# Practice English Speaking&Listening with: Proof: U = (3/2)PV or U = (3/2)nRT | Thermodynamics | Physics | Khan Academy

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

the internal energy of a system.

And it's really everything thrown in there.

It's the kinetic energy of the molecules.

It has the potential energy if the molecules are vibrating.

It has the chemical energy of the bonds.

It has the potential energy of electrons that want to get

some place.

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

an introductory chemistry, physics, or thermodynamics

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

that's an ideal gas.

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

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

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

be the kinetic energy of each of these particles.

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

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

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

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

volume, or temperature.

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

internal energy.

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

what's the change in internal energy.

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

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

or done by the system.

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

how do we know how much internal energy we even have

in a system?

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

experiment.

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

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

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

And something tells me that I might have already done this

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

think I related exactly to internal energy.

So I'll do that here.

Let's say my system is this cube.

And let's say the dimensions of the cube

are x in every direction.

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

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

And let's say I have n particles in my

system, capital N.

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

it straightforward.

I have N particles.

So they're all doing what they will.

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

simplification.

But I think it's reasonable.

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

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

possible random direction.

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

sides, that's what causes the pressure.

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

cetera, in all random directions.

Now, for the sake of simplicity of our mathematics,

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

time, I'm going to make an assumption.

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

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

parallel to each of the axes.

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

guess we could say, left to right.

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

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

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

it makes our math a lot simpler.

And if you actually were to do the statistical mechanics

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

would actually end up getting the same result.

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

oversimplification.

There is some infinitesimally small chance that we actually

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

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

it's such a small probability.

But this could actually be our system.

And this system would generate pressure.

And it makes our math a lot simpler.

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

So let's take a sideways view.

Let's take a sideways view right here.

And let's just study one particle.

Maybe I should have done it in green.

But let's say I have one particle.

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

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

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

particle exert on this wall right here?

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

The area of this wall is x times x.

So it's x squared area.

How much force is being exerted by this particle?

Well, let's think about it this way.

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

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

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

We know that force is equal to mass times acceleration.

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

mass times, change in velocity over change in time.

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

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

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

We could put that inside of the change.

So it's delta mv over change in time.

And this is just change in momentum, right?

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

So that's another way to write force.

So what's the change in momentum going

to be for this particle?

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

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

Its momentum is equal to mv.

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

ricochet straight back.

And what's its momentum going to be?

Well, it's going to have the same mass

and the same velocity.

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

Nothing is lost to heat or whatever else.

But the velocity is in the other direction.

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

velocity has switched directions.

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

off with a momentum of minus mv, what's

my change in momentum?

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

well, it's the difference between these two,

which is just 2mv.

Now, that doesn't give me the force.

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

So how often does this happen?

How frequently?

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

We're going to hit this wall.

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

off of that wall, and then come back

here and hit it again.

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

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

collisions?

Well, the particle has to travel x going back.

It's going to collide.

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

This distance is x.

Let me do that in a different color.

This distance right here is x.

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

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

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

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

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

Distance is equal to rate times time.

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

of time we took.

This is just our basic motion formula.

Our delta T, the distance we have to

travel is back and forth.

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

Well, our rate is our velocity.

Divided by v.

There you go.

So this is our delta T right here.

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

incident momentum.

Because we ricocheted back with the same magnitude, but

negative momentum.

So that's our change in momentum.

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

It's the total distance we have to travel between

collisions of this wall, divided by our velocity.

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

the reciprocal of this-- so this is just fraction

math-- v over 2x.

And what is this equal to?

The 2's cancel out.

So that is equal to mv squared, over x.

Interesting.

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

me for a second.

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

this-- force from one particle on this wall.

Now, what was the area?

We wrote it up here.

The pressure is equal to the force per area.

So this is the force of that particle.

So that's mv squared over x, divided by

the area of the wall.

Well, what's the area of the wall?

The area of the wall here, each sideis x.

And so if we draw the wall there, it's x times x.

It's x squared.

So divided by the area of the wall, is x squared.

And what does this equal?

This is equal to mv squared over x cubed.

You can just say, this is times 1 over x squared, when

this all becomes x cubed.

This is just fraction math.

So now we have an interesting thing.

The pressure due to this one particle-- let's just call

this from this one particle-- is equal to m v

squared over x cubed.

Now, what's x cubed?

That's the volume of our container.

Over the volume.

I'll do that in a big V, right?

So let's see if we can relate this to something else that's

interesting.

So that means that the pressure being exerted by this

one particle-- well, actually let me just take another step.

So this is one particle on this wall, right?

This is from one particle on this wall.

