Brian Bi

What is the pressure needed to contain a 1 molar electron gas at room temperature?

I'm not sure how to work out the thermodynamics, but I can do the calculation at low temperatures (i.e. where the system will be found very close to its ground state).

For ease of calculation, we will take the container to be spherical. Say the container has volume \(V\) and there are \(N\) electrons inside.

From basic electrostatics we know that the ground state of this system will have all the electrons at the surface. A straightforward calculation gives the surface area of the container as \(A = 4\pi\left(\frac{3V}{4\pi}\right)^{2/3}\). So we get a surface charge density of \[\sigma = Ne/A = \frac{Ne}{4\pi} \left(\frac{3V}{4\pi}\right)^{-2/3}.\]

A surface with net charge density feels an electrostatic pressure of \(\frac{\epsilon_0}{2}E^2\) where \(E\) is the magnitude of the electric field just outside the surface. By Gauss's law and the vanishing of the electric field in the container's interior this is \[P = \frac{\epsilon_0}{2}E^2 = \frac{\epsilon_0}{2}(\sigma/\epsilon_0)^2 = \frac{\sigma^2}{2\epsilon_0}\]

Putting it all together \[P = \frac{N^2 e^2}{32\pi^2\epsilon_0} \left(\frac{3V}{4\pi}\right)^{-4/3}\] Putting \(N = 6.022 \times 10^{23}\) and \(V\) = 1 dm3 we obtain the result

\(P \sim 2.25 \times 10^{23}\) pascals

This is about 10 million times greater than the pressure at the core of the sun.

A curious feature is that the pressure is not related to the particle density \(N/V\) in any straightforward way. This is because the electrostatic repulsion between electrons is a long-ranged force and the expressions for the virial coefficients diverge.