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Topic: Mean Free Path  (Read 4959 times)

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Offline zeshkani

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Mean Free Path
« on: August 31, 2007, 09:18:28 PM »
ok today in lecture the prof calculated the mean free path og N2 at 1atm and 25C  and got 70nm
but when i tired to do that i get a total different answer

lambda= (kT)/sigma*P*2^1/2
i believe k is boltzmann constant

so with all the numbers in
lambda= (1.38065e^-23*298.15K)/ (.43nm^2)*1*2^1/2
         =  6.7692e^-21  and thats no way the right answer
so my question is em i doing this calculation wrong, or there is something more too it


Offline Yggdrasil

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Re: Mean Free Path
« Reply #1 on: September 01, 2007, 12:31:05 AM »
Always include units in your calculation.

Offline zeshkani

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Re: Mean Free Path
« Reply #2 on: September 01, 2007, 12:35:58 AM »
i figured it out, is was just using p in atm but it needs to be in Pa and the cross section was .52 which i also used like that but this was wrong, its (.52e^-2)^2m
its so easy know, next time i wont just start a new topic, it just took me some time to figure it out

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