# Enthalpy of a Van der Waals gas continuation

In my previous question Enthalpy of a Van der Waals gas, I got the expression of the enthalpy generalised, but I am still having issues finishing it since it might have some quite hard calculus...

For my Van der Waals gas with constant temperature we got to the following expression by the already mentioned question: $$dH=TdS+VdP=T(-\frac{\partial P}{\partial V}dP)+VdP=(V-\frac{\partial P}{\partial V}T)dP$$ while the Van der Waals equation is $$(P+\frac{an^2}{V^2})(V-nb)=P(V-nb)+\frac{an^2}{V}-\frac{an^3 b}{V^2}=nRT$$ and solving for $$P$$ to get the $$\frac{\partial P}{\partial V}$$ we get $$P=\frac{nRT-\frac{an^2 V-an^3 b}{V^2}}{V-nb}=\frac{nRTV^2-an^3 V-an^3 b}{(V-nb)V^2}$$ and so we can find the partial derivative term $$\frac{\partial P}{\partial V}=\frac{(2nRTV-ab^3)(V^2 (V-nb))-(nRTV^2-an^3 V-an^3 b)(3V^2-2nbV)}{(V^3-nbV^2)^2}$$ which seems already extremely large, am I on the right path or is there an easier way?

• It would be a good idea to indicate explicitly which variables are constant in every partial derivative. Nov 9, 2023 at 23:21
• One way that is for sure easier is to find the internal energy U and then enthalpy is just H = U + PV Nov 10, 2023 at 1:44
• @ErJio how would that be done though? I'm looking for $\Delta H$ not $H$, and I wouldn't know how to find the $\Delta U$ Nov 10, 2023 at 9:16
• @Ulshy $P = \frac{nRT}{V-bn} - \frac{an^2}{V^2}$ so $\frac{\partial P}{\partial T}T - P = (\frac{nR}{V-bn})T - (\frac{nRT}{V-bn} - \frac{an^2}{V^2}) = \frac{an^2}{V^2}$ which is easy to integrate Nov 11, 2023 at 9:23
• @Ulshy No worries. First, we both have the same expression for $P$ however you combined it into a single term. You can derive my form from $(P+\frac{an^2}{V^2})(V-bn) = nRT$ by dividing both sides by $(V-bn)$ then subtracting both sides by $\frac{an^2}{V^2}$. Second, your expression for $\frac{\partial P}{\partial V}$ will not simplify into $\frac{an^2}{V^2}$. Since you've expressed $\mathrm dH$ in terms of $\mathrm dP$ you have to integrate by $P$ which involves expressing $V$ as a function of $P$ (not easy). This is why I suggested using internal energy, since it involves simple expressions. Nov 11, 2023 at 12:45

Your equation is wrong. It should read $$dH=\left(V-T\left(\frac{\partial V}{\partial T}\right)_P\right)dP$$ You may have to integrate this numerically because of the non-linearity with of the VDW equation with respect to V.
If you know the initial and final states of the gas, it is easier to get the enthalpy change for a VDW gas by working with the internal energy change $$\Delta U$$ than $$\Delta H$$. This is because the internal energy change for a VDW gas can be expressed analytically: $$\Delta U=nC_v\Delta T+nR\ln{\left(\frac{V_2-nb}{V_1-nb}\right)}$$For a VDW gas, Cv is independent of V. From the definition of $$\Delta H$$, $$\Delta H=\Delta U+\Delta (PV)$$
• Sorry for responding late, I didn't see this. The internal energy seems different from the one in my book but on any case I don't know how to express $\Delta(PV)=\Delta P V+\Delta V P$, I'm guessing I can write $P=P(V)$ in terms of the volume and integrate and likewise with $V=V(P)$, or else I don't know how Nov 14, 2023 at 7:22
• The initial and final volume, temperature, $a$ and $b$, and the mass, so yeah we can find the initial and final pressure with the Van der Waals equation, isn't it? Nov 14, 2023 at 17:44
• $\Delta (PV)=P_2V_2-P_1V_1$, and you know all four of these. Nov 15, 2023 at 11:47