Hi. It’s Mr. Andersen and this AP Physics essentials video 130. It is on kinetic theory
and temperature. Temperature is a macroscopic value. In other words if I look to the left
I have some water with a thermometer in it. We could say the temperature is around 25
degrees celsius. As I add some warm water to it, what is happening to the temperature?
It is increasing. It is a higher value and now it is around 35 degrees celsius. But that
is macroscopic. It is what we are measuring in the world that we live in. But hopefully
you know the temperature is caused by microscopic interactions. So we have all of these molecules
that are bouncing around. And so some have slow velocity. Some have fast velocities.
But the average kinetic energy of all of those molecules is equal to the temperature. The
more they are moving the more their velocities are the greater the temperature is. And so
wouldn’t it be wonderful if we had an equation that allowed us to move from the microscopic
to the macroscopic world. Well, temperature has to be on the left, kinetic energy has
to be on the right. This bar on the top means that it is the average. So we could put in
our formula for kinetic energy, 1/2 m v squared. Now this velocity is the average velocity
of the molecules. We call that the root mean square. And then on the left side we have
to put in a constant. So that is the Boltzmann’s constant. So three halves Boltzmann’s constant
times temperature. This has to be measured in kelvin is equal to one half m v squared.
And so this allows us to move from the microscopic world to the macroscopic world. Because temperature
is the average kinetic energy of all of the molecules. And lots of times you will represent
that using a distribution where we put the speed on the bottom and then the number of
molecules on the left. So we get a nice distribution that looks like that. So if I were to say
where is that average velocity? The tendency is to want to put the average right here but
it is actually going to be, since it goes to the right, it is going to be a little bit
off center like that. So here is our formula again. What is this? This is going to be the
root mean square, which is going to be the average velocity of all of those molecules.
And so what we can do is if we know that value, know the temperature and Boltzmann’s constant
we can figure out the kinetic energy of even one molecule. And so temperature, remember,
is going to be molecules bouncing around. On the left I have some cold water. On the
right I have some warmer hot water. I put some dye in it so you can see the molecules
interacting with the dye. I have done some time lapse here so you can see it moving around.
So you can see on the right side there is more molecular motion. More of those molecules
are moving around so we get greater distribution of the dye. And so we use a thermometer to
measure how fast those molecules are going. So with the celsius scale 0 is going to be
freezing. But what happens if we slow those molecules to a stop? That is going to be absolute
zero and we use the Kelvin scale to do that. And so remember if you are ever converting
a temperature to Kelvin, which you will have to do with any of these problems, all you
do is simply add 273 to it. So that would be 288 Kelvin on the left, 338 Kelvin on the
right. And so here is a way that we could apply that formula in the easiest sense. What
is the average kinetic energy of a gas molecule at 25 degrees celsius? What is neat is that
it does not matter what that gas molecule is. So we know kinetic energy is three-halves
k T. That is what we are asked for in this problem. I would look up the Boltzmann’s
constant. That is going to be 1.38 times 10 to the negative 23 joules per Kelvin. And
do not forget to convert 25 degrees celsius to Kelvin. So that would be 298 Kelvin. And
so we could figure out the kinetic energy of 1 gas molecule is going to be 6.2 times
10 to the negative 22 joules. And see how powerful that equation is. It allows us to
go from just a temperature to this microscopic value, the amount of kinetic energy of a single
gas molecule. So if we look at the Maxwell-Boltzmann distribution we are going to have a bunch
of gas moving around. They are each going to have different velocities. Some high some
low. So we get a distribution that looks like that. The largest value, most probable velocity,
but to the right side of that we are going to have what is called the root mean square
velocity. That is going to be the average velocity of all the molecules that are moving
around. So here is the root mean square. So how do we calculate that? Well here is that
formula again. If I want to solve for the velocity, so if I want to solve for just this
value right here, my root mean square is equal to the root of 3 k T divided by m. m is going
to be the mass of the molecule itself. And so this could be a quantitative problem that
you would solve. Find the root mean square of a nitrogen molecule at 0 degrees celsius.
So how would you solve this problem? Well we know the Boltzmann’s constant, we know
the temperature, since it is at 0 degrees celsius. We then have to find the mass of
a nitrogen molecule. So you would have to use a little bit of chemistry. And so we could
use the periodic table to figure out the mass of a nitrogen molecule is equal to 4.65 times
10 to the negative 26. We plug all of those values in here and then we would find the
root mean square of nitrogen. But in AP Physics they do not want you to solve in quantitatively.
They want you to be able to read a distribution like this. So let me pose the same question.
What is the root mean square of a nitrogen molecule at 0 degrees celsius? So reading
this graph I see that nitrogen is going to be the red. You can see that nice distribution
like this. We are reading velocities on the bottom. And so if I were to approximate off
to the right here, I would say it is going to be a little over 500 meters per second.
And so you can read the distribution to figure it out. And so did you learn to qualitatively
connect the microscopic to macroscopic world? I hope so. And I hope that was helpful.