Physics
Thermodynamics
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Carnot cycle and Carnot engine
Introduction to the Carnot cycle and Carnot heat engine
Discussion and questions for this video
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 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 welldefined 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 covereddome 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 quasistatic 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 Paxis.
 That's the Vaxis.
 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.
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.
I'm still pretty confused about the 2 adiabiatic processes in the Carnot cycle. If there is no exchange of heat during the move from B to C and from D to A, and the temperature is constant during that move, how is it that the temperature all of a sudden jumps from T2 back to T1? Please don't answer "because it is an adiabatic process", that doesn't help me...
The temperature is not constant. In the adiabatic expansion, the work done by the gas is supplied by the energy in the gas, and the temperature then decreases.
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.
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.
If you compress a gas it will always get warmer, likewise it will cool down if you expand it. That's because an expanding gas has to "use" energy to "push the surrounding gas away to make room for itself" (at least that's how my simple brain views it). Reallife example: Canned spray deodorant will always feel very cold on your skin. That's because the gas inside the can was compressed. If you release it, it will expand and cool down.
You can also compress air with a tire pump, it will feel warmer.
You can also compress air with a tire pump, it will feel warmer.
No heat definitely does not mean no change in temperature. Heat is just one way of transferring energy. You can, for instance, change the temperature of on object just by doing work on it. Try rubbing some objects together  they'll get warmer even though you're only doing work on them. Temperature simply reflects the internal energy , not the heat.
What's the main purpose of the Carnot Cycle though??
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.
It is the theoretical underpinnings of our modern day engines. This may help if you are a reader: http://en.wikipedia.org/wiki/Carnot_cycle
Ok, like calgeko, I am confused why adiabatic expansion in carnot cycle decreases temp. if we apply the principle of PV=nRT, it obviouslt states that V varies directly with T. meaning V increase (expansion) means T increase. Pls. kindly help me with this. Thanks.
In an adiabatic expansion, heat does not go in or out. But that does not mean that energy cannot go in or out. Energy can be transferred through the flow of heat, yes, but it can also be transferred out mechanically. That's what happens in adiabatic expansion. The gas does work to make the volume expand. In doing the work, the gas loses energy, and the result is lower temperature.
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.
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.
in this comment. i cant understand the part u told about adiabatic compression, that is, 3rd paragraph to be exact. can u explain me? thanks!
This one helps. Thanks! :)
From B to C , how come its follows a ln curve even though temperature isn't constant ?
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.
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.
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?
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?
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.
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.
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 ?
Energy that was used up from adiabatic process was partly regained in the isothermal process where exchange of energy is possible
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
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.
At 15:25 when we start adding pebbles to the system to move from state D to state A was it an adiabatic process?
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.
B to C and D to A are both adiabatic processes.
A to B and C to D are both isothermic processes.
I'm a bit confused. if D is back to conditions we have in point A. I'm assuming that both Temp must be the same (seeing that it seems to be an adiabatic process) but there is no indication that temp from D is back to initial temp in point A. I actually assume point A has temp of T1 whereas point D has temp of T2. am I missing anything?
I couldn't exactly get what really happened in the case (not named, in the video) just after the 'C' case. When we have a gas at temp T2 put on a reservoir at T2, why and how will the heat be transferred to the reservoir(as the arrow depicts, at 12:59), when both are at same temp.??
This hypothetical reservoir is infinitely large and can take in any amount of heat without raising its temperature. When pebbles are added after the C point and the pressure increases, there is a tendency for the temperature (hence the total kinetic energy) to also rise. The reservoir allows the system to bleed off this excess heat into the cold reservoir, or sink, and thus allows the system to retain the temperature of T2 from C to D. The system is therefore isothermic between C and D.
In the adiabatic process from B to C, temperature and pressure are reducing. Wouldn´t the volume be increasing because of the piston moving up (as a result of "pebbles" being removed off of it) and the gas expanding?
Sal's drawing is a little misleading. From B to C, volume is being held constant by definition so the drawing should be closer to a straight line (assuming it will increase or decrease in a real example). As temperature decrease so will the pressure.
