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In reference [1], page 4173, the following relation is provided: $$Z=P V /(N k T)=\mu /(k T)-A /(N k T)$$ where $Z$ is compressibility and $A$ is the Helmholtz free energy. This means pressure is: $$P =\mu \frac{N}{V}-\frac{A}{V}$$ My question is, how can I derive this from definition of pressure which is $P=-\frac{\partial A}{\partial V}$ ?

[1] Gil-Villegas, Alejandro, et al. "Statistical associating fluid theory for chain molecules with attractive potentials of variable range." The Journal of chemical physics 106.10 (1997): 4168-4186.

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A better starting point is $A\equiv U-TS=-PV+\mu N$, the definition of the Helmholtz free energy. This can be immediately rearranged to provide

$$P=\frac{\mu N}{V}-\frac{A}{V},$$ QED.

In addition, from the fundamental relation $dU=T\,dS-P\,dV+\mu\,dN$, we have $dA=-S\,dT-P\,dV+\mu\,dN$, with $$P=-\left(\frac{\partial A}{\partial V}\right)_{T,N}.$$

Make sense? Or are you specifically seeking a way to show the former starting from the latter?

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  • $\begingroup$ To be more precise, your starting point is the definition of the Helmholtz free energy and the Euler equation resulting from being a homogeneous function of degree 1 of its extensive variables. $\endgroup$
    – GiorgioP-DoomsdayClockIsAt-90
    Commented Oct 28, 2022 at 9:39
  • $\begingroup$ Right. We recognize that the first equation consists of conjugate pairs, each containing an intensive and an extensive variable. The energy scales up with the extensive variables, implying the fundamental relation (and also the Gibbs–Duhem relation, for consistency). $\endgroup$
    – Chemomechanics
    Commented Oct 28, 2022 at 16:35
  • $\begingroup$ @Chemomechanics The first approach clarified it for me. Thanks $\endgroup$
    – Ali
    Commented Oct 29, 2022 at 20:57

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