Now, of all the particles-- we have N particles in our cube--

what fraction of them are going to be

bouncing off of this wall?

That are going to be doing the exact same

thing as this particle?

Well, I just said.

1/3 are going to be going in this direction.

1/3 are going to be going up and down.

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

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

to be doing exactly what this particle is going to be doing.

This is the pressure from one particle.

If I wanted the pressure from all of the particles on that

wall-- so the total pressure on that wall is going to be

from N over 3 of the particles.

The other particles aren't bouncing off that wall.

So we don't have to worry about them.

So if we want the total pressure on that wall-- I'll

just write, pressure sub on the wall.

Total pressure on the wall is going to be the pressure from

one particle, mv squared, over our volume, times the total

number of particles hitting the wall.

The total number of particles is N divided by 3, because

only 3 will be going in that direction.

So, the total pressure on that wall is equal to mv squared,

over our volume of our container, times the total

particles divided by 3.

Let's see if we can manipulate this thing a little bit.

So if we multiply both sides by-- let's see what we can do.

If we multiply both sides by 3v, we get pv times 3 is equal

to mv squared, times N, where N is the number of particles.

Let's divide both sides by N.

So we get 3pv over-- actually, no, let me leave the N there.

Let's divide both sides of this equation by 2.

So we get, what do we get?

We get 3/2 pv is equal to-- now this is interesting.

It's equal to N, the number of particles we have, times mv

squared over 2.

Remember, I just divided this equation right

here by 2 to get this.

And I did this for a very particular reason.

What is mv squared over 2?

mv squared over 2 is the kinetic energy of that little

particle we started off with.

That's the formula for kinetic energy.

Kinetic energy is equal to mv squared over 2.

So this is the kinetic energy of one particle.

Now, we're multiplying that times the total number of

particles we have, times N.

So N times the kinetic energy of one particle is going to be

the kinetic energy of all the particles.

And, of course, we also made another assumption.

I should state that I assumed that all the particles are

moving with the same velocity and have the same mass.

In a real situation, the particles might have very

different velocities.

But this was one of our simplifying assumptions.

So, we just assumed they all have that.

So, if I multiply N times that-- this statement right

here-- is the kinetic energy of the system.

Now, we're almost there.

In fact, we are there.

We just established that the kinetic energy of the system

is equal to 3/2 times the pressure, times the volume of

the system.

Now, what is the kinetic energy of the system?

It's the internal energy.

Because we said all the energy in the system, because it's a

simple ideal monoatomic gas, all of the energy in the

system is in kinetic energy.

So we could say the internal energy of the system is equal

to-- that's just the total kinetic energy of the system--

it's equal to 3/2 times our total pressure, times our

total volume.

Now you might say, hey, Sal, you just figured out the

pressure on this side.

What about the pressure on that side, and that side, and

that side, or on every side of the cube?

Well, the pressure on every side of the

cube is the same value.

So all we have to do is find in terms of the pressure on

one side, and that's essentially the pressure of

the system.

So what else can we do with that?

Well, we know that pv is equal to nRT, our ideal gas formula.

pv is equal to nRT, where this is the number of moles of gas.

And this is the ideal gas constant.

This is our temperature in kelvin.

So if we make that replacement, we'll say that

internal energy can also be written as 3/2 times the

number of moles we have, times the ideal gas constant, times

our temperature.

Now, I did a lot of work, and it's a little bit mathy.

But these results are, one, interesting.

Because now you have a direct relationship.

If you know the pressure and the volume, you know what the

actual internal energy, or the total kinetic energy,

of the system is.

Or, if you know what the temperature and the number of

molecules you have are, you also know what the internal

energy of the system is.

And there's a couple of key takeaways I want you to have.

If the temperature does not change in our ideal situation

here-- if delta T is equal to 0-- if this doesn't change,

the number particles aren't going to change.

Then our internal energy does not change as well.

So if we say that there is some change in internal

energy, and I'll use this in future proofs, we could say

that that's equal to 3/2 times nR times-- well, the only

thing that can change, not the number molecules or the ideal

gas constant-- times the change in T.

Or, it could also be written as 3/2 times the change in pv.

We don't know if either of these are constant.

So we have to say the change in the product.

Anyway, this was a little bit mathy.

And I apologize for it.

But hopefully, it gives you a little bit more sense that

this really is just the sum of all the kinetic energy.

We related it to some of these macro state variables, like

pressure, volume, and time.

And now, since I've done the video on it, we can actually

use this result in future proofs.

Or at least you won't complain too much if I do.

Anyway, see you in the next video.

The Description of Proof: U = (3/2)PV or U = (3/2)nRT | Thermodynamics | Physics | Khan Academy