Gas Distribution: Temperature, MFP & More

[Glenn G]

Diffusion of gases seems to follow normal distribution. I imagine deviations from the mean would depend on temperature, mean free path and speed of gas molecules. Any other?

Cheers,
Glenn

 

[muscaria]

Diffusion of gases seems to follow normal distribution."
Glenn

What do you mean by this? Diffusion in single component gas due to density gradients? 2 component mixture? Or are you talking about the maxwell-boltzmann distribution?

 

[Glenn G]

Sorry I meant that if you modeled the displacement over time of a body of gas and plot where all they have got to in terms of displacement from their original position that it follow the normal distribution (the mean displacement should be zero I'd imagine because they are moving in random directions) is any of this correct?

 

[muscaria]

Sorry I meant that if you modeled the displacement over time of a body of gas and plot where all they have got to in terms of displacement from their original position that it follow the normal distribution (the mean displacement should be zero I'd imagine because they are moving in random directions) is any of this correct?

Well if the body of gas is in equilibrium, then the gas will be homogeneously distributed i.e density even everywhere. So that's where it is trying to get to and doesn't depend on the initial configuration of the gas as long as you wait long enough for equilibrium to happen at a given temperature for fixed volume say. The temperature and pressure determine the mean energy and the temperature alone determines the distribution about this mean value due to statistical fluctuation. For a non interacting gas this corresponds to kinetic energy and thus determines the average speed and the distribution of speeds about this average, which is a gaussian distribution.
Does that help or am I not getting something?

 

[Cutter Ketch]

I believe he means that in a gas at equilibrium the probability for the net displacement of any single molecule over a fixed time is normally distributed. (I can't say that it is always true, but I suppose the central limit theorem makes it hard for it not to be true)

I believe the mean displacement in time depends only on the mean size of the steps (mean free path) and the time of the steps (mean free time). These vary with pressure (shorter paths) and temperature (faster molecules means shorter time between collisions). I suspect any stickiness (chemical bonding when the collision energy happens to be low) would also increase the time.

 

[muscaria]

I believe he means that in a gas at equilibrium the probability for the net displacement of any single molecule over a fixed time is normally distributed.

Ah yes OK, I probably should have been able to get what was meant but it was the

a body of gas and plot where all they have got to

that threw me off.
Yes for a single particle when the fixed time is less than mean free time or if the walls of the container are at infinity.
The main point I was trying to make earlier hints at why this is:

For a non interacting gas this corresponds to kinetic energy and thus determines the average speed and the distribution of speeds about this average, which is a gaussian distribution.

I was making reference to the Maxwell-Boltzmann distribution i.e that the speed of a particle is normally distributed.. which then leads to a normal distribution of displacement for fixed time.

(I can't say that it is always true, but I suppose the central limit theorem makes it hard for it not to be true)

Here the normal distribution arises from the fact that the energy of a non-interacting gas is purely kinetic and therefore quadratic in the velocities. So the probability of measuring a particle with energy $\epsilon=mv^2/2$ is proportional to
$$exp\left[-\frac{mv_x^2+mv_y^2+mv_z^2}{2k_BT}\right]$$
where $v=\sqrt{v_x^2+v_y^2+v_z^2}$.

I imagine deviations from the mean would depend on temperature, mean free path and speed of gas molecules. Any other?

As with all statistical mechanics observables, the spread from the average occurs due to energy fluctuating between the system and heat bath during thermal equilibrium and it is the temperature $\textit{alone}$ which determines this fluctuation (pressure doesn't play any role for instance). Increasing the energy of a system through work (pressure, E-M fields etc..) shifts energy levels upwards, whereas increasing energy by adding heat and raising the temperature doesn't change the energy levels but shifts the occupation towards higher energy levels: work changes the energy levels and leaves distribution unchanged, heat changes the distribution and leaves the energy levels unchanged.
So the deviation from the mean you are asking about, should be entirely fixed by the temperature of the system - for a given particle mass that is.. You can see this is the case from the probability distribution above. You can find everything you want to know from the Boltzmann distribution, e.g. the mean speed (and therefore mean displacement), the spread from the mean etc..

 

[Glenn G]

Thanks for that very insightful. The fact that work changes energy levels but leaves distribution unchanged whereas heating does the reverse is unexpected for me. Good stuff.