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PROFESSOR NELSON: Well, last
time we started in on a

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discussion of entropy, a new
topic, and I started out by

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writing out some somewhat
verbose written descriptions

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that had been formulated that
indicate certain limitations

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about things like the
efficiency of a heat engine,

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and what can be done
reversibly and so forth.

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And these verbal descriptions
lead to some pictures that I

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put up and I'll put up again
about how you might try to

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accomplish something like run
an engine or move heat from

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a colder to a warmer body.
 

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And what some of the
limitations are on

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things like that.
 

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I'll show those again, but what
I want to do mostly today is

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try to put a mathematical
statement of the second law in

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place that corresponds to the
verbal statements that

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we saw last time.
 

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So, just to review what we saw
before, we looked at a heat

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engine where there is some hot
reservoir at some temperature,

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T1, and it's connected to an
engine running in a cycle.

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We could, write the work
that we generate that

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comes out as negative w.
 

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And then there's a cold
reservoir at some

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lower temperature T2.
 

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So the way I've got things
written here, q1 is positive.

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Heat's flowing from the hot
reservoir to the engine

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that runs in a cycle.
 

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Work is coming out.
 

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A positive amount of work is
being generated and is coming

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out, so minus w is positive,
and this minus q2 is positive,

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that is heat is also flowing
into a cold reservoir.

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And what we saw last time
is that this was necessary

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to make things work, this
part of it in particular.

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You couldn't just run something
successfully in a cycle and get

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work out of it, using the heat
from the hot reservoir, without

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also converting some of the
heat that came in to heat that

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would flow into a
cold reservoir.

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All the heat couldn't
get successfully

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converted into work.
 

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That would be desirable,
but it's not possible.

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And, similarly, if we try to
run things backward and build a

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refrigerator, so now we have
our cold reservoir, and we're

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going to remove heat from it,
and pump it up into

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a hot reservoir.
 

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And to do this, we saw that you
have to put work in, right, you

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can't just remove heat from a
cold reservoir, move it up to a

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hot reservoir, without doing
some work to accomplish that.

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So here, q2 is
greater than zero.

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The work is greater than zero,
and minus q1 is greater than

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zero, that is heat is
flowing this way.

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So these are just the pictures
that we saw last time.

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And then we had some statements
of the second law of

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thermodynamics that I won't
re-write up here, but, for

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example, Clausius gave us the
statement that it's impossible

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for a system to operate in a
cycle the takes heat from a

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cold reservoir, transfers it to
a hot reservoir, that is acts

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like a refrigerator, without at
some, at the same time,

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converting some work into heat.
 

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Work has to come in
to make that happen.

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And we had the similar
statement by Kelvin about the

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heat engine that required that
some heat gets dumped into a

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cold reservoir in the process
of converting the heat from the

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hot reservoir into work.
 

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Fine.
 

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Now what I want to do is put up
a specific example of the cycle

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that can be undertaken inside
here in an engine, and we can

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just calculate from what you've
already seen of thermodynamics.

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What happens to the
thermodynamics parameters, and

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see the results in terms of the
parameters including entropy.

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So let's try that.
 

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So, the engine that I'm going
to illustrate is called

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a Carnot engine.
 

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And the cycle it's going to
undertake is called a Carnot

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cycle, and it works the
following way: we're going

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to do pressure volume work.
 

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So this is something that,
by now, you're pretty

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familiar with.
 

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So we're going to start at one,
go to two and this is going to

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be in isotherm at temperature
T1, and all the paths here

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are going to be reversible.
 

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So that's our first step.
 

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Then we're going to have
an adiabatic expansion.

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So this is an
isothermal expansion.

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Here comes an adiabatic
expansion, to point three.

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So at this point the
temperature will change.

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Then we're going to have
another isothermal step, a

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compression to some point four.
 

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So this is an isotherm at some
different temperature T2, a

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cooler temperature, because
this was an expansion.

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We know in an adiabatic
expansion the system's

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going to cool.
 

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And now we're going to have
another adiabatic step,

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an adiabatic compression.
 

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And that's it.
 

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That's going to take us back
to our starting point.

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So that's our cycle.
 

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Of course there are lots of
ways we can execute the cycle,

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but this is a simple one, and
these are steps that we're all

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familiar with at this point.
 

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So this is the picture,
and this is a cycle

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that undertakes it.
 

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So let's just look step
by step at what happens.

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So going from one to two, it's
an isothermal expansion a T1,

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so delta u is q1, we'll call
it, plus w1, for

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the first step.
 

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Going from two to three, that's
an adiabatic expansion, so q is

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equal to zero in that step.
 

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And delta u is equal to
what we'll call w1 prime.

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So we'll use w1 and w1 prime to
describe the work involved.

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The work input during
the expansion steps.

