Carnot cycle and Carnot engine
Introduction to the Carnot cycle and Carnot heat engine
Carnot cycle and Carnot engine
- 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.
- 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.
- 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.
- 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.
- 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.
- 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.
- 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.
- 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.
Be specific, and indicate a time in the video:
At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?
Have something that's not a question about this content?
This discussion area is not meant for answering homework questions.
You can also compress air with a tire pump, it will feel warmer.
In the equation U = Q - W (forgive the lack of delta's) adiabatic and isothermal are the edge cases. Isothermal means U=0, adiabatic means Q=0. In the first case, all work is at the expense of heat entering (or leaving) the system. I the second case, work is a "cash out" of internal energy, hence (for an ideal gas) temperature.
Isothermal expansion and compression , Adiabatic expansion and compression.
But why do we need a two adiabatic states? Work is being done during the isothermal expansion and compression so What's the point of isolating the system for adiabatic expansion and compression? Can't the system directly be taken to the cold reservoir after keeping it on the hot reservoir?
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.
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?
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.
The problem with applying PV=nRT the way you are doing above is that you are implcitly assuming P is constant when you say that V has to vary directly with T. IF P is constant, you are correct. But in an adiabatic expansion, P is not constant. It goes down.
Think about the opposite of an adiabatic expansion: a compression. If I tell you I have an insulated piston, and I am going to compress the gas in it by pushing on the piston, your intuition will probably tell you that the pressure and volume are both going to go up, right? Try it with a bicycle pump; you will be able to feel the pump get hot as you try to pump up a tire that is under high pressure.
If it gets hot when you compress, it should get cooler when you decompress, right? So that's adiabatic expansion: lower pressure and lower temperature.
The first law of thermo confirms this. delta U = Q - W. Q is zero in adiabatic expansion. So delta U = - W. In expansion, W is a positive number, so delta U is a negative number. Delta U represents the change in internal energy. In an ideal gas, internal energy is all kinetic energy, so negative delta U means kinetic energy has decreased, which means temperature has decreased.
B to C and D to A are both adiabatic processes.
A to B and C to D are both isothermic processes.
I suppose that this is wrong and only a spoken mistake, because the move to the right o in the positive direction on the V axes can only mean more volume and lower pressure. Lower volume does not normally go hand in hand with lower pressure. So, although Khan probably meant more volume and lower pressure he said lower volume and lover pressure. Please confirm;)
But that's the only slight mistake, because the rest of the video is quite alright, just the wrong word use "down". But, you do see what he's trying to mean, right?
How do we know that it is a smaller amount of heat?
Basically what does a mere (V,p) with arrows diagram tell us?
Is it just the work done to/by the system or are we able to deduce also the isothermic/adiabatic processes between the points??
I guess we need the T:s as well to be able to say that?
Carnot considered a heat engine in his famous theorems. The cell simply uses a different form of energy (electrochemical potential energy) to get work (the creation of ATP via ATP synthase) out of the system.
And this is because when the temperature of a system is at lower state you need less force to do work to the system.
heat(total) = work + heat(leaving the engine)
A process with no change in volume is called isochoric or isovolumetric. No work gets done because the volume does not change.
I was hoping to please clarify:This is only because it keeps the temperature of the piston low right? Not because it is colder than the piston
isobaric, where the pressure does not change and maximum work gets done
isochoric, where the volume does not change, and no work gets done
adiabatic, where no heat comes or goes, so the internal energy change (ie the temperature change) determines the work
In a way, an isothermal process is the opposite of an adiabatic one. In an isothermal process, you are allowing heat to come and go so that you are at the same temperature as your surroundings. You have to do an isothermal change slowly and with easy flow of energy into or out of the system. Adiabatic is the opposite. You are not letting any heat come in or go out. If your system gets very far from the temperature of the surroundings, still no heat can flow in or out to restore thermal equilibrium. The way you do that is to insulate your system as much as possible. You also do an adiabatic change quickly, sort of like you are trying to get away with it before the heat can leak in or out.
If this is done wouldn't this mean that work(the moving from B) would be done on system B additional to the work that comes from removing the rocks.
2)why is is it only PV diagram why not PT diagram
3)why cant v keep the cylinder first on the cold reservoir and then keep it on the hot reservoir?
adiabetic process the volume decreases then the curve in the PV diagramm should not bent to the right but to the left or why not?
I'm not very good at English but I hope this helped.
i) A to B and C to D are isothermal processes.
ii) B to C and D to A are adiabatic processes.
dW(i.e. small change in workdone) = P.dV
we have two equations, isothermal equation; PV=constant and ideal gas equation; PV=nRT.
Now in P.dV, we can write P as (constant/V), and this constant equals nRT as told earlier.
Now, dW= (nRT/V).dV
Integrating on both sides, W= nRT(lnV), where ln means natural logarithm.
now when gas goes from volume V1 to V2, W=nRT(lnV2 - lnV1)
or, W= nRT[ln(V2/V1)]. Remeber this result. Now you can put the given values and obtain the answer!
All I do is just using the formula Kc=QL/(QH-QL) for part (a) and K=QL/W for part (b)? Is that a correct method to do it? Anyone can help me with this?
PV^g = constant,
different constants for different adiabatics and where the exponent, g, on the volume, is the ratio of the specific heat at constant pressure, cp, to the specific heat at constant volume, cv, i.e., g = cp/cv. Since cp > cv, g > 1 and adiabatic curves are steeper than isothermal curves in a P V diagram.
How could the gas have magically done work without any change energy?
Am i correct that if it is Adiabetic the temperature in the system will change? (heat will be added or removed from a "reservoir"?) And if its a Isotherm the tempertature will not change? And that a system cannot be Adiabetic and Isothermic at the same time?
can I say that for adiabatic:
since Q=0 then DU=-w
since U=3/2 PV then DU=(3/2P2V2)-(3/2P1V1)
so I can calculate the work done by the system in a adiabatic expansion knowing initial and final pressures and volumes? Does that correspond also to the area under the adiabatic curve in the PV graph?
1) when we reached to state D we will again remove the reservoir , again adiabatic process will occur ? what i meant to say is that , again the WHOLE process will work?
2) why we use carnot engine ? what are the uses of carnot engine?
i hope you understand the questions which i have asked you .
limit that no heat is given out, ALL of the heat can be converted to useful work.
2) It's very useful in showing what the efficiency limit is for any heat machine. No process operating with the help of two reservoirs (With temperatures T1 and T2 in this video), can be more efficient than a Carnot Engine/cycle operating under the same conditions.
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