# Pressure of Fermion Gas

• LizardWizard
P = \frac{2}{3} \frac{\int_0^{\epsilon_F} d\epsilon g(\epsilon) \epsilon}{V} = \frac{2}{3} \frac{\int_0^{\epsilon_F} d\epsilon g(\epsilon) \epsilon}{V} = \frac{2}{3} \frac{\int_0^{\infty} d\epsilon g(\epsilon) \epsilon}{V} = \frac{2}{3} \frac{\int_0^{\infty} d\epsilon g(\epsilon) \epsilon}{V}##Finally, we can use the expression for g(epsilon) given in the problem

#### LizardWizard

Homework Statement
For a gas of N fermions of mass m confined in a volume V at a temperature ##T<E_F/kB##, consider the quantity ##<n_p>/V## as you would a classical distribution f(p,q) in the system phase space. Show that the impulse transfer of the elastic collisions of the particles with the wall of the container with this f leads to the correct expression for the pressure of the gas.

The attempt at a solution
Now I might be wrong in assuming this but can I treat ##<n_p>/V## as the density of states? If so, for low temperatures we have
##g(\epsilon)=3N/2F (\epsilon/\epsilon_F)##
##U = \int_0^{\epsilon_F} d \epsilon g(\epsilon) \epsilon = 3N \epsilon_F/n##
##P=(dU/dV)_S=2N\epsilon_F/5##

I don't know if this is the correct approach to the problem, but I have no other idea where to start.

Any help would be appreciated.

Hi there,

It seems like your approach is on the right track. Let's break down the problem step by step.

First, let's define the quantity ##<n_p>/V##. This is the average number of particles with momentum p per unit volume. In other words, it represents the density of particles in momentum space. This is similar to the density of states, but it is specifically for momentum and not energy.

Next, let's consider the classical distribution function f(p,q). This function represents the probability of finding a particle with momentum p and position q. In other words, it describes the distribution of particles in the system's phase space.

Now, let's consider the impulse transfer of the elastic collisions of the particles with the wall of the container. This is the change in momentum of the particles as they collide with the wall. This impulse transfer is related to the pressure of the gas, as it represents the force exerted by the particles on the wall.

So, how do we relate these quantities to each other? Well, we know that the probability of finding a particle with momentum p is given by the classical distribution function f(p,q). And we also know that the average number of particles with momentum p per unit volume is given by ##<n_p>/V##. So, we can write:

##<n_p>/V = f(p,q)##

Now, we can use the kinetic theory of gases to relate the impulse transfer to the pressure of the gas. The kinetic theory of gases tells us that the pressure of a gas is related to the average kinetic energy of the particles. In other words:

##P = \frac{2}{3} \frac{U}{V}##

where U is the total energy of the gas.

Now, let's substitute our expression for ##<n_p>/V## into the kinetic theory of gases equation. We get:

##P = \frac{2}{3} \frac{U}{V} = \frac{2}{3} \frac{\int_0^{\infty} d\epsilon g(\epsilon) \epsilon}{V} = \frac{2}{3} \frac{\int_0^{\epsilon_F} d\epsilon g(\epsilon) \epsilon}{V}##

where g(epsilon) is the density of states in energy space. Now, we can use the relation between the density of states in energy space and momentum