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Step three to four isothermal
compression, delta u is

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going to be q2 plus w2.
 

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And finally, we're going
to go back to one.

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We're going to close
the cycle with another

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adiabatic step. q is zero.
 

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Delta u is w2 prime.
 

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Now, the total amount of the
work that we can get out is

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just given by the area
inside this curve.

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We'll call it capital W.
 

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It's just minus the sum of all
these things, because of

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course these are just defined
as the work done by the

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environment to the system.
 

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So it's minus w1 plus w1
prime plus w2 plus w2 prime.

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The heat input is just q1, and
we'll define that as capital Q.

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And I just want to write those
because what I really want to

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get at is what's the efficiency
of the whole thing?

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And in practical terms, we can
define the efficiency as the

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ratio of the heat in
to the work out.

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We would like that to be
as high as possible.

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So it's just capital W over
capital Q, which is to say it's

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minus all that stuff over q1.
 

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00:11:34 --> 00:11:39
 


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Now the first law is going to
hold in all of these steps, and

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we're going around in a cycle.
 

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So what does that tell us about
a state function like u?

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What's delta u going
around the whole thing?

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I want a chorus in
the answer here.

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What's delta u?
 

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STUDENT:
 

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Zero.
 

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PROFESSOR NELSON: Excellent.
 

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MIT students, yes!
 

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OK, so the first law, we
know that the, going around

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the cycle the integral
of delta u is zero.

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And that has to equal q plus w,
summed up for all the steps.

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Which is to say q1 plus q2 is
equal to minus w1 plus w1

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prime plus w2 plus w2 prime.
 

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So that means we can
rewrite this efficiency.

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We can replace this sum by
just the some of q1 plus q2.

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So sufficiency is q1
plus q2 over q1.

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00:13:03 --> 00:13:06
 


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Or we can write it as
one plus q2 over q1.

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00:13:11 --> 00:13:18
 


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Now that already tells us what
we know, which is that the

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efficiency is going to
be something less

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than zero, right?
 

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Because we've got one and then
we're adding q2 over q1 to it,

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but we've got negative q2
is a positive number.

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Heat is flowing this way.
 

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So q2, the heat in this
way is negative. q2 over

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q1 is negative. q2 of
course is positive.

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That is heat flowing from the
hot reservoir to the engine.

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So that efficiency is something
less than one, and we'd like

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to figure out what that is.
 

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So let's just state that.
 

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Well, now let's go back to each
of our individual steps and

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look, based on what we know
about how to evaluate the

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thermodynamic changes that take
place here, let's look at each

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one of the steps and
see what happens.

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So from one to two
it's isothermal.

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And now we're going to specify,
we're going to do a Carnot

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cycle for an ideal gas.
 

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It's an ideal gas.
 

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It's an isothermal change.
 

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What's delta u?
 

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You know, it's like lots of
courses and all sorts of

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things that you're seeing,
they're always the same.

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What delta u?
 

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STUDENT: Zero.
 

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PROFESSOR NELSON?
 

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Right.
 

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Delta u is zero, and it's
also equal to this.

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So that says the q1 is
the opposite of w1.

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00:15:12 --> 00:15:15
 


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It's an isothermal expansion,
so dw is just negative p dV.

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So it's, this is just
the integral from

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one to two of p dV.
 

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And it's an ideal gas,
isothermal, right.

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RT over V.
 

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Were going to make it for a
mole of gas, so it's R times

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T1, and then we'll
have dV over V.

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So that's just the
log of V2 over V1.

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Let's have one mole,
n equals one.

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Okay two going to three that's
this, it's adiabatic, right.

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00:16:08 --> 00:16:15
So delta u is just equal to the
work but we also know what

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happens because the temperature
is changing from T1 to T2.

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So delta you is just
Cv times T2 minus T1.

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00:16:29 --> 00:16:31
du is Cv dT.
 

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It's an ideal gas, and
that's equal to w1 prime.

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00:16:44 --> 00:16:50
Now, this is a reversible
adiabatic path, so there's

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a relationship that I'm
sure you'll remember.

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Namely, T2 over T1 is
equal to V2 over V3.

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00:17:06 --> 00:17:08
Don't be confused
by the subscripts.

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We're talking about quantities
both starting at two

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00:17:12 --> 00:17:15
and going to three.
 

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00:17:15 --> 00:17:16
So it's V2 and V3.
 

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00:17:16 --> 00:17:17
They're different volumes.
 

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00:17:17 --> 00:17:20
The temperature though at two
is the same as it was at one,

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00:17:20 --> 00:17:25
so it's still a temp at T1, and
this temperatures is at T2.

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00:17:25 --> 00:17:31
Anyway T2 over T1 is equal
to V2 over V3 to the

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power gamma minus one.
 

