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August 9, 2006, 11:48
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Mean Free Path of water vapor?
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#1 |
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Guest
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Can anyone tell me the mean free path of water vapor at 293.15K, and partial vapor pressure of 0.023062atm ?
(Remember that at 293.15K, P_vapor = 0.023062 atmospheres and P_air = 0.976938 atmospheres is the saturation condition, i.e. 100% Relative Humidity.) |
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August 12, 2006, 12:05
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Re: Mean Free Path of water vapor?
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#2 |
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Guest
Posts: n/a
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Go to HyperPhysics web site and you will be able to calculate it online based on the parameters you enter. Remember that the path will be limited by collision with of all kinds of molecules (i.e. water, O2, N2, etc.). My geuss is it will be something on the order of 10e-7 m.
Angen |
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