How Do You Derive Pressure from the Grand Partition Function?

[Rawrzz]

Can someone take a look at picture and show me how to derive the pressure from the grand partition function ?

 

[unscientific]

Grand potential $\Phi_G = -PV$
Grand potential $Z = e^{-\beta \Phi_G}$

Thus, $PV = kT ln Z$

 

[Rawrzz]

Unscietific,

I can't use the fact that the grand potential equals -PV because my goal is to prove that the grand potential in terms of the partition function is equivalent to (-PV).

I know that those sums on the left side must equal (PV/KT) but I don't know the details of how to show it.

 

[unscientific]

Unscietific,

I can't use the fact that the grand potential equals -PV because my goal is to prove that the grand potential in terms of the partition function is equivalent to (-PV).

I know that those sums on the left side must equal (PV/KT) but I don't know the details of how to show it.

The grand partition function is sum of all states $Z_G = \sum_i e^{\beta(\mu N_i - E_i)}$ and Probability is i-th state over all possible states: $P_i = \frac{e^{\beta(\mu N_i - E_i)}}{Z_G}$.
$$S = -k\sum_i P_i ln P_i = \frac{U - \mu N + kT ln (Z_G)}{T}$$
Rearranging,
$$-kT ln (Z_G) = \Phi_G = U - TS - \mu N = F - \mu N$$

Now we must find $F$ and $\mu$.

Starting with partition function of an ideal gas: $Z_N = \frac{1}{N!}(\frac{V}{\lambda_{th}^3})^N$, what is $F$?

Using the below relation, how do you find $\mu$?

$$dF = -pdV - SdT + \mu dN$$
Putting these together, can you find $\Phi_G$?