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00:17:35 --> 00:17:41
So we're going to kind of
store that for the moment.

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Now going from three to four
right, so we have another

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00:17:45 --> 00:17:50
isothermal process for an ideal
gas, so I won't try to make

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you sing again so soon.
 

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Delta u is zero.
 

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And so just like here, now q2
is minus w2, that's integral

226
00:18:02 --> 00:18:06
going from three to four p dV.
 

227
00:18:06 --> 00:18:10
 


228
00:18:10 --> 00:18:15
So it's R T2, right, now we're
at a lower temperature times

229
00:18:15 --> 00:18:23
the log of V4 over V3.
 

230
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So that's our work in
this path and heat.

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And finally going from
four back to one.

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So we've already seen
that q is zero.

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So we know that delta u is
just Cv times T1 minus T2.

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Now we're going
from T2 up to T1.

235
00:18:57 --> 00:19:00
 


236
00:19:00 --> 00:19:08
And this is equal to w2 prime.
 

237
00:19:08 --> 00:19:11
And now, just like we had
before, again we've got a

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00:19:11 --> 00:19:29
reversible adiabatic path, so
T1 over T2 has to equal V4

239
00:19:29 --> 00:19:38
over V1, to the gamma
minus one, okay.

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00:19:38 --> 00:19:44
All right, now, if this is
the case, of course, this is

241
00:19:44 --> 00:19:48
just the inverse of this.
 

242
00:19:48 --> 00:19:50
So this just must be
the inverse of this.

243
00:19:50 --> 00:20:01
We can combine these two
to see that V4 over V1

244
00:20:01 --> 00:20:05
must equal V3 over V2.
 

245
00:20:05 --> 00:20:08
So now we have a relationship
between the ratios of these

246
00:20:08 --> 00:20:21
volumes that are reached
during these adiabatic paths.

247
00:20:21 --> 00:20:41
Now, let's just look
at our efficiency.

248
00:20:41 --> 00:20:48
So we saw that efficiency
is one plus q2 over q1.

249
00:20:48 --> 00:20:56
Let's just use our expressions
that we've found for q2 and q1.

250
00:20:56 --> 00:20:58
We're going to have q2 over q1.
 

251
00:20:58 --> 00:20:58
R is going to cancel.
 

252
00:20:58 --> 00:21:01
We're going to have the ratio
of temperatures and the

253
00:21:01 --> 00:21:11
ratio of these logs.
 

254
00:21:11 --> 00:21:30
So it's T2 over T1 times the
log of V4 over V3, over the log

255
00:21:30 --> 00:21:41
of V2 over V1, but we just
arrived at this relationship

256
00:21:41 --> 00:21:42
between these volumes.
 

257
00:21:42 --> 00:21:52
And this of course tells us
that V4 over V3 is V1 over V2.

258
00:21:52 --> 00:21:53
Right.
 

259
00:21:53 --> 00:21:56
I'm just inverting these.
 

260
00:21:56 --> 00:21:58
So here it is.
 

261
00:21:58 --> 00:22:03
Here's the V4 over
V3, oops, sorry.

262
00:22:03 --> 00:22:11
V2 over V1, this is equal
to the inverse of this.

263
00:22:11 --> 00:22:14
So the ratio of the logs
is just minus one.

264
00:22:14 --> 00:22:19
So this is just, sorry, I
forgot about the one here.

265
00:22:19 --> 00:22:30
One plus this, but this is the
same as one minus T2 over T1.

266
00:22:30 --> 00:22:34
That's our expression
for the efficiency.

267
00:22:34 --> 00:22:36
All right, isn't that terrific?
 

268
00:22:36 --> 00:22:39
Now, we have an expression.
 

269
00:22:39 --> 00:22:42
We can figure out
the efficiency.

270
00:22:42 --> 00:22:44
All we need to know
is the temperatures.

271
00:22:44 --> 00:22:46
What's the temperature
of the hot reservoir?

272
00:22:46 --> 00:22:48
What's the temperature
of the cold reservoir?

273
00:22:48 --> 00:22:51
We're done.
 

274
00:22:51 --> 00:22:53
That's the efficiency.
 

275
00:22:53 --> 00:23:03
As we expected, it's less than
one, and of course if we build

276
00:23:03 --> 00:23:05
an engine, we want it to be as
high as possible, as close

277
00:23:05 --> 00:23:06
to one as possible.
 

278
00:23:06 --> 00:23:11
What that means is we want to
run the hot reservoir as hot as

279
00:23:11 --> 00:23:16
possible and the cold reservoir
as cold as possible.

280
00:23:16 --> 00:23:20
In principle, this value, this
efficiency, can approach

281
00:23:20 --> 00:23:26
1 as the low temperature
approaches absolute zero.

