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Topic: Diffusion coefficient for an ideal gas  (Read 10319 times)

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Offline Hunt

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Diffusion coefficient for an ideal gas
« on: July 19, 2008, 06:43:51 PM »
Does anyone know of a good reference that shows how to find the expression for the diffusion coefficient Do of an ideal gas ?

It's easy to show that Do is a linear function of the mean free path , Lambda. But I cant figure out how to get the exact expression :
Do= Lambda x c(bar) / 3

c(bar) = average velocity of the gas

Offline enahs

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Re: Diffusion coefficient for an ideal gas
« Reply #1 on: July 20, 2008, 11:03:14 AM »
I believe it is just defined that way, as it makes the averaging of the total flux consistent with the most basic relationship between flux and the spatial gradient in transported properties.

In other words, the Diffusion Coefficient is nothing more then the diffusion proportionally constant in the basic expression of the flux defined to work with the appropriate units and values for this specific derivation.

Offline Hunt

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Re: Diffusion coefficient for an ideal gas
« Reply #2 on: July 20, 2008, 11:26:07 AM »
Yes ofcourse , Enah ..but why is the 1/3 factor added ? This is not part of dimensional analysis. A simple derivation shows that a system that undergoes steady-state diffusion, the particle flux
F = - [ <c> Lambda / 2 ] dN / dx

This is an approximate expression that validates fick's empirical 1st law for diffusion. However, this way the diffusion coefficient becomes :

Do = <c> Lambda / 2 , contrary to what all chemistry books admit. It's the 1/3 fraction that doesnt make sense to me.

Offline enahs

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Re: Diffusion coefficient for an ideal gas
« Reply #3 on: July 20, 2008, 11:50:23 AM »
It is assumed in the derivation that the particles moves from the planes located at ±λ to the flux plane directly along the x axis. But the particle trajectory is not aligned with the x axis, and so the particle will not reach the flux plane after traveling one mean free path. It is at this point, collisions with other particles can occur, resulting in post-collision trajectories away from the flux plane, therefore these particles will not contribute to the flux.  If you include those trajectories you are required to take the orientational average of the mean free path. This average results in a reduction in the total flux by a factor of 2/3 (1/2 * 2/3 = 1/3).


The reduction by 2/3 is just and estimate based on standard probability density functions of the particle hitting another particle before it reaches the other plane.
Feel free to do that math. It is not really hard, just tedious and I have no desire to try and type it here.


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Re: Diffusion coefficient for an ideal gas
« Reply #4 on: July 21, 2008, 03:57:40 AM »
Thanks for your input.

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