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PROFESSOR: As you can probably
tell from the volume of

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my voice, I'm not Moungi.
 

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I'm Bob Field, and I will be
lecturing today as a special

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booby prize, because
Bawendi is so wonderful.

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Last time you heard
about work and heat.

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So there's p v work and it's
given by the integral minus

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p external dv or just the
integral from one to two of dw.

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And this little slash
here means an in

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inexact differential.
 

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The reason for inexact doesn't
mean it's a crummy measurement,

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it means that it's path
dependent, and so the value of

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this integral depends on how
you get from one to two. w

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greater than zero means that
work is done on the system, and

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it's really important to keep
track of the signs

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of these things.
 

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For heat, heat is defined, or
the energy unit for heat,

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calorie, is defined as the
amount of heat needed to raise

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one gram of water from 14.5
degrees c to 15.5 degrees c.

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That great deal of specificity
implies that heat is also

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path-dependent and again we
have the convention that if

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heat is added to the
system, the quantity

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is greater than zero.
 

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OK, so that's a quick statement
of what happened last time.

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Today we're going to talk
about heat capacity.

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The first law, a process that
takes a gas from p1, V1, T to

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p2, V2, T examined
for various paths.

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And what you'll find is that
the maximum work out is obtain

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for a reversible path.
 

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We'll then look at the
quantity, internal energy,

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which we define through the
first law, and we think

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of it as a function of
two variables T and V.

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And whenever we say something
like that, we'll be able to

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write a differential, the
partial of u, with respect to T

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at constant V, dT, plus the
partial of u with respect

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to V at constant T dV.
 

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So, if we write something
like this, it implies

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this kind of equation.
 

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And the main points of the
latter part of the lecture is

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what are these quantities?
 

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How do we measure them?
 

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How do we understand them?
 

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This one turns out to be the
heat capacity, and this one

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turns out to be something
that we measure in the

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Joule-free expansion.
 

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So, that's the menu for today,
at least that's what Moungi

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assigned me to cover, and
I'll do the best I can.

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So, let's talk about
heat capacity.

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Heat capacity relates the
amount of heat that you add to

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the system to the change in
temperature, and this

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is the relationship.
 

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The heat-added,
temperature, and this is a

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proportionality constant.
 

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Heat capacity depends on path.
 

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So, here are two kinds
of experiments.

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Here we have a fixed volume,
and we have a little candle,

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and we're adding heat, and
when we add heat, the

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pressure does what?
 

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STUDENT:
 

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PROFESSOR: It increases
and the temperature.

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OK, we're you know,
this is MIT.

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You guys know something.
 

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You should yell it out,
not just whisper it.

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You have the confidence -- I
mean maybe up the street we

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whisper, but here we know it.
 

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And, so here is a different
kind of system where we have a

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constant external pressure.
 

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We add heat to the system.
 

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And so here the
volume can change.

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The temperature can change.
 

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And so as we add heat here, the
temperature goes up and the

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volume -- that was even
quieter than last time.

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OK, but you know it.
 

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You understand it.
 

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Now, the coefficient that
relates the amount of heat in

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to the temperature change is
obviously going to be different

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for these two cases.
 

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in this case, where we have a
fixed volume, we say dq is

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equal that Cv, heat capacity
dT, where the v specifies

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what's constant for the path.
 

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And so this is one
path, constant volume.

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This has a particular value.
 

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Over here, we have dq=Cp dT,
the heat, the proportionality

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between heat and temperature
rise is given by this, the

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constant pressure
heat capacity.

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They're different quantities.
 

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If we want to know the total
heat added to the system, we

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can measure it, which is the
straightforward thing, but

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sometimes you want to calculate
in advance, or sometimes you

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want to calculate
it on an exam.

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So, we do an integral over a
path, for the heat capacity

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along that path, dT.
 

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So, for Cp and Cv, these are
often quantities that are

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measured as a function of
temperature, and one could, in

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fact, calculate this integral.
 

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For most problems on exams,
those quantities are constant,

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independent of temperature.
 

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But there's no guarantee
that they will be.

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Now I get to tell a fun story.
 

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The relationship between heat
and work was initially proposed

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in the 1940's by Joule.
 

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Now, there are two stories
about Joule and how he

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came to this insight.
 

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One is that he observed when
people were machining cannon

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barrels, a lot of heat was
generated, and there was

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a lot of work done.
 

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And so maybe the work
generated the heat.

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Or there was a relationship
between work and heat.

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Well, this is ridiculous
because people were making

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cannons long before 1840.
 

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The other story, which is
probably just as untrue, is

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that Joule, on his honeymoon,
took his wife to a mountain

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resort, and they were sitting
at the bottom of a waterfall,

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and he had this wonderful idea.
 

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And that was the water at the
bottom of the waterfall is

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probably going to be hotter
than the water at the

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top of the waterfall.
 

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So he grabbed his thermometer,
and went and made a couple of

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measurements and discovered the
first law of thermodynamics.

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In fact, the water at the
bottom of the waterfall is

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hotter than the water at
the top of the waterfall.

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And the idea was that gravity
did work on the water and

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falling, and that work led
to the generation of heat.

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00:09:23 --> 00:09:27
If you think about this, you
probably can come up with

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several other ideas for how
the water at the bottom might

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be warmer than at the top.
 

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I mean it's flowing fast at the
top, and it's sort of pooled at

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00:09:35 --> 00:09:38
the bottom, and there is
sunlight and all sorts of

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00:09:38 --> 00:09:43
things that could make this
coincidental as opposed to

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an insightful observation.
 

