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

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

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 you 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 on our PV diagram we got back there you 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 try to define some heat some heat related state variable let's say I call it heat content and I defined change in heat content is 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 was 10 well I added some heat here in this in this state 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 heat then I 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 with this 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 the internal if this is 0 if this right here is zero then the amount of the amount of heat we add to the system is the amount of work we did our internal energy is definitely zero 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 a heat of 10 here or whatever my heat and mystical variable I made up just now when we go around it would then be 10 plus W we go around again it would be 10 plus 2w + 10 + 3 w so it's state 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 the system not a valid state variable ignore it all now we know that Q 1 we added more heat here than we took away so there is some net heat added but that we can maybe you know there's something interesting here we added it from aha at a higher temperature and here we took less heat away at a lower temperature so maybe we can define another another state variable that can have the result that when we go around the cycle we do get back to our same value and 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 chain and 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 an intuition about what this actually means and kind of our minds but let's see if at least at least this is a valid state variable if as we go around the Carnot cycle whether our change in Delta s is zero right to be a legitimate state variable we have some value for s here maybe it's a hundred I don't know we once we go back around the Carnot cycle it should be a hundred again or a delta s should be zero 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 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're 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 zero so it's plus zero then we went from C to D that we were at a new temperature we were on a new ISO term we're 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 will 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 zero right these zeroes are zeros over you know changing temperature but this just zero so this thing should be equal to zero 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 change in our Mystikal new candidate state variable s is as we go around the Carnot cycle is equal to q1 over t1 plus q2 over t2 will seek to q2 is negative so what is the amount of what is q1 can we calculate a q1 well as we go as we are 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 it's the area under this curve it's the area under this curve not just the area in the cycle it'd 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 q1 is equal to the work done as we went from A to B and work remember can just be written as pressure 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 for a small change in volume and we're going to integrate it all over the little sums right this this DV is this little change in volume right there ty pressure that makes a little rectangle and then we sum up all the rectangles so we sum up all of the rectangles from our initial volume which is VA to our final volume which is V B 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 our pressure 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 what's the heat was being taken out of the system so we're going to go from we're going to go from where was our starting point VC and we go to V VD now how can we evaluate these integrals well we do we've done this before in a previous video we use the fact both in Q both of these circumstances when we go to a to B and when we go from C to D both of these circumstances occur on isotherms right so the only things that are changing our pressure and volume temperature is not changing and so if we go back to our ideal gas equation that PV is equal to n RT we can just rewrite this by dividing both sides by V as P is equal to n R T over V and use 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 V a to V B of n R T over V DV and q2 is equal to the integral from VC to VD of n R T over V DV I'm going to do to two integrals in parallel just so that you kind of see that we're essentially solving the same thing okay so how can we solve this well we know in both of these cases were moving along an isotherm that our temperature our temperatures are constant and actually we know what the temperatures are on the 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 t 2 it was kept that way by our reservoir right t2 when we moved from C to D and t1 we move from A to B those were our temperatures and they're constant fair enough so let's we can take so n is constant R is definitely a constant and is this the 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 n R T 1 over the integral from VA to VB times 1 over V DV and q2 we can write as n R t2 times the integral from VC to V D 1 over V DV all right now this integral is pretty straightforward to evaluate the antiderivative of 1 over V is the natural log of V so we get Q 1 is equal to n R t1 times the natural log of V evaluated at VB minus it evaluated at VA and Q 2 let me just solve this whole equation right here so this is equal to what this is equal to the natural log of VB minus the natural log of VA which is the same thing as the natural log of VB over VA times n R t1 and all that's equal to Q 1 now the same logic Q 2 is going to be equal to what Q 2 is going to be equal to n R t2 now the only difference with this integral is where I had VB now I have VD sorry so now then it becomes natural log of VD and where I had VA now I have VC so over V CE alright now what was our original question that we were dealing with we said this is a legitimate state variable if our chortles if this if the change in this whatever this value s is we go around the cycle is equal to zero that means it didn't change so these two things when you sum them have to equal zero q1 over t1 plus q2 over t2 so let's add them so q1 over t1 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 cute 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 n our n R times the natural log of VB over VA that's that right there and then plus q2 over t2 which is just n R times the natural log of VD over VA this is sorry over V over VC this is VC this is a VA here all right now let's see what we can do this is equal to almost there homestretch 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 a natural log of a B so this is equal times the natural log of VB over V V a times VD over VC all right now this is so this is our change in our s state variable that we're playing with right now now what is this equal to now what if we divide the numerator and then or well let's say we multiply let me think of the best way to to say this so let's 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 by VC over VD or let me multiply its numerator denominator by VC over VD so I can rewrite this as the natural log of VB over VA / right instead of multiplying it times this I can divide it by this is reciprocal VC over VD VD so I just rewrote it I just did a little bit of fraction math that's all I did instead of multiplying it times this I divided by its reciprocal now now you see why the previous video I did was done what is this equal to in 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 one if that's equal to one then what's the natural log of one so our change in our mystical S state variable is n R times the natural log of one what's the natural log of one e to the what power is equal to one 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 that deals with heat if we define change in s change in s is equal to 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 it kind of a micro a microstate level but at least we've stumbled upon some property of something where ever if s is 10 here if s is 10 we go and 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 we come back here our change in s is going to be 0 again or sorry is 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 of what it actually means we'll leave that to a future video