If a gas expands adiabatically, why does its temperature decrease just because its volume increases? (We're not keeping the pressure constant, so volume is not directly proportional to temperature)
If you imagine the particles inside a container, the walls of the container are moving outward. That means every time a particle collides with the wall it loses some momentum and bounce off the wall slower than it hit the wall. Slower particles = lower temperature.
how come the temperature is not constant in a adiabatic process as dq = ncdt
and for a adiabatic process dq = 0 which means dt must be constant
and for a adiabatic process dq = 0 which means dt must be constant
Because there's also work being done, one way or another. While heat doesn't flow in or out during an adiabatic process, work either becomes heat or internal heat becomes work (depending on direction of process).
When expanding, some of the internal heat is becoming work as it expands, pushing something out of the way for it's new larger, footprint. When compressing, something outside is doing work on the system to get it in a smaller volume. Without the ability for heat to flow elsewhere, these both result in an internal temperature change.
When expanding, some of the internal heat is becoming work as it expands, pushing something out of the way for it's new larger, footprint. When compressing, something outside is doing work on the system to get it in a smaller volume. Without the ability for heat to flow elsewhere, these both result in an internal temperature change.
Although I intuitively feel that Q1Q2=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 Q1Q2=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?
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.
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.
why can not the whole heat be converted into work...?? if we think of isothermal process then from first law of thermodynamics, dQ =dW.... or Q=W....!here the whole heat is conveted into work....why so...?
The best you can do in theory is for Q to be equal to W. In the real world, at least a little bit of the Q is going to leak away rather than be converted into W, because of friction and similar effects.
if we look at the case of common heat engine,also whole heat cant be converted into work...if this happens then temperature of source will be equal to that of working substance....there will be no flow of heat....!! engine will stop working or its efficiency will become 0....!am i right..?
I have a question and im sorry if this is slightly off topic but i am just curious. In biology class I just learned about cellular respiration . can i say that the generation of ATP through ATP synthase and the proton gradient across the mitochrondrial inner membrane is like the carnot engine? i didnt understand this video entirely but if the carnot engine is similar to the way ATP is made, it would simplify a lot for me. any help is much appreciated!
Thermodynamics certainly comes into play in biological processes. In the case of cellular respiration, the laws of thermodynamics plays a huge factor! The electrochemical potential gradient that is created by the ETC is restored by allowing those protons back through, while ATP synthase harnesses the potential energy and turns it into chemical potential energy by creating ATP from ADP and Pi. The second law says that over time, chemical potentials tend to equilibrate in a closed system.
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.
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.
Suppose 100 moles of a monatomic ideal gas undergo an isothermal expesion from 1.00m3 to 4.00m3 .How much work is done by by the gas during this expansion ie. Wkj
For an isothermal process, PV=constant as in PV=nRT, 'nRT' is constant when temperature is constant. And we know that workdone is integration of P.dV, so we write it as:
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!
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!
During each cycle a reversible engine absorbs 2600j of heat from a high temperature reservoir and performs 2300j work, what is the efficiency of the engine ie. E=?
2300/2600 *100% = 88%
I understand what happened in the Carnot cycle, and I understand that work done by the system is the are of that Carnot loop in the PV diagram. My only question is, if gas did work, where did that energy go? Since we ended up back at point A with the same mass on piston, height of piston, pressure of gas, and volume of gas, there was no change in the sum of gravitational potential energy of the rocks and internal energy of the gas.
How could the gas have magically done work without any change energy?
How could the gas have magically done work without any change energy?
can an adiabatic system exchange energy with the surroundings? i mean it does not exchange heat but it can give out work so what is the correct answer ?
Yes of course if it can do work that's giving energy
what is exactly carnot engine?
How are the Rankine Cycle and the Carnot Cycle related? And is the Carnot Cycle related to the Otto Cycle? or am I just confusing all the cycles?
If this is the wrong place to ask this, I apologize, but what is a _state variable_?
