Partition function of a gas

[rayman123]

Homework Statement

The partition function of a given gas can be written
$$z=(\frac{V-Nb}{N})^{N}(\frac{mk_{B}T}{2\pi\hbar^2})^{\frac{3N}{2}}e^{{\frac{N^2a^2}{Vk_{B}T}}$$

Homework Equations

$$lnz= Nln(\frac{V-Nb}{N})+\frac{3N}{2}ln(\frac{mk_{B}T}{2\pi\hbar^2})+\frac{N^2a^2}{Vk_{B}T}}$$

The Attempt at a Solution

mean energy
$$U=-\frac{\partial lnz}{\partial \beta}= \frac{3N}{2}{\frac{1}{\frac{mk_{B}T}{2\pi\hbar^2}}\cdot\frac{mk_{B}T}{2\pi\hbar^2}$$
taking $$\beta=\frac{1}{k_{B}T}$$
I get
$$\frac{3}{2}Nk_{B}T$$
is this correct?
it seems like the minus from the formula for the mean energy is missing...it the whole calculation correct?

Homework Statement

calculate the pressure of the gas

now i have
$$\frac{1}{\beta}\frac{\partial lnz}{\partial V}= -\frac{N_{b}}{V-N_{B}}$$


can someone show me how to calculate it correctly?

 

[BigJDubs]

Homework Equationslnz= Nln(\frac{V-Nb}{N})+\frac{3N}{2}ln(\frac{mk_{B}T}{2\pi\hbar^2})+\frac{N^2a^2}{Vk_{B}T}}The Attempt at a SolutionPressure of the gas is given byP=-\frac{1}{\beta}\frac{\partial lnz}{\partial V}Substituting the expression for lnZ in the above equation we get,P=-\frac{N}{V-Nb}\left[ln\left(\frac{V-Nb}{N}\right)+\frac{3}{2}ln\left(\frac{mk_{B}T}{2\pi\hbar^2}\right)+\frac{N^2a^2}{Vk_{B}T}\right]Therefore, the pressure of the gas is given by,P=-\frac{N}{V-Nb}\left[ln\left(\frac{V-Nb}{N}\right)+\frac{3}{2}ln\left(\frac{mk_{B}T}{2\pi\hbar^2}\right)+\frac{N^2a^2}{Vk_{B}T}\right]