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Proof: S (or entropy) is a valid state variable

Proof that S (or entropy) is a valid state variable. Created by Sal Khan.

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  • blobby green style avatar for user raiderchia10
    I got kind of lost when he started using the integrals, could someone explain what he was trying to prove and how he proved it in simpler terms?
    (14 votes)
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    • blobby green style avatar for user afrykman
      The integral was just the mathematical way to write the heat, Q, in terms of the volume, V, because that's what he had the proof in terms of from the last video. He couldn't prove that Q1/T1 + Q2/T2 = 0, because he doesn't know what Q1 and Q2 are, but he knows for an isothermal (constant temperature) process that the heat, Q, is equal to the integral of pressure, P, over the change in volume, V. So then he ends up with delta(S) equaling something in terms of volumes, which, form the previous video, he can then use to show that the change in S is equal to zero.
      (16 votes)
  • mr pink red style avatar for user Rainman
    what is state variable and state function?
    (8 votes)
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    • blobby green style avatar for user C Hart
      A state variable means it will always be the same AT THAT POINT. So if you choose P1 and V1 as your point for a certain system of ideal gas, no matter how you change P and V, when you get back to P1 and V1, the U will be the same as it was at the beginning (U1).

      Think if it this way: On a mountain top, there are always the same Latitude and Longitude, not matter if you hike straight up, or cycle around the mountain, or parachute in. Those coordinates DEFINE your location on the mountain top. Is "tiredness" or "effort" a variable that is always the same for someone on top of the mountain? No. If you crawled up the mountain, you would be more tired than the person next to you who parachuted in. Is your elevation a state variable for a mountain top? Yes. Is your thirstiness? No.

      Likewise P, V, T, S, and U are always the same at the same point on a PV diagram, no matter what wacky path gets you to that point. But the amount of W or Q needed depends heavily on what path you took to that PV point (iso-volume is no work, etc).
      (17 votes)
  • purple pi purple style avatar for user Potatocar
    It's often emphasized how important it is that the process is reversible. I think I have an idea of what reversible and irreversible processes are, but I can't seem to understand why so many things apply to reversible processes only. Could someone explain that?
    (9 votes)
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    • male robot donald style avatar for user Zak
      Because reversible processes are ideal, and simplifications of real world problems. In the real world we would have to account for changes in kinetic and potential energy along with friction (and more). That would make these problems much more difficult to solve! So reversible is just an assumption created to make the problems more simple in an introductory course and that's why everything seems to apply just to reversible processes in these videos.
      (6 votes)
  • piceratops ultimate style avatar for user Arunabha
    Has entropy been proven?
    (0 votes)
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  • blobby green style avatar for user neelima993
    why dont u take Q2 negetive?wouldn't it mean Q2=-p.del(V)
    (5 votes)
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  • leafers tree style avatar for user jwfisher90
    Around Sal says that in the second isothermic process (from C to D) you take away less heat than you needed to add in the first isothermic process (from A to B). Can anyone please help clarify why this is the case and why the heat added and taken are not the same?
    (3 votes)
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  • leaf grey style avatar for user Shlomo Fingerer
    I have no idea what's going on, as to be expected. I don't have the background. I am watching these videos, so that when I do take Chemistry, I will have heard the terms before. I know you guys are all taking Chem. now, tell me, will my strategy work?
    (1 vote)
    Default Khan Academy avatar avatar for user
    • Well, I am taking Chemistry for sure.
      Actually Sal hasn't covered chemistry that well. It might be better for you leave this for a while, if you are not understanding. You can always come back later
      Get started with calculus. Start from basic limits, and then work your way up. He's explained calculus better than chemistry. Both of them are very important, calculus more so. Ask questions regarding what you don't get. You will get answers. Don't think calculus is too hard. Don't ever give up on it.
      here, watch this videos of this person twice your age - http://www.khanacademy.org/stories/bruce
      If he can do it, then why can' you. If you don't get something, then try searching for it on the internet. For chemistry, you can go to academicearth.org and listen to the various lectures. You won't get points for it, but atleast you'll get a better understanding.
      (9 votes)
  • male robot hal style avatar for user KhanEagle
    I don't Know Calculus so this means can i skip it , Or it is really necessary ?
    (2 votes)
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  • blobby green style avatar for user cormac.stopes
    Why is it that we can say this derivation applies in general, and not only in Carnot engines?
    (4 votes)
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  • marcimus pink style avatar for user Bridget
    How does one complete this entropy problem using algebra based physics?
    (3 votes)
    Default Khan Academy avatar avatar for user

