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In the video where I first
introduced the concept of

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entropy, I just tried
something out.

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I defined my change in entropy
as being equal to the heat

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added to a system, divided by
the temperature at which it

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was added to the system.

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And then I tested to
see if this was

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a valid state variable.

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And when I did that, I looked
at the Carnot cycle.

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And this is a bit of a review.

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Never hurts to review.

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Let me draw the PV
diagram here.

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We saw that we start at this
state here, and then we

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proceed isothermically.

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We removed little pebbles
off the piston.

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So we increased the volume
and lowered the pressure.

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Then we proceed adiabatically,
where we isolated things and

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we moved like that.

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That was adiabatically.

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Then at this other isotherm,
we added the pebbles back.

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And then we isolated
the system again.

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So adiabatically, we continued
to add more pebbles, and we

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got back to our original
state.

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And I did a couple of videos
where I show that if you take

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the heat added here-- so this is
all being done at some high

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temperature, T1.

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This is being done at some
low temperature, T2.

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There's some heat being added
here, Q1, and that there's

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some heat being released
here, Q2.

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And since these are adiabatic,
there's no transfer of heat to

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and from the system.

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And when I looked at this, and
when I looked at the Carnot

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cycle, and I used this
definition of entropy, I saw

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that the total change in S, when
I go from this point all

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the way around and got back, the
change in S, was equal to

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Q1 over T1 plus Q2 over T2.

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And then I actually showed you
that this was equal to 0,

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which is exactly the result
that I wanted to see.

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Because in order for this to be
a state variable, in order

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for S to be a state variable, it
should not be dependent on

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how I got there.

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It should only be dependent
on my state variables.

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So even if I go on some crazy
path, at the end of the day,

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it should get back to 0.

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But I did something, I guess,
a little bit-- what I did

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wasn't a proof that this is
always a valid state variable.

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It was only a proof that it's
a valid state variable if we

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look at the Carnot cycle.

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But it turns out that it was
only valid because the Carnot

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cycle was reversible.

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And this is a subtle but super
important point, and I really

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should've clarified this
on the first video.

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I guess I was too caught up
showing the proof of the

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Carnot cycle to put the
reversibility there.

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And before I even show you why
it has to be reversable, let

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me just review what
reversibility means.

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Now, we know that in order to
even define a path here, the

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system has to be pretty close to
equilibrium the whole time.

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That's the whole reason why
throughout these videos, I've

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been drawing this piston, you
know, have the gases down

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here, and then always-- instead
of having one big

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weight on top that I took off
or took on, because it would

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throw the system out of
equilibrium-- I did it in

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really small increments.

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I just moved grains of sand, so
that the system was always

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really close to equilibrium.

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And that's called quasistatic.

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and I've defined that before.

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And that means that you're
always in kind of a

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quasi-equilibrium.

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So your state variables
are always defined.

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But that, by itself, does not
give you reversability.

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You have to be quasistatic and
frictionless in order to be

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reversible.

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Now, what do we mean
by frictionless?

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Well, I think you know what
frictionless means.

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Is that like you see in this
system right here, if I make

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this piston a little bit bigger,
that when this piston

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rubs against the side of this
wall, in kind of our real

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world, there's always a little
bit of friction.

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Those molecules start bumping
against each other, and then

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they start making them vibrate,
so they transfer some

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kinetic energy.

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From just by rubbing into each
other, they start generating

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some kinetic energy,
or some heat.

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So you normally have some heat
generated from friction.

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Now, if you have some heat
generated from friction, when

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I remove a pebble-- first all,
when I remove that first

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pebble, it might not
even do anything.

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Because it might not even
overcome-- you can kind of

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view it as the force
of friction.

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But let's say I remove some
pebbles, and this thing moves

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up a little bit.

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But because some of the, I guess
you could say, the force

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differential, the pressure
differential between the

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pebbles and the gas inside, and
the pressure of the gas,

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was used to generate heat as
opposed to work, when I add

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the pebbles back, if I have
friction, I'm not going to get

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back to the same point
that I was before.

