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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 at state C.
- So this volume and that volume.
- So volume at C 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, 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 math-y.
- But I think the fun part about it, even, it will
- 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 to 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 0.
- So if we go back to our original definitions-- let me
- show you that here.
- Right here, at the step and 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 definition, almost, or our first law of
- thermodynamics, we know that the 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, maybe it's a very small change in volume, while
- the pressure is constant.
- But if we're doing a quasi-static process, we can
- write this.
- Pressure, you can view it as kind of constant for that very
- small change in volume.
- So that's what we have there.
- Now, if it's adiabatic, we know that this is 0.
- And if that's 0, we can add P delta V to both sides of this
- equation, and we will get that-- this is only true if it
- were adiabatic-- that delta U, our change in internal energy,
- plus our pressure times our change in
- volume, is equal to 0.
- And 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 proved to you that U, the internal
- energy at any point in time-- let me write it here.
- The internal energy at any point in time is equal to 3/2
- times n times R times T.
- Which is also equal to 3/2 times PV.
- Now, if I have a change in internal energy, what can
- change on this side?
- Something must have changed.
- Well, 3/2 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 U could be rewritten as delta-- let me do
- it in a different color-- delta U could be written as
- 3/2 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 do something with this P here.
- Well, we'll just resort to our ideal gas equation.
- Because we're dealing with an ideal gas, we might as well.
- PV is equal to nRT.
- This should be emblazoned in your mind, at this point.
- So if we want to solve for P, we get O is
- equal to nRT 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 thing.
- So that means that 3/2 nR delta T plus P-- P is this
- thing-- plus nRT over V times delta V is equal to 0.
- Interesting.
- So what can we do further here?
- And I'll kind of tell you where I'm going with this.
- So that tells me, 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, you know, over each little small
- increments, they're going to add up to zero.
- So 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 temperature here.
- So don't try to even imagine, when we do the integral, that
- we should think of it in some terms 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, 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 divide it by nRT, let's divide it by nRT.
- And we have to do it to both sides of the equation. nRT.
- Well, on this term, the n's cancel out, the R cancels out.
- Over here, this nRT cancels out with this nRT.
- And what are we left with?
- We're left with 3/2-- we have this 1 over T left-- times 1
- over T delta T plus 1 over V delta V is equal to-- well,
- zero divided by anything is just equal to 0.
- Now we're going to integrate over a bunch of really small
- delta T's and delta V's.
- So let me just change those to our calculus terminology.
- We're going to do an infinite sum over infinitesimally small
- changes in delta T and delta V.
- So I'll rewrite this as 3/2 1 over T dt plus 1 over V dv is
- equal to 0.
- Remember, this just means a very, very
- small change in volume.
- This is a very, very, very, small change, an
- infinitesimally small change, in temperature.
- And now I want to do the total change in temperature.
- 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 start to
- temperature finish.
- And this will be going from our volume
- start to volume finish.
- Fair enough.
- Let's do these integrals.
- This tends to show up a lot in thermodynamics, these
- antiderivatives.
- The antiderivative of 1 over T is natural log of T.
- So this is equal to 3/2 times the natural log of T.
- We're going to evaluate it at the final temperature and then
- the starting temperature, 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 final
- velocity, and 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?
- I mean, we could integrate both sides-- well, 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 we could rewrite this green part as-- so it's 3/2 times
- the natural log of TF minus the natural log of TS, which
- is just, using our log properties, the 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, for the same reason, the natural log of VF over the
- natural log of VS. When you evaluate this, it's the
- natural log of VF minus the natural log of VS, which can
- be simplified this to this, just from our logarithmic
- properties.
- So this equals 0.
- And now we can-- this coefficient out front, we can
- use our logarithmic properties.
- Instead of putting a 3/2 natural log of this, we can
- rewrite this as the natural log of TF over TS to the 3/2.
- 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-- The natural
- log of TF over TS to the 3/2 power, times the natural log
- of VF over VS. And this is a fatiguing proof.
- All right.
- And all that is going to be equal to 0.
- Now what can we say?
- Well, we're saying that e to the 0 power-- the natural log
- is log base e-- e to the 0 power is equal to this thing.
- So this thing must be 1.
- E to the 0 power is 1.
- So we can say-- we're almost there-- that TF, our final
- temperature over our starting temperature to the 3/2 power,
- times our final volume over our starting
- volume is equal to 1.
- Now let's take this result that we worked reasonably hard
- to produce.
- 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 our PV equals nRT formulas.
- Although this was kind of PV-- this is internal energy at any
- point is equal to 3/2 times PV.
- And then we integrated over all the changes, and we said,
- look, this is adiabatic.
- So the total change-- the sum of all of our change in
- internal energy and work done by the system has to be 0,
- 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 T2.
- Watch the Carnot cycle video, if you forgot that.
- This was the VB All of these things up here were at
- temperature 1.
- All of the things down here were at temperature 2 So we're
- at temperature 1 up here, and temperature 2 down
- here, volume C.
- So let's look at that.
- So on that right part, that right process, our final
- temperature was temperature 2.
- So let me write it down.
- Temperature 2.
- Our initial temperature was temperature 1, where we
- started off at point B to the 3/2.
- 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 1.
- Neat.
- That's the result we got from this adiabatic process.
- We got that formula saying, 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, what's your final temperature?
- Don't want to get you dizzy going up.
- Well 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 3/2 power is times-- let me
- write our form formula there.
- Our final volume is the volume at A.
- That's where we moved to.
- And we moved from our volume at D.
- And this is going to be equal to 1.
- OK.
- We're almost there, if your eyes are
- beginning to glaze over.
- But this is interesting.
- And if anything, it's a little bit of fun mathematics to wake
- you up in the morning.
- So let's see.
- 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 this equation
- right here.
- So obviously if we take the reciprocal of this-- and we
- could just say, this is T2 over T1 to the minus 3/2
- power, which is the same thing as T1 over
- T2, to the 3/2 power.
- Right?
- That's just the reciprocal.
- And we're taking both sides to the negative 1 power, so we're
- going to have to take this to the negative 1 power.
- VB over VC.
- And when you take the reciprocal of 1, that just
- equals 1, That still equals 1.
- Which this also equals, so we could say, that equals this
- thing over here.
- So that is equal to T1 over T2, to the 3/2 power,
- times VA over VD.
- Now, these things are equal to each other.
- We can get rid of the 1.
- These two-- actually, let me just erase some of this.
- I don't want to make it say not equal to.
- They're equal to each other.
- They both equal 1.
- So they both equal each other.
- This thing and this thing are the same thing.
- T1 over T2 to the 3/2, T1 over T2 to the 3/2.
- 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 arithmetic to get there.
- Let's just cross-multiply.
- And you get VB times VD is equal to VC times VA.
- Now if we divide both sides by VBVC-- actually, let's do it
- the other way.
- Let's divide both sides by VDVA.
- 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 0.
- 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 of the work done by the system.
- When you add them up, you get to 0.
- Then we use that result from a few videos ago, where we said
- the internal energy at any point is 3/2 times nRT.
- So the change in internal energy is that times delta T,
- because that's the only thing that can change.
- We used 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 0.
- And we ended up with this formula over here, and then we
- just applied it to our two adiabatic processes.
- And 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.