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Topic: Gas diffusion with a porous pot  (Read 16406 times)

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Offline confusedstud

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Gas diffusion with a porous pot
« on: October 20, 2012, 08:38:45 AM »
In a question there was a experimental set up like this: http://www.chemist.sg/kpt/Velocities%20of%20Some%20Common%20Gases/files/assets/seo/page8_images/0002.jpg

There is air inside and outside there is hydrogen gas. We learned that the hydrogen will diffuse into the pot at a faster rate until the concentration in the pot and outside is the same as compared to the air diffusing out. As a result, the water level will drop from the initial position of water. After a while the water level will rise back up to the same position as the air inside the pot will slowly diffuse out as well.

But will the water level rise to the same position? I'm not very sure as the pressure of hydrogen and air is not the same (my guess here) so shouldn't the pressure of air be different from a mixture of air and hydrogen? Or if the pressure is the same why is that so?

Thanks!
« Last Edit: October 20, 2012, 08:47:19 AM by Arkcon »

Offline confusedstud

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Re: Gas diffusion with a porous pot
« Reply #1 on: October 20, 2012, 10:06:44 AM »
I was thinking about this and if the gases diffuse equally by the volume, the number of moles of molecules inside and outside the porous pots will be the same. But what is the pressure dependent on? Is it the number of moles or the Mr of the gases present in it?

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