00:01
In this problem, on the topic of thermal properties of matter, we are given the molar mass of oxygen or the o2 molecule as 32 grams per mole.
00:11
And we are asked to calculate various quantities regarding oxygen.
00:16
The first of which is the average translational kinetic energy of an oxygen molecule at a given temperature of 300 kelvin.
00:25
So to find this translational kinetic energy, we know this is a half.
00:31
Times the mass of this molecule m times the average velocity squared.
00:40
And remember, we can write this as 3 over 2 times the boltman's constant k times the temperature of the molecule t.
00:51
So we can write this again as 3 over 2 times 1 .38 times 10 to the minus 23.
01:04
Which is in a joule per molecule kelvin multiplied by the temperature which is given as 300 kelvin.
01:19
This gives us the average translational kinetic energy for the oxygen molecule to be 6 .21 times 10 to the minus 21 joules.
01:34
So with the average translational velocity of the oxygen molecule, we can calculate its average kinetic energy.
01:45
Next, for part b, we want to calculate the average value of the square of its speed.
01:53
So to get this, we first need to find the mass of one molecule.
02:00
The mass of one molecule of the oxygen is its molar mass, divided by a vagabergative.
02:09
Number n or n a this is equal to the given molar mass which is 32 times 10 to the minus 3 so that we can use the unit kilogram per mole which is s i units divided by avogadro's number which is 6 .022 times 10 to the power 20 the molecules in every mole so calculating we get the mass of the oxygen o2 molecule to be 5 .314 times 10 to the minus 26 kilograms.
03:05
So every molecule of this o2 measures 5 .314 times 10 to the minus 6 kilograms.
03:13
Now that we have the mass and we also know from part a what the kinetic energy is, we said a half m v squared was 6 .21 times 10 to the minus 21 joules.
03:42
So from here we can find, as the question asks for the square of the speed, so v squared, we can solve a v squared to be, if we rearrange this equation 2, in terms of to 6 .21 times 10 to the minus 21 joules over m.
04:09
And we've just calculated the mass.
04:12
So this is 2 into 6 .21 times 10 to the minus 21 joules divided by the mass of the o2 molecule, which is 5 .314 times 10 to the minus 26 kg.
04:31
So we get the square of the speed.
04:34
To be 2 .34 times 10 to the power 5, and this is meter squared per second squared.
04:45
And so that's the answer for the square of the speed, or the average speed of the oxygen molecule.
04:53
Now for part c, we are asked to calculate the root mean square speed of this molecule.
05:02
Now the root mean square speed is vrms, and we know this to be, the square root v squared rms so the root mean square is simply the square root of the square of the quantity which is the square root of v squared which we calculated previously as to 0 .34 times 10 to the power of 4 not 10 to the power of 4 but rather 10 to the power 5 as we calculated above in si units of meter squared per second squared.
05:53
In calculating, we get the root mean square velocity vrms of the oxygen molecule to be 484 meters per second.
06:04
So that's the solution.
06:06
We have the root mean square velocity, the root mean square speed of this molecule.
06:12
Now, in part d, we are asked to calculate the momentum of an oxygen molecule.
06:17
That is traveling at the speed.
06:21
Now we have the speed, we are asked to calculate the momentum.
06:25
We know the momentum p is equal to the mass of the molecule m times the rms speed, so vrms.
06:42
Now the mass of this molecule we calculated earlier to be 5 .314 times 10 to the minus 26 kg times vrm.
06:56
Rms of 484 meters per second we calculated in part c and we get the momentum of the molecule traveling at this speed to be 2 .57 times 10 to the minus 23 kilogram meter per second and that's the momentum of this molecule.
07:27
Now we go on to part e of the problem which asks us to find the or says that if we suppose an oxygen molecule is traveling at this speed and bounces back and forth between opposite sides of a cubicle vessel with a given side length, we want to know the average force that the molecule will exert on one of the walls of the container.
07:52
Okay, so we know the time between collisions with one wall, we'll call t, is equal to the distance 0 .2 meters divided by the speed in which the molecule is traveling, which is vrms, which is 0 .2 meters divided by 484 meters per second.
08:19
And we get the time between collisions to be 4 .13 times 10 to the minus 4 seconds...