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:

  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 [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.


1 Answer 1


I am really puzzled by the lack of a clear answer to this question, even after reading the references you cited.

The real answer (your ref. [5] goes quite near to it, but it misses the point as well) is simply that it is not possible to obtain it if the only information is the van der Waals equation of state! This is quite well explained in Callen's book on thermodynamics. A strong indication comes from the fact that, even in the perfect gas, the equation of state alone does not distinguish among mono- di- or poly-atomic perfect gases.

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.

Also, arguments based on assuming that the limiting behavior at large volumes is the same as a perfect gas are flawed. Perfect gas is the limit of real gases at large volume (low pressure) and high temperature. Van der Waals equation of state is routinely used for describing fluids even close to their triple point. Extending so much the validity of the asymptotic regime is equivalent to postulate the thermal equation of state, and it is not proof.

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.


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