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I've talked a lot about the
general idea that in order to

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have a state variable like,
say, U, which is internal

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energy, at any point in this
PV diagram, that state

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variable should be that value.

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So for example, if at this
point, U is equal to 5, and I

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go do this whole Carnot cycle,
when I come back to state A, U

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should still be equal to 5.

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It should not have changed.

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It's not dependent upon what
we did to get there.

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So if we did some kind of crazy
path on our PV diagram,

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we got back there, U should
always be the same.

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That's what it means to
be a state variable.

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It's only dependent upon its
position in this PV diagram.

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It's only dependent on its
state, not how you got there.

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And because of that, heat is
something that we can't really

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use as a state variable.

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For example, if I tried to
define some heat-related state

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variable, let's say I call it
heat content, and I defined

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change in heat content as equal
to the amount of heat

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added to the system.

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Well, if we go back to our
Carnot cycle here, let's say

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that my heat content
here was 10.

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Well, I added some heat here,
in this process here.

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Nothing happened, because this
was adiabatic from B to C.

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Then from C to D, I took
out some heat.

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But I took out less heated
than was added here.

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And then here, nothing was
done with regard to heat.

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So I did add some heat
to the system.

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The net heat that I added to
the system as I went around

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the cycle-- in this case, Q
would be equal to Q1 minus Q2.

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And we know that this number
is larger than this.

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The net amount of heat we added
to the system was the

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amount of work we did on the
system, because the internal

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energy didn't change.

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So if this is 0, then the amount
of heat we add to the

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system is the amount
of work we did.

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Our internal energy is
definitely 0 as we go all the

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way around.

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We did this shaded portion of
work-- I showed you that

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several videos ago, that the
area inside of our little

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cycle is the amount
of work we did.

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And so that's also the
net amount of heat

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added to the system.

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So if we added that amount of
heat to the system, if we

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started off at the heat of 10
here, or whatever my heat

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content mystical variable I made
up just now, when we go

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around, it would then
be 10 plus W.

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If we go around again,
it would be 10 plus

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2W and 10 plus 3W.

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So it can't be a legitimate
state variable, because it's

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completely dependent on what
we did to get there.

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And if we keep going around the
cycle, we can increase it,

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even though we get to
the same point.

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So this is an illegitimate state
variable, where I define

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the change in our little made up
heat content to be equal to

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the heat added to system.

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Not a valid state variable.

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Ignore it all.

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Now we know that Q1, we added
more heat here than we took

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away, so there was something
net heat added.

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But there's something
interesting here.

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We added it at a higher
temperature.

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And here we took less heat away
at a lower temperature.

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So maybe we can define another
state variable that can have

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the result that when we go
around the cycle, we do get

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back to our same value.

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Now let me just-- we're
just experimenting.

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Although I know where this
experiment will go.

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I wouldn't have been doing
it if I didn't.

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So let's say I define a
new state variable S.

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And I define a change in S.

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So I say, a change in S-- I'm
just making up a definition--

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is equal to the heat added to
the system divided by the

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temperature at which it was
added to the system.

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Now, I don't know what
this means just yet.

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In future videos, maybe we'll
get intuition about what this

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actually means in kind
of our minds.

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But let's see if at least this
is a valid state variable.

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If, as we go around the Carnot
cycle, whether our change in

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delta S is 0, right?

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To be a legitimate state
variable, we have some value

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for S here.

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Maybe it's 100.

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I don't know.

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Once we go back around the
Carnot cycle, it should be 100

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again, or our delta
S should be 0.

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So what's the delta S?

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So the delta S, as we go around
the whole cycle-- let

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me write delta S-- let
me do another color.

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Delta S.

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As we are go around, I'll say
c for the Carnot cycle.

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As we go around the Carnot
cycle, is going to be equal

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to-- well, when we went from A
to B, we were at a constant

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temperature, and we added Q1.

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So it's Q1, and we were
at temperature T1.

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Fair enough.

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Then when we went from B
to C, it was adiabatic.

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We added or took away no heat.

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So this value, Q over
T, would just be 0.

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So it's plus 0.

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Then we went from C to D.

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We were at a new temperature,
we were on a new isotherm.

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We were at T2.

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And we took away, or I won't put
the sign here, let's just

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say we added Q2 heat.

