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I've already told you multiple
times that big, uppercase U is

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the internal energy
of a system.

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And it's really everything
thrown in there.

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It's the kinetic energy
of the molecules.

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It has the potential energy if
the molecules are vibrating.

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It has the chemical energy
of the bonds.

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It has the potential energy of
electrons that want to get

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some place.

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But, for our sake, and
especially if we're kind of in

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an introductory chemistry,
physics, or thermodynamics

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course, let's just assume that
we're talking about a system

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that's an ideal gas.

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And even better, it's a kind
of a monoatomic ideal gas.

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So everything in on my system
are just individual atoms. So

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in that case, the only energy in
the system is all going to

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be the kinetic energy of each
of these particles.

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So what I want to do in this
video-- it's going to get a

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little bit mathy, but I think
it'll be satisfying for those

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of you who stick with it-- is
to relate how much internal

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energy there really is in a
system of a certain pressure,

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volume, or temperature.

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So we want to relate pressure,
volume, or temperature to

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internal energy.

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Notice all the videos we've done
up until now, I just said

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what's the change in
internal energy.

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And we related that to the heat
put into or taken out of

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a system, or the work
done, or done to,

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or done by the system.

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But now, let's just say before
we do any work or any heat,

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how do we know how much internal
energy we even have

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in a system?

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And to do this, let's do a
little bit of a thought

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experiment.

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There is a bit of a
simplification I'll make here.

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But I think you'll find it OK,
or reasonably satisfying.

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So let's say-- let me just
draw it-- I have a cube.

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And something tells me that I
might have already done this

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pseudo-proof in the physics play
list. Although, I don't

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think I related exactly
to internal energy.

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So I'll do that here.

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Let's say my system
is this cube.

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And let's say the dimensions
of the cube

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are x in every direction.

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So it's x high, x wide,
and x deep.

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So its volume is, of course,
x to the third.

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And let's say I have
n particles in my

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system, capital N.

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I could have written lowercase
n moles, but let's just keep

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it straightforward.

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I have N particles.

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So they're all doing
what they will.

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Now, this is where I'm going
to make the gross

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simplification.

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But I think it's reasonable.

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So in a normal system, every
particle, and we've done this

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before, is just bouncing off
in every which way, every

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possible random direction.

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And that's what, when they
ricochet off of each of the

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sides, that's what causes
the pressure.

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And they're always bumping into
each other, et cetera, et

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cetera, in all random
directions.

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Now, for the sake of simplicity
of our mathematics,

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and just to be able to do it
in a reasonable amount of

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time, I'm going to make
an assumption.

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I'm going to make an assumption
that 1/3 of the

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particles are going-- well, 1/3
of the particles are going

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parallel to each of the axes.

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So 1/3 of the particles are
going in this direction, I

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guess we could say,
left to right.

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1/3 of the particles are
going up and down.

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And then 1/3 of the particles
are going forward and back.

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Now, we know that this isn't
what's going in reality, but

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it makes our math
a lot simpler.

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And if you actually were to do
the statistical mechanics

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behind all of the particles
going in every which way, you

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would actually end up getting
the same result.

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Now, with that said, I'm
saying it's a gross

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oversimplification.

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There is some infinitesimally
small chance that we actually

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do fall onto a system where
this is already the case.

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And we'll talk a little bit
later about entropy and why

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it's such a small probability.

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But this could actually
be our system.

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And this system would
generate pressure.

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And it makes our math
a lot simpler.

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So with that said, let's
study this system.

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So let's take a sideways view.

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Let's take a sideways
view right here.

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And let's just study
one particle.

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Maybe I should have
done it in green.

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But let's say I have
one particle.

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It has some mass, m, and
some velocity, v.

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And this is one of the capital
N particles in my system.

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But what I'm curious is how
much pressure does this

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particle exert on this
wall right here?

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We know what the area of
this wall is, right?

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The area of this wall
is x times x.

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So it's x squared area.

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How much force is being exerted
by this particle?

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Well, let's think about
it this way.

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It's going forward, or left
to right just like this.

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And the force will be exerted
when it changes its momentum.

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I'll do a little bit of review
of kinetics right here.

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We know that force is equal to
mass times acceleration.

