00:02
So we're given that the temperature 20 degrees celsius, which is 293 degrees kelvin, and that the pressure is 750 tor, which we can just leave in tor.
00:20
So we'll be only taking ratios.
00:24
We have the mean free path for argonne is 9 .9 times 10 to the minus 6 centimeters.
00:32
And for di - nitrogen, 27 .3.
00:37
5 times 10 to the minus 6 centimeters.
00:44
So we want to find first the ratio of the two diameters.
00:50
So r is diameter of argon over diameter of di -nitrogen.
00:58
So we can use the formula for the mean -free path.
01:01
It's just 1 over pi square root 2, d squared, n over v.
01:11
So here, let's the only thing that changes for the two atoms or molecules is d.
01:17
So we can just take a ratio.
01:20
So the wavelength for argon over wavelength for di -nitrogen is just d squared dinitrogen, because here it was down on bottom over d squared for argon.
01:37
So our ratio is just square root of the free path ratio, which is 1 .7 when we plug in our numbers here.
01:58
We want to know how does this mean free path change as we change temperature or pressure...