Calculating Mean Free Path & Avg. Separation of Oxygen Molecules

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SUMMARY

The mean free path for oxygen molecules at 300K and atmospheric pressure (105 Pa) is calculated using the formula λ = kT / (4π√2 * r² * p), where the diameter of an oxygen molecule is 0.29 nm. At this condition, the mean free path is approximately 1.11 x 10^-9 m. At an altitude of 300 km, where the pressure is 10^-11 atmospheres, the mean free path increases significantly to about 1.11 x 10^9 m. The discussion highlights the need to understand the average molecular separation to compare the ratio of mean free path to molecular separation at different altitudes.

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Homework Statement



(a)What is the mean free path for oxygen molecules at 300K and atmospheric
pressure (105 Pa) and the average frequency of collisions for a particular molecule?
(The diameter of an oxygen molecule is 0.29 nm).

(b)What is the mean free path of oxygen molecules at an altitude of 300 km, where
the pressure is only 10-11 atmospheres? (Assume the temperature is 300 K). Comment
on the implications of your result for simulating conditions at these altitudes in the
laboratory.

(c) Compare the ratio of λ to the average molecular separation for oxygen at sea level
and at an altitude of 300 km.

Homework Equations



\lambda = \frac{kT}{4*\pi*\sqrt{2}*r^{2}p}

The Attempt at a Solution



I've done parts (b) and (c) I think, and got answers of 1.11*10^-9m and 1.11*10^9m for the mean free paths. I have no idea how to do part (c) though. Just thinking about it, you would expect the ratio to stay roughly 1, since the average molecular separation and mean free path should increase alongside one another. How do you get an estimate of the average separation though?!

Thanks.
 
Physics news on Phys.org
Can you get the number of molecules per m3, given temperature and pressure? That's helpful in determining the average molecular separation.
 

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