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We know acceleration can be
written as, which is equal to

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mass times, change in velocity
over change in time.

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And, of course, we know that
this could be rewritten as

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this is equal to-- mass is a
constant and shouldn't change

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for the physics we deal
with-- so it's delta.

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We could put that inside
of the change.

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So it's delta mv over
change in time.

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And this is just change
in momentum, right?

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So this is equal to change in
momentum over change in time.

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So that's another way
to write force.

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So what's the change
in momentum going

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to be for this particle?

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Well, it's going to bump
into this wall.

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In this direction, right now,
it has some momentum.

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Its momentum is equal to mv.

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And it's going to bump into this
wall, and then going to

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ricochet straight back.

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And what's its momentum
going to be?

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Well, it's going to
have the same mass

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and the same velocity.

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We'll assume it's a completely
elastic collision.

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Nothing is lost to heat
or whatever else.

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But the velocity is in
the other direction.

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So the new momentum is going
to be minus mv, because the

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velocity has switched
directions.

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Now, if I come in with a
momentum of mv, and I ricochet

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off with a momentum of
minus mv, what's

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my change in momentum?

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My change in momentum, off of
that ricochet, is equal to--

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well, it's the difference
between these two,

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which is just 2mv.

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Now, that doesn't give
me the force.

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I need to know the change in
momentum per unit of time.

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So how often does this happen?

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How frequently?

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Well, it's going to happen
every time we come here.

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We're going to hit this wall.

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Then the particle is going to
have to travel here, bounce

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off of that wall, and
then come back

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here and hit it again.

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So that's how frequently
it's going to happen.

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So how long of an interval do
we have to wait between the

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collisions?

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Well, the particle has to
travel x going back.

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It's going to collide.

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It's going to have to travel
x to the left.

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This distance is x.

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Let me do that in a
different color.

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This distance right here is x.

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It's going to have to
travel x to go back.

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Then it's going to have
to travel x back.

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So it's going to have to
travel 2x distance.

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And how long will it take it
to travel 2x distance?

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Well, the time, delta T, is
equal to, we know this.

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Distance is equal to
rate times time.

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Or if we do distance divided by
rate, we'll get the amount

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of time we took.

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This is just our basic
motion formula.

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Our delta T, the distance
we have to

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travel is back and forth.

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So it's 2 x's, divided
by-- what's our rate?

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Well, our rate is
our velocity.

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Divided by v.

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There you go.

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So this is our delta
T right here.

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So our change in momentum per
time is equal to 2 times our

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incident momentum.

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Because we ricocheted back with
the same magnitude, but

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negative momentum.

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So that's our change
in momentum.

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And then our change in time
is this value over here.

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It's the total distance we
have to travel between

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collisions of this wall, divided
by our velocity.

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So it is, 2x divided by v, which
is equal to 2mv times

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the reciprocal of this-- so
this is just fraction

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math-- v over 2x.

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And what is this equal to?

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The 2's cancel out.

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So that is equal to mv
squared, over x.

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Interesting.

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We're getting someplace
interesting already.

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And if it doesn't seem too
interesting, just hang on with

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me for a second.

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Now, this is the force being
applied by one particle, is

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this-- force from one particle
on this wall.

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Now, what was the area?

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00:09:03,840 --> 00:09:05,090
We care about the pressure.

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00:09:05,090 --> 00:09:12,200


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00:09:12,200 --> 00:09:14,610
We wrote it up here.

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The pressure is equal to
the force per area.

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00:09:16,890 --> 00:09:21,360


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So this is the force
of that particle.

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So that's mv squared
over x, divided by

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00:09:28,570 --> 00:09:29,840
the area of the wall.

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00:09:29,840 --> 00:09:31,730
Well, what's the area
of the wall?

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00:09:31,730 --> 00:09:35,200
The area of the wall here,
each sideis x.

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00:09:35,200 --> 00:09:37,610
And so if we draw the wall
there, it's x times x.

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It's x squared.

213
00:09:39,360 --> 00:09:43,390
So divided by the area of
the wall, is x squared.

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00:09:43,390 --> 00:09:44,250
And what does this equal?

215
00:09:44,250 --> 00:09:52,370
This is equal to mv squared
over x cubed.

