Fermi/Bose gases

Question

Knowing that $$\Omega = -\frac{2}{3}\frac{gVm^{\frac{3}{2}}}{2^{\frac{1}{2}}\pi^{2}h^{3}} \int_0^{\infty}\frac{z^{\frac{3}{2}}dz}{z^{\frac{z-\mu}{T}}\pm1} $$ is the thermodynamic potential for Bose Fermi particle gas and $$ \Omega = -\frac{2}{3} E,$$ where E is total energy of gas. The final equation is $$PV = \frac{2}{3}E,$$where we know that $$\Omega = -PV$$ My question is how to prove that $\Omega=-PV$. I found one suggestion : to use Euler's theorem for homogeneous functions.

Answer 1

Use this fact: From $$d\Omega = -SdT-PdV-Nd\mu$$ we know that $$-P=\left( \frac{\partial \Omega}{\partial V} \right)_{T,\ \mu}$$ You don't need Euler's theorem.