Enthalpy of a Van der Waals gas

Question

I was thinking about how would one find the enthalpy in terms of $V,T$ but not on the usual ideal gas, but instead on a Van Der Waals gas with constant temperature, that is, a gas that satisfies: $$(P+\frac{an^2}{V^2})(V-nb)=nRT$$ with $a,b$ constants. We all know that the enthalpy is defined as $$H=U+PV\Rightarrow dH=dU+VdP+PdV$$ and since we are taking constant temperature (isothermic) to make our lifes easier, and $dU=TdS-PdV$ we get $$dH=dU+PdV+VdP=TdS+VdP$$ since and by definition of the entropy in terms of pressure and temperature $dS=C_p \frac{dT}{T}+\frac{nR}{P}dP=\frac{nR}{P}dP$ (since temperature is constant) we can get $$dH=T(\frac{nR}{P}dP)+VdP=(\frac{nRT}{P}+V)dP$$ but am I supposed to substitute one of the variables $n,V,T$ with the Van Der Waals? On any case I could get a very long expression and I don't think it is possible, how can I enhance this?

Answer 1

Your equation for dS is for an ideal gas. For a real gas, it reads: $$dS=\frac{C_P}{T}dT-\left(\frac{\partial V}{\partial T}\right)_PdP$$

Answer 2

Expand the enthalpy in $T$ and $V$:

$$\begin{align}\require{cancel}dH&=\left(\frac{\partial H}{\partial T}\right)_V\cancelto{0}{dT}+\left(\frac{\partial H}{\partial V}\right)_TdV\\&=\left[T\left(\frac{\partial S}{\partial V}\right)_T+V\left(\frac{\partial P}{\partial V}\right)_T\right]dV\\&=\left[T\left(\frac{\partial P}{\partial T}\right)_V-K\right]dV\\&=\left[-T\left(\frac{\partial V}{\partial T}\right)_P\left(\frac{\partial P}{\partial V}\right)_T-K\right]dV\\&=\left(\alpha T K-K\right)dV\\&=K(\alpha T-1)\,dV,\end{align}$$

where $K$ is the isothermal bulk modulus and $\alpha$ is the constant-pressure thermal expansion coefficient and where a Maxwell relation and the triple product rule were applied. This relation holds for all closed systems at constant temperature. (For the ideal gas, for which $\alpha=1/T$, it is identically zero.)