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We're going to actually
solve for it later.

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We added Q2 heat.

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We'lll see that it's actually
a negative value.

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And then finally, when we
went from D to A, it

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was adiabatic again.

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So no transfer of heat.

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So plus 0, right?

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The 0's are 0's over, you know,
changing temperature,

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but this is just 0.

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So this thing should be equal to
0 in order for this to be a

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valid state variable.

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So let's figure out what
this value is.

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What Q1-- so our change in our
mystical new candidate state

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variable, S, as we go around the
Carnot cycle, is equal to

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Q1 over T1, plus Q2 over T2.

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And we'll see, Q2 is negative.

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So what is Q1?

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Can we calculate Q1?

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Well, as we're on this top
isotherm, our temperature

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doesn't change, our internal
energy doesn't change.

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So if your internal energy
does not change, if your

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internal energy is 0, then the
heat added to the system is

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equal to the work done
by the system.

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So its the area under
this curve.

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Not just the area
in the cycle.

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It would be the whole area
under the curve.

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So what's the whole area
under the curve?

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Well, so let me do a
little aside here.

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So Q1 is equal to the work done
as we went from A to B.

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And work, remember, can just be
written as pressure times

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change in volume.

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We're going to do a little
calculus here, so I'll write

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dV for a small change
in volume.

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And we're going to integrate
it all over the

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little sums, right?

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This dV is this little
change in volume

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right there times pressure.

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That makes a little rectangle.

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And then we sum up all the
rectangles from our initial

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volume, which is VA, to our
final volume, which is VB.

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And then, what's Q2 going
to be equal to?

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Well, Q2 is going to be
essentially the same thing.

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It's going to be the sum of the
work done by our system,

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which in this case is going to
be negative, because work was

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done to our system as we
go from here to here.

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Right?

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That's when Q2 was operating,
the heat was being taken out

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of the system.

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So we're going to go from--
where was our starting point?

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VC and we go to VD.

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Now, how can we evaluate
these integrals?

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Well, we've done this before
in a previous video.

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We use both of these
circumstances-- When we go to

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A to B, and we go from C
to D-- both of these

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circumstances occur on
isotherms, right?

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So the only things that
are changing are

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pressure and volume.

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Temperature is not changing.

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And so if we go back to our
ideal gas equation-- PV is

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equal to nRT-- we can just
rewrite this by dividing both

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sides by V, as P is equal
to nRT over V.

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And substitute that back in
for P, in both cases.

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That is P as a function of V.

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We now have the equation
of the curve.

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And we're taking the area
under it in both cases.

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So Q1 is equal to the integral
from VA to VB

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of nRT over V, dv.

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And Q2 is equal to the integral
from VC to VD of nRT

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over V, dv I'm going to do two
integrals in parallel, just so

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that you kind of see that we're
essentially solving the

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same thing.

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OK.

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So how can we solve this?

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Well, we know in both of
these cases, we're

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moving along an isotherm.

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That our temperatures
are constant.

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And actually, we know the
temperatures are.

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When we're moving from VA to
VB, our temperature is T1.

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It was kept that way
by our reservoir.

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When we moved from VC to VD,
our temperature was T2.

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It was kept that way by
our reservoir, right?

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T2, when we move from C to D.

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And T1, we move from A to B
Those were our temperatures.

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And they're constant.

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Fair enough.

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So we can take-- so
n is constant.

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R is definitely a constant.

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n is just a number of molecules
we have. And then

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our temperature is also
constant, so we can take it

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out of the integral.

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So we can rewrite Q1 is equal to
nRT1 over the integral from

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VA to VB times 1/V dv, and Q2 we
can write as nRT2 times the

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00:09:49,430 --> 00:09:56,000
integral from VC
to VD, 1/V dv.

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All right.

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Now this integral is fairly
straightforward to evaluate.

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The antiderivative of 1/V
is the natural log of V.

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00:10:02,660 --> 00:10:11,470
So we get Q1 is equal to nRT1
times the natural log of V

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00:10:11,470 --> 00:10:16,050
evaluated at VB, minus
it evaluated at VA.

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And Q2-- well, let me just
solve this whole equation

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00:10:19,650 --> 00:10:20,580
right here.

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So this is equal to what?

