The change of Gibbs energy at constant temperature and species numbers, $\Delta G$, is given by an integral $\int_{p_1}^{p_2}V\,{\mathrm d}p$. For the ideal gas law, $p\,V=n\,RT$, this comes down to $$\int_{p_1}^{p_2}\frac{1}{p}\,{\mathrm d}p=\ln\frac{p_2}{p_1}.$$ We find this logarithm in many formulas in chemistry.

I tried and failed to find a general formula for $\Delta G$ for the Van der Waals gas (see this question on the physics board).

It makes me wonder: Is there even another case of a chemical situation, other than the ideal gas, where $\int_{p_1}^{p_2}V(p)\,{\mathrm d}p$ or $\int_{V_1}^{V_2}p(V)\,{\mathrm d}V$ is known analytically?

My questions are mostly motivated by the desire to understand the functional dependencies of the chemical potential $\mu(T)$, that is essentially given by the Gibbs energy. For the ideal gas and any constant $c$, we see that a state change from e.g. the pressure $c\,p_1$ to another pressure $c\,p_2$ doesn't actually affect the Gibbs energy. The constant factors out in $\frac{1}{p}\,{\mathrm d}p$, resp. $\ln\frac{p_2}{p_1}$. However, this is a mere feature of the gas law with $V\propto \frac{1}{p}$, i.e. it likely comes from the ideal gas law being a model of particles without interaction with each other.

  • $\begingroup$ Hi, I think this might be a rather lousy solution as it does not explicitly address the van der Waals state equation, but Atkins' Physical Chemistry talks a little about the fugacity of a real gas, how it can be derived from the compression factor $Z = pV_m/RT$, and how it can be used to find $\Delta G$ for the process you are asking about. Because I'm too lazy to type it out in a proper answer, I've just uploaded the relevant section from the 9th ed here, you can take a look if you want: drive.google.com/file/d/0B0GATVY1ly6cOXMtX1JubTlhdm8/… $\endgroup$ Aug 19, 2015 at 11:27
  • $\begingroup$ I tried the integration and I see what you mean, I couldn't make $V_m$ into a function of $p$ so I have no idea how to evaluate the integral $\int V_m\,\mathrm{d}p$. Not sure if what I wrote above is useful because it also involves an integral which is probably equally difficult to carry out. Hopefully someone else will have some idea. $\endgroup$ Aug 19, 2015 at 11:32
  • $\begingroup$ @orthocresol: No, I don't think so. The problem with this, as I understand it, is that the fugacity is exactly introduced to take into account that some gases $\Delta G$#s do diverge from the $pV=nRT$-model. That's ad hoc and begins with what I want to end up with. It's not like there the real volume $V_m$ in your link has a theoretical model, an equation of state that's different from $=\frac{nRT}{p}$ in a specified sense. I'm asking for a material theory, giving $V(p)$ that's different from $\propto \frac{1}{p}$ and still let's me compute $\int V(p)\,{\mathrm d}p$. $\endgroup$
    – Nikolaj-K
    Aug 19, 2015 at 11:35
  • $\begingroup$ Oh I see, you want something specific... Try the virial equation of state: $pV_m = RT(1 + B'p + C'p^2 + \cdots)$ I think that one is integrable. Or do you find that it's still too close to the ideal gas law? The most common form in which the virial expansion is written is $pV_m = RT(1 + B/V_m + C/V_m^2 + \cdots)$ but I suspect that's not going to work. $\endgroup$ Aug 19, 2015 at 11:40
  • $\begingroup$ @orthocresol: Before you post you think it's integrable, why not try it? This virial equation of state doesn't contain positive powers of the pressure, in any case. $\endgroup$
    – Nikolaj-K
    Aug 19, 2015 at 12:09

1 Answer 1


With regard to the Van der Waals equation, have you considered integrating by parts: $V\mathrm dp=\mathrm d(pV)-p\mathrm dV$. That can certainly be integrated analytically.

The answer will be in terms of $V$, but that’s not a big limitation.


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