# Is there a thermodynamical function $\Xi$ such that $d \Xi = V d P/H$?

Question

Is there a thermodynamical function $\Xi$ such that generally $$d \Xi = \frac{V d P}{H}$$ where $V$ is the volume, $P$ the pressure, and H the enthalpy $H=U + PV$? If such a function generally does not exist, which conditions must the thermodynamical system (and its state equation) fulfill for $\Xi$ to exist?

We may assume the usual thermodynamical identities such as $dU=-PdV + T dS$.

Attempt at solution and comments

A naive effort to solve this is to simply assume that we can express $P=P(V,H)$ and then integrate $$\frac{\partial \Xi}{\partial H}|_{V=const.} = \frac{V}{H} \frac{\partial P}{\partial H}|_{V=const.}$$ $$\frac{\partial \Xi}{\partial V}|_{H=const.} = \frac{V}{H} \frac{\partial P}{\partial V}|_{H=const.}$$ However, this leads to the integrability condition $$\frac{\partial P}{\partial H}|_{V=const.}=-\frac{V}{H} \frac{\partial P}{\partial V}|_{H=const.}$$ and a nice formal solution for $\Xi$. However, does every physically reasonable thermodynamical system fulfill the integrability condition above?

Solution for ideal gas

The trivial example of an ideal gas with $K$ degrees of freedom allows for and expression of $P$ as $$P=\frac{2}{K+2} \frac{H}{V}$$ which can be easily seen to fulfill the integrability conditions above and we obtain $$\Xi = \frac{1}{K+2}\log(\frac{H}{H_0} \frac{V_0}{V})$$ where $H_0, V_0$ are integration constants.

Since the pressure $p$ is intensive and the volume $V$ and enthalpy $H$ are extensive variables the function $p=p(V,H)$ is homogeneous degree $0$ so you always have $$H \frac{\partial p}{\partial H} + V \frac{\partial p}{\partial V} = 0$$