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Carnot cycle and Carnot engine

Introduction to the Carnot cycle and Carnot heat engine. Created by Sal Khan.

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  • blobby green style avatar for user ilovejingle
    for the D to A process, it's adiabatic and u add pebbles to reach state A, but how do u know it will reach state A exactly instead of somewhere else. i mean how do you make sure it will be the same volume after u add the pebbles
    (7 votes)
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    • leaf green style avatar for user Gordon N. Fleming
      One has to choose the state, D, very carefully. In other words, the isothermal compression from state C must proceed only until the system reaches the intersection of the isothermal curve for temperature, T2, and the adiabatic curve coming down from state A. That intersection defines the state, D.
      (9 votes)
  • old spice man green style avatar for user Michael Tardrew
    Is this how a reverse cycle air-conditioner works? With the inside being a reservoir and the out side being the other reservoir so for cooling the outside half is compressed allowing the heat to be drawn out side and for heating the inside is compressed allowing the heat to flow into the room?
    (6 votes)
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    • purple pi purple style avatar for user APDahlen
      Hello Michael,

      Correct:

      Summer: the evaporator is inside the house

      Winter: the evaporator is outside the house

      Most systems will use the same coils for summer and winter but will include valves (solenoids) to reverse the connections to the compressor.

      At home I have on of these heat pumps. On a cold winter day the outside coil is seriously cold. You would not want to stand in front of it!

      Regards,

      APD
      (4 votes)
  • blobby green style avatar for user konkarmilla
    At when we start adding pebbles to the system to move from state D to state A was it an adiabatic process?
    (2 votes)
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  • blobby green style avatar for user Jordan Yusufali
    From B to C , how come its follows a ln curve even though temperature isn't constant ?

    (2 votes)
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    • piceratops tree style avatar for user h.j.l.hendrickx
      The curve from A to B is an isotherm, meaning it's temperature is contstant.
      The curve from B to C is adiabatic, which is a different curve. (you can see that is doesn't follow the isotherm.)
      That it is a curve is because the system can not jump from one volume to another.

      Hope this helped.
      (10 votes)
  • piceratops sapling style avatar for user Vaishnavi Gaikwad
    Although I intuitively feel that Q1-Q2=W,I still doubt this looking at the PV chart. The area under curve AB =Q1, similarly area under curve CD=Q2 but it is not evident from graph that Q1-Q2=W since curve CD is slightly shifted towards right than AD. How then can we substract? Is there a proof that the substraction yeilds work done?
    (4 votes)
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    • blobby green style avatar for user ambientsilence
      Look at stages 2 (going from B to C) and 4 (D->A) of the cycle. I claim that the work done by the system in stage 2 is the same as the work done to the system in stage 4.

      Since these processes are adiabatic we have Q=0 and thus:

      delta_U = W

      Now note that A and B lie on the same isotherm so they have same internal energy. Recall the video where Sal proved that U=3/2nrT. Let's call this internal energy U_1. By the same argument the internal energy at C and D are the same. Let's call this U2.

      So we have

      U_2 - U_1 = W (stage 2)
      U_1 - U_2 = W (stage 4)

      proving that the total work for these two stages combined is indeed 0.
      (3 votes)
  • leaf green style avatar for user Mehul Sharma
    Doesn't an adiabatic process violate the 1st law of thermodynamics ?? If the system is isolated and cant exchange heat with its surroundings and if the KE is used up in doing work , how does the system regain that energy ?? Is it converted to Potential Energy ?
    (3 votes)
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    • leaf blue style avatar for user Chris Pedersen
      Adiabatic means that the heat exchange, Q, is 0, not that the internal energy change is 0 (internal energy change is 0 for the isothermal processes). If the process is adiabatic, Q=0, so you can work out the change in internal energy (and therefore temperature) using the 1st law:

      dU=dQ-dW
      dU=-dW

      So the energy for the work done comes from the KE of the particles being used up like you say, and the gas just has a lower internal energy (=lower temperature) at the end of the process.
      (4 votes)
  • leaf red style avatar for user elliekat94
    What's the main purpose of the Carnot Cycle though??
    (2 votes)
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    • male robot johnny style avatar for user surya.bond.prakash
      originally carnot gone into developing this theorem to find the answer to two questions
      1)is the work from the fuel is unbounded
      2)can we increase the efficiency of steam engine by replacing the working fluid

      for the first question we know that the work from a fuel is not unbounded,but limited by the heat content of the system
      for the second question,from the equation of carnot efficiency the engine efficiency cannot be increased by replacing the working fluid.

