# Ideal gas

- Page ID
- 5212

Recall the canonical partition function expression for the ideal gas:

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

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

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

Thus, the free energy is

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

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

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

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

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