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00:09:52,370 --> 00:09:54,790
You can just say, this is times
1 over x squared, when

217
00:09:54,790 --> 00:09:55,760
this all becomes x cubed.

218
00:09:55,760 --> 00:09:57,530
This is just fraction math.

219
00:09:57,530 --> 00:09:59,120
So now we have an interesting
thing.

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00:09:59,120 --> 00:10:07,160
The pressure due to this one
particle-- let's just call

221
00:10:07,160 --> 00:10:14,220
this from this one particle--
is equal to m v

222
00:10:14,220 --> 00:10:16,750
squared over x cubed.

223
00:10:16,750 --> 00:10:18,650
Now, what's x cubed?

224
00:10:18,650 --> 00:10:20,960
That's the volume of
our container.

225
00:10:20,960 --> 00:10:22,210
Over the volume.

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00:10:22,210 --> 00:10:27,120
I'll do that in a
big V, right?

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00:10:27,120 --> 00:10:29,130
So let's see if we can relate
this to something else that's

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00:10:29,130 --> 00:10:29,630
interesting.

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So that means that the pressure
being exerted by this

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00:10:32,820 --> 00:10:36,220
one particle-- well, actually
let me just take another step.

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So this is one particle
on this wall, right?

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00:10:39,280 --> 00:10:40,900
This is from one particle
on this wall.

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00:10:40,900 --> 00:10:45,890
Now, of all the particles-- we
have N particles in our cube--

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00:10:45,890 --> 00:10:47,860
what fraction of them
are going to be

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bouncing off of this wall?

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00:10:49,170 --> 00:10:50,730
That are going to be doing
the exact same

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00:10:50,730 --> 00:10:52,950
thing as this particle?

238
00:10:52,950 --> 00:10:53,820
Well, I just said.

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00:10:53,820 --> 00:10:55,380
1/3 are going to be going
in this direction.

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00:10:55,380 --> 00:10:56,820
1/3 are going to be
going up and down.

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And 1/3 are going to go
be going in and out.

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So if I have N total particles,
N over 3 are going

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00:11:02,420 --> 00:11:06,195
to be doing exactly what this
particle is going to be doing.

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00:11:06,195 --> 00:11:08,740


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00:11:08,740 --> 00:11:10,350
This is the pressure
from one particle.

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00:11:10,350 --> 00:11:12,850
If I wanted the pressure from
all of the particles on that

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00:11:12,850 --> 00:11:15,960
wall-- so the total pressure
on that wall is going to be

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00:11:15,960 --> 00:11:18,000
from N over 3 of
the particles.

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00:11:18,000 --> 00:11:19,990
The other particles aren't
bouncing off that wall.

250
00:11:19,990 --> 00:11:21,700
So we don't have to
worry about them.

251
00:11:21,700 --> 00:11:26,620
So if we want the total pressure
on that wall-- I'll

252
00:11:26,620 --> 00:11:29,150
just write, pressure
sub on the wall.

253
00:11:29,150 --> 00:11:31,060
Total pressure on the wall is
going to be the pressure from

254
00:11:31,060 --> 00:11:37,400
one particle, mv squared, over
our volume, times the total

255
00:11:37,400 --> 00:11:40,950
number of particles
hitting the wall.

256
00:11:40,950 --> 00:11:44,750
The total number of particles
is N divided by 3, because

257
00:11:44,750 --> 00:11:47,270
only 3 will be going
in that direction.

258
00:11:47,270 --> 00:11:50,800
So, the total pressure on that
wall is equal to mv squared,

259
00:11:50,800 --> 00:11:52,790
over our volume of our
container, times the total

260
00:11:52,790 --> 00:11:54,115
particles divided by 3.

261
00:11:54,115 --> 00:11:57,610
Let's see if we can manipulate
this thing a little bit.

262
00:11:57,610 --> 00:12:02,420
So if we multiply both sides
by-- let's see what we can do.

263
00:12:02,420 --> 00:12:14,390
If we multiply both sides by 3v,
we get pv times 3 is equal

264
00:12:14,390 --> 00:12:21,730
to mv squared, times N, where N
is the number of particles.

265
00:12:21,730 --> 00:12:24,580
Let's divide both sides by N.

266
00:12:24,580 --> 00:12:34,270
So we get 3pv over-- actually,
no, let me leave the N there.