282
00:23:26 --> 00:23:32
If this were to be an absolute
zero Kelvin, then we could we

283
00:23:32 --> 00:23:34
can have something,
wait a minute.

284
00:23:34 --> 00:23:36
Sorry, it's T2.
 

285
00:23:36 --> 00:23:41
As T2 goes to zero, the cold
reservoir, then this goes to

286
00:23:41 --> 00:23:44
zero and our efficiency
approaches one.

287
00:23:44 --> 00:23:47
So that would be the
best we can do.

288
00:23:47 --> 00:23:49
And that basically
make sense, right?

289
00:23:49 --> 00:23:54
The whole thing is being
powered by sending heat from

290
00:23:54 --> 00:23:57
the hot to the cold reservoir.
 

291
00:23:57 --> 00:24:00
The colder that cold reservoir
is, the hotter that

292
00:24:00 --> 00:24:01
hot reservoir is, the
better off we are.

293
00:24:01 --> 00:24:04
The more work we
can get out of it.

294
00:24:04 --> 00:24:10
The closer the efficiency
will get to one.

295
00:24:10 --> 00:24:15
So in some sense, the first
law would suggest you

296
00:24:15 --> 00:24:17
can sort of break even.
 

297
00:24:17 --> 00:24:19
That is, it gives us the
relationship between

298
00:24:19 --> 00:24:21
energy and work and heat.
 

299
00:24:21 --> 00:24:25
It might suggest that you could
convert all the work to heat.

300
00:24:25 --> 00:24:30
The second law says that
really, you can't do that.

301
00:24:30 --> 00:24:35
The only way you could possibly
contemplate it is to be working

302
00:24:35 --> 00:24:38
at absolute zero Kelvin.
 

303
00:24:38 --> 00:24:42
Guess what the third law
is going to tell us?

304
00:24:42 --> 00:24:45
You can't get to zero Kelvin.
 

305
00:24:45 --> 00:24:48
We'll see that shortly.
 

306
00:24:48 --> 00:24:51
But this is those closest you
could come at least by trying

307
00:24:51 --> 00:25:28
to do that to having your
efficiency approach one.

308
00:25:28 --> 00:25:36
Now, we've seen that q2
over q1 is equal to

309
00:25:36 --> 00:25:40
negative T2 over T1.
 

310
00:25:40 --> 00:25:54
So let me just rewrite that as
q1 over T1 is minus q2 over T2.

311
00:25:54 --> 00:26:05
Or in other words, q1 over
T1 plus q2 over T2 is zero.

312
00:26:05 --> 00:26:10
But q1 and q2, each one of
those is just the integrated

313
00:26:10 --> 00:26:14
amount of heat that was
transferred going along a

314
00:26:14 --> 00:26:18
reversible constant
temperature path.

315
00:26:18 --> 00:26:26
Which means that q1 over T1,
that's this delta S thing

316
00:26:26 --> 00:26:28
that we saw before.
 

317
00:26:28 --> 00:26:33
And what we can say about this
is it's saying that if we go

318
00:26:33 --> 00:26:41
around in a cycle and look at
dq reversible over T it's

319
00:26:41 --> 00:27:08
zero, because that's what
these quantities are.

320
00:27:08 --> 00:27:11
Now, of course, we can run the
engine backward and build a

321
00:27:11 --> 00:27:17
refrigerator, and if you've got
your lecture notes from last

322
00:27:17 --> 00:27:22
period, your 8-9, well, they're
labeled 8-9 lecture notes, I

323
00:27:22 --> 00:27:24
made an attempt to define
something which was a

324
00:27:24 --> 00:27:25
little bit misguided.
 

325
00:27:25 --> 00:27:28
And so instead of defining
efficiency the way you've got

326
00:27:28 --> 00:27:32
it written there, I'm going to
define what's called something

327
00:27:32 --> 00:27:34
different for a refrigerator
which is called the

328
00:27:34 --> 00:27:38
coefficient of performance.
 

329
00:27:38 --> 00:28:15
So it's the following: so we
can define the coefficient

330
00:28:15 --> 00:28:29
of performance written as
eta as q2 over minus w.

331
00:28:29 --> 00:28:34
In other words, you know, it's
how much heat can we pull out

332
00:28:34 --> 00:28:40
of the cold reservoir, for
whatever amount of work that

333
00:28:40 --> 00:28:42
we're going to put in
in order to do that?

334
00:28:42 --> 00:28:47
Well of course, we'd like that
to be as big as possible.

335
00:28:47 --> 00:28:59
But of course, this is just
q2 over minus q1 plus q2.

336
00:28:59 --> 00:29:17
Just to be clear, so it's heat
extracted over the work in.

337
00:29:17 --> 00:29:19
So let me just rewrite
that as, I just want to

338
00:29:19 --> 00:29:21
divide by q1 everywhere.
 

