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In the book "Physical Chemistry An Advanced Treatsie VIIIA" Eyring, Henderson and Jost use the thermodynamic relation:

\begin{equation} \left(\frac{\partial\rho}{\partial\mu}\right)=\rho\left(\frac{\partial\rho}{\partial P}\right) \end{equation} $\rho$ is the number density ($N/V$), $\mu$ is the chemical potential and $P$ is the pressure. This relation is used in the grand canonical ensemble. I am having a hard time in deriving this relation. I have tried deriving the expression using Maxwell relations derived from the Landau grand potential, but I have not been successful so far. Any help in deriving this expression would be greatly appreciated.

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Let me try: let $\Omega(\mu,V,T)$ be the grand-canonical potential, then the thermodynamic quantities conjugated to $\mu$ and $V$ (which are particle number $N$ and pressure $P$ respectively), are given by the relations $$ N = \rho V = - \frac{\partial \Omega}{\partial \mu}; \;\;\;\; P = - \frac{\partial \Omega}{\partial V}. $$ Now we can write in a smart way $\partial \rho/\partial \mu$: $$ \frac{\partial \rho}{\partial \mu} = \frac{\partial \rho}{\partial P}\frac{\partial P}{\partial \mu}. $$ Now the proof is complete once you are able to show that $\partial P / \partial \mu = \rho$, but this is very simple using the equations for $N$ and $P$ above: $$ \frac{\partial P}{\partial \mu} = \frac{\partial }{\partial \mu} \left( - \frac{\partial \Omega}{\partial V} \right) = \frac{\partial }{\partial V} \left( - \frac{\partial \Omega}{\partial \mu} \right) = \frac{\partial }{\partial V} \left( \rho V \right) = \rho; $$ where I have simply swapped the derivatives.

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