267
00:12:34,270 --> 00:12:41,240
Let's divide both sides
of this equation by 2.

268
00:12:41,240 --> 00:12:43,520
So we get, what do we get?

269
00:12:43,520 --> 00:12:49,320
We get 3/2 pv is equal to--
now this is interesting.

270
00:12:49,320 --> 00:12:54,780
It's equal to N, the number of
particles we have, times mv

271
00:12:54,780 --> 00:12:57,920
squared over 2.

272
00:12:57,920 --> 00:12:59,610
Remember, I just divided
this equation right

273
00:12:59,610 --> 00:13:00,650
here by 2 to get this.

274
00:13:00,650 --> 00:13:02,280
And I did this for a very
particular reason.

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00:13:02,280 --> 00:13:05,260
What is mv squared over 2?

276
00:13:05,260 --> 00:13:09,910
mv squared over 2 is the kinetic
energy of that little

277
00:13:09,910 --> 00:13:11,140
particle we started off with.

278
00:13:11,140 --> 00:13:12,900
That's the formula for
kinetic energy.

279
00:13:12,900 --> 00:13:20,070
Kinetic energy is equal
to mv squared over 2.

280
00:13:20,070 --> 00:13:21,950
So this is the kinetic energy
of one particle.

281
00:13:21,950 --> 00:13:28,570


282
00:13:28,570 --> 00:13:30,890
Now, we're multiplying that
times the total number of

283
00:13:30,890 --> 00:13:33,280
particles we have, times N.

284
00:13:33,280 --> 00:13:35,890
So N times the kinetic energy of
one particle is going to be

285
00:13:35,890 --> 00:13:37,490
the kinetic energy of
all the particles.

286
00:13:37,490 --> 00:13:39,430
And, of course, we also made
another assumption.

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00:13:39,430 --> 00:13:41,160
I should state that I assumed
that all the particles are

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00:13:41,160 --> 00:13:43,510
moving with the same velocity
and have the same mass.

289
00:13:43,510 --> 00:13:45,880
In a real situation, the
particles might have very

290
00:13:45,880 --> 00:13:47,060
different velocities.

291
00:13:47,060 --> 00:13:48,980
But this was one of our
simplifying assumptions.

292
00:13:48,980 --> 00:13:50,850
So, we just assumed they
all have that.

293
00:13:50,850 --> 00:13:53,890
So, if I multiply N times that--
this statement right

294
00:13:53,890 --> 00:13:56,285
here-- is the kinetic energy
of the system.

295
00:13:56,285 --> 00:14:01,700


296
00:14:01,700 --> 00:14:03,140
Now, we're almost there.

297
00:14:03,140 --> 00:14:04,480
In fact, we are there.

298
00:14:04,480 --> 00:14:08,610
We just established that the
kinetic energy of the system

299
00:14:08,610 --> 00:14:12,730
is equal to 3/2 times the
pressure, times the volume of

300
00:14:12,730 --> 00:14:13,650
the system.

301
00:14:13,650 --> 00:14:16,270
Now, what is the kinetic
energy of the system?

302
00:14:16,270 --> 00:14:17,320
It's the internal energy.

303
00:14:17,320 --> 00:14:19,290
Because we said all the energy
in the system, because it's a

304
00:14:19,290 --> 00:14:22,900
simple ideal monoatomic gas,
all of the energy in the

305
00:14:22,900 --> 00:14:25,770
system is in kinetic energy.

306
00:14:25,770 --> 00:14:30,980
So we could say the internal
energy of the system is equal

307
00:14:30,980 --> 00:14:33,370
to-- that's just the total
kinetic energy of the system--

308
00:14:33,370 --> 00:14:38,020
it's equal to 3/2 times our
total pressure, times our

309
00:14:38,020 --> 00:14:38,620
total volume.

310
00:14:38,620 --> 00:14:41,070
Now you might say, hey, Sal,
you just figured out the

311
00:14:41,070 --> 00:14:42,130
pressure on this side.

312
00:14:42,130 --> 00:14:43,960
What about the pressure on that
side, and that side, and

313
00:14:43,960 --> 00:14:45,730
that side, or on every
side of the cube?

