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the goal in this video is to essentially prove a pretty simple result and that's that the ratio between the volumes let me write this down that the ratio between the volume at state B and the volume at state a so the ratio of that volume to that volume is equal to in our Carnot cycle is equal to the ratio between the volume of state C so this volume and that volume so volume it's e to the volume at D so this is what I'm about to embark to prove a fairly simple result that maybe is even if you look at this it's that looks about right so if you're happy just knowing that you don't have to watch the rest of the video but if you are curious how we get there I encourage you to watch it although it gets a little bit mathy but I think the fun part about it even you know one satisfy you that this is true but the other thing is we'll be able to study adiabatic processes a little bit more so just kind of launching off of that the whole proof revolves around this step right here and this step right here when we go from D to a so by definition an adiabatic process is one where there's no transfer of heat so our heat transfer in an adiabatic process is zero so if we go back to our original definition so let me show you that here right here at this step in this step we have no transfer of heat so if we go back to there adiabatic we're completely isolated from the rest of the world so there's nothing to transfer heat to or from so if we go to our our definition almost or our first law of thermodynamics we know that the change in internal energy change in internal energy is equal to the heat applied to the system minus the work done by the system and the work done by the system is equal to the pressure of the system times some change in volume at least you know maybe it's a very small change in volume while the pressure is constant but if we're doing a a quasi-static process we can write this pressure can you can view it as kind of constant for that little very small change in volume so that's what we have there now if it's adiabatic we know that this is zero we know that this right there is zero and if that's zero we can add P Delta V to both sides of this equation and we will get then this is only true if we're adiabatic the Delta U our change in internal energy plus our pressure times our change in volume is equal to zero let's see if we can do this somehow we can do something with this equation to get to that result that I'm trying to get to so a few videos ago I prove to you that you the internal energy at any point in time let me write it here the internal energy at any point of time is equal to let me is equal to three-halves times n times R times T which is also equal to three-halves times P V now if I have a change in internal energy what can change on this side something must have changed well three-halves can't change n can't change we're not going to change the number of molecules we have the universal gas constant can't change so the temperature must change so there you have it you have Delta you could be rewritten as Delta let me do it in a different color Delta you could be written as 3 halves n times R times our change in temperature and that's why I keep saying in this especially when we're dealing with the situation where all of the internal energy is essentially kinetic energy that if you don't have a change in temperature you're not going to have a change in internal energy likewise if you don't have a change in internal energy you're not going to have a change in temperature so let me put this aside right here I'm going to substitute it back there but let's see if we can if we can do something with this with this P here well we'll just resort to our our ideal gas equation because we're dealing with an ideal gas we might as well PV is equal to NRT this is should be should be emblazoned in your mind at this point so if we want to solve for P we get P is equal to n R T over V fair enough so let's put both of these things aside and substitute them into this formula so Delta U is equal to this so that means that three-halves n r delta T plus P so plus P P is this thing plus n R T over V times Delta V times Delta V that's the Delta V is equal to 0 interesting so what can we do for the year and I'll kind of tell you where I'm going with this so this tells me if I over my change in internal energy over a very small delta T this tells me my work done by the system over a very small Delta V and we're saying that we you know over at each little small increments they're going to add up to zero what I want to do is Inc so this is let me just go back to the original graph so this is over a very small Delta V right there let me do it in a more vibrant color a very small change as we go from there to let's say there we're going to have some change in our volume and you don't see the top the temperature here so don't try to even imagine when we do the integral that we should think of it in some times of area but we're going to integrate over the change in temperature as well the temperature changes a little bit from there to there so what I want to do this is over very this is right here over a small change I want to integrate eventually over all of the changes that occur during our adiabatic process so let's see if we can simplify this before I break out the calculus so if we divide both sides by NRT what do we get so let's divided by n RT let's divide it by n RT and we have to do to both sides of this equation and RT well on this term the ends cancels out the R cancels out over here this NRT cancels out with this NRT and what are we left with we're left with three-halves we have this 1 over T left times 1 over T delta T plus plus 1 over V Delta V is equal to zero divided by anything is just is equal to zero is equal to zero now we're going to integrate over a bunch of really small Delta T's and and end and Delta V so let me just change those to our calculus terminology we're going to set do an infinite sum over an infinitesimally small changes in delta T and Delta V so I'll rewrite this as 3 halves 1 over T DT plus 1 over V DV is equal to 0 remember this just means a very very small change in in volume this is a very very very small change infinitesimally small change in temperature and now I want to do the total change in temperature and the total I want to integrate over the total change in temperature and the total change in volume so let's do that so I want to go from always temperature temperature start to temperature finish the temperature finish and this will be going from our volume start volume start to volume finish fair enough let's do these integrals we've just this tends to show up a lot in thermodynamics these these anti derivatives the antiderivative over 1 over T is natural log of T so this is equal to three-halves times the natural log of T we're going to evaluate it at the final temperature and then the starting temperature plus plus the natural log the antiderivative of 1 over V is just the natural log of V plus the natural log of V evaluated from our Farmar well from our final velocity we're going to subtract out the starting velocity this is just the calculus here and this is going to be equal to 0 right now I mean we could integrate both sides of this equal if every infinitesimal change is equal to the sum is equal to 