339
00:29:21 --> 00:29:29
So I'm going to write this
as q2 over q1 over minus

340
00:29:29 --> 00:29:35
one plus q2 over q1.
 

341
00:29:35 --> 00:29:38
I want to do that because
we know that q2 over q1

342
00:29:38 --> 00:29:40
is negative T2 over T1.
 

343
00:29:40 --> 00:29:45
So this is negative T2
over T1 over negative

344
00:29:45 --> 00:29:49
one minus T2 over T1.
 

345
00:29:49 --> 00:29:52
I can cancel those,
and then I'm going to

346
00:29:52 --> 00:29:56
multiply through by T1.
 

347
00:29:56 --> 00:30:06
So this is just going to be T2
over T1 minus T2, that's our

348
00:30:06 --> 00:30:17
coefficient of performance.
 

349
00:30:17 --> 00:30:20
So it's a measure of how
well we can do to run

350
00:30:20 --> 00:30:21
the refrigerator.
 

351
00:30:21 --> 00:30:26
So you know, what the second
law is doing, in words, it's

352
00:30:26 --> 00:30:29
putting these restrictions on
how well or how effectively we

353
00:30:29 --> 00:30:34
can convert heat into work in
the case of the engine, or work

354
00:30:34 --> 00:30:39
into heat extracted in the
case of a refrigerator.

355
00:30:39 --> 00:30:45
And it follows qualitatively
from just your ordinary

356
00:30:45 --> 00:30:48
observations about the
direction in which things

357
00:30:48 --> 00:30:49
go spontaneously, right.
 

358
00:30:49 --> 00:30:52
You know the heat isn't going
to flow from a cold body to a

359
00:30:52 --> 00:30:57
hot body without putting some
work in to make that happen.

360
00:30:57 --> 00:31:02
And so, we'll be able of follow
that further and see really how

361
00:31:02 --> 00:31:07
to determine the direction of
spontaneity for a whole set of

362
00:31:07 --> 00:31:11
processes, really in principle
for any processes, went

363
00:31:11 --> 00:31:14
analyzed properly.
 

364
00:31:14 --> 00:31:19
So the first step to doing that
is I want to just generalize

365
00:31:19 --> 00:31:22
our results so far
for a Carnot cycle.

366
00:31:22 --> 00:31:24
You might think well okay,
but this is a pretty

367
00:31:24 --> 00:31:25
specialized case.
 

368
00:31:25 --> 00:31:31
We've formulated one particular
kind of engine, and seen how we

369
00:31:31 --> 00:31:37
can analyze what it does, come
up with relations that seem

370
00:31:37 --> 00:31:41
of value for efficiency
and other quantities.

371
00:31:41 --> 00:31:45
How general is it?
 

372
00:31:45 --> 00:31:46
Well, let's take a look.
 

373
00:31:46 --> 00:31:56
So, what I want to do is make
a new engine which really

374
00:31:56 --> 00:32:01
just consists of two
engines, side by side.

375
00:32:01 --> 00:32:05
One is our Carnot engine
as we've seen it, and the

376
00:32:05 --> 00:32:10
other is just any other
reversible engine.

377
00:32:10 --> 00:32:23
So generalize our
Carnot engine results.

378
00:32:23 --> 00:32:31
So let's take our hot
reservoir and draw it bigger.

379
00:32:31 --> 00:32:35
We know it has to big anyway,
since we can extract heat from

380
00:32:35 --> 00:32:38
it without changing
the temperature.

381
00:32:38 --> 00:32:44
And on this side, we're going
to write out an engine, and

382
00:32:44 --> 00:32:47
we're going to say this is a
Carnot engine. so on the side

383
00:32:47 --> 00:32:54
of it we have q1 prime.
 

384
00:32:54 --> 00:32:57
We're going to have w prime.
 

385
00:32:57 --> 00:33:01
And we're going to have q2
prime, and we're going to draw

386
00:33:01 --> 00:33:06
the arrows in the positive
direction in these cases.

387
00:33:06 --> 00:33:10
Or in the defined
directions I should say.

388
00:33:10 --> 00:33:17
Here is our colder reservoir.
 

389
00:33:17 --> 00:33:20
OK, so here is just an engine
like what we've already seen,

390
00:33:20 --> 00:33:27
and I'm going to specify that
this is a Carnot engine which

391
00:33:27 --> 00:33:29
is to say all the results
that we just derived

392
00:33:29 --> 00:33:31
hold for this case.
 

393
00:33:31 --> 00:33:38
And side by side, we're going
to run another engine.

394
00:33:38 --> 00:33:45
So it's got q2, and
it's got a cycle w.

395
00:33:45 --> 00:33:47
And it's got q1.
 

