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Topic: Diffusion Coefficient Problem  (Read 3538 times)

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Offline GMAN20

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Diffusion Coefficient Problem
« on: February 25, 2014, 07:56:37 PM »
The diffusion of H2 gas through a 0.005-cm-thick piece of Pd foil with an area of 0.750 cm2 is measured. On one side of the foil, a volume of gas maintained at 298 K and 1 atm is applied, while a vacuum is applied to the other side of the foil. After 24 hours, the volume of H2 has decreased by 15.2 cm2. What is the diffusion coefficient of H2 gas in Pd?

So far I have tried using the equation for mutual diffusion D=(3/8)*(RT/2piμ)^(1/2)*(1/(r1+r2)2tot

I am having trouble is determining the reduced mass (μ). I understand it is kg/mol opposed to g/mol but which molar mass do I use or do I add them together?

Offline Corribus

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Re: Diffusion Coefficient Problem
« Reply #1 on: February 26, 2014, 12:05:43 PM »
The most obvious way to solve this problem would be to use the diffusion equation (Fick's 2nd Law). However from the wording here it is unclear whether the steady state applies, because the concentration on one side of the film seems to be changing. If steady state cannot be assumed, this will require some fairly complex differential equation solutions.  Is this the exact wording of the problem? (Minor thing, the units on your volume of H2 is cm2, which can't be right.)
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline GMAN20

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Re: Diffusion Coefficient Problem
« Reply #2 on: February 26, 2014, 07:21:59 PM »
That is the exact wording of the problem and I do not believe steady state can be assumed.. Seems like I might be SOL  ???

Offline Corribus

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Re: Diffusion Coefficient Problem
« Reply #3 on: February 26, 2014, 09:37:52 PM »
I have a book on diffusion which solves the diffusion equation for this kind of scenario. However I still maintain there is some ambiguity in the wording.  What is the initial concentration of gas, for instance? It says the volume of gas is maintained at 1 atm on one side of the film. Does this mean the starting pressure of gas is 1 atm? Or is it unknown? Also, is it maintained at 1 atm during the whole experiment? (Probably not, since they say 15.2 cm2 is lost.) They why use "maintained"? Perhaps the overall pressure is, but that would sure be a strange way to do an experiment to measure the diffusion constant. Either way, it seems the starting amount should be a critical piece of information.... since the diffusion flux is proportional to the concentration.  15.2 lost from a starting pressure of 16 is a lot different than 15.2 lost from a starting pressure of 1000.  Hmm, I will give it some more thought. I actually deal with gas diffusion through films a lot in my own research, so it's rather ironic that this one has me a bit stumped.
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline Enthalpy

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Re: Diffusion Coefficient Problem
« Reply #4 on: February 27, 2014, 12:40:10 PM »
I understand it simply. "Maintained at 1 atm" and "vacuum", while the initial volume dimishes. That's for instance a piston - or a hermetic bag, since 1 atm is for free. And a perfect steady state.

So, knowing the pressure difference over the known distance, and the voume that passes through the known area, you get the diffusion coefficient.

What's the bizarre formula for "mutual diffusion"? It looks rather like kinetic theory for gas mixes.

The only subtlety about hydrogen through palladium is whether atomic hydrogen diffuses, rather than molecular - but one can define a diffusion coefficient as if hydrogen just stayed molecular.

Offline GMAN20

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Re: Diffusion Coefficient Problem
« Reply #5 on: February 27, 2014, 12:51:15 PM »
I figured it out.. My professor threw the problem out but I found the "long" version

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