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what will happen when pores are much smaller than means free path of gases ?
so here is the question ...
One side of glass cylinder is initially filled with H2 at pressure P and other side with N2 at pressure 2P . This glass cylinder is closed at both ends and a glass filter in middle of cylinder divides into two compartments of equal volume . Describe what happens starting at time t = 0 at constant temperature and assuming that pores of filter have a diameter much smaller than mean free path of gases
i know H2 and N2 will not react ....so maybe this question is about rate of effusion i.e. grahams law = r1/r2 = P1/P2 X square root of (M2/M1)
my only doubt is what does the line - ' pores of filter have a diameter much smaller than mean free path of gases ' mean ?
i know mean free path is distance between two successive collision
please help ??
1 Answer
- az_lenderLv 72 years agoFavorite Answer
The line about the pores being much smaller than the mean free path is what enables Graham's Law to be applied. If the pores were quite large, the diffusion of the high-pressure nitrogen into the low-pressure hydrogen would be nearly immediate, and hydrogen would be going the other direction, too.
So yes, the problem is just about the rate of effusion through the porous filter.