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### Course: Thermal physics (Essentials) - Class 11th > Unit 3

Lesson 7: Efficiency of a Carnot engine# Carnot cycle and Carnot engine

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

## Want to join the conversation?

- What's the main purpose of the Carnot Cycle though??(24 votes)
- Sadi Carnot introduced the Carnot cycle in an analysis of the efficiency of heat engines in the early 19th century. He showed that efficiency was lost whenever heat engines deviated from being in thermal equilibrium and that any heat engine operating between a maximum temperature, T1, and a minimum temperature, T2, could not have greater efficiency than a Carnot cycle operating between the same temperatures. So in this sense the Carnot cycle is the theoretical ideal.(47 votes)

- 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?(9 votes)- 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.(16 votes)

- From B to C , how come its follows a ln curve even though temperature isn't constant ?

(8 votes)- 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)

- 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)
- 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)

- 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)
- 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)

- At15:25when we start adding pebbles to the system to move from state D to state A was it an adiabatic process?(2 votes)
- The short answer is yes.

B to C and D to A are both adiabatic processes.

A to B and C to D are both isothermic processes.(10 votes)

- 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)
- 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)

- 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)
- 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)

- Why is pv work done by the system -ve in chemistry and +ve in physics?(3 votes)
- 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)

- Can space be considered a reservoir? if so how come the earth and all the other planets didn't match its temperature?(2 votes)
- One aspect of a reservoir is being able to transfer heat to and from it, space being almost empty of matter has no significant ability to receive or provide heat.(2 votes)

## 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.