0 the sum of all of the infinitesimal changes are also going to be equal to 0 so this is still equal to 0 see what we can do here so that we could rewrite this green part as so it's three-halves times the natural log of TF minus the natural log of TS which is just using our log properties natural log of TF over the natural log of TS right when you evaluate you get natural log of TF minus the natural log of TS that's the same thing as this plus plus for the same reason the natural log of VF over the natural log of V s when you evaluate this as the natural log of VF minus the natural log of V S which can be simplified this to this from just from our logarithmic properties so that equals zero and now we can this coefficient out front we can use our logarithmic properties instead of putting a three half natural log of this we can rewrite this as the natural log of TF over TS to the three halves now we can keep doing our logarithm properties you take the log of something plus the log of something that's equal to the log of their product so this is equal to I'll switch colors this is equal to the natural log the natural log of TF over TS to the three half power times the natural log of VF over V s and this is a fatiguing proof all right and all of that is going to be equal to zero now what can we say well we're saying that e to the zero power write the natural log is log base E e to the zero power is equal to this thing so this thing must be 1 e to the zero power is one so we can say we're almost there we can say that TF TFR final temperature over our starting temperature to the three half power times our final volume over our starting volume is equal to 1 now let's take this result that we we worked reasonably hard to produce let's take this result remember all of this we just said we're dealing with an adiabatic process and we started from the principle of just what the definition of internal energy is and then we substitute it with kind of our PV equals NRT formulas although this was kind of PV you know the is the internal energy at any point is equal to three halves times PV and then we got and then we integrated over all of the changes and we said look this is adiabatic so the total change has to the sum of all of our all of our change in internal energy and work done by the system has to be zero then we use the property of log to get to this result now let's do these for both of these adiabatic processes over here so the first one we could do is this one where we go from volume B at t1 to volume C at T to watch the Carnot video Carnot cycle video if you forgot that this was VB all of these things up here were at temperature one all of the things down here we're at temperature two so we're at temperature one up here and temperature two down here at volume C so let's look at that so on that right part that write process our final temperature was temperature two so let me write it down temperature - our initial temperature was temperature one where we started off at point B to the three-halves times what was our final volume our final volume was our volume at C divided by the volume at B and that's going to be equal to one neat that was what that's the result we got from this adiabatic process we got that formula saying that this is adiabatic we did a bunch of math and then we just substitute for our initial and final volumes and temperatures let's do it the same way but let's go from D to a so when you go from D to a when you go from D to a what's your final temperature let me don't want to get you dizzy going up and what your final temperature we're going from D to a so our final temperature is t1 and our final volume is the volume at a go back down so our final temperature is t1 our initial temperature is t2 we're going from D to a to the three halves power is x times let me write our formula there are final our final volume is a volume at a that's where we moved to from and we moved from our volume at D and this is going to be equal to one okay we're almost there if you're getting if your eyes are beginning to glaze over but this is interesting and if anything it's a little bit of fun mathematics to to wake you up in the morning so let's see if we can almost relate these two things we could set them equal to each other but it's not quite satisfying yet let's take the reciprocal of both sides of let's take the reciprocal of both sides of this equation right here so obviously if we take the reciprocal of this any you know this is we could just say this is t2 over t1 to the minus three halves power which is the same thing as which is the same thing as t1 over t2 to the three halves power right that's just the reciprocal and we're taking it at both sides to the negative one power so we're going to take this to the negative one power VB over VC and when you take the reciprocal of one that just equals one that's still equals one and which is this also equals so we can say that equals this thing over here so that is equal to t1 over t2 to the three-halves power times VA over VD now these things are equal to each other we can get rid of the one these two actually let me just erase let me erase some of this I want to make it say not equal to they're equal to each other they both equal one so they both equal each other these both equal each other well this thing and this thing are the same thing t1 over t2 to the three-halves T 1 over T to the three-halves so let's just divide both sides by that those cancel out and what are we left with I think you can see the finish line the finish line is near we have VB over VC is equal to VA over VD now that's not quite the result we wanted but it takes a little bit of simple rhythmic arithmetic to get there let's just cross multiply and you get VB times VD is equal to VC times VA now if we multiply if we divide both sides by by let's divide both sides by VB VC so if we divide both sides by VB VC actually let's do it the other way let's divide both sides by VD VA VD V a VD V a what do we get these cancel out and these cancel out and we are left with VB over VA is equal to VC over VD all that work for a nice and simple result but that's better than doing a lot of work for a hairy and monstrous result so that's what we set out to prove that VB over VA is equal to VC over VD and we got it all from the notion that we're dealing with adiabatic process that our change in heat is zero and we just went to our formula or our definition of our change in internal energy the first law of thermodynamics that if we have any change in internal energy it must be equal to the amount of work done by the system or at least a negative done of the work done by the system when you add them up you get to zero then we use that result from a few videos ago where we said three the internal energy at any point is three halves times NRT so the change in internal energy is that times delta T because that's the only thing that can change we use PV equal NRT and then we just integrated along all of the little changes in temperature and volume as we moved along this line as we moved along the line we took the integral we said that had to be equal to zero and we ended up with this formula over here and then we just applied it to our two adiabatic processes when we went from B to C and we went from D to a and we got these results and we got to our finish line see you in the next video