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## Class 11 Physics (India)

### Course: Class 11 Physics (India) > Unit 16

Lesson 2: Work from PV graph and internal energy- Work from expansion
- PV-diagrams and expansion work
- Work from PV diagrams
- What are PV diagrams?
- Proof: U = (3/2)PV or U = (3/2)nRT
- Internal energy
- Heat capacity at constant volume and pressure
- Applications of first law
- Applications of first law - cyclic processes

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# Proof: U = (3/2)PV or U = (3/2)nRT

Conceptual proof that the internal energy of an ideal gas system is 3/2 PV. Created by Sal Khan.

## Want to join the conversation?

- Why do we count the number of particles striking a wall as 1/3 instead of 1/6 (the six faces of a cube)?(11 votes)
- actually he is saying 1/3 particles stricking two parllel walls which we have only 3 parllel walls(21 votes)

- i'm still confused, why do we have to divide the particles by 3? :s(5 votes)
- Daniel, the particles are divided by 3 because you have 1/3 of the particles going in one direction (up and down), 1/3 of the particles going in another direction (side to side), and 1/3 of the particles going in another direction (forward or backward). For example, suppose I have 9 particles in the cube; that means N = 9. That means that 3 are going up and down, 3 are going to the sides, and 3 are going forward and backward. So, if I asked you to talk about the pressure in one direction, you would talk about only 3 of the particles, which is N/3 (=9/3=3).

Hopefully, that helps!(15 votes)

- Even though people have tried explaining the time being 2x / v, it is still confusing for me.

Force is defined as follows: "If a force F is applied to a particle for a time interval Δt, the momentum of the particle changes by an amount Δ P = F * t". So if the time, t, is how long the force was applied for, then why include the time that the particle is not applying a force to the wall? Which is what is done when you include the time taken to travel 2x. It is not applying a force to the wall during that time.

Let's make a macroscopic example of this situation. An elastic ball is rolling from one wall in a buiding to another with mass, m, and velocity, v. It rolls back and forth one time by 'bouncing' off the wall elastically as it reaches the wall on the other side. The time of the bounce is 1 second, in other words the time the ball has to apply a force to the wall and have its momentum changed. It takes 10 second for the ball to reach the wall and come back to its initial position.

Calculating the force applied to the wall as shown in the video means F = m* delta v / delta t, where delta t is 10 seconds (the time it takes the ball to go to and from the wall), and delta v is 2v because the difference is the velocity vector direction has changed so you get delta v = v - (-v) = 2v. So F = m * 2v / 10 = m*v/5

If I calculate it according to how I originally understand how force equals the change in momentum I get the following:

F = m * delta p / delta t, where delta t is the 1 second the ball is in contact with the wall during the 'bounce' and delta p is the same as above: 2v. We get F = m * 2v / 1 = 2*mv

Clearly the method shown in the video gives a much smaller force than when considering time as only the time when the object is applying the force to the wall. What am I misunderstanding here?(7 votes)- In the video, it is safe to assume that the force is applied over a time period of 2x/v, because within that time period the particle hits each wall once. Imagine with your bouncy ball, you had it bouncing between 2 buildings. It would hit both buildings every 10 seconds, so it would apply the force of those two collisions every ten seconds. This may seem confusing on a small scale, but on a large scale with many particles, that averages out to about the same number of collisions per second, which is really what we're trying to calculate with this.(7 votes)

- U=(3/2)PV is ONLY true for an IDEAL gas. Is this correct?(7 votes)
- Yes. That's why he can say that the total internal energy is accounted for by kinetic energy only.(8 votes)

- At14:19it's said that all of the internal energy of the system is in kinetic energy because it's an ideal monoatomic gas. Even in a monoatomic gas, such as He, every atom has potential and kinetic energy "stored" within it due to the electrons moving about the nucleus. Why isn't this energy also included as part of the internal energy?(3 votes)
- He is not an ideal gas. An ideal gas is hypothetical(5 votes)

- At6:19shouldn't the change in momentum for the molecule that got ricochet off the wall be minus 2mv (because change in momentum for the molecule= final momentum - initial momentum that is equal to minus 2mv). Did Sal considered momentum imparted to the wall that is equal to plus 2mv?(2 votes)
- Yes u are right Sal considered that momentum imparted to wall by particle.when we calculate force as rate of change of momentum , we take force exerted by molecule on the wall not the force exerted by wall on the molecule. force exerted by wall on the molecule would be -2mv/t but force exerted by molecule on the wall will be 2mv/t because they are action reaction forces and will be equal in magnitude but opposite in direction.(6 votes)

- I thought Total Thermal Energy---not Internal Energy, U---is the "Kinetic energy of all particles."

Or is the "Total Thermal Energy" the same as "Internal Energy," U, for Ideal gases, since there are no potential energies of attraction between gaseous species?(2 votes)- Yes, for ideal gas, thermal energy IS the total internal energy, because there are assumed to be no other forms of energy present (eg, as you said, there is no PE)(2 votes)

- isn't the change in momentum equal to -2mv? my physics teacher told me that so i'm a little bit confused right now.(2 votes)
- Your point is somehow correct. just consider the particle is moving from
**right to left**instead of thinking it going from left to right. then you will get**2mv**. Hope it helps.(2 votes)

- F=change in p/change in T

T should be the impact time

But why did you take delta T ... (2X/V) to be the time taken for the particle to go and return??(2 votes)- This means that every 2x/v time, the particle hits both walls. If T was simply the time it took for the particle to impact the wall, that would represent the single particle hitting a side of the container constantly, without bouncing back.(1 vote)

- Why do we count the number of particles striking a wall as 1/3?(2 votes)
- Since there are three dimensions of a cube, 1/3 would go in the x direction, 1/3 would go to the y direction, while the remaining 1/3 would go to the z direction. Hope this helps with your understanding!(1 vote)

## Video transcript

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. We're getting someplace
interesting already. 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 care about the pressure. 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.