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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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• 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?
• 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.
• what is state variable and state function?
• 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).
• 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?
• 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.
• Has entropy been proven?
• well yes, entropy has been proven. It is part of the second "law" of thermodynamics... laws are rigorously tested and have not been disproven since formulated
(1 vote)
• 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?
• The second isotherm is at a lower temperature and pressure, so it takes less work to compress it, and therefore less heat has to leave the system after the compression to keep the system at the same temperature (on that same lower isotherm).
• 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)
• I don't Know Calculus so this means can i skip it , Or it is really necessary ?
• better u learn it...as it's an application based concept...it shall help u in ol d subjects..!
• Why is it that we can say this derivation applies in general, and not only in Carnot engines?
• How does one complete this entropy problem using algebra based physics?
• at i dont get why heat isnt a state variable. if we went around the cycle, why heat at A would be 10 + W?
• Posted 4 years ago
because Heat isn't an intrinsic property of the system, therefore you are not able to evaluate the heat without talking of exchange of energy with the enviroment, In fact it's a measure of the flux of energy. So you are able to find the value of heat only if you have transition of energy. So it dipends on the path, think about this example: you have two ways to get to the floor one to the floor two, the elevator or the staircase. The height between the floor two and one is a state variable, so we don't care about how do we get to the second floor ( the path), because the height between these two floors is the same.
Instead what you spend in terms of heat it's different, because you follow two different paths ( the elevator or the staircase).
Ok, now, also if I'm not sure of my answer, I'm going to answer your second question.
heat, in the video, is equal to work because the internal energy of your system doesn't change, so the net heat is equal to the work done by the system. Morover you have to add to this value 10J, because it's the heat content in your system at state A: in fact if you keep going around the carnot cycle and eventually you return at state A, the value of your heat content won't be equal to 10J + w, but 10J + 2w, because Q is equal to 2w and in fact you have to add a value of heat equal to the energy transferred thanks to the work to keep the temperature constant, avoiding its reduction( because it is an isothermic process > and so I have to replace the energy which i take away from the system thanks to work, adding the same value of heat). Furthermore you have to add 10 j to 2w because you start from state A.
If I make some mistakes, check them
(1 vote)