0
$\begingroup$

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 $f(\mu,T)=0$?

$\endgroup$
3
  • 1
    $\begingroup$ 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. $\endgroup$
    – Christophe
    Commented Mar 16, 2018 at 16:44
  • $\begingroup$ Why don't you start from the definition of the grand canonical partition function and work it out? $\endgroup$ Commented Mar 16, 2018 at 17:21
  • $\begingroup$ See here: nyu.edu/classes/tuckerman/stat.mechII/lectures/lecture_5/… $\endgroup$
    – valerio
    Commented Mar 17, 2018 at 11:55

2 Answers 2

1
$\begingroup$

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 $$ \ln Z = \frac{PV}{kT} $$

$\endgroup$
1
$\begingroup$

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

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.