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Current time:0:00Total duration:15:38

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 I defined my change in entropy z-- 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 what I did that I looked at the Carnot cycle I looked at the Carnot cycle and just as a bit of a review never hurts to review let me draw the PV diagram here we saw that okay we started this state here and then we proceeded isothermally we removed little pebbles off the piston so we increase the volume and lower the pressure then we proceeded adiabatically where we isolated things where we isolated things we move like that that was adiabatic lee then at this other isotherm we added the pebbles back so we added some pebbles back and then we kept and then we isolated the system again so adiabatically we continue 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 that there's some heat being added here q1 and then 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 when I looked at the Carnot cycle and I used this definition of entropy I saw that the change in s 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 q1 over t1 plus q2 over t2 and then I actually showed you that this was equal to zero 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 should only be dependent on my state variables so even if I go on some easy path at the end of the day it should get back to zero but I did something I guess a little bit it 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 is it was only valid because the Carnot cycle was reversible and this is a subtle but super important point and I really should have 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 and and and before I even show you why it has to be over so let me just review what reversibility means now we know that it 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 had the gasses down here and then I'll have 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 and so that the system was always really close to equilibrium and that's called quasi static and I've defined that before quasi static quasi static 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 a reversibility you have to be quasi static and frictionless and frictionless in order to be reversible now what do we mean by frictionless well I I think you know what frictionless means is that like this e in this system right here if I make this piston a little bit bigger that when this 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 when I remove a pebble first of all when I that first pebble it might not even do anything because it might not even overcome the you can kind of view it as the force of friction let's say I remove some pebbles and this thing moves up a little bit it 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 D and the nd 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 remove ten pebbles and add the ten val couples 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 I my my my my piston won't move as much as you would expect if it was frictionless so this is key this is a key assumption for reversibility now the Carnot cycle by definition is reversible it's reversible and that's why no one could actually really implement an engine that that 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 potential more efficient engine you could have a 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 is because it's it's frictionless any engineer who makes engines can tell you wow if I could just remove all of the friction from my system I will be get I would get a lot more efficient now with that said so I've I've told you that look this has 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 just defined it Q divided by T on an irreversible system so let's say that I have let me draw another PV diagram I have another PV diagram I'm going to do a very almost a very I would say 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 a 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 when this thing moves a little bit of heat is generated and this and when it moves in either direction some heat is generated so we couldn't 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 let's call this t1 and let's just start removing pebbles let's just start removing pebbles and we will move along an isotherm will move an ice 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 friction this has friction I'm not going to get quite as far along the isotherm then if I if I didn't have friction if this was a frictionless system I would have gotten a little bit further along the isotherm so the number of rocks isn't going to be the same as if 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 had the same number of rocks or different were actually 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 2 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 Delta internal energy is a state variable 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 what's our change in internal energy as we expand our Delta u of expansion is equal to was equal to the heat added to the system right it's equal to the heat added to the system minus the work done by the system minus the work done by the system so minus 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 plus some heat added by the friction what's what's that I was on some random website on the others we're off off the screen and all of a sudden that cartoon sound started up I didn't had 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 by 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 this is that the system is creating this heat itself as it expands is this friction right there fair enough so since this is the one variation when now that we're not dealing with a 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 is that is leaves the system that 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 minus Q released let's just say this is a positive number as if I'm releasing it it's going to be a minus right there and I want to just do that just to make hopefully make things a little bit clearer plus the work performed on the system I'm assuming that work is always positive so for 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 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 that some of these has to be equal to zero because this is just a state variable so if the sum of the all of this has to be equal to zero let's sum this up so this gets us to Q a minus Q R so the heat accepted minus the heat released the WS cancel out plus plus let me see right here plus two times the heat of friction in either direction all of that has to be equal to zero let's see we can what we can do is we can we can rewrite this as the heat accepted minus the heat released is equal to minus two times the amount of heat generated from friction and then if we just switch these around we'll get the heat released - the heat accepted is equal to well I just wanted to get all positive numbers two times the heat of friction now why did I do all of this because I wanted to do an experiment with a with an irreversible system and this was a very simple experiment with the irreversible system now we said that Delta s Delta s which in a long time ago I defined as Q divided by T and in this video I said it had to be reversible I wanted to show you right now that what if this is what if I didn't make the 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 your Delta s here would be you just divide everything by T you would just divide everything by T because our temperature was constant the entire time we were just on a reservoir you'll see that this usually this is your going to be your Delta s this is your total change in that your I guess you could say your net heat added to the system so this is let me say this is this is the heat added to the system let me do this way heat added to the system divided by the temperature at which it was added divided by the temperature yeah exactly the divided by the temperature was added which is a positive number even though we got to the exact same place on the state diagram so in an irreversible 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 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 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 and its path might have gone like that that's it necessarily been assuming it's quasi static 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 okay what would over what would have taken for a reversible system to go from this state to this state and then we maybe a reversible system would have done something like this I'm sorry I want to make it a smooth curve maybe a reversible system might have done something like that and that and this change this heat added in there 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 say call this s final sorry 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 was added not just heat to any system it just happen to work when I did it and I should have been clear about it when I first explained it that it that it worked only because it was a Carnot cycle which is which is reversible