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Current time:0:00Total duration:20:53

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 a ceiling that's movable the gas and we're thinking of mono atomic ideal gases in here they're exerting some pressure onto this 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 well you know 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 well-defined 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 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 the 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 adiabatic Li what is adiabatic Li mean if we did if we remove 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 in 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 of reservoirs you know let's say I had a a cup of water in a stadium and the stadium is the air conditioner in the stadium is at 60 degrees well no matter what I do to that water if I could 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 reservoirs temperature decrease if it's throwing off heat well it would but it's so much larger that it's it's impact isn't noticeable for example if I put a cup of boiling water into into a super large covered domed stadium the water will get colder to the the ambient temperature of the stadium the stadium will get warmer but it'll be so marginally warmer that 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 let me get my pen correctly okay so our volume has gotten larger by roughly this mount 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 only because of this reservoir I want to make that clear and also just to review this is a quasi-static process because I'm doing it very slowly the system is in equilibrium the whole time so let's draw let's draw what we have so far on our famous PV diagram so this is this is P axis that's the V axis right label them this is P this is V let me call this let me 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 you do it like that that's state a and it moves to state B and notice I kept the temperature constant what did we learn and 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 you're 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 stage B I'll move over here to state B and the whole time this was at a constant temperature t1 t1 t1 now we've done a bunch of videos now we said okay 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 under 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 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 changes 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 I 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 out so if 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 along this isotherm well we moved along 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 get the put the the work done by the system 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 somebody to these rocks so it was giving energy away it was giving energy outside of the system so what how did it how did it maintain its internal energy well someone had to give it some energy and it was given that energy by this by this reservoir so let's say and the way the convention for doing this is it's to say that it was given let me write this down it was given some energy q1 we just say we just put this down an hour so that some energy went into the system here fair enough now now let's take this state B and remove the reservoir so now 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 some even more rocks than we had in biess maybe I only have one rock left I only have one rock left and obviously the floor the 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 okay let me fill that in and then I just use the blue this is 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 my we'll have gone down there they're going to bump into each into the walls less and because I removed I removed the reservoir what's going to happen to the temperature I'm going to 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 hour continues down here and I'll say a adiabatic adiabatic now since I'm moving from one temperature to another this is now at t2 this is now 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 now I would have kept moving along this hyperbola but now that we didn't keep our tension 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 it like that so let's say I have another everything actually let me it should actually move 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 got kept going down and my volume would have kept going down so this move from B to C for to state C will look like this from B to state C let me do it in another color let me do it in the orange color of this arrow so that it'll look like this and now we are at state C state C now this was adiabatic adiabatic which means there is no exchange of heat so I don't have to figure out what how much heat got transferred into the system now there's something interesting here we still did do we'd 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 where well the main thing is is what what was reduced by that work energy and well if you think well I to future video our entry are our internal energy was reduced right because our temperature went down so our internal energy went down we'll talk more about that in a future video so now that we're at state C and we're at temperature t2 let's let's put back another sink here let's put back another sink but this sink what it's going to have is a reservoir so let me put two things right here so this sink 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 isothermal 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 swap 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 a let me make it let me make it blue because it's going to be because here what's happening I'm now adding pebbles in I'm compressing the gas the gas if we were 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 I this is t2 remember this reservoir is a 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 there's heat being 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 so I also say and that amount of heat is q2 q2 so I move along this so this right this is right here I'm moving along another isotherm I'm moving along this isotherm until I get to 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 you can make the argument that we moved along an isotherm our temperature did not change from C to D we know that our 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 it was at temperature t2 right because we had this reservoir here it stayed the same if your temperature stayed 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 the heat 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 other the force times distance went the other way so given that work was done to the system that he added to the system was negative right we're just applying the same thing heat if our if our internal energy is zero 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 the system gave away heat will say put that with Q - Q - and where did it give that heat it gave it to this reservoir that we put here this kind of cold reservoir you can almost view it as a well I'll just it's accepting the heat okay we're almost there now let's say we remove this reservoir from under our system again so it's it's completely isolated from everything else at least in terms of heat and what we do is we start adding so state D it still wasn't you know 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 to my pebble color so we start adding more pebbles again to get it back to state a so that's this process right here that's this process let me do a different color this is let's say this this is green so as we add pebbles that's this movement right here this we're moving from one isotherm up to another isotherm at a higher temperature and remember this whole time we went in this clockwise direction so a couple of interesting things are going on here because we're assuming an ideal scenario nothing was lost to friction this piston just you know moves up and down no heat loss due to that what we can say is is that we've achieved we have 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 is what we started with so what we've done here is completed a cycle in this particular cycle it's an important one it's called Carnot cycle it's named after a French engineer who is trying to just optimize engines in the early 1800s so it's a Carnot cycle 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 it what they're talking about 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 we relate it to kind of more modern notions of it now something 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 hey Sal this doesn't seem like a great engine I have to move pebbles and all of that and you're right it's not 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 heatsink transferred some heat to the system and then the system transferred a smaller amount of heat back to the ambient back to the other back to the other reservoir right so this in this 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 we or the area inside of this cycle so this is the work done right this is the work done by our Carnot engine and if you the way you think about it is this the area 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 could write this Carnot engine like this we could write it like this it's taking its starting so you have a reservoir at t1 at t1 and then you have your engine right here and then it's connected adventure so it's takes is takes q1 and from this reservoir it does some work right the work is represented by the amount of the work right here is 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 that 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 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 the 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 thermodynamics class I encourage you to even you know go off on your own and do this yourself kind of you can almost you know take a pencil and paper and and kind of redo this video that I just did because 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 regards to entropy will be a little bit more intuitive and not too confusing