314
00:14:45,730 --> 00:14:46,850
Well, the pressure on
every side of the

315
00:14:46,850 --> 00:14:48,250
cube is the same value.

316
00:14:48,250 --> 00:14:50,530
So all we have to do is find
in terms of the pressure on

317
00:14:50,530 --> 00:14:52,420
one side, and that's essentially
the pressure of

318
00:14:52,420 --> 00:14:54,480
the system.

319
00:14:54,480 --> 00:14:55,590
So what else can we
do with that?

320
00:14:55,590 --> 00:15:00,700
Well, we know that pv is equal
to nRT, our ideal gas formula.

321
00:15:00,700 --> 00:15:06,475
pv is equal to nRT, where this
is the number of moles of gas.

322
00:15:06,475 --> 00:15:08,410
And this is the ideal
gas constant.

323
00:15:08,410 --> 00:15:10,240
This is our temperature
in kelvin.

324
00:15:10,240 --> 00:15:12,880
So if we make that replacement,
we'll say that

325
00:15:12,880 --> 00:15:17,140
internal energy can also be
written as 3/2 times the

326
00:15:17,140 --> 00:15:20,500
number of moles we have, times
the ideal gas constant, times

327
00:15:20,500 --> 00:15:22,290
our temperature.

328
00:15:22,290 --> 00:15:24,560
Now, I did a lot of work, and
it's a little bit mathy.

329
00:15:24,560 --> 00:15:27,780
But these results are,
one, interesting.

330
00:15:27,780 --> 00:15:30,140
Because now you have a
direct relationship.

331
00:15:30,140 --> 00:15:32,860
If you know the pressure and the
volume, you know what the

332
00:15:32,860 --> 00:15:37,530
actual internal energy, or
the total kinetic energy,

333
00:15:37,530 --> 00:15:38,660
of the system is.

334
00:15:38,660 --> 00:15:41,350
Or, if you know what the
temperature and the number of

335
00:15:41,350 --> 00:15:44,320
molecules you have are, you also
know what the internal

336
00:15:44,320 --> 00:15:45,630
energy of the system is.

337
00:15:45,630 --> 00:15:48,520
And there's a couple of key
takeaways I want you to have.

338
00:15:48,520 --> 00:15:52,070
If the temperature does not
change in our ideal situation

339
00:15:52,070 --> 00:15:56,930
here-- if delta T is equal to
0-- if this doesn't change,

340
00:15:56,930 --> 00:15:58,680
the number particles aren't
going to change.

341
00:15:58,680 --> 00:16:04,750
Then our internal energy does
not change as well.

342
00:16:04,750 --> 00:16:07,030
So if we say that there is
some change in internal

343
00:16:07,030 --> 00:16:10,790
energy, and I'll use this in
future proofs, we could say

344
00:16:10,790 --> 00:16:16,360
that that's equal to 3/2 times
nR times-- well, the only

345
00:16:16,360 --> 00:16:18,850
thing that can change, not the
number molecules or the ideal

346
00:16:18,850 --> 00:16:21,350
gas constant-- times
the change in T.

347
00:16:21,350 --> 00:16:27,000
Or, it could also be written as
3/2 times the change in pv.

348
00:16:27,000 --> 00:16:28,450
We don't know if either
of these are constant.

349
00:16:28,450 --> 00:16:30,520
So we have to say the change
in the product.

350
00:16:30,520 --> 00:16:32,780
Anyway, this was a
little bit mathy.

351
00:16:32,780 --> 00:16:34,050
And I apologize for it.

352
00:16:34,050 --> 00:16:36,840
But hopefully, it gives you a
little bit more sense that

353
00:16:36,840 --> 00:16:39,300
this really is just the sum
of all the kinetic energy.

354
00:16:39,300 --> 00:16:42,420
We related it to some of these
macro state variables, like

355
00:16:42,420 --> 00:16:44,590
pressure, volume, and time.

356
00:16:44,590 --> 00:16:47,640
And now, since I've done the
video on it, we can actually

357
00:16:47,640 --> 00:16:50,990
use this result in
future proofs.

358
00:16:50,990 --> 00:16:53,220
Or at least you won't complain
too much if I do.

359
00:16:53,220 --> 00:16:54,470
Anyway, see you in
the next video.

360
00:16:54,470 --> 00:00:00,000