I was searching for an expression for the bulk modulus $B$ for a gas that obeys the van der Waals equation ("van der Waals gas" from now on). Since for an ideal gas we have that $B=\gamma P$ (where $\gamma$ is the adiabatic index and $P$ the pressure), I thought that I could arrive to something similar for a van der Waals gas.
Something I thought was just starting from the definition (considering that the temperature is constant): $$B=-V\frac{\partial P}{\partial V}.$$
Then I solve for $P$ in the van der Waals equation to get $$P=\frac{nRT}{V-nb}-a\frac{n^2}{V^2}$$ and just substitute it and calculate its derivative with respect to $V$:
$$\begin{align*}B &= -V\frac{\partial }{\partial V}\left(\frac{nRT}{V-nb}-a\frac{n^2}{V^2}\right) \\[2mm] B &=\frac{VnRT}{(V-bn)^2}-\frac{2an^2}{V^2}.\end{align*}$$
But I don't like this expression; it's not compact and it depends on a lot of variables. Also, this is not the method you use to get to $B=\gamma P$ for an ideal gas. In the deduction of $B=\gamma P$ that I saw (this one), the first law of thermodynamics and the expression for the internal energy of an ideal gas are used. So my question is: is there a way to use the first law and the expression for the internal energy of a van der Waals gas ($U=nc_VT-\frac{an^2}{V}$, deduced here) to get to a simpler expression for $B$? Or is there any other way to get to a simpler expression for $B$?
Edit: I just realized that $B$ is different depending on whether the process is isothermal or adiabatic. The expression I found for $B$ is for an isothermal process and I want an expression for an adiabatic process (also, if you can find a simpler expression for an isothermal process that would be great).
