Deriving Gas Equation of State

The equation of state can be found from the partition function using the thermodynamic relation between pressure ($P$) and the partition function ($Q$). The relevant formula is:

$P = k_B T \left( \frac{\partial \ln Q}{\partial V} \right)_{N,T}$

where $k_B$ is the Boltzmann constant, $T$ is the temperature, $V$ is the volume, and $N$ is the number of particles.

Partition Function Analysis

The given approximate partition function is:

$Q(N,V,T) = \frac{1}{N!} \left( \frac{2\pi mk_B T}{h^2} \right)^{3N/2} (V-Nb)^N$

To use the formula for pressure, we first need $\ln Q$:

$\ln Q = \ln \left( \frac{1}{N!} \right) + \frac{3N}{2} \ln \left( \frac{2\pi mk_B T}{h^2} \right) + N \ln(V-Nb)$

Calculating Pressure

Now, we differentiate $\ln Q$ with respect to $V$, keeping $N$ and $T$ constant:

• The first two terms are independent of $V$, so their derivative is zero.
• The derivative of the third term, $N \ln(V-Nb)$, with respect to $V$ is:
$ \left( \frac{\partial [N \ln(V-Nb)]}{\partial V} \right)_{N,T} = N \cdot \frac{1}{V-Nb} \cdot \frac{\partial (V-Nb)}{\partial V} $ $ = N \cdot \frac{1}{V-Nb} \cdot 1 = \frac{N}{V-Nb} $

So, the partial derivative is:

$ \left( \frac{\partial \ln Q}{\partial V} \right)_{N,T} = \frac{N}{V-Nb} $

Final Equation of State

Substitute this result back into the pressure formula:

$ P = k_B T \left( \frac{N}{V-Nb} \right) $

Rearranging this equation gives the equation of state:

$ P(V-Nb) = Nk_B T $

This matches the first option.