Video transcript

I've talked a lot about the general idea that in order to have a state variable like, say, U, which is internal energy, at any point in this PV diagram, that state variable should be that value. So for example, if at this point, U is equal to 5, and I go do this whole Carnot cycle, when I come back to state A, U should still be equal to 5. It should not have changed. It's not dependent upon what we did to get there. So if we did some kind of crazy path on our PV diagram, we got back there, U should always be the same. That's what it means to be a state variable. It's only dependent upon its position in this PV diagram. It's only dependent on its state, not how you got there. And because of that, heat is something that we can't really use as a state variable. For example, if I tried to define some heat-related state variable, let's say I call it heat content, and I defined change in heat content as equal to the amount of heat added to the system. Well, if we go back to our Carnot cycle here, let's say that my heat content here was 10. Well, I added some heat here, in this process here. Nothing happened, because this was adiabatic from B to C. Then from C to D, I took out some heat. But I took out less heated than was added here. And then here, nothing was done with regard to heat. So I did add some heat to the system. The net heat that I added to the system as I went around the cycle-- in this case, Q would be equal to Q1 minus Q2. And we know that this number is larger than this. The net amount of heat we added to the system was the amount of work we did on the system, because the internal energy didn't change. So if this is 0, then the amount of heat we add to the system is the amount of work we did. Our internal energy is definitely 0 as we go all the way around. We did this shaded portion of work-- I showed you that several videos ago, that the area inside of our little cycle is the amount of work we did. And so that's also the net amount of heat added to the system. So if we added that amount of heat to the system, if we started off at the heat of 10 here, or whatever my heat content mystical variable I made up just now, when we go around, it would then be 10 plus W. If we go around again, it would be 10 plus 2W and 10 plus 3W. So it can't be a legitimate state variable, because it's completely dependent on what we did to get there. And if we keep going around the cycle, we can increase it, even though we get to the same point. So this is an illegitimate state variable, where I define the change in our little made up heat content to be equal to the heat added to system. Not a valid state variable. Ignore it all. Now we know that Q1, we added more heat here than we took away, so there was something net heat added. But there's something interesting here. We added it at a higher temperature. And here we took less heat away at a lower temperature. So maybe we can define another state variable that can have the result that when we go around the cycle, we do get back to our same value. Now let me just-- we're just experimenting. Although I know where this experiment will go. I wouldn't have been doing it if I didn't. So let's say I define a new state variable S. And I define a change in S. So I say, a change in S-- I'm just making up a definition-- is equal to the heat added to the system divided by the temperature at which it was added to the system. Now, I don't know what this means just yet. In future videos, maybe we'll get intuition about what this actually means in kind of our minds. But let's see if at least this is a valid state variable. If, as we go around the Carnot cycle, whether our change in delta S is 0, right? To be a legitimate state variable, we have some value for S here. Maybe it's 100. I don't know. Once we go back around the Carnot cycle, it should be 100 again, or our delta S should be 0. So what's the delta S? So the delta S, as we go around the whole cycle-- let me write delta S-- let me do another color. Delta S. As we are go around, I'll say c for the Carnot cycle. As we go around the Carnot cycle, is going to be equal to-- well, when we went from A to B, we were at a constant temperature, and we added Q1. So it's Q1, and we were at temperature T1. Fair enough. Then when we went from B to C, it was adiabatic. We added or took away no heat. So this value, Q over T, would just be 0. So it's plus 0. Then we went from C to D. We were at a new temperature, we were on a new isotherm. We were at T2. And we took away, or I won't put the sign here, let's just say we added Q2 heat. We're going to actually solve for it later. We added Q2 heat. We'lll see that it's actually a negative value. And then finally, when we went from D to A, it was adiabatic again. So no transfer of heat. So plus 0, right? The 0's are 0's over, you know, changing temperature, but this is just 0. So this thing should be equal to 0 in order for this to be a valid state variable. So let's figure out what this value is. What Q1-- so our change in our mystical new candidate state variable, S, as we go around the Carnot cycle, is equal to Q1 over T1, plus Q2 over T2. And we'll see, Q2 is negative. So what is Q1? Can we calculate Q1? Well, as we're on this top isotherm, our temperature doesn't change, our internal energy doesn't change. So if your internal energy does not change, if your internal energy is 0, then the heat added to the system is equal to the work done by the system. So its the area under this curve. Not just the area in the cycle. It would be the whole area under the curve. So what's the whole area under the curve? Well, so let me do a little aside here. So Q1 is equal to the work done as we went from A to B. And work, remember, can just be written as pressure times change in volume. We're going to do a little calculus here, so I'll write dV for a small change in volume. And we're going to integrate it all over the little sums, right? This dV is this little change in volume right there times pressure. That makes a little rectangle. And then we sum up all the rectangles from our initial volume, which is VA, to our final volume, which is VB. And then, what's Q2 going to