# speed distribution and mean free path of gases at pressure

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?

• What are you defining $c$ to be? – Spaderdabomb Feb 21 '16 at 21:40
• @Spaderdabomb c is the speed or average magnitude – Jackson Hart Feb 21 '16 at 21:41

• 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. – Spaderdabomb Feb 21 '16 at 22:05