Boltzmann Distribution with two gasses

1. Jan 9, 2012

edpell

What is the Boltzmann distribution look like if we have a mix of two gasses with very different molecular weights? Say hydrogen at 2 amu and xenon at 132 amu. Do they have equal average momentum and hence the hydrogen has an average velocity 66 times the velocity of the xenon and an energy .5*m*v^2 that is 66 times higher than the xenon? Or do they have equal average energy? But shouldn't they both have just .5*kT of energy per degree of freedom?

2. Jan 9, 2012

klimatos

Since the two gases are mixed, at equilibrium they will have the same temperature and hence the same mean kinetic energy of translation. They will not have the same momentum. Molecular momentum is a function of the molecular velocity, but energy of translation is a function of the means of the squares of the velocities. The Maxwell distribution curves of their respective kinetic energies of translation will be different as well. The curve for hydrogen will be lower and broader. The mean total KE per molecule will be 3/2 kT for each gas.

In short, the KE means will be the same, the distributions will be different.

3. Jan 9, 2012

Curl

The Boltzmann distribution describes a statistic (mean speed) in the microcanonical ensemble. As such, particles of different kinds are non-interacting so in a mixture of two gasses, each species can be treated independently.

If you want a combined statistic for the entire gas, just add up the 2 statistics in the end using the appropriate mathematics.

4. Jan 10, 2012

Rap

If you really mean momentum, and not absolute value of momentum, then they will have the same average momentum per particle, which will be zero. Since the absolute value of momentum is not a conserved quantity, the gases will not have the same average absolute value of momentum. Energy is conserved, so they will have the same average energy per particle.