A state variable is a property of a system that only depends on the CURRENT STATE of system and NOT on WHICH DIFFERENT PROCESSES YOU DID TO ACHIEVE THE STATE.Means that for a given state of the system a state variable can have only 1 value. To give you an example let the system have some temperature and some pressure. So its volume is also FIXED i.e. it can have one and only one value. The same is true for temperature, and internal energy. Another thing to note is that if your system has some state variable for a particular state say A then no matter what you do if you go back to state A you will find that ultimately the value of the state variable didn't change at all. i.e you get to exactly the same situation you were in
ok so the engine gets energy from heat provided does work and thus releases heat while doing work. so if my engine doesnt give out heat while doing work then its 100% efficient. but thats not possible. am i right?
Yes you are right thats not possible beacuase of friction due, to friction a large of heat is radiated out!
Is the net work done in cycle is zero ? please tell me that.....?
yes. This is due to couple. couple is nothing but a pair of equal and opposite forces acting on an object. When we peddle the couple acts and therefore the net force is zero. work is the product of force and displacement. Since force is zero work done will also be zero.
In my book (atkins) it says that the integral of (dq_reversible/T) = zero for a carnot cycle. The passage is used to explain to us (students) that entropy is a state function. The book makes good sense. Why however is Q total for the carnot cycle not equal to that q_reversible that we were originally intending to find? Is it because the isothermal steps are not reversible?
this may sound a stupid question but how is A to B an isothermal process when the work is done by the system?
Why does the temperature increase or decrease when the volume changes? Wouldn't the pressure simply change as well to compensate and leave the temperature constant?
PV = nRT. When volume changes, pressure AND temperature can both change.
At 7:00, why is there a change in temperature when there should be no change in temperature in an isothermic process?
I don't see a change in temperature. Is it possible you confused the labeling of the PV diagram? The T1 written on the curve states that the temperature of the system is constantly T1 throughout the process from A to B. What changes is the volume of the system which increases, and the pressure of the system which decreases.
I am confused as to how Q1  Q2 = W. I mean the system got heat from the first reservoir and then gave heat to the second reservoir and converted some of that heat to work, then how does the system return to it's first state to complete the cycle? Where does that energy come from if the system gave some of that energy as work?
I do not understand how Q can change when the temperature is constant, as in the first step of the carnotcycle. Because Q = mcdT and m is not changing and c is a constant, or have I got i wrong? Is c the factor that makes Q change?
The specific heat formula, Q = mcdT, in which m is the mass of the system and c is the specific heat, applies most straightforwardly to solids and liquids which do not expand very much, compared to gasses, when they are heated. For a gas there is a specific heat for heating at a constant volume, cv, and a specific heat for heating at constant pressure, cp. The latter, cp, is always the larger specific heat since the heat absorbed both changes the temperature and does some work as the volume expands at constant pressure. For ideal gasses the difference is, cp  cv = R, the universal gas constant (if the mass is measured in moles). Now an isothermal process can be thought of as equivalent to a sequence of very tiny volume changes at constant pressure followed by very tiny pressure changes at constant volume, each pair returning to the chosen fixed temperature from which it started. Depending on the direction one is moving along the isothermal, one of the changes in a pair will absorb heat into the system while the other change in the pair will release heat from the system. But since the corresponding specific heats have different values the net effect will be either to absorb heat along the isothermal (increasing volume direction) or to release heat along the isothermal (decreasing volume direction). In this way heat absorption or release at constant temperature is seen to be compatible with the specific heat formula.
so if we move from left to right work is done by the system?
Yes. The volume of the cylinder was increased by the particles inside the cylinder, so the system DID work.
Between C and D, shouldn't Q2 be an arrow pointed away from the isotherm to indicate that heat was transferred into the reservoir from the system?
Okay, let's see if I got this right:
The example right in the beginning is NOT an adiabatic process because the reservoir is adding heat to the canister.
It is, in fact an isothermal process.
The work done by the canister in increasing the volume has reduced the temperature and therefore, the internal energy.
Of course, that isn't the end of the story. The reservoir has nullified this effect by adding an amount of heat to the system equal to the work the canister didhence the change in internal energy is 0.
This is corroborated by the fact that delta T is 0 (isothermal process)
No isothermal process is adiabatic.
Sorry, this is so long. I just have to make sure I've got it right.
Thank you.
The example right in the beginning is NOT an adiabatic process because the reservoir is adding heat to the canister.