      But now in common we know that diesel engine is more efficient than petrol engine??
      please explain this phenomenon when the working fluid changes it does have an impact on engine efficiency?
      (3 votes)
  • leafers tree style avatar for user #Sonu Mathew
    Why is pv work done by the system -ve in chemistry and +ve in physics?
    (3 votes)
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    • female robot ada style avatar for user Rodrigo Campos
      That's more of a semantics difference than an actual distinction in definition. You can say ∫PdV is +(work done by the system) or -(work done on the system). It doesn't depend on whether you're studying physics or chemistry, as well. I've had classes with teachers of different disciplines using the same convention, its something kind of personal haha. My tip: use the one you feel more confortable with and stick to it, while keeping in mind what I wrote on the second sentence "You can say ∫PdV is the +(work done by the system) or -(work done on the system)".
      (2 votes)
  • male robot hal style avatar for user Jerrod Hess
    What equation does the adiabatic curve follow (specifically)?
    I know that PV = nRT.
    I'm guessing that the temperature is a function of pressure AND volume T(P,V), but what is the relationship?
    (2 votes)
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    • blobby green style avatar for user Siddharth Suhas Pawar
      P(V)^n=Constant where n is called the adiabatic exponent IF the process is ADIABATIC For an adiabatic process
      n=cp/cv. where cp is the specific heat at constant pressure and cv is the specific heat at constant volume....... Notice that when n=0 ,u get a const pressure line (horizontal). ,when n =1 u get a isothermal curve (rectangular hyperbola) and when n= infinity u get a constant volume curve (vertical line on PV graph.)... Now for adiabatic process ,lets say for system of air, the adiabatic exponent equals 1.40 whixh is >1 and hence the adiabatic curve is MORE sloping down than isothermal curve for air.
      (2 votes)
  • duskpin ultimate style avatar for user kassinjack
    Can space be considered a reservoir? if so how come the earth and all the other planets didn't match its temperature?
    (2 votes)
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Video transcript

