- Macrostates and microstates
- Quasistatic and reversible processes
- First law of thermodynamics / internal energy
- More on internal energy
- What is the first law of thermodynamics?
- Work from expansion
- PV-diagrams and expansion work
- What are PV diagrams?
- Proof: U = (3/2)PV or U = (3/2)nRT
- Work done by isothermic process
- Carnot cycle and Carnot engine
- Proof: Volume ratios in a Carnot cycle
- Proof: S (or entropy) is a valid state variable
- Thermodynamic entropy definition clarification
- Reconciling thermodynamic and state definitions of entropy
- Entropy intuition
- Maxwell's demon
- More on entropy
- Efficiency of a Carnot engine
- Carnot efficiency 2: Reversing the cycle
- Carnot efficiency 3: Proving that it is the most efficient
Conceptual proof that the internal energy of an ideal gas system is 3/2 PV. Created by Sal Khan.
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- Why do we count the number of particles striking a wall as 1/3 instead of 1/6 (the six faces of a cube)?(12 votes)
- actually he is saying 1/3 particles stricking two parllel walls which we have only 3 parllel walls(22 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!(17 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)
- 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)
- According to the newtons second law force exerted by a particle is equal to the change in momementom of a particle divided by the time in which momemtom changed.Then why Sal is taking time as the time taken by the partice to come back after collision here.(2 votes)
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.