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## Physics library

### Unit 10: Lesson 1

Temperature, kinetic theory, and the ideal gas law- Thermodynamics part 1: Molecular theory of gases
- Thermodynamics part 2: Ideal gas law
- Thermodynamics part 3: Kelvin scale and Ideal gas law example
- Thermodynamics part 4: Moles and the ideal gas law
- Thermodynamics part 5: Molar ideal gas law problem
- What is the ideal gas law?
- The Maxwell–Boltzmann distribution
- What is the Maxwell-Boltzmann distribution?

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# Thermodynamics part 2: Ideal gas law

To begin, Sal solves a constant temperature problem using PV=PV. Then he relates temperature to kinetic energy of a gas. In the second half of the video, he derives the ideal gas law. Created by Sal Khan.

## Video transcript

Welcome back. In the last video, I told you
that pressure times volume is a constant. That if you increase the
pressure-- or if you increase the volume, you're going to
decrease the pressure. Hopefully, you got an
intuitive sense why. Or likewise, if you squeezed the
balloon, or the box, and there are no openings there,
then the pressure within the box would increase. With that said, let's see if we
can do a couple of fairly typical problems that
you'll see. So let's say that I have
a box, or a balloon, or something, and it has a volume,
and so let me call this the initial volume. My initial volume is 50 cubic
meters, and my initial pressure is 500 pascals. Just so you remember,
what's a pascal? That's 500 newtons
per meter cubed. I take that box, or balloon, or
whatever, and I compress it down to 20 meters cubed. So I compress it, so I squeeze
it-- that was the first example that I gave last time. It was the same container,
and I squeeze it down to 20 meters cubed. What's going to be
the new pressure? You should immediately have an
intuition-- what happens when you squeeze a balloon? It becomes harder to do it. What's going to be
the new pressure? It's definitely going to be
higher-- when you decrease the volume, the pressure increases
are inversely related. The pressure's going to go up,
and let's see if we can calculate it. We know that P1 times v1 is
equal to some constant, and since we have no aggregate
change in energy-- I'm just telling you that the box is
squeezed, I'm not telling you whether it did any work, or
anything like that-- the same constant is going to be equal to
the new pressure times the new volume, which is equal
to P2 times V2. You could just have the general
relationship: P1 times V1 is equal to P2 times V2,
assuming that no work was done, and there was no exchange
of energy from outside of the system. In most of these cases, when you
see this on an exam, that is the case. The old pressure was 500 pascals
times 50 meters cubed. One thing to keep in mind,
because this equivalence is not equal, and we're not saying
it has to equal some necessary absolute number--
for example, we don't know exactly what this K is, although
we could figure it out right now-- as long as
you're using one unit for pressure on this side, and one
unit for volume on this side, you just have to use
the same units. We could have done this same
exact problem the exact same way, if instead of meters cubed,
they said liters, as long as we had liters here. You just have to make sure
you're using the same units on both sides. In this case, we have 500
pascals as the pressure, and the volume is 50 meters cubed. That's going to be equal to the
new pressure, P2, times the new volume, 20
meters cubed. Let's see what we can do: we can
divide both sides by 10, so we can take the 10 out of
there, and we could divide both sides by 2, so that
becomes a 250. We we get 250 times 5 is equal
to P2, and so P2 is equal to 1250 pascals, and if we kept
with the units, you would have seen that. When I decreased the volume
by roughly 60%, I have the pressure actually increased by
2 1/2, so that gels with what we talked about before. Let's add another variable into
this mix-- let's talk about temperature. Like pressure, volume, work, and
a lot of concepts that we talk about in physics,
temperature is something that you probably are at least
reasonably familiar with. How do you view temperature? A high temperature means
something is hot, and a low temperature means something is
cold, and I think that also gives you intuition that a
higher temperature object has more energy. The sun has more energy than an
ice cube-- I think that's fair enough. I think you also have the sense
that-- what would have more energy? A 100 degree cup of tea, or a
100 degree barrel of tea. I want to make them equivalent
in terms of what they're holding. I think you have a sense. Even though they're the same
temperature, they're both pretty warm-- let's say this
is 100 degrees Celsius, so they're both boiling-- that the
barrel, because there's more of it, is going to
have more energy. It's equally hot, and there's
just more molecules there. That's what temperature is. Temperature, in general, is a
measure roughly equal to some constant times the kinetic
energy-- the average kinetic energy-- per molecule. So the average kinetic energy
of the system divided by the total number of molecules we
have. Another way we could talk about is, temperature is essentially energy per molecule. So something that has a lot of
molecules, where N is the number of molecules. Another way we could view this
is that the kinetic energy of the system is going to be
equal to the number of molecules times the
temperature. This is just a constant-- times
1 over K, but we don't even know what this is, so we
could say that's still a constant-- so the kinetic energy
of the system is going to be equal to some constant
times the number of particles times temperature. We don't know what this
is, and we're going to figure this out later. This is another interesting
concept. We said that pressure times
volume is proportional to the kinetic energy of the system--
the aggregate, if you take all of the molecules and combine
their kinetic energies. These aren't the same K's-- I
could put another constant here and call that K1. And we also know that the
kinetic energy of the system is equal to some other constant
times the number of molecules I have times
the temperature. If you think about it, you could
also say that this is proportional to this, and this
is proportional to this. You could say that pressure
times volume is proportional to the number-- and these are
all different proportional constants, and we'll figure
out this exact constant later-- so we could say that
pressure times volume is proportional to molecules we
have, times temperature. And we said that we can
view temperature as energy per molecule. Another way we could say is
that if this constant is constant, which is by
definition, and the number of molecules is constant-- we
have PV over temperature. Pressure times volume over
temperature is going to be equal to something times the
number of molecules, so we could say that's some other
constant, like k4. This is another interesting
thing to think about: we said pressure times volume is equal
to pressure times volume, and now we added temperature
into the mix. We could say P1 times V1
over T1 is equal to P2 times V2 over T2. Does this make sense to you? What happens if I have another
box, and I have my particles bouncing around like always. I have some volume, and some
amount of pressure-- what happens when the temperature
goes up? What am I saying? I'm saying that the average
kinetic energy per molecule is going to go up, so they're going
to bounce against the walls more. If they bounce against the walls
more, the pressure's going to go up, assuming
volume stays flat. Another way you could think
about it-- let's say the temperature goes up, and the
pressure stays flat. So what did I have to do? I just said if the temperature
goes up, the average kinetic energy of each molecule--
they'll bounce more. In order to make them bounce
against the sides of the walls as often, I'd have to
increase the volume. If you hold pressure constant,
the only way you can do that is by increasing the volume
while you increase the temperature. Let's keep this in mind, and we
will use this to solve some pretty typical problems
in the next video.