Let's start with this classic system that I keep referring to in our thermodynamics videos. I have a cylinder. It's got a little piston on the top of it, or it's got a ceiling that's movable. The gas, and we're thinking of monoatomic ideal gases in here, they're exerting pressure onto this ceiling. And the reason why the ceiling isn't moving all the way up, is because I've placed a bunch of rocks on the top to offset the force per area of the actual gas. And I start this gas when it's in equilibrium. I can define its macrostates. It has some volume. It has some pressure that's being offset by these rocks. And it has some well-defined temperature. Now, what I'm going to do is, I'm going to place this system here-- I'm going to place it on top of a reservoir. And I talked about what a reservoir was either in the last video or a couple of videos ago. You can view it as an infinitely large object, if you will, of a certain temperature. So if I put it next to-- if I put our system next to this reservoir-- and let's say I start removing pebbles from our system. We learned a couple of videos ago that if we did it adiabatically-- what does adiabatically mean? If we removed these pebbles in isolation, without any reservoir around, the volume would increase, the pressure would go down, and actually the temperature would decrease, as well. We showed that a couple of videos ago. So by putting this big reservoir there that's a lot larger than our actual canister, this will keep the temperature in our canister at T1. You can kind of view a reservoir as-- say I had a cup of water in a stadium. And the air conditioner in the stadium is at 60 degrees. Well, no matter what I do to that water, I could put it in the microwave and warm it up, but if I put it back in that stadium, that stadium is going to keep that water at 60 degrees. And you might say, oh, won't the reservoir's temperature decrease if it's throwing off heat? Well, it would, but it's so much larger that its impact isn't noticeable. For example, if I put a cup of boiling water into a super large covered-dome stadium, the water will get colder to the ambient temperature of the stadium. The stadium will get warmer, but it will be so marginally warmer than you won't even notice it. So you can kind of view that as a reservoir. And theoretically, this is infinitely large. So the effect of this is, as we remove these little rocks, we're going to keep the temperature constant. And remember, if we're keeping the temperature constant, we're also keeping the internal energy constant, because we're not changing the kinetic energy of the particles. So let me see what happens. So I keep doing that. And so I get to a point-- let me see-- where my volume has increased-- so let me delete some of my rocks here. Delete some of the rocks. So some of the rocks are gone. And now my overall volume is going to be larger. Let me move this up a little bit. And then let me color this in black. Oh, whoops. Let me color this in, just to give an idea. So our volume has gotten a bit larger. And let me get my pen correctly. So our volume has gotten larger by roughly this amount. We have the same number of particles. They're going to bump into the ceiling a little less frequently, so my pressure would have gone down. But because I kept this reservoir here, because this reservoir was here the whole time during this process, the temperature stayed at T1. And that was only because of this reservoir. And I want to make that clear. And also, just as review, this is a quasi-static process, because I'm doing it very slowly. The system is in equilibrium the whole time. So let's draw what we have so far on our famous PV diagram. So this is the P-axis. That's the V-axis. You label them. This is P. This is V. Let me call this-- I'm going to do it in a good color. This is state A of the system. This is state B of the system. So state A starts at some pressure and volume-- I'll do it like that. That's state A. And it moves to state B. And notice, I kept the temperature constant. And what did we learn in, I think it was one or two videos ago? Well, we're at a constant temperature, so we're going to move along an isotherm, which is just a rectangular hyperbola. Because when your temperature is constant, your pressure times your volume is going to equal a constant number. And I went over that before. So we're going to move over-- our path is going to look something like this, and I'll move here, to state B. I'll move over here to state B. And the whole time, this was at a constant temperature T1. Now, we've done a bunch of videos now. We said, OK, how much work was done on this system? Well, the work done on the system is the area under this curve. So some positive work was-- not done on the system, sorry. How much work was done by the system? We're moving in this direction. I should put the direction there. We're moving from left to right. The amount of work done by the system is pressure times volume. We've seen that multiple times. So you take this area of the curve, and you have the work done by the system from A to B. Right? So let's call that work from A to B. Now, that's fair and everything, but what I want to think more about, is how much heat was transferred by my reservoir? Remember, we said, if this reservoir wasn't there, the temperature of my canister would have gone down as I expanded its volume, and as the pressure went down. So how much heat came into it? Well, let's go back to our basic internal energy formula. Change in internal energy is equal to heat applied to the system minus the work done by the system Now, what is the change in internal energy in this scenario? Well, it was at a constant temperature the whole time, right? And since we're dealing with a very simple ideal gas, all of our internal energy is due to kinetic energy, which temperature is a measure of. So, temperature didn't change. Our average kinetic energy didn't change, which means our kinetic energy didn't change. So our internal energy did not change while we moved from left to right along this isotherm. So we could say our internal energy is zero. And that is equal to the heat added to the system minus the