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Because friction is always
resisting the movement.

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So in order for something to be
reversible, when I remove a

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couple of pebbles, if I removed
ten pebbles and add

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the ten pebbles back, I should
be at the exact same state.

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But as, you know, you can just
do the thought experiment.

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If there's friction, I won't
be at the exact same state.

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My piston won't move as much as
you would expect if it was

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frictionless.

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So this is a key assumption
for reversability.

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Now, the Carnot cycle, by
definition, is reversible.

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And that's why no one could
actually really implement an

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engine that fully does
the Carnot cycle.

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And we even showed that, that
the Carnot cycle is the most

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efficient potential engine.

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That if anyone made a more
efficient engine, you could

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have a perpetual motion
machine, or a

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perpetual energy machine.

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And the reason why the Carnot
engine is the most efficient

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engine-- and there's no secret
here-- it's because it's

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frictionless.

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Any engineer who makes engines
could tell you, wow, if I

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could just remove all of the
friction from my system, I

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would get a lot more
efficient.

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Now, with that said-- so I've
told you that look.

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The definition doesn't have to
be-- it happened to work, Q

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divided by T, because
I was dealing with

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a reversible system.

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And just to hit the point home,
let me show you that it

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would not have worked if I had
just defined it Q divided by T

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on an irreversible system.

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So let's say that I have-- let
me draw another PV diagram.

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And I'm going to do almost
a very simple thought

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experiment.

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Now I'm going to have an
irreversible system.

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And I start here at some
point on my PV diagram.

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And you know, this could be some
type of cylinder, and it

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has a piston on top, and I have
my rocks, like always.

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But this time, there's a
little bit of friction.

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When this thing moves, a
little bit of heat is

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generated in this.

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And when it moves in
either direction,

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some heat is generated.

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So we can call that the
heat from friction.

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When it moves either up
or when it moves down.

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So let's do something.

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Let's stick this on a big
reservoir, like we tend to do.

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So it's an isothermal system.

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So let's call this T1.

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And let's just start
removing pebbles.

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And we'll move along
an isotherm.

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Maybe to that point there.

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And then we're going to-- and
I want to make a very

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important point here.

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Because this is has friction,
I'm not going to get quite as

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far along the isotherm than
if I didn't have friction.

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If this was a frictionless
system, I would've gotten a

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little bit further along
the isotherm.

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So the number of rocks isn't
going to be the same as it was

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frictionless.

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But let's just say I
move from here to

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here on the PV diagram.

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And then we go and add a bunch
of rocks back, and we want to

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go all the way back.

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And I'm not even saying whether
we have the same

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number of rocks or
different rocks.

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You're probably going to have to
add a few more rocks to get

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back to this point.

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But the idea here is, is that
we've gotten back to the same

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point on the state diagram.

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So our delta U total should be
equal to 0, which is equal to

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the delta U of expansion.

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So the delta U of expansion is
the delta u to go that way,

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delta U of expansion, plus the
delta U of contraction, which

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is the delta U of going
back like this.

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Those have to be equal to
0 by definition, right?

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Because internal energy is a
state variable, and if we get

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to that same point, our delta
U has to be equal to 0.

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So what's our delta
U of expansion?

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What's our change in internal
energy as we expand?

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Our delta U of expansion is
equal to the heat added to the

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system minus the work
done by the system.

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And we know how much
work was done, this

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whole area right there.

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And then plus the heat added
by the friction.

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There's some heat added
by the friction.

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Let me do that in brown.

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What's that?

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I was on some random website off
the screen, and all of a

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sudden that cartoon
sound started up.

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I have no idea what that was.

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But anyway, where was I?

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So I said our change in internal
energy from expansion

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is going to be the heat added
to the system from our

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reservoir minus the work done
by the system, as we expand,

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plus the heat added to the
system or generated by the

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system-- I guess you could say,
it's not being added.

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The system is creating this heat
itself, as it expands.