It is, in fact an isothermal process.
The work done by the canister in increasing the volume has reduced the temperature and therefore, the internal energy.
Of course, that isn't the end of the story. The reservoir has nullified this effect by adding an amount of heat to the system equal to the work the canister didhence the change in internal energy is 0.
This is corroborated by the fact that delta T is 0 (isothermal process)
No isothermal process is adiabatic.
Sorry, this is so long. I just have to make sure I've got it right.
Thank you.
No isothermal process is adiabatic. That's correct.
I thought this was isothermal not adiabatic
According to graph given in the vid:
i) A to B and C to D are isothermal processes.
ii) B to C and D to A are adiabatic processes.
i) A to B and C to D are isothermal processes.
ii) B to C and D to A are adiabatic processes.
It's both, depending on what part of the cycle you are in.
I am wondering since change in internal energy U = 0 if T=const. Is that also true if there is a phase change because then we would not need to have a temperature change but we would still have a change in latent energy which i thought was the internal energy.
Thank you.
Thank you.
ah true thank you!!
The assumption that U = O while T is constant rests on the concept of an ideal gas. Obviously, these principles no longer apply if we're not dealing with an ideal gas. Outside of the ideal gas theory, U is dependent on other factors, as well as T.
At @ 15:50 Sal draws the final adiabatic curve from D to A. Is this curve exactly the same shape, length etc as the adiabatic curve from B to C? Thanks!
No, it is not exactly the same shape, length, etc. But it is the same kind of curve and that kind of curve is of the form,
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.
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.
I got a question!! Usually this is done with a cylinder with a piston.
Suppose there is a some pump of volume 50.
But there is no piston fit into the pump.
Let's say there is a volume of 40 occupied with molecules
When the volume occupied by the molecules changes to 50, and the pressure remains the same then is there any work done even if there is no piston.?
Suppose there is a some pump of volume 50.
But there is no piston fit into the pump.
Let's say there is a volume of 40 occupied with molecules
When the volume occupied by the molecules changes to 50, and the pressure remains the same then is there any work done even if there is no piston.?
If there is no piston, then there is no pressure, and no way to define the volume.
No pressure means no work.
No pressure means no work.
At 12:32 Sal says" its kind of a cold reservoir"
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
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
At 17:50 sal says " the system transfers a smaller amount of heat back to the ambient/other reservoir"
How do we know that it is a smaller amount of heat?
How do we know that it is a smaller amount of heat?
To make ice, a freezer that is a reverse Carnot engine extracts 42kJ as heat as 15 degree celcius during each cycle, with coefficient of performance 5.7. The room temperature is 30.3 degree celcius. How much (a) energy per cycle is delivered as heat to the room and (b) work per cycle is required to run the freezer?
All I do is just using the formula Kc=QL/(QHQL) for part (a) and K=QL/W for part (b)? Is that a correct method to do it? Anyone can help me with this?
All I do is just using the formula Kc=QL/(QHQL) for part (a) and K=QL/W for part (b)? Is that a correct method to do it? Anyone can help me with this?
Sal, is putting on or off the pebbles to depict compression and expansion of vol respectively..But, what happens actually when we have a flexible piston and a container that is firstly put on a source(at T1), then on an insulating stand, further on a sink and again back on insulating stand?
how to interpret work done on the system in an isothermic process. Suppose, I pull up the piston of the system connected to a sink at temperature T. Is it the work on the system?
If so, 0=Q+W => Q=W. What does "W" imply?
If so, 0=Q+W => Q=W. What does "W" imply?
It means that negative work was done on the system, and the system has to lose thermal energy to account for that.
One last thing to get confirmed is that in a carnot cycle two adiabatic processes and two isothermic processes take place. Is that right ?
If an completely isolated system(That doesn't exchange temperature with the surroundings), does work, then it does less work than it should (or less than it would when it would exchange the temperature with surroundings), because when the temperature decreases the, pressure and volume of the final state also decreases. Is that correct?
How is net work done on the surroundings when the process seems to follow the same processes from A to C and back. The system does more work going from A to C than the surroundings do from C to A?
why is adiabiatic curve steeper than the isothermic curve??
