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## Homework Statement

I have to find the mean Energy $<E>$ and pressure of a system of N fermions with spin 1/2. The energy per particle is

\begin{equation}

\varepsilon = \frac{p^2}{2m}.

\endu{equation}

## Homework Equations

The relevant equations are the degeneracy of the system:

\begin{equation}

g(\varepsilon)=\frac{4V (2m)^{3/2}\pi\varepsilon^{1/2}}{h^3}

\end{equation}

the mean occupation numbers of fermions

\begin{equation}

<n_\varepsilon>_{FD} = \frac{1}{e^{\beta(\varepsilon - \mu)}+1}

\end{equation}

## The Attempt at a Solution

I calculated the mean energy doing

\begin{equation*}

\begin{split}

<E> &= \int_{0}^{\infty} \varepsilon g(\varepsilon) <n_\varepsilon>_{FD} d\varepsilon\\

&= \frac{4V (2mkT)^{3/2}\pi}{h^3} kT \int_{0}^{\infty} \frac{x^{3/2}}{z^{-1}e^x+1}dx.

\end{split}

\end{equation*},

and finally obtained:

\begin{equation}

<E> = \frac{V}{\lambda^3} kT \sum_{l=1}^{\infty} (-1)^{l+1}\frac{z^l}{l^{5/2}}

\end{equation}

where z is the fugacity.

\begin{equation*}

z=\exp{\beta \mu}

\end{equation*}

Then I calculated the pressure making use of the grand partition function

\begin{equation}

\mathcal{Z} = \prod_\varepsilon 1+e^{-\beta(\varepsilon - \mu)}

\end{equation}

and doing this

\begin{equation*}

\begin{split}

\frac{PV}{kT} &= \ln \mathcal{Z} = \sum_{\varepsilon} \ln \left(1+e^{-\beta(\varepsilon - \mu)}\right)\\

&= \frac{4V (2m)^{3/2}\pi}{h^3} \int_{0}^{\infty} \varepsilon^{1/2} \ln \left(1+e^{-\beta(\varepsilon - \mu)}\right) d\varepsilon\\

&= \frac{4V (2m)^{3/2}\pi}{h^3} \left[\frac{2}{3} \varepsilon^{3/2} \ln \left(1+e^{-\beta(\varepsilon - \mu)}\right)\Big|_0^{\infty} - \frac{2\beta}{3} \int_{0}^{\infty} \frac{\varepsilon^{3/2}}{e^{\beta(\varepsilon - \mu)}+1} d\varepsilon \right]\\

&= \frac{8V (2m)^{3/2}\pi}{3h^3} \beta \int_{0}^{\infty} \frac{\varepsilon^{3/2}}{e^{\beta(\varepsilon - \mu)}+1} d\varepsilon

\end{split}

\end{equation*}

I finally obtained that the pressure is

\begin{equation}

P = \frac{2}{3} \frac{1}{\lambda^3} kT \sum_{l=1}^{\infty} (-1)^{l+1}\frac{z^l}{l^{5/2}}

\end{equation}

Now, the problem is that, if I want to obtain the pressure at room temperature, I calculate using the canonical ensemble the chemical potential, and approximate the series

\begin{equation}

\mu = \frac{1}{\beta} \ln \left(\frac{N \lambda^3}{V}\right),

\end{equation}

\begin{equation}

\sum_{l=1}^{\infty} (-1)^{l+1}\frac{z^l}{l^{5/2}} \approx z

\end{equation}

I get that the pressure is:

\begin{equation}

P = \frac {2} {3} \frac{NkT}{V}

\end{equation}

That result really bothers me because of that factor of 2/3. I did the same process to obtain the pressure and energy for bosons and obtained the correct result.

I thought that the factor may arise because I am considering electrons and the spin may do a contibution.

If anyone could help me with this it would be very appreciated!

Thank you