Mean free path and speed as a function of pressure and temp

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At 18,000 ft, the atmospheric pressure is approximately half of sea level pressure, and the temperature decreases by 70°C. This change affects the speed distribution of air molecules, which can be analyzed using the Maxwell-Boltzmann distribution formula. For the mean free path, the relationship indicates that a decrease in pressure would indeed increase the mean free path, but the lower temperature must also be considered. The calculations suggest that the mean free path increases, as indicated by the ratio of the initial and final conditions. Overall, the temperature and pressure changes significantly influence both the speed distribution and mean free path of air molecules.
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Homework Statement



I know that at 18,000 ft. above mean sea level, the atm. pressure ~half of what it is at seal level (760 Torr). The temperature also decreases by 70C.

a. How would this change the speed distribution of the air molecules quanitatively?
b.How would it change the mean free path of the air molecules quantitatively?

Homework Equations


For a) all I can seem to find is this:
$$ c=\int_0^\infty vP(v)dv=(8kT/\pi m)^{1/2}$$

But since I am interested in the speed distribution, should I use:

$$P(v)=4\pi [\frac{m}{2 \pi kT}]^{3/2} (v^2) exp(\frac{-mv^2}{2kt})$$

For b) all I can seem to find is this:

$$\lambda_{mfp}=\frac{kt}{\pi d^2 P (2)^{1/2}}$$

Since the pressure is half, that would seem to double the mfp correct? and then i need to take into account the change in temp?

Thus the change would be

$$\frac{T1}{T2} \frac{P2}{P1}$$
is this correct?

The Attempt at a Solution



For b) If I plug the numbers in I get like 0.66 which to me, might be correct. This is the ratio of mfp_1 to mfp_2 (lower pressure and temperature). Thus 0.66 * mfp_1 / mfp_2 = 0.65 so mfp_2 is larger (which I would expect)

I'm not sure if my approach to b) is correct, and for a), I am not very sure how to solve it. Do I just plug the temperatures into the equation and see the difference, noting that v also depends on temperature?
 
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b: correct.

a: I'm not sure what else you can do quantitatively apart from writing down the formula. The distribution changes as given by this formula.
 

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