work done by the system. Right? So if you just-- we put the work done by the system on the other side, and then switch the sides, you get heat added to the system is equal to the work done by the system. And that makes sense. The system was doing some work this entire time, so it was giving energy to-- well, you know, it was giving essentially maybe some potential energy to these rocks. So it was giving energy away. It was giving energy outside of the system. So how did it maintain its internal energy? Well, someone had to give it some energy. And it was given that energy by this reservoir. So let's say, and the convention for doing this is to say, that it was given-- let me write this down. It was given some energy Q1. We just say, we just put this downward arrow to say that some energy went into the system here. Fair enough. Now let's take this state B and remove the reservoir, and completely isolate ourselves. So there's no way that heat can be transferred to and from our system. And let's keep removing some rocks. So if we keep removing some rocks, where do we get to? Let me go down here. So let's say we remove a bunch of more rocks. So let me erase even more rocks than we had in B. Maybe I only have one rock left. And obviously, the overall volume would have increased. So let me make our piston go up like that, and I can make our piston is maybe a lot higher now. And let me just fill in the rest of our, just so that we don't have some empty space there. So if I fill that in right there-- OK Let me fill that in. And then I just use the blue-- I should be talking about thermodynamics, not drawing. But you get the idea. And then I have some more-- you know, I shouldn't add particles. But my volume has increased a good bit. My pressure will have gone down, they're going to bump into the walls less. And because I removed the reservoir, what's going to happen to the temperature? My temperature is going to go down. This was an adiabatic process. So an adiabatic just means we did it in isolation. There was no exchange of heat from one system to another. So let me just-- this arrow continues down here. I'll say adiabatic. Now, since I'm moving from one temperature to another, this is at T2. So I will have moved to another isotherm. This is the isotherm for T1. If I keep my temperature constant, I move along this hyperbola. And I would have kept moving along this hyperbola. But now that we didn't keep our temperature constant, we now move like this. We move to another isotherm. So let's say I have another isotherm at T2. It looks something like this. So let me draw like that. So let's say I have another-- it should actually curve up a little bit. So let's say, everything at temperature T2, depending on its pressure and volume, is someplace along this curve that asymptotes up like that, and then goes to the right like that. Now, I would have moved down to this isotherm, and my pressure would have kept going down, and my volume would have kept going down. So this move, from B to state C, will look like this. Let me do to it in another color. Let me do it in the orange color of this arrow. So it will look like this. And now we're at state C. Now, this was adiabatic. Which means, there is no exchange of heat. So I don't have to figure out how much heat got transferred into the system. Now, there's something interesting here. We still did do some work. We can take the area under this curve. And we're going to leave it to a future video to think about where that work energy-- well, the main thing is, is what was reduced by that work energy. And, well, if you think [UNINTELLIGIBLE] to leave it to future video. Our internal energy was reduced, right? Because our temperature went down. So our internal energy went down. We'll talk more about that in the future video. So now that we're at state C, and we're at temperature T2. Let's put back another sink here. But this sink, what it's going to have is a reservoir. So let me put two things right here. So I'm going to add-- let me erase some of these blocks in black. So now I'm going to add blocks back. I'm going to add little pebbles back into it. But I'm going to do it as an isothermic process. I'm going to do it with a reservoir here. But this reservoir here, it's not going to be the same reservoir that I put up there. I swapped that one out. I got rid of any reservoir when I went from B to C. And now I'm going to swap in a new reservoir. Actually, let me make it blue. Because it's going to be-- Because here's what's happening. I'm now adding pebbles in. I'm compressing the gas. If this was an adiabatic process, the gas would want to heat up. So what I'm doing is, I need to put a reservoir to keep it at T2, to keep it along this isotherm. So this is T2. Remember, this reservoir is kind of a cold reservoir. It keeps the temperature down. As opposed to here. This was a hot reservoir. It kept the temperature up. So you can imagine. The heat generated in the system, or internal energy being generated in the system-- well, no, I shouldn't say that. The temperature of the system will want to go up, but it's being released, because it's able to transfer that heat into our new reservoir. And that amount of heat is Q2. So I move along this. This is right here. I'm moving along another isotherm, I'm moving along this isotherm. Until I get to state D. We're almost there. This is state D. So state D will be someplace here, along this isotherm right here. Maybe this is state D. And once again, you can make the argument that we moved along an isotherm Our temperature did not change from C to D. We know that our internal energy went down from B to C, because we did some work. But from C to D, our temperature stayed the same. It was at temperature-- let me write it down-- T2, right? Because we had this reservoir here. It stayed the same. If your temperature stays the same, then your internal energy stays the same. At least for the system we're dealing with, because it's a very simple gas. It's actually the system you'll deal with most of the time, in an intro thermodynamics course. So. Given our internal energy didn't change, we can apply the same argument that the heat added