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There's this friction
right there.

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Fair enough.

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So this is the one variation.

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Now that we're not dealing with
the reversible process,

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we have this friction.

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Now what's our change in energy
from contraction?

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So our change in internal energy
from contraction is

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going to be the heat that leaves
the system, that has to

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go back into the reservoir
as we contract.

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Because otherwise, if we didn't
have the reservoir, the

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temperature would go up.

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But we want to release heat.

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So we want to say, heat
released-- and

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I'm going to do something.

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Let's just assume that all
of the Q's are positive.

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So if I'm releasing heat, it's
going to be a minus Q release.

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Let's just say that this is a
positive number, and if I'm

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releasing it, it's going to
be a minus right there.

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And I want to just do that, just
to hopefully make things

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little bit clearer.

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Plus the work performed
on the system.

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I'm assuming that work is always
positive, so if we're

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doing work, it'll
be minus work.

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If work is being done to
us, it'll be plus work.

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But in this situation
as well, we're still

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adding heat from friction.

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Or heat from friction is still
being generated in the system.

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This is still positive.

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In either direction, when we
move upwards or downwards, the

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system is generating friction.

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Now, we always said, we went all
the way here, we went all

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the way back.

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So the sum of these has to be
equal to 0, because this is a

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state variable.

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So if the sum of all of this
has to be equal to 0, let's

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sum this up.

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So this gets us to
Qa minus Qr.

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So the heat accepted minus
the heat released.

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The W's cancel out.

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Plus-- let me see right here--
plus 2 times the heat of

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00:11:41,870 --> 00:11:43,250
friction in either direction.

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All of that has to
be equal to 0.

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Let's see.

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00:11:47,290 --> 00:11:51,120
What we can do is, we can
rewrite this as the heat

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00:11:51,120 --> 00:11:56,210
accepted minus the heat released
is equal to minus 2

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00:11:56,210 --> 00:11:59,760
times the amount of heat
generated from friction.

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00:11:59,760 --> 00:12:03,240
And then if we just switch these
around, we'll get the

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00:12:03,240 --> 00:12:08,070
heat released minus the heat
accepted is equal to-- well, I

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00:12:08,070 --> 00:12:10,580
just wanted to get all positive
numbers-- 2 times the

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00:12:10,580 --> 00:12:11,560
heat of friction.

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00:12:11,560 --> 00:12:13,200
Now why did I do all of this?

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Because I wanted to do
an experiment with an

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00:12:15,450 --> 00:12:18,810
irreversible system, and this
was a very simple experiment

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with an irreversible system.

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00:12:20,580 --> 00:12:25,580
Now, we said that delta S,
which a long time ago I

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defined as Q divided by T-- and
in this video, I said it

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had to be reversible.

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And I wanted to show you right
now that what if I didn't make

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the constraint that this
has to be reversible?

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00:12:36,830 --> 00:12:39,000
Because if this doesn't have to
be reversible, and I just

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use this definition right here,
you'll see that your

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delta S here would be-- you just
divide everything by T--

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because our temperature was
constant the entire time, we

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were just on a reservoir--
you'll see that this is going

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to be your delta s.

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This is your total change in
the, I guess you could say,

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00:13:03,240 --> 00:13:05,950
your net heat added
to the system.

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So this is, let me say,
this is the heat

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added to the system.

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00:13:11,012 --> 00:13:13,290
Let me do it this way.

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Heat added to the system,
divided by the temperature at

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00:13:19,200 --> 00:13:20,655
which it was added.

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00:13:20,655 --> 00:13:24,840


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Which is a positive number.

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00:13:26,290 --> 00:13:28,410
Even though we got to
the exact same place

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00:13:28,410 --> 00:13:30,060
on this date diagram.

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00:13:30,060 --> 00:13:32,780
So in an reversible system,
this wouldn't be

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a valid state variable.

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00:13:34,220 --> 00:13:37,830
So it's only a valid state
variable if it's reversible.