Because temperature changes during adiabatic processes, so that means you have to move from one isotherm to another, and the only way to do that is to be steeper.
What is the effect of the Work W done by the cycle? In what form is it wasted if there is no friction?
Basically why is Q1 > Q2?
Basically why is Q1 > Q2?
how are isothermal and adiabatic processes related to each other? during the state of thermal equilibrium(dT=0),there is no flow of heat too(dQ=0)...
explain....!
explain....!
The way I think about the relationship is this: you can draw a series of isotherm curves that each represent the PV combinations at a certain temperature. If you want to stay at the same temperature, you travel along an isotherm. If you want to move from one temperature to another, you have to move from one isotherm curve to another. There are three ways you can do that:
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.
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 that (V,p)coordinate diagram or pressure as a function of volume diagram is given with arrows, what can one deduce about it without further knowledge.
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?
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?
I don't understand why the net amount of heat is >0. I would think that Q1 and Q2 cancel eachother out completely.
Think about it this way. As Sal mentioned, the net internal energy as the system is always the same at point A (assuming no energy loss). However, we know the system performed work (Since work performed is the integral of P(V), and the area under the curve on the path from A to C and the different path from C to A are not equal). Since Delta U = Q  W, U = 0, and we know W is positive, Q has to be positive. Therefore, the net heat added to the system is positive.
I wonder why the videos after Thermodynamics 5 aren't in Physics playlist on youtube? Before using the actual website I used to watch video on youtube directly. I think that would be much more conveniently if they were added there as well. But still many thanks for the lessons! :)
@10:12 Khan says that his pressure would go down at T2 but his volume would go down also.
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;)
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;)
i noticed the same. Yes, I'm sure it was a small error
I do, thank You.
Hmm...I see what you mean, he moves his arrow saying volume goes down, whereas actually, what he's showing is that its increasing.
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?
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?
we learned that in an adiabetic process pressure, volume and temperature go down  should' nt then the corresponding curve fall vertically (no increase in volume)?
In an adiabatic process, no heat goes in or out. That means you move from one isotherm to another as pressure or volume changes. There's no reason you have to move straight up or down.
A process with no change in volume is called isochoric or isovolumetric. No work gets done because the volume does not change.
A process with no change in volume is called isochoric or isovolumetric. No work gets done because the volume does not change.
I was not clear enough im my question yersterday. If in an
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?
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?
The things that determine the lines in the diagram are Pressure and Volume. If the line would bend to the left, this would indicate that if the volume becomes smaller the pressure becomes smaller as well, which is not the case. (think of the pistons Sal so often uses to explain features in thermodynamics). So in the case of a cilinder with a piston, if the curve would bend the to left this would mean that you would DECREASE the volume and with this DECREASE the amount of bumps all the gas particles would deliver to the walls of the cilinder(pressure). This leads to a contradiction.
I'm not very good at English but I hope this helped.
I'm not very good at English but I hope this helped.
why are we taking change in internal energy _*ZERO*_ ???
1)its given that in insothermal process Q=W In what way does the gas do the work?
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?
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?
The gas expands. The expansion does work.
You can draw a PT diagram, too. It will convey the same information. PV is sort of easier to think about.
You can draw a PT diagram, too. It will convey the same information. PV is sort of easier to think about.
How can B go to C in an adiabatic process when B originally came from A in a isothermic progress, does this mean B was just 'pulled out' of the giant reservoir keeping it isothermic.
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.
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.
How is work of adding or removing the pebbles accounted for? Without this work, nothing gets done.
In physics Work is not the normal work. Its practically something humans will never find useful in future ('sigh").....I mean how much work did Peter do in lifting the weight (100kg) and in keeping it back on the stand ? Hes doin lots of work but since displacement is 0 , WORK IS 0. Imagine the poor thing's plight!
Why does the volume continue to expand during the "adiabatic expansion" if no additional heat is added? And likewise, why does the volume continue to fall during the "adiabatic compression" after the heat sink is removed?
The volume expands until the interior pressure falls to match the exterior pressure.
On the return cycle, the volume contracts due to outside pressure until the interior pressure rises to meet it.
