# Caloric equation of state for the Van der Waals gas

I'm currently trying to reproduce a specific derivation of the caloric equation of state for the Van der Waals gas, which I saw a couple of months ago in a Thermodynamics lecture. I'm well aware of the fact that there are already multiple derivations on this site and online (see , , , , ), but none of them are what I'm looking for.

The things that I remember from the derivation:

1. The inner energy was a function of $$T$$ and $$v$$.
2. We used the fact that $$s$$ is a state variable.
3. After using the first law we expanded all differentials in some variables.
4. With this we obtained $$u(T)=f(T)-a/v$$ and then proceeded to find $$f(T)$$ by using the fact that $$C_V = (\delta Q/\partial T)_V = (\partial u/\partial T)_V$$.

What I've tried so far: Using the first law we obtain $$Tds=du+pdv$$. We can express $$du$$ and $$dv$$ in terms of $$T,p$$, which leads to the equation $$ds = \frac{1}{T}\left(\frac{\partial u}{\partial p}\Bigg|_{T} + p \frac{\partial v}{\partial p}\Bigg|_{T}\right)dp+\frac{1}{T}\left(\frac{\partial u}{\partial T}\Bigg|_{p}+ p\frac{\partial v}{\partial T}\Bigg|_{p}\right)dT.$$ Using the fact that $$s$$ is a state variable we get the relation $$-\frac{\partial v}{\partial p}\Bigg|_{T} = \frac{1}{T}\left(\frac{\partial u}{\partial T}\Bigg|_{p}+p \frac{\partial v}{\partial T}\Bigg|_{p}\right).$$ This simplifies the first equation a bit but isn't really helpful since it's impossible to acutally take the derivative of $$v$$ with respect to any variable due to the form of the Van der Waals equation.

Since I couldn't figure anything out with this I tried the following as another approach: \begin{align} ds &= \frac{1}{T} \left(\frac{\partial u}{\partial v}\Bigg|_{T}dv + \frac{\partial u}{\partial T}\Bigg|_{v}dT\right) + \frac{p}{T}dv \\ & = C_V \frac{dT}{T}+ \frac{1}{T}\left(\frac{\partial u}{\partial v}\Bigg|_{T}+p \right)dv.\end{align} This approach seems to similar to the one from , but I'm 100% sure that we didn't use the Helmholtz equation as suggested by @juanrga.

TL;DR Could someone help me derive $$u(T)=C_VT-a/v$$ (caloric equaiton of state) for the Van der Waals gas without using any fancy equations, but rather fundamentals like the fact that $$s$$ is a state variable, etc.

From the formal point of view, the root of the problem is trivially: to have $$U(V,T)$$ or any equivalent information, one has to integrate the pressure. Knowing $$P(T,V,N)$$, we know the partial derivative of the Helmholtz free energy $$F(T,V,N)$$ with respect to $$V$$ at fixed $$T,N$$. The integration of $$P = - \left. \frac{\partial{F}}{\partial{V}} \right|_{T,N}$$ can provide $$F$$ only within an arbitrary function of $$N$$ and $$T$$. Which is equivalent to know almost nothing about the thermal equation of state.
The real answer is that knowledge of the equation of state $$P=P(T,V,N)$$ alone does not allow to obtain the caloric equation of state. Some additional information must be added, for example, information about the constant volume specific heat as a function of (T,V,N). In any case, it is an independent additional information.