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00:13:37,830 --> 00:13:40,670
Now, does that mean that you
can only talk about entropy

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00:13:40,670 --> 00:13:43,950
for reversible reactions?

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00:13:43,950 --> 00:13:44,740
No.

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00:13:44,740 --> 00:13:46,300
You can talk about entropy
for anything.

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00:13:46,300 --> 00:13:50,050
But what you do is-- and this
is another important point.

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00:13:50,050 --> 00:13:57,630
So let's say that I have some
irreversible reaction that

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00:13:57,630 --> 00:14:00,730
goes from here to here.

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00:14:00,730 --> 00:14:03,110
And I want to figure out
its change in entropy.

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00:14:03,110 --> 00:14:03,370
Right?

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00:14:03,370 --> 00:14:05,410
And it might have done all
sorts of crazy things.

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00:14:05,410 --> 00:14:08,000
It's an irreversible reaction,
its path might

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00:14:08,000 --> 00:14:08,630
have gone like that.

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00:14:08,630 --> 00:14:11,410
That's assuming it's
quasistatic, if we can even

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00:14:11,410 --> 00:14:13,020
look at its path like that.

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00:14:13,020 --> 00:14:15,700
If we wanted to figure out its
change in entropy, though, we

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00:14:15,700 --> 00:14:18,470
wouldn't worry about the heat
that was added to it and the

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00:14:18,470 --> 00:14:20,130
different temperatures at
which it was added.

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00:14:20,130 --> 00:14:21,300
We wouldn't worry about that.

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00:14:21,300 --> 00:14:22,380
We would just say, OK.

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00:14:22,380 --> 00:14:26,900
What would it have taken for a
reversible system to go from

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00:14:26,900 --> 00:14:29,100
this state to this state?

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00:14:29,100 --> 00:14:31,290
And then maybe a reversible
system would have done

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00:14:31,290 --> 00:14:32,490
something like this.

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00:14:32,490 --> 00:14:34,710
Sorry, I want to make
it a smooth curve.

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00:14:34,710 --> 00:14:37,060
Maybe a reversible system
might have done

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00:14:37,060 --> 00:14:38,180
something like that.

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00:14:38,180 --> 00:14:43,540
And this change, this heat added
by the reversible system

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00:14:43,540 --> 00:14:47,160
divided by the temperature for
the reversible system, would

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00:14:47,160 --> 00:14:49,090
be the change in entropy.

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00:14:49,090 --> 00:14:56,280
And this change in entropy--
we could call this S final,

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00:14:56,280 --> 00:15:00,540
and this is S initial, it's
going to be the same for both

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00:15:00,540 --> 00:15:04,060
systems. It's just, we don't use
the irreversible system to

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00:15:04,060 --> 00:15:05,310
figure out our entropy.

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00:15:05,310 --> 00:15:08,750
We would use the reversible heat
and temperature to figure

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00:15:08,750 --> 00:15:10,460
out the actual change.

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00:15:10,460 --> 00:15:12,300
Hopefully that clarifies
something else.

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00:15:12,300 --> 00:15:15,370
It's, on some level, a subtle
point, but on some level it's

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00:15:15,370 --> 00:15:16,140
super important.

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00:15:16,140 --> 00:15:18,160
Because you can't-- the
thermodynamic definition of

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00:15:18,160 --> 00:15:20,360
entropy has to be this.

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00:15:20,360 --> 00:15:23,560
It has to be heat added to a
reversible system divided by

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00:15:23,560 --> 00:15:24,820
the temperature that
was added.

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00:15:24,820 --> 00:15:26,750
Not just heat to any system.

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00:15:26,750 --> 00:15:28,870
It just happened to work when
I did it, and I should have

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00:15:28,870 --> 00:15:32,540
been clearer about it when I
first explained it, that it

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00:15:32,540 --> 00:15:35,790
worked only because it was
a Carnot cycle, which is

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00:15:35,790 --> 00:15:37,040
reversible.

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00:15:37,040 --> 00:00:00,000