# Ideal gas

Recall the canonical partition function expression for the ideal gas:

$Q(N,V,T) = {1 \over N!} \left[{V \over h^3}\left({2\pi m \over \beta}\right)^{3/2}\right]^{N}$

Define the thermal wavelength $$\lambda (\beta)$$ as $\lambda(\beta) = \left({\beta h^2 \over 2 \pi m}\right)^{1/2}$ which has a quantum mechanical meaning as the width of the free particle distribution function. Here it serves as a useful parameter, since the canonical partition can be expressed as

$Q(N,V,T) = {1 \over N!}\left({V \over \lambda^3}\right)^N$

The grand canonical partition function follows directly from $$Q(N,V,T)$$:

${\cal Z}(\zeta,V,T) = \sum_{N=0}^{\infty}{1 \over N!}\left({V\zeta \over \lambda^3}\right)^N = e^{V\zeta/\lambda^3}$

Thus, the free energy is

${PV \over kT} = \ln {\cal Z} = {V \zeta \over \lambda^3}$

In order to obtain the equation of state, we first compute the average particle number $$\langle N \rangle$$

$\langle N \rangle = \zeta {\partial \over \partial \zeta}\ln {\cal Z}= {V \zeta \over \lambda^3}$

Thus, eliminating $$\zeta$$ in favor of $$\langle N \rangle$$ in the equation of state gives

$PV = \langle N \rangle kT$

as expected. Similarly, the average energy is given by

$E = -\left({\partial \ln {\cal Z}\over \partial \beta}\right )_{\zeta V} = {3V\zeta \over \lambda^4}{\partial \lambda \over \partial \beta} ={3 \over 2}\langle N \rangle kT$

where the fugacity has been eliminated in favor of the average particle number. Finally, the entropy

$S(\mu,V,T) = k\ln {\cal Z}(\mu,V,T) - k\beta\left({\partial\ln {\cal Z} (\mu, V, T) \over \partial \beta} \right)_{\mu, V} = {5 \over 2}\langle N \rangle k + \langle N \rangle k\ln \left[{V\lambda^3 \over \langle N \rangle} \right]$

which is the Sackur-Tetrode equation derived in the context of the canonical and microcanonical ensembles.