Entropy and Pressure

Derive pressure and chemical potential of the ideal gas from entropy.
Remember that for a three dimensional system of $N$ billiards of mass $m$ in a volume $V$ , we have found $Ω_{q} \propto V^{N}$ and $Ω_{p} \propto m (2 m E)^{\frac{3 N - 2}{2}}$ .

Entropy of the Billiards Model

S (E, V, N) = N k_{B} \ln V - \frac{3}{2} N k_{B} \ln N + \frac{3}{2} N k_{B} \ln E + C (N)

Temperature:

T = - \frac{\partial S / \partial E}{\partial S / \partial μ} \Rightarrow \frac{1}{T} = \frac{\partial S}{\partial E} = \frac{3 N k_{B}}{2 E} \Rightarrow E = \frac{3}{2} N k_{B} T

Pressure

Moving a piston: $W = P Δ V$ , so $d E = - P d V$ .
Setting $d S = 0$ :

\frac{\partial S}{\partial E} (- P d V) + \frac{\partial S}{\partial V} d V = 0

P_{eq} = T \frac{\partial S}{\partial V}

For the Billiards model, $\frac{\partial S}{\partial V} = \frac{N k_{B}}{V}$ , so:

P V = N k_{B} T

Chemical Potential

Setting $d S = 0$ , $d N = 1$ :

d S = \frac{\partial S}{\partial E} (- μ) + \frac{\partial S}{\partial N} d N = 0

μ_{eq} = T \frac{\partial S}{\partial N}