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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?

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? Shouldn't the average speed also depend on the pressure? However, the equation I show seems to just depend on kT after integrating. Could someone help me figure out how these would quantitatively change at the different pressure and thus temp?

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  • $\begingroup$ What are you defining $c$ to be? $\endgroup$ – Spaderdabomb Feb 21 '16 at 21:40
  • $\begingroup$ @Spaderdabomb c is the speed or average magnitude $\endgroup$ – Jackson Hart Feb 21 '16 at 21:41
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Think about what is physically going on. We know that the speed should be affected by the temperature because the temperature is a measure of the internal energy of the system. As for pressure though, think of a bunch of particles bouncing around in a box. Now if we shrink that box adiabatically so we impart no new energy to the system, but have the same number of particles, the average energy of each particle has not changed. Therefore, they will be bouncing around at the same speed.

It's true that they will be colliding with each other much more often, so the mean free path is expected to decrease. This is simply because more particles in less volume = less space to freely move around without colliding. But bouncing off walls and each other will not change the internal energy of the particle, thus the average speed of the particles should remain constant when changing pressure adiabatically.

Note that a change in temperature should not affect the mean free path. This is because the particles are simply bouncing around faster - but there is still the same amount of particles and the same amount of volume. However, increasing the temperature would increase the collision rate, or the mean free time.

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  • $\begingroup$ Ok, thanks for the response. I need to know how this would change the distribution quantitatively. How would you figure this out? For the mfp, I just find that T1/T2 P2/P1 $\endgroup$ – Jackson Hart Feb 21 '16 at 22:02
  • $\begingroup$ My equation for mfp has kT in it though? $\endgroup$ – Jackson Hart Feb 21 '16 at 22:03
  • $\begingroup$ In your equation for mean free path, before you do anything, all I see is a $P$ in the denominator. Therefore if the pressure halves, then the mean free path should double. You may be overcomplicating by involving temperature. It may be indirectly affected by temperature if a change in temperature changes the pressure. But it is inherently a change in pressure that will change your mean free path. $\endgroup$ – Spaderdabomb Feb 21 '16 at 22:05
  • $\begingroup$ Could you comment more on part a? How will it affect the speed distribution? The distribution equation I have has v in it as well, which also depends on temperature. $\endgroup$ – Jackson Hart Feb 21 '16 at 22:06
  • $\begingroup$ At hyperphysics.phy-astr.gsu.edu/hbase/kinetic/menfre.html it shows in the calculator at the bottom, that the temperature and pressure will both affect the mfp $\endgroup$ – Jackson Hart Feb 21 '16 at 22:10

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