00:01
So we have a 10 centimeter diameter, 20 centimeter long cylinder.
00:06
It contains two times 10 to the 22nd molecules of or atoms of argon.
00:14
So argon is just monotomic at a temperature of 50 degrees celsius.
00:20
So and the first question is, is what is the number done to the gas? well, that means we just need to divide the number of molecules or atoms we have.
00:28
Since it's argon, we could just call it and say they're atoms.
00:31
Divided by the volume, and the volume, you know, is just the volume of a cylinder.
00:37
And we wind up with 1 .27 times 10 to the 25th atoms per cubic meter.
00:45
Then they ask for the root mean square speed, so that the, that is a square root of three times boltsman constant times temperature divided by the molecular, the weight of one of one argon atom.
01:03
Temperature we can convert into, we need to convert into calvin.
01:07
And the weight of one argon atom, so we know that a mole of argon weighs 40 kilograms, or 40 grams.
01:17
And so we divide that by avogadro's number, and we get 6 .64 times 10 to the minus 26 kilograms.
01:30
So we have that.
01:32
We can plug that in here.
01:33
We can plug this into here.
01:35
And boltzman's constant.
01:36
And tanking through this calculation, we get 449 meters per second, which again, seems reasonable given the information we have in the chapter.
01:48
So then we want to know what is vmax, the rms value of vx max, the rms value of the x component of velocity.
01:55
Well, that's just the rms value of the speed divided by square to three.
02:04
And that winds up being 259 meters per second.
02:12
Now, we're asked for the rate of atoms colliding with one end of the cylinder.
02:22
Well, we can see here that we have that that is the one half times the number density, times the area that we're looking at where they're hitting, and times the velocity in the x direction.
02:36
And we're assuming that the velocity along the length of the cylinder is x.
02:42
Again, we call it whatever we want.
02:45
Again, x, y, or z, if we would have then needed vz, v, sub z, which would have been the same value.
02:56
So x, y, and z are all the same...