210
00:10:22,320 --> 00:10:24,710
This is equal to a natural
log of VB minus the

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00:10:24,710 --> 00:10:26,160
natural log of VA.

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00:10:26,160 --> 00:10:33,390
Which is the same thing as the
natural log of VB over VA

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00:10:33,390 --> 00:10:36,590
times nRT1.

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00:10:36,590 --> 00:10:38,850
And all that is equal to Q2.

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00:10:38,850 --> 00:10:42,690
Now, the same logic, Q2, is
going to be equal to what?

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00:10:42,690 --> 00:10:48,220
Q2 is going to be
equal to nRT2.

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00:10:48,220 --> 00:10:50,090
Now the only difference with
this integral is where I had

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VB, now I have VD.

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Sorry.

220
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So then it becomes natural
log of VD.

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And where I had VA, now
I have VC, So over VC.

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00:11:01,300 --> 00:11:02,290
All right.

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00:11:02,290 --> 00:11:04,870
Now what was our original
question that we

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00:11:04,870 --> 00:11:05,730
were dealing with?

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00:11:05,730 --> 00:11:09,020
We said, this is a legitimate
state variable.

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If the change in this-- whatever
this value S as we go

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around the cycle-- is
equal to 0, that

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means it didn't change.

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So these two things, when you
sum them, have to equal 0.

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00:11:19,400 --> 00:11:21,800
Q1 over T1, plus Q2 over T2.

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00:11:21,800 --> 00:11:23,860
So let's add them.

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00:11:23,860 --> 00:11:29,820
So Q1 over T1 is equal
to that over T1.

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That cancels out.

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Q2 over T2 is equal
to that over T2.

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That cancels out.

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So our change in our mystical
state variable, as we go

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around the Carnot cycle,
is equal to Q1 over T1,

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plus Q2 over T2.

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00:11:47,730 --> 00:11:56,770
Which is equal to nR times the
natural log of VB over VA.

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00:11:56,770 --> 00:11:58,760
That's that, right there.

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00:11:58,760 --> 00:12:06,070
And then plus Q2 over T2, which
is just nR times the

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00:12:06,070 --> 00:12:15,120
natural log of VD over V.

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00:12:15,120 --> 00:12:16,310
This is VC.

244
00:12:16,310 --> 00:12:18,460
This is a VA here.

245
00:12:18,460 --> 00:12:19,360
All right.

246
00:12:19,360 --> 00:12:20,140
Now let's see what we can do.

247
00:12:20,140 --> 00:12:22,570
This is equal to--
almost there.

248
00:12:22,570 --> 00:12:23,720
Home stretch.

249
00:12:23,720 --> 00:12:24,820
nR.

250
00:12:24,820 --> 00:12:26,930
We can factor out an nR.

251
00:12:26,930 --> 00:12:30,130
And then the natural log of A
plus the natural log of B is

252
00:12:30,130 --> 00:12:32,080
just the same thing as the
natural log of AB.

253
00:12:32,080 --> 00:12:40,590
So this is equal to times the
natural log of VB over VA,

254
00:12:40,590 --> 00:12:45,620
times VD over VC.

255
00:12:45,620 --> 00:12:47,440
All right.

256
00:12:47,440 --> 00:12:51,280
So this is our change in our
S, state verbal that we're

257
00:12:51,280 --> 00:12:52,660
playing with right now.

258
00:12:52,660 --> 00:12:53,910
Now what is this equal to?

259
00:12:53,910 --> 00:12:59,890


260
00:12:59,890 --> 00:13:02,030
Let me think of the best
way to say this.

261
00:13:02,030 --> 00:13:06,030
So let's divide the numerator
and the denominator by VC.

262
00:13:06,030 --> 00:13:10,420
So let me take this expression
here, and divide its numerator

263
00:13:10,420 --> 00:13:14,590
and its denominator,
essentially, by VC over VD.

264
00:13:14,590 --> 00:13:16,260
Or let me multiply
its numerator and

265
00:13:16,260 --> 00:13:17,720
denominator VC over VD.

266
00:13:17,720 --> 00:13:24,260
So I can rewrite this as the
natural log of VB over VA,

267
00:13:24,260 --> 00:13:26,000
divided by-- right?

