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 [1], [2], [3], [4], [5]), but none of them are what I'm looking for.
The things that I remember from the derivation:
- The inner energy was a function of $T$ and $v$.
- We used the fact that $s$ is a state variable.
- After using the first law we expanded all differentials in some variables.
- 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 [1], 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.