State equation in grand canonical ensemble

I have just found a way to prove that $ln Z = frac{PV}{kT}$ as follows: the grand canonical potential is defined, in thermodynamics, to be $ Xi = U - TS - mu N $. Using Euler relation (Callen, eq. (3.6), also known as Euler integrals here in Wikipedia) $ U = TS - PV + mu N $ then $$ Xi = -PV. $$

On the other hand the grand canonical potential can be obtained from the grand canonical partition function as $$ Xi = -kT ln Z .$$

Now it is trivial that $$ ln Z = frac{PV}{kT} $$

Let us consider a grand canonical ensemble of $M$ identical systems such that the total number of particles in the ensemble is $Mbar{N}$ and a total energy of $Mbar{E}$. Let $n_{r,s}$ be the number of systems having energy $E_{s}$ and number of particles $N_{r}$. Putting these statements mathematically, we have

$sum_{r,s} n_{r,s}= M$

$sum_{r,s} n_{r,s} N_{r}=Mbar{N}$

$sum_{r,s} n_{r,s} E_{s}=M bar{E}$

The above part lays the groundwork. The solutions begins here.

Let $P_{r,s}$ be the probability that a given member of the ensemble has an energy $E_{s}$ and number of particles $n_{r}$. From basic considerations of the grand canonical ensemble, we have

$P_{r,s}=frac{exp(-alpha N_{r}-beta E_{s})}{sum_{r,s}exp(-alpha N_{r}-beta E_{s})}$ .............. (1)

Now, the denominator is the grand canonical partition function $Z_{G}$

$Z_{G}=sum_{r,s}exp(-alpha N_{r}-beta E_{s})$ .............. (2)

Therefore,

$P_{r,s}=frac{exp(-alpha N_{r}-beta E_{s})}{Z_{G}}$ .............. (3)

Now, from the definition of entropy, we have

$S=-k_{B}<ln(P_{r,s})>$ .............. (4)

$=> S=-k_{B}sum_{r,s} P_{r,s} ln(P_{r,s}) $ .............. (5)

Now, we substitute the expression of $P_{r,s}$ obtained in equation (3) inside the $ln$. We leave the $P_{r,s}$ outside the $ln$ unchanged. (This makes the calculations easier as we will see shortly).

$S=-k_{B} sum_{r,s} P_{r,s} ln[frac{exp(-alpha N_{r}-beta E_{s})}{Z_G}]$ .............. (6)

$=>S=-k_{B} sum_{r,s} P_{r,s}[-alpha N_{r}-beta E_{s}-ln(Z_{G})]$ .............. (7)

$=>S=k_{B} alpha sum_{r,s} P_{r,s}N_{r}+k_{B} beta sum_{r,s} P_{r,s} E_{s}+k_{B}ln(Z_{G}) sum_{r,s} P_{r,s} $ .............. (8)

From normalization of probabilities,

$sum_{r,s} P_{r,s}=1$.

From the basic definition of mean, we have

$sum_{r,s} P_{r,s}N_{r}=bar{N}$

and

$sum_{r,s} P_{r,s}E_{s}=bar{E}$

Therefore, equation (8) can be written as

$S=k_{B} alpha bar{N}+k_{B} beta bar{E}+k_{B}ln(Z_{G}) $ .............. (9)

Now, using the first law and second law of thermodynamics,

$dbar{E}=TdS-PdV+mu dbar{N}$ .............. (10)

Since $S$, $V$, and $N$ are extensive quantities, using Euler's homogeneous function theorem, we get

$bar{E}=TS-PV+mubar{N}$ .............. (11)

Rearranging the terms of (11) and using the fact that $beta=frac{1}{k_{B}T}$, we get

$S=k_{B}betabar{E}+k_{B}beta PV - k_{B}beta mu bar{N} $ .............. (12)

Comparing (9) and (12), we get

$k_{B}beta PV =k_{B}ln(Z_{G})$

$=>beta PV=ln(Z_{G})$

$=>frac{PV}{k_{B}T}=ln(Z_{G})$ .............. (13)

[Hence proved]

Note : The key here is to bridge statistical mechanics, and thermodynamics. Entropy is often a useful tool for this purpose as we saw here.