\d P= \left(\frac{\partial P}{\partial T}\right)\_V\d T + \left(\frac{\partial P}{\partial V}\right)\_T\d V.\tag2$$


Substitute $(2)$ into $(1)$ and rearrange the differentials:


$$\left[1-\left(\frac{\partial V}{\partial P}\right)\_T\left(\frac{\partial P}{\partial V}\right)\_T\right]\d V = \left[\left(\frac{\partial V}{\partial T}\right)\_P+\left(\frac{\partial V}{\partial P}\right)\_T\left(\frac{\partial P}{\partial T}\right)\_V\right]\d T.$$


This equality will hold in general if and only if the corresponding coefficients of the differentials are both equal to zero. We are interested in the coefficient of $\d T$, namely


$$\left(\frac{\partial V}{\partial T}\right)\_P = -\left(\frac{\partial V}{\partial P}\right)\_T\left(\frac{\partial P}{\partial T}\right)\_V.$$


Assuming the partial derivative of $V$ with repect to $P$ is non-zero we can solve for $\left(\frac{\partial P}{\partial T}\right)\_V$.


$$\left(\frac{\partial P}{\partial T}\right)\_V = -\frac{\left(\frac{\partial V}{\partial T}\right)\_P}{\left(\frac{\partial V}{\partial P}\right)\_T}$$
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This result is identical to what [Chester Miller](https://chemistry.stackexchange.com/users/15095/chester-miller) reached.



> 
>  **Hint:** multiply by $1$ and apply definitions of expansion coefficients to finish the derivation.
> 
> 
> 


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$$dV=\left(\frac{\partial V}{\partial T}\right)\_pdT+\left(\frac{\partial V}{\partial p}\right)\_Tdp$$So at constant volume dV = 0 and $$\left(\frac{\partial p}{\partial T}\right)\_V=-\frac{\left(\frac{\partial V}{\partial T}\right)\_p}{\left(\frac{\partial V}{\partial p}\right)\_T}$$


The rest is straightforward.


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