On the return cycle, the volume contracts due to outside pressure until the interior pressure rises to meet it.
Why are the adiabatic parts of the PV diagram of the Carnot Cycle shaped like a rectangular parabola as well? .. They do not occur at a constant temperature so they need not be shaped that way right?
how does he know (19:34) that there is more heat going in the systm than heat going out?
Since work=heat, and from step C to A (that the system is receiving heat) the work is less than from step A to C.
And this is because when the temperature of a system is at lower state you need less force to do work to 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.
The reservoir (unlimited space) has heat to always put into the system (mentioned @7:27). More importantly, the carnot engine is doing work on the gas which goes into the overall conservation of energy;
heat(total) = work + heat(leaving the engine)
heat(total) = work + heat(leaving the engine)
Why is it that in most cases the PV diagram is a rectangular hyperbola ?? Like a downward curve ?? Could it also be like an upward curve , or a diagonal straight line maybe ?? Thanks !!! :)
it cnt be an upward curve , ur decreasing the volume it will have to go down
I'm still confused. What is a Carnot cycle, then?
The Carnot Cycle has been mathematically proven to be the most efficient method for converting thermal energy into work and vise versa.
20:52 I cannot understand why the point B to C can happen because of adiabatic situation. Their temperature changed T1 to T2, I guess. What happened between B to C?
Remember that physically we allowed the gas to expand by decreasing the restraining force (pebbles). If you think about the gas laws PV/T = constant and since we did not keep T constant, the net result is V increase, P decrease, T decrease.
also, in from a more abstract perspective, we jumped from one of those allimportant isotherm to another.
From point B to C, we decreased pressure so the volume went up. Temperature decreased because it's an adiabatic situation (no reservoir).
How can cooling and heating things ever be adiabatic??
System changes in the real world can't be perfectly adiabatic, but if you insulate something very well, you can isolate the system enough so that you can analyze it as adiabatic.
I understand that for isothermal w=nRT(lnV2lnV1)
can I say that for adiabatic:
since Q=0 then DU=w
since U=3/2 PV then DU=(3/2P2V2)(3/2P1V1)
so: w=(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?
Thanks,
G.
can I say that for adiabatic:
since Q=0 then DU=w
since U=3/2 PV then DU=(3/2P2V2)(3/2P1V1)
so: w=(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?
Thanks,
G.
Yes you can. watch the previous video again. :)
from a to b system has a conducting wall because reservoirs transfer heat to the cylinder.but from b to c it is an adiabatic process meaning that the cylinder has a diathermic wall(no heat can enter or exit)how is this possible?
How carnot engine works ?
I have a formula for Work done per cycle of the Carnot cycle:
W = η R TH ln (V3/V4)
Where η is efficiency & TH is Temp hot.
I'm not sure how this was derived; any ideas? Thanks
Nic
W = η R TH ln (V3/V4)
Where η is efficiency & TH is Temp hot.
I'm not sure how this was derived; any ideas? Thanks
Nic
Can anyone please explain to me what an Adiabetic system is?
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?
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?
Ok I have heard the term adiabatic before. What is adiabatic pressure or compression? I work at a ASU Air separating unit  cryogenics. I understood it was similar to lets say water hammer. But you got me think more on the macro and micro level of that system. Lets Use 99% oxygen at 160degree F traveling town a 5 inch pipe with a gauge pressure of 56psi. How would you draw up and explain the adiabatic pressure if that system flow was interrupted by a sudden stop of flow  a valve closing quickly
Why does the Temperature goes down, in the process of BC??
hey i ve two questions regarding carnot cycle
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 .
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 .
Carnot Engine has no particular use....except in theory. It's just a module to show that maybe, when things become more advanced and efficient to the
limit that no heat is given out, ALL of the heat can be converted to useful work.
limit that no heat is given out, ALL of the heat can be converted to useful work.
1) Yes, A>D is an adiabatic process. And yes, after you have completed a whole cycle (A>B>C>D>A), you're back at the start, with everything being exactly the same as it was before you began, so you can continue on and on. (If this was not what you meant, please rephrase, I had trouble understanding what you wrote.)
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.
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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