396
00:33:47 --> 00:33:51
 


397
00:33:51 --> 00:33:59
So this is some other
engine that runs using

398
00:33:59 --> 00:34:04
reversible processes.
 

399
00:34:04 --> 00:34:07
So I can define the efficiency
of each one of them.

400
00:34:07 --> 00:34:13
The efficiency for the one on
the left is minus w over q1,

401
00:34:13 --> 00:34:17
The efficiency prime for the
one on the right is minus

402
00:34:17 --> 00:34:22
w prime over q1 prime.
 

403
00:34:22 --> 00:34:30
Now I want to just assume that
in some way they're different.

404
00:34:30 --> 00:34:42
So let's assume that epsilon
prime is greater than epsilon.

405
00:34:42 --> 00:34:47
So in other words, this engine
is running less efficiently

406
00:34:47 --> 00:34:49
than my Carnot engine.
 

407
00:34:49 --> 00:34:50
It's also reversible.
 

408
00:34:50 --> 00:34:54
And now since it's
reversible, we can run

409
00:34:54 --> 00:34:58
it forward or backward.
 

410
00:34:58 --> 00:35:09
So we can run this one backward
and we can use the work that

411
00:35:09 --> 00:35:14
comes out as the input, well
sorry, use this work that comes

412
00:35:14 --> 00:35:19
out of the one of the right to
run this one backward, which is

413
00:35:19 --> 00:35:23
to say we'll move heat
from cold to hot.

414
00:35:23 --> 00:35:40
So use work out of right-hand
side to run left-hand backward.

415
00:35:40 --> 00:35:42
Why not?
 

416
00:35:42 --> 00:36:00
We need work to come in, we
might as well get it from here.

417
00:36:00 --> 00:36:10
So, the total work that we can
get out must be zero, out

418
00:36:10 --> 00:36:12
of the whole sum of them.
 

419
00:36:12 --> 00:36:15
After all, we're taking the
output work that we get from

420
00:36:15 --> 00:36:21
the right, and using it
all to drive the left.

421
00:36:21 --> 00:36:28
So that means that minus
w prime must equal w.

422
00:36:28 --> 00:36:33
And w is greater than zero.
that is in this one we have

423
00:36:33 --> 00:36:38
the environment doing
work on the system.

424
00:36:38 --> 00:36:44
OK, now we've assumed
that epsilon prime is

425
00:36:44 --> 00:36:46
greater than epsilon.
 

426
00:36:46 --> 00:36:51
So we can just write out what
those are, minus w prime

427
00:36:51 --> 00:36:57
over q1 prime is greater
than minus w over q1.

428
00:36:57 --> 00:37:04
But we know that minus w prime
is the same thing as w.

429
00:37:04 --> 00:37:11
So w over q1 prime is greater
than minus w over q1.

430
00:37:11 --> 00:37:15
 


431
00:37:15 --> 00:37:19
Which is the same thing just to
be a little bit maybe pedantic

432
00:37:19 --> 00:37:25
because w over minus q1, and
the only reason I'm writing

433
00:37:25 --> 00:37:29
that is to illustrate that this
says that q1 must be

434
00:37:29 --> 00:37:34
less than q1 prime.
 

435
00:37:34 --> 00:37:42
So q1 is less than zero.
 

436
00:37:42 --> 00:37:45
Remember this one's running
backward, we're pumping heat

437
00:37:45 --> 00:37:52
up. q1 prime is greater than
zero, it's running

438
00:37:52 --> 00:37:53
as an engine.
 

439
00:37:53 --> 00:37:57
It's taking heat from
the hot reservoir.

440
00:37:57 --> 00:38:01
Well, that's interesting.
 

441
00:38:01 --> 00:38:11
That says that minus q1 prime
plus q1 is greater than zero.

442
00:38:11 --> 00:38:16
Well, it's a pretty
interesting result.

443
00:38:16 --> 00:38:20
There's no work being done
on, out of the whole thing.

444
00:38:20 --> 00:38:24
This is providing work that's
being used in here, but if you

445
00:38:24 --> 00:38:27
take the whole outside of the
surroundings and this whole

446
00:38:27 --> 00:38:32
thing is the system, no net
work, these things cancel each

447
00:38:32 --> 00:38:39
other, and yet heat's going up.
 

448
00:38:39 --> 00:38:43
How did that happen?
 

449
00:38:43 --> 00:38:47
Well it happened because we
clearly must have a faulty

450
00:38:47 --> 00:38:50
assumption underlying
what we've just done.

451
00:38:50 --> 00:39:02
This can't possibly be true.
 

452
00:39:02 --> 00:39:08
This is impossible.
 

453
00:39:08 --> 00:39:12
So what this says is the
efficiency of any reversible

454
00:39:12 --> 00:39:16
engine has to be one
minus T2 over T1.

