Calculating Mean Free Path Ratios in a Divided Ideal Gas System

Tachyonprince
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I have a box with a wall in mid dividing it in 2 sections, and the wall has a hole of diameter d. There is ideal gas in both sections at 150 K in one section and at 300 K in another. How am I supposed to calculate ratio of mean free paths in 2 sections.

My attempt: L ~ Volume / Number of particles
=> L ~ Temperature / Pressure

Now, Assuming pressure to be same on both sections, ratio must be half. But that is incorrect. Why? Correct answer is 0.7.
 
on Phys.org
Hello your majesty, :welcome:

And read the PF guidelines -- they apply to royalty, too..

But what the heck, I'll risk my neck (perhaps I can be knighted posthumously... :rolleyes: ):

Tachyonprince said:
L ~ Volume / Number of particles
Can't be right: this only applies if target particles aren't moving
 
BvU said:
Can't be right: this only applies if target particles aren't moving
No, that refers to the precise expression involving the average velocity of particles. The moving target is corrected for by multiplying the average velocity by a factor of √2. The proportionality to V/N is unaffected.
 
I agree. Unfortunate Wrong way to put it on my part
 

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