00:01
Okay, in this problem we have air at standard temperature pressure.
00:05
We're given its mean free path.
00:07
It's about 9 times 10 to negative 8 meters.
00:09
And we were asked to find the collision frequency for these molecules in this ideal gas.
00:17
So this may be perceived as cheating, but in problem 49, we derived an expression for the frequency, which is f equals 16 times the pressure, times the radius of our particles squared times the square root of pi over mkt this was derived in the previous problem 49 and so from that we can actually do get everything if we assume that the mass of air which we've done in a number of problems in this chapter is about 29 times the mass of a proton and we have standard pressure temperature we can we can do this rather easily we also can assume that the radius of our air particles is about 1 .5 times 10 to negative 10.
01:15
This was also, this has also been done numerous times in this chapter, so i think it's fair game.
01:22
So just using this result, we can get this frequency, plugging in our numbers.
01:33
We have 16 times standard temperature, standard pressure, 1 .013, 10 to the negative 10 to the 5, pascal, that's standard pressure...