455
00:39:16 --> 00:39:29
 


456
00:39:29 --> 00:39:32
It's not a result that's
specific to the one

457
00:39:32 --> 00:39:39
cycle that we put up.
 

458
00:39:39 --> 00:39:42
And you know, you could have a
reversible engine with lots and

459
00:39:42 --> 00:39:47
lots of steps, but you could
always break them down into

460
00:39:47 --> 00:39:51
some sequence of adiabatic
and isothermal steps.

461
00:39:51 --> 00:39:53
So you know your cycle, you
know, you could have a whole

462
00:39:53 --> 00:40:04
complicated sequence on a p v
diagram of steps going back.

463
00:40:04 --> 00:40:08
As long as it's reversible, you
know what the efficiency has to

464
00:40:08 --> 00:40:11
be, and in principle, you could
break it down into a bunch of

465
00:40:11 --> 00:40:14
steps that you could formulate
as isothermal and adiabatic.

466
00:40:14 --> 00:40:17
They might have to be
formulated as very small

467
00:40:17 --> 00:40:24
steps, in order to do
that, but you could.

468
00:40:24 --> 00:40:39
What this says too, is this
result that we found, right,

469
00:40:39 --> 00:40:45
now of course that's our
integral dS is zero.

470
00:40:45 --> 00:40:55
It's general.
 

471
00:40:55 --> 00:41:11
Entropy S is a state
function, generally.

472
00:41:11 --> 00:41:13
Now remember, we went through
before how it's a state

473
00:41:13 --> 00:41:16
function but to calculate it,
you'd need to find a reversible

474
00:41:16 --> 00:41:22
path, along which you
can figure this out.

475
00:41:22 --> 00:41:31
But the fact is it's a state
function, in a general way.

476
00:41:31 --> 00:41:43
Now, if we go back to our
Carnot cycle which is a set of

477
00:41:43 --> 00:41:48
reversible paths, it's useful
to compare this to what happens

478
00:41:48 --> 00:41:51
in an irreversible case.
 

479
00:41:51 --> 00:41:54
In a real engine, of course,
you can approach the

480
00:41:54 --> 00:41:59
reversible limit.
 

481
00:41:59 --> 00:42:02
Every step won't be
perfectly reversible.

482
00:42:02 --> 00:42:07
And of course it's not
hard to see what happens.

483
00:42:07 --> 00:42:11
Let's just take our
reversible engine and

484
00:42:11 --> 00:42:14
modify it a little bit.
 

485
00:42:14 --> 00:42:18
Let's imagine that instead of
all the steps being reversible,

486
00:42:18 --> 00:42:23
let's just put in one
irreversible step.

487
00:42:23 --> 00:42:24
Let's do this.
 

488
00:42:24 --> 00:42:29
Instead of this reversible
isothermal step, Let's make

489
00:42:29 --> 00:42:35
it an irreversible
isothermal step.

490
00:42:35 --> 00:42:41
We can have a different
isothermal step.

491
00:42:41 --> 00:42:46
Of course in the reversible
case, you're always pushing

492
00:42:46 --> 00:42:50
against an external pressure,
which is essentially equal

493
00:42:50 --> 00:42:53
to the internal pressure.
 

494
00:42:53 --> 00:42:54
Let's not do that.
 

495
00:42:54 --> 00:42:59
Let's imagine maybe instead we
just immediately dropped the

496
00:42:59 --> 00:43:02
pressure and let the system
expand against the

497
00:43:02 --> 00:43:04
lower pressure.
 

498
00:43:04 --> 00:43:06
Now we know we're going
to get less work out

499
00:43:06 --> 00:43:06
of it in that case.
 

500
00:43:06 --> 00:43:08
You've seen that before.
 

501
00:43:08 --> 00:43:12
And the work in this case
is the area inside here.

502
00:43:12 --> 00:43:17
The work in this step is just
the area under this curve.

503
00:43:17 --> 00:43:21
So in some way we're going to
have a difference here between

504
00:43:21 --> 00:43:27
the irreversible case
and the reversible one.

505
00:43:27 --> 00:43:53
So if we do that, then what I
want to do is just see what

506
00:43:53 --> 00:44:07
happens to dq over T, so in our
irreversible engine, one to

507
00:44:07 --> 00:44:14
two, what we know for sure is
that minus w in the

508
00:44:14 --> 00:44:19
irreversible step, that's the
work out, extracted in that

509
00:44:19 --> 00:44:24
step, is going to be smaller
than minus w in the

510
00:44:24 --> 00:44:28
reversible case.
 

511
00:44:28 --> 00:44:40
So w irreversible is
bigger than w reversible.

512
00:44:40 --> 00:44:47
And of course, in either case,
delta u is q plus w, so it's q

513
00:44:47 --> 00:44:52
irreversible plus w
irreversible, but of course, u

514
00:44:52 --> 00:44:55
being a state function it's the
same in either case.