to the system is equal to the work done by the system. Right? Same math as we did up here. Now, in this case, the work wasn't done by the system. The work was done to the system. We compressed this piston. The force times distance went the other way. So given that work was done to the system, the heat added to the system was negative, right? We're just applying the same thing. If our internal energy is 0, the heat added to the system is equal to the work done by the system. The work done by the system is negative. Work was done to it. So the heat added to the system would be negative. Or another way to think about it is that the system gave away heat. We put that with Q2. And where did it give that heat? It gave it to this reservoir that we put here, this kind of cold reservoir. You could almost view it as a-- well, it's accepting the heat. OK. We're almost there. Now, let's say we remove this reservoir from under our system again, so it's completely isolated from everything else, at least in terms of heat. And what we do is, we start adding-- so state D, we still had a few less pebbles. But we start adding more pebbles again. We start adding more pebbles to get it to state A. So let me change my pebble color. So we start adding more pebbles again to get it to state A. So that's this process right here. Let me do a different color. Let's say this is green. So as we add pebbles, that's this movement right here. We're moving from one isotherm up to another isotherm at a higher temperature. And remember, this whole time we went this clockwise direction. So a couple of interesting things are going on here. Because we're assuming an ideal scenario, nothing was lost of friction. This piston just moves up and down. No heat loss due to that. What we can say is that we've achieved-- we are back at our original internal energy. In fact, this is one of the properties of a state variable, is that if we're at the same point on the PV diagram, the same exact point, we have the same state variable. So now we have the same pressure, volume, temperature, and internal energy as what we started with. So we've done here is completed a cycle. And this particular cycle, it's an important one, it's called the Carnot cycle. It's named after a French engineer who was trying to just optimize engines in the early 1800s. So Carnot cycle. And we're going to study this a lot in the next few videos to really make sure we understand entropy correctly. Because in a lot of chemistry classes, they'll throw entropy at you. Oh, it's measure of disorder. But you really don't know what they're talking about, or how can you quantify it, or measure it anyway. And we really need to deal with the Carnot cycle in order to understand where the first concepts of entropy really came from, and then relate it to kind of more modern notions of it. Now, a system that completes a Carnot cycle is called a Carnot engine. So our little piston here that's moving up and down, we can consider this a Carnot engine. You might say, oh, Sal, this doesn't seem like a great engine. I have to move pebbles and all of that. And you're right. You wouldn't actually implement an engine this way. But it's a useful engine, or it's a useful theoretical construct, in order for understanding how heat is transferred in an engine. I mean, if you think about what's happening here, is this first heat sink transferred some heat to the system, and then the system transferred a smaller amount of heat back to the other reservoir. Right? So this system was transferring heat from one reservoir to another reservoir. From a hotter reservoir to a colder reservoir. And in the process, it was also doing some work. And what was the work that it did? Well, it's the area under this curve, or the area inside of this cycle. So this is the work done by our Carnot engine. And the way you think about it is, when you're going in the rightward direction with increasing volume, it's the area under the curve is the work done by the system. And then when you move in the leftward direction with decreasing volume, you subtract out the work done to the system, and then you're left with just the area in the curve. So we can write this Carnot engine like this. It's taking, it's starting-- so you have a reservoir at T1. And then you have your engine, right here. And then it's connected-- so this takes Q1 in from this reservoir. It does some work, all right? The work is represented by the amount of-- the work right here is the area inside of our cycle. And then it transfers Q2, or essentially the remainder from Q1, into our cold reservoir. So T2. So it transfers Q2 there. So the work we did is really the difference between Q1 and Q2, right? You say, hey. If I have more heat coming in than I'm letting out, where did the rest of that heat go? It went to work. Literally. So Q1 minus Q2 is equal to the amount of work we did. And actually, this is a good time to emphasize again that heat and work are not a state variable. A state variable has to be the exact same value when we complete a cycle. Now, we see here that we completed a cycle, and we had a net amount of work done, or a net amount of heat added to the system. So we could just keep going around the cycle, and keep having heat added to the system. So there is no inherent heat state variable right here. You can't say what the value of heat is at this point in time. All you could say is what amount of heat was added or taken away from the system, or you can only say the amount of work that was done to, or done by, the system. Anyway, I want to leave you there right now. We're going to study this a lot more. But the real important thing is, and if you never want to get confused in a thermodynamics class, I encourage you to even go off on your own, and do this yourself. Kind of-- you can almost take a pencil and paper, and redo this video that I just did. Because it's essential that you understand the Carnot engine, understand this adiabatic process, understand what isotherms are. Because if you understand that, then a lot of what we're about to do in the next few videos with regard to entropy will be a little bit more intuitive, and not too confusing.