State equation in grand canonical ensemble

My teacher told us that $$\ln Z = \frac{PV}{kT}$$ is the equation of state for an ideal gas, being $Z$ the grand canonical partition function and $k$ the Boltzmann constant. Where does this formula come from? (we then used this formula for studying Bose-Einstein and Fermi-Dirac systems).

I have tried the following: in the grand canonical ensemble I know that $\Xi = -kT \ln Z$ is the grand canonical potential. Also, from thermodynamics, I know that the grand canonical potential differential is $d\Xi = -S~dT - P~dV - N~d\mu$ so, integrating, $\Xi = -PV + f(\mu,T)$ where $f$ is some unknown function. Equating this two facts about $\Xi$ gives $$\ln Z = \frac{PV - f(\mu,T)}{kT}$$ but how can be shown that $f(\mu,T)=0$?

• You can argue that the grand canonical potential $\Xi$ is extensive. It is indeed the case for $PV/k_BT$ while $\mu$ and $T$, and therefore $f(\mu,T)$, are intensive variables. Commented Mar 16, 2018 at 16:44
• Why don't you start from the definition of the grand canonical partition function and work it out? Commented Mar 16, 2018 at 17:21
• Commented Mar 17, 2018 at 11:55

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 $$M\bar{N}$$ and a total energy of $$M\bar{E}$$. 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

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

$$\sum_{r,s} n_{r,s} N_{r}=M\bar{N}$$

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

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

$$S=-k_{B}$$ .............. (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,

$$d\bar{E}=TdS-PdV+\mu d\bar{N}$$ .............. (10)

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

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

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

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