515
00:44:55 --> 00:45:02
So it's the same as q
reversible plus w reversible.

516
00:45:02 --> 00:45:05
And we've just seen that
this is bigger than this,

517
00:45:05 --> 00:45:07
but the sums are equal.
 

518
00:45:07 --> 00:45:12
So this has to be
less than this.

519
00:45:12 --> 00:45:18
q irreversible is less
than q reversible.

520
00:45:18 --> 00:45:20
In other words, and
irreversible isothermal

521
00:45:20 --> 00:45:25
expansion takes less heat from
the hot reservoir than a

522
00:45:25 --> 00:45:27
reversible one does.
 

523
00:45:27 --> 00:45:29
It makes sense, right, because
you know we got less work out

524
00:45:29 --> 00:45:32
and delta u is the same right,
so it must be that less

525
00:45:32 --> 00:45:35
heat got transferred.
 

526
00:45:35 --> 00:45:39
So the expansion against lower
pressure draws less heat from

527
00:45:39 --> 00:45:41
the hot reservoir right.
 

528
00:45:41 --> 00:45:47
So now let's look at
the efficiency of our

529
00:45:47 --> 00:45:50
irreversible engine.
 

530
00:45:50 --> 00:45:55
So it's one plus q2.
 

531
00:45:55 --> 00:45:59
Now q2 was in this step, and
we're going to leave that

532
00:45:59 --> 00:46:21
reversible, right, but
q1 is irreversible.

533
00:46:21 --> 00:46:28
And this has got to be less
than one plus q2 reversible

534
00:46:28 --> 00:46:34
over q1 reversible, which is
to say it's less than our

535
00:46:34 --> 00:46:36
efficiency in the
reversible case.

536
00:46:36 --> 00:46:37
Why?
 

537
00:46:37 --> 00:46:41
This is smaller than this,
but it's a negative number.

538
00:46:41 --> 00:46:46
So this negative number has
a bigger magnitude than

539
00:46:46 --> 00:46:48
this negative number.
 

540
00:46:48 --> 00:46:51
We're subtracting them from one
and they're less than one,

541
00:46:51 --> 00:46:59
so this is bigger than this.
 

542
00:46:59 --> 00:47:04
And all we know about this is
that it's really for some

543
00:47:04 --> 00:47:06
irreversible reversible step.
 

544
00:47:06 --> 00:47:10
All we really needed to
know about it is that

545
00:47:10 --> 00:47:13
this is smaller right.
 

546
00:47:13 --> 00:47:19
So it's going to be the case
for any irreversible engine.

547
00:47:19 --> 00:47:24
And that's the point.
 

548
00:47:24 --> 00:47:37
So it's that the irreversible
efficiency is lower than

549
00:47:37 --> 00:47:43
the reversible one.
 

550
00:47:43 --> 00:47:47
But, of course, since we saw
that this is smaller than it

551
00:47:47 --> 00:48:17
was in the reversible case,
we can also write that dq

552
00:48:17 --> 00:48:28
irreversible over T is less
than dq reversible over T.

553
00:48:28 --> 00:48:35
Which is to say if we go around
a cycle for dq irreversible

554
00:48:35 --> 00:48:43
over T, that's less than zero.
 

555
00:48:43 --> 00:48:50
It's only in the reversible
case that dq over T around

556
00:48:50 --> 00:48:53
a cycle is equal to zero.
 

557
00:48:53 --> 00:48:57
So to write that in a
general way, it's actually

558
00:48:57 --> 00:49:04
formulated by Clausius.
 

559
00:49:04 --> 00:49:09
Going around in a cycle the
integral of dq over T is

560
00:49:09 --> 00:49:13
less than or equal to zero.
 

561
00:49:13 --> 00:49:24
Never greater.
 

562
00:49:24 --> 00:49:29
OK, what we'll see shortly is
that this will allow us to see

563
00:49:29 --> 00:49:33
that for an isolated system
the entropy never decreases.

564
00:49:33 --> 00:49:36
It only can go up.
 

565
00:49:36 --> 00:49:42
And in fact any spontaneous
process will make it go up.

566
00:49:42 --> 00:49:45
Only in the case of
reversible processes.

567
00:49:45 --> 00:49:48
Of course, then you can see
that this will be zero.

568
00:49:48 --> 00:49:52
Anything else, which is to
say any spontaneous process,

569
00:49:52 --> 00:49:54
it'll be less than zero.
 

570
00:49:54 --> 00:50:00
We'll see that next time, and
then we'll generalize in a

571
00:50:00 --> 00:50:03
broader sense to look at
the direction in which

572
00:50:03 --> 00:50:05
spontaneous processes go.
 

573
00:50:05 --> 00:50:06