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Thermodynamic entropy definition clarification

Video transcript
In the video where I first introduced the concept of entropy, I just tried something out. I defined my change in entropy as being equal to the heat added to a system, divided by the temperature at which it was added to the system. And then I tested to see if this was a valid state variable. And when I did that, I looked at the Carnot cycle. And this is a bit of a review. Never hurts to review. Let me draw the PV diagram here. We saw that we start at this state here, and then we proceed isothermically. We removed little pebbles off the piston. So we increased the volume and lowered the pressure. Then we proceed adiabatically, where we isolated things and we moved like that. That was adiabatically. Then at this other isotherm, we added the pebbles back. And then we isolated the system again. So adiabatically, we continued to add more pebbles, and we got back to our original state. And I did a couple of videos where I show that if you take the heat added here-- so this is all being done at some high temperature, T1. This is being done at some low temperature, T2. There's some heat being added here, Q1, and that there's some heat being released here, Q2. And since these are adiabatic, there's no transfer of heat to and from the system. And when I looked at this, and when I looked at the Carnot cycle, and I used this definition of entropy, I saw that the total change in S, when I go from this point all the way around and got back, the change in S, was equal to Q1 over T1 plus Q2 over T2. And then I actually showed you that this was equal to 0, which is exactly the result that I wanted to see. Because in order for this to be a state variable, in order for S to be a state variable, it should not be dependent on how I got there. It should only be dependent on my state variables. So even if I go on some crazy path, at the end of the day, it should get back to 0. But I did something, I guess, a little bit-- what I did wasn't a proof that this is always a valid state variable. It was only a proof that it's a valid state variable if we look at the Carnot cycle. But it turns out that it was only valid because the Carnot cycle was reversible. And this is a subtle but super important point, and I really should've clarified this on the first video. I guess I was too caught up showing the proof of the Carnot cycle to put the reversibility there. And before I even show you why it has to be reversable, let me just review what reversibility means. Now, we know that in order to even define a path here, the system has to be pretty close to equilibrium the whole time. That's the whole reason why throughout these videos, I've been drawing this piston, you know, have the gases down here, and then always-- instead of having one big weight on top that I took off or took on, because it would throw the system out of equilibrium-- I did it in really small increments. I just moved grains of sand, so that the system was always really close to equilibrium. And that's called quasistatic. and I've defined that before. And that means that you're always in kind of a quasi-equilibrium. So your state variables are always defined. But that, by itself, does not give you reversability. You have to be quasistatic and frictionless in order to be reversible. Now, what do we mean by frictionless? Well, I think you know what frictionless means. Is that like you see in this system right here, if I make this piston a little bit bigger, that when this piston rubs against the side of this wall, in kind of our real world, there's always a little bit of friction. Those molecules start bumping against each other, and then they start making them vibrate, so they transfer some kinetic energy. From just by rubbing into each other, they start generating some kinetic energy, or some heat. So you normally have some heat generated from friction. Now, if you have some heat generated from friction, when I remove a pebble-- first all, when I remove that first pebble, it might not even do anything. Because it might not even overcome-- you can kind of view it as the force of friction. But let's say I remove some pebbles, and this thing moves up a little bit. But because some of the, I guess you could say, the force differential, the pressure differential between the pebbles and the gas inside, and the pressure of the gas, was used to generate heat as opposed to work, when I add the pebbles back, if I have friction, I'm not going to get back to the same point that I was before. Because friction is always resisting the movement. So in order for something to be reversible, when I remove a couple of pebbles, if I removed ten pebbles and add the ten pebbles back, I should be at the exact same state. But as, you know, you can just do the thought experiment. If there's friction, I won't be at the exact same state. My piston won't move as much as you would expect if it was frictionless. So this is a key assumption for reversability. Now, the Carnot cycle, by definition, is reversible. And that's why no one could actually really implement an engine that fully does the Carnot cycle. And we even showed that, that the Carnot cycle is the most efficient potential engine. That if anyone made a more efficient engine, you could have a perpetual motion machine, or a perpetual energy machine. And the reason why the Carnot engine is the most efficient engine-- and there's no secret here-- it's because it's frictionless. Any engineer who makes engines could tell you, wow, if I could just remove all of the friction from my system, I would get a lot more efficient. Now, with that said-- so I've told you that look. The definition doesn't have to be-- it happened to work, Q divided by T, because I was dealing with a reversible system. And just to hit the point home, let me show you that it would not have worked if I had just defined it Q divided by T on an irreversible system. So let's say that I have-- let me draw another PV diagram. And I'm going to do almost a very simple thought experiment. Now I'm going to have an irreversible system. And I start here at some point on my PV diagram. And you know, this could be some type of cylinder, and it has a piston on top, and I have my rocks, like always. But this time, there's a little bit of friction. When this thing moves, a little bit of heat is generated in this. And when it moves in either direction, some heat is generated. So we can call that the heat from friction. When it moves either up or when it moves down. So let's do something. Let's stick this on a big reservoir, like we tend to do. So it's an isothermal system. So let's call this T1. And let's just start removing pebbles. And we'll move along an isotherm. Maybe to that point there. And then we're going to-- and I want to make a very important point here. Because this is has friction, I'm not going to get quite as far along the isotherm