268
00:13:26,000 --> 00:13:28,330
Instead of multiplying it times
this, I can divide it by

269
00:13:28,330 --> 00:13:31,500
this reciprocal, VC over VD.

270
00:13:31,500 --> 00:13:34,960


271
00:13:34,960 --> 00:13:35,980
So I just rewrote it.

272
00:13:35,980 --> 00:13:38,030
I just did a little bit
of fraction math.

273
00:13:38,030 --> 00:13:38,590
That's all it is.

274
00:13:38,590 --> 00:13:40,690
Instead of multiplying it times
this, I divided by its

275
00:13:40,690 --> 00:13:42,030
reciprocal.

276
00:13:42,030 --> 00:13:47,530
Now you see why the previous
video I did was done.

277
00:13:47,530 --> 00:13:49,610
What is this equal to?

278
00:13:49,610 --> 00:13:54,230
On the previous video, I showed
you that VB over VA is

279
00:13:54,230 --> 00:13:56,340
equal to VC over VD.

280
00:13:56,340 --> 00:14:00,630
We did that big convoluted hairy
proof to prove this.

281
00:14:00,630 --> 00:14:04,190
And now that we proved it, we
can use this to know that this

282
00:14:04,190 --> 00:14:06,270
quantity is equal to
this quantity.

283
00:14:06,270 --> 00:14:08,920
So if you divide something by
itself, they're equal to each

284
00:14:08,920 --> 00:14:11,770
other, this is equal to 1.

285
00:14:11,770 --> 00:14:15,570
If that's equal to 1, then
what's the natural log of 1?

286
00:14:15,570 --> 00:14:20,780
So our change in our mystical
S state variable is nR times

287
00:14:20,780 --> 00:14:22,280
the natural log of 1.

288
00:14:22,280 --> 00:14:23,200
What's the natural log of 1?

289
00:14:23,200 --> 00:14:25,210
E to the what power
is equal to 1?

290
00:14:25,210 --> 00:14:27,690
e to the 0 is equal to 1.

291
00:14:27,690 --> 00:14:30,760
n times R times 0-- I don't
care how big or

292
00:14:30,760 --> 00:14:32,220
whatever-- this is 0.

293
00:14:32,220 --> 00:14:33,990
So it equals 0.

294
00:14:33,990 --> 00:14:34,960
So there we have it.

295
00:14:34,960 --> 00:14:38,260
We've stumbled upon a legitimate
state variable that

296
00:14:38,260 --> 00:14:40,770
deals with heat.

297
00:14:40,770 --> 00:14:49,600
If we define change in S is
equal to the heat added to the

298
00:14:49,600 --> 00:14:52,810
system, divided by the
temperature at which the heat

299
00:14:52,810 --> 00:14:54,890
was added to the system,
this is a

300
00:14:54,890 --> 00:14:57,100
legitimate state variable.

301
00:14:57,100 --> 00:14:59,350
Now, we don't have much
intuition about what it really

302
00:14:59,350 --> 00:15:02,860
means at kind of a micro
state level.

303
00:15:02,860 --> 00:15:04,230
But at least we've
stumbled upon

304
00:15:04,230 --> 00:15:06,210
some property of something.

305
00:15:06,210 --> 00:15:10,530
If S is 10 here, and we go
around here, our change in S

306
00:15:10,530 --> 00:15:12,750
will be 0, S is 10 again.

307
00:15:12,750 --> 00:15:15,270
If S is, I don't know, let's
say S is 15 here, and we go

308
00:15:15,270 --> 00:15:18,170
around some crazy cycle and we
come back here, our change in

309
00:15:18,170 --> 00:15:19,860
S is going to be 0 again.

310
00:15:19,860 --> 00:15:21,420
Or sorry, it's going
to be 15 again.

311
00:15:21,420 --> 00:15:24,750
So we didn't have-- our change
in S will be 0, so our s

312
00:15:24,750 --> 00:15:26,470
itself will be 15 again.

313
00:15:26,470 --> 00:15:29,410
So S is a legitimate state
variable, but we don't have a

314
00:15:29,410 --> 00:15:31,795
good sense of what it
actually means.

315
00:15:31,795 --> 00:15:34,570


316
00:15:34,570 --> 00:15:37,290
We'll leave that to
a future video.

317
00:15:37,290 --> 00:00:00,000