be equal to? Well, Q2 is going to be essentially the same thing. It's going to be the sum of the work done by our system, which in this case is going to be negative, because work was done to our system as we go from here to here. Right? That's when Q2 was operating, the heat was being taken out of the system. So we're going to go from-- where was our starting point? VC and we go to VD. Now, how can we evaluate these integrals? Well, we've done this before in a previous video. We use both of these circumstances-- When we go to A to B, and we go from C to D-- both of these circumstances occur on isotherms, right? So the only things that are changing are pressure and volume. Temperature is not changing. And so if we go back to our ideal gas equation-- PV is equal to nRT-- we can just rewrite this by dividing both sides by V, as P is equal to nRT over V. And substitute that back in for P, in both cases. That is P as a function of V. We now have the equation of the curve. And we're taking the area under it in both cases. So Q1 is equal to the integral from VA to VB of nRT over V, dv. And Q2 is equal to the integral from VC to VD of nRT over V, dv I'm going to do two integrals in parallel, just so that you kind of see that we're essentially solving the same thing. OK. So how can we solve this? Well, we know in both of these cases, we're moving along an isotherm. That our temperatures are constant. And actually, we know the temperatures are. When we're moving from VA to VB, our temperature is T1. It was kept that way by our reservoir. When we moved from VC to VD, our temperature was T2. It was kept that way by our reservoir, right? T2, when we move from C to D. And T1, we move from A to B Those were our temperatures. And they're constant. Fair enough. So we can take-- so n is constant. R is definitely a constant. n is just a number of molecules we have. And then our temperature is also constant, so we can take it out of the integral. So we can rewrite Q1 is equal to nRT1 over the integral from VA to VB times 1/V dv, and Q2 we can write as nRT2 times the integral from VC to VD, 1/V dv. All right. Now this integral is fairly straightforward to evaluate. The antiderivative of 1/V is the natural log of V. So we get Q1 is equal to nRT1 times the natural log of V evaluated at VB, minus it evaluated at VA. And Q2-- well, let me just solve this whole equation right here. So this is equal to what? This is equal to a natural log of VB minus the natural log of VA. Which is the same thing as the natural log of VB over VA times nRT1. And all that is equal to Q2. Now, the same logic, Q2, is going to be equal to what? Q2 is going to be equal to nRT2. Now the only difference with this integral is where I had VB, now I have VD. Sorry. So then it becomes natural log of VD. And where I had VA, now I have VC, So over VC. All right. Now what was our original question that we were dealing with? We said, this is a legitimate state variable. If the change in this-- whatever this value S as we go around the cycle-- is equal to 0, that means it didn't change. So these two things, when you sum them, have to equal 0. Q1 over T1, plus Q2 over T2. So let's add them. So Q1 over T1 is equal to that over T1. That cancels out. Q2 over T2 is equal to that over T2. That cancels out. So our change in our mystical state variable, as we go around the Carnot cycle, is equal to Q1 over T1, plus Q2 over T2. Which is equal to nR times the natural log of VB over VA. That's that, right there. And then plus Q2 over T2, which is just nR times the natural log of VD over V. This is VC. This is a VA here. All right. Now let's see what we can do. This is equal to-- almost there. Home stretch. nR. We can factor out an nR. And then the natural log of A plus the natural log of B is just the same thing as the natural log of AB. So this is equal to times the natural log of VB over VA, times VD over VC. All right. So this is our change in our S, state variable that we're playing with right now. Now what is this equal to? Let me think of the best way to say this. So let's divide the numerator and the denominator by VC. So let me take this expression here, and divide its numerator and its denominator, essentially, by VC over VD. Or let me multiply its numerator and denominator VC over VD. So I can rewrite this as the natural log of VB over VA, divided by-- right? Instead of multiplying it times this, I can divide it by this reciprocal, VC over VD. So I just rewrote it. I just did a little bit of fraction math. That's all it is. Instead of multiplying it times this, I divided by its reciprocal. Now you see why the previous video I did was done. What is this equal to? On the previous video, I showed you that VB over VA is equal to VC over VD. We did that big convoluted hairy proof to prove this. And now that we proved it, we can use this to know that this quantity is equal to this quantity. So if you divide something by itself, they're equal to each other, this is equal to 1. If that's equal to 1, then what's the natural log of 1? So our change in our mystical S state variable is nR times the natural log of 1. What's the natural log of 1? E to the what power is equal to 1? e to the 0 is equal to 1. n times R times 0-- I don't care how big or whatever-- this is 0. So it equals 0. So there we have it. We've stumbled upon a legitimate state variable that deals with heat. If we define change in S is equal to the heat added to the system, divided by the temperature at which the heat was added to the system, this is a legitimate state variable. Now, we don't have much intuition about what it really means at kind of a micro state level. But at least we've stumbled upon some property of something. If S is 10 here, and we go around here, our change in S will be 0, S is 10 again. If S is, I don't know, let's say S is 15 here, and we go around some crazy cycle and we come back here, our change in S is going to be 0 again. Or sorry, it's going to be 15 again. So we didn't have-- our change in S will be 0, so our s itself will be 15 again. So S is a legitimate state variable, but we don't have a good sense of what it actually means. We'll leave that to a future video.