than if I didn't have friction. If this was a frictionless system, I would've gotten a little bit further along the isotherm. So the number of rocks isn't going to be the same as it was frictionless. But let's just say I move from here to here on the PV diagram. And then we go and add a bunch of rocks back, and we want to go all the way back. And I'm not even saying whether we have the same number of rocks or different rocks. You're probably going to have to add a few more rocks to get back to this point. But the idea here is, is that we've gotten back to the same point on the state diagram. So our delta U total should be equal to 0, which is equal to the delta U of expansion. So the delta U of expansion is the delta u to go that way, delta U of expansion, plus the delta U of contraction, which is the delta U of going back like this. Those have to be equal to 0 by definition, right? Because internal energy is a state variable, and if we get to that same point, our delta U has to be equal to 0. So what's our delta U of expansion? What's our change in internal energy as we expand? Our delta U of expansion is equal to the heat added to the system minus the work done by the system. And we know how much work was done, this whole area right there. And then plus the heat added by the friction. There's some heat added by the friction. Let me do that in brown. What's that? I was on some random website off the screen, and all of a sudden that cartoon sound started up. I have no idea what that was. But anyway, where was I? So I said our change in internal energy from expansion is going to be the heat added to the system from our reservoir minus the work done by the system, as we expand, plus the heat added to the system or generated by the system-- I guess you could say, it's not being added. The system is creating this heat itself, as it expands. There's this friction right there. Fair enough. So this is the one variation. Now that we're not dealing with the reversible process, we have this friction. Now what's our change in energy from contraction? So our change in internal energy from contraction is going to be the heat that leaves the system, that has to go back into the reservoir as we contract. Because otherwise, if we didn't have the reservoir, the temperature would go up. But we want to release heat. So we want to say, heat released-- and I'm going to do something. Let's just assume that all of the Q's are positive. So if I'm releasing heat, it's going to be a minus Q release. Let's just say that this is a positive number, and if I'm releasing it, it's going to be a minus right there. And I want to just do that, just to hopefully make things little bit clearer. Plus the work performed on the system. I'm assuming that work is always positive, so if we're doing work, it'll be minus work. If work is being done to us, it'll be plus work. But in this situation as well, we're still adding heat from friction. Or heat from friction is still being generated in the system. This is still positive. In either direction, when we move upwards or downwards, the system is generating friction. Now, we always said, we went all the way here, we went all the way back. So the sum of these has to be equal to 0, because this is a state variable. So if the sum of all of this has to be equal to 0, let's sum this up. So this gets us to Qa minus Qr. So the heat accepted minus the heat released. The W's cancel out. Plus-- let me see right here-- plus 2 times the heat of friction in either direction. All of that has to be equal to 0. Let's see. What we can do is, we can rewrite this as the heat accepted minus the heat released is equal to minus 2 times the amount of heat generated from friction. And then if we just switch these around, we'll get the heat released minus the heat accepted is equal to-- well, I just wanted to get all positive numbers-- 2 times the heat of friction. Now why did I do all of this? Because I wanted to do an experiment with an irreversible system, and this was a very simple experiment with an irreversible system. Now, we said that delta S, which a long time ago I defined as Q divided by T-- and in this video, I said it had to be reversible. And I wanted to show you right now that what if I didn't make the constraint that this has to be reversible? Because if this doesn't have to be reversible, and I just use this definition right here, you'll see that your delta S here would be-- you just divide everything by T-- because our temperature was constant the entire time, we were just on a reservoir-- you'll see that this is going to be your delta s. This is your total change in the, I guess you could say, your net heat added to the system. So this is, let me say, this is the heat added to the system. Let me do it this way. Heat added to the system, divided by the temperature at which it was added. Which is a positive number. Even though we got to the exact same place on this date diagram. So in an reversible system, this wouldn't be a valid state variable. So it's only a valid state variable if it's reversible. Now, does that mean that you can only talk about entropy for reversible reactions? No. You can talk about entropy for anything. But what you do is-- and this is another important point. So let's say that I have some irreversible reaction that goes from here to here. And I want to figure out its change in entropy. Right? And it might have done all sorts of crazy things. It's an irreversible reaction, its path might have gone like that. That's assuming it's quasistatic, if we can even look at its path like that. If we wanted to figure out its change in entropy, though, we wouldn't worry about the heat that was added to it and the different temperatures at which it was added. We wouldn't worry about that. We would just say, OK. What would it have taken for a reversible system to go from this state to this state? And then maybe a reversible system would have done something like this. Sorry, I want to make it a smooth curve. Maybe a reversible system might have done something like that. And this change, this heat added by the reversible system divided by the temperature for the reversible system, would be the change in entropy. And this change in entropy-- we could call this S final, and this is S initial, it's going to be the same for both systems. It's just, we don't use the irreversible system to figure out our entropy. We would use the reversible heat and temperature to figure out the actual change. Hopefully that clarifies something else. It's, on some level, a subtle point, but on some level it's super important. Because you can't-- the thermodynamic definition of entropy has to be this. It has to be heat added to a reversible system divided by the temperature that was added. Not just heat to any system. It just happened to work when I did it, and I should have been clearer about it when I first explained it, that it worked only because it was a Carnot cycle, which is reversible.