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But it's a good story,
 

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Joule decided that there must
be a direct relationship

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between work and heat.
 

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They are the same quantity.
 

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They are both forms of energy.
 

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00:10:00 --> 00:10:04
And so, we have this cartoon.
 

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00:10:04 --> 00:10:09
Again, we have an open beaker
and a candle, and we're putting

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00:10:09 --> 00:10:15
only heat into this beaker, and
the temperature goes

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00:10:15 --> 00:10:18
from T1 to T2.
 

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00:10:18 --> 00:10:25
And delta T is given by the
heat, which has to do with how

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00:10:25 --> 00:10:29
much of the candle burnt,
divided by the constant

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00:10:29 --> 00:10:32
pressure heat capacity.
 

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00:10:32 --> 00:10:35
Now you could do a similar sort
of thing, but instead of having

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a candle, you have a paddle
wheel, and the paddle wheel

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00:10:40 --> 00:10:46
is spun by a weight that's
dropping from here to here.

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00:10:46 --> 00:10:51
So, this weight is being spun--
now I'm really fantastic at

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00:10:51 --> 00:10:55
drawing this mechanical device,
but you can imagine that

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00:10:55 --> 00:10:59
dropping a weight can cause
a paddle wheel to turn.

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And we know about gravity,
and we know about work.

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00:11:04 --> 00:11:12
So, if you're doing physics,
work is the integral of

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00:11:12 --> 00:11:17
force, dx, x1 to x2.
 

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Right, that's work.
 

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00:11:19 --> 00:11:24
Now, gravity, well force,
is equal to mass times

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00:11:24 --> 00:11:26
acceleration, and the
acceleration due

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00:11:26 --> 00:11:29
to gravity is g.
 

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The force, as this weight drops
is constant, and so the work is

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just going to be m g
h, where this is h.

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00:11:41 --> 00:11:45
So, it's a trivial matter, by
looking at what is the weight,

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and how far does it drop,
to say OK, how much work is

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done by the paddle wheel.
 

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And this is in an isolated,
in an adiabatic container.

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All the energy that is inserted
into this, which might be

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00:12:02 --> 00:12:08
turbulence initially, becomes
heat, or becomes -- it

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00:12:08 --> 00:12:10
raises the temperature.
 

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And so, again, we see a
temperature increase, and we

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know the work, and the
temperature increase, it's a

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constant pressure thing.
 

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And so we now have
two observations.

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The same temperature increase,
work and heat, and we have a

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relationship between
heat and work.

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00:12:37 --> 00:12:43
The first law of thermodynamics
is -- I'm fine.

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I don't -- I'm all set,
I only use this chalk.

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00:12:47 --> 00:12:56
I brought it with me.
 

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00:12:56 --> 00:12:58
The first law of
thermodynamics, written

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concisely is dw plus dq, two
inexact differentials,

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integrated over any
closed path, is zero.

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This is an abstract and
powerful mathematical

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statement of the first
law of thermodynamics.

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00:13:21 --> 00:13:27
There are much better or
more appealing expressions.

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00:13:27 --> 00:13:34
One is, du, u is called the
internal energy or just the

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00:13:34 --> 00:13:42
energy, is equal to dq plus dw.
 

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00:13:42 --> 00:13:45
OK, notice we have two
inexact differentials

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00:13:45 --> 00:13:49
and exact differentials.
 

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00:13:49 --> 00:13:51
This is a condition.
 

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00:13:51 --> 00:13:55
If you have a quantity which
is constant over any closed

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path, that quantity is a
thermodynamics state function.

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00:14:00 --> 00:14:03
So, this observation is
equivalent to saying that

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00:14:03 --> 00:14:09
there must be something
that is path independent.

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00:14:09 --> 00:14:11
Therefore, we don't have
a cross through the d.

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00:14:11 --> 00:14:16
And the first law says, well
heat and work are different

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00:14:16 --> 00:14:20
forms of energy, and we can add
them, and the path dependence

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00:14:20 --> 00:14:24
of these two things is somehow
cancelled in the fact that we

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have this internal energy.
 

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So, we can also write delta u
as integral from 1 to 2 of du.

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00:14:38 --> 00:14:46
That's u2 minus u1,
and it's q plus w.

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00:14:46 --> 00:14:50
So, these two quantities,
again, are path dependent.

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00:14:50 --> 00:14:52
This is not.
 

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00:14:52 --> 00:14:54
That's the first law
of thermodynamics.

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00:14:54 --> 00:14:59
It seems trivial, but
it's really important.

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00:14:59 --> 00:15:02
It's almost as important as the
second law of thermodynamics,

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00:15:02 --> 00:15:07
which you'll see
in a week or so.

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00:15:07 --> 00:15:10
There is a corollary to this.
 

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00:15:10 --> 00:15:15
If we say we have a system,
and it's in its surroundings.

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00:15:15 --> 00:15:23
We can talk about du for the
system well, that's q plus w.

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00:15:23 --> 00:15:27
And we can say du for
the surroundings.

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00:15:27 --> 00:15:31
Well the system got its work
from the surroundings.

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00:15:31 --> 00:15:35
It either had work, got its
heat from the surroundings,

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00:15:35 --> 00:15:39
or it got worked on
by the surroundings.

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00:15:39 --> 00:15:45
And so, we can immediately
write this and then we can

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00:15:45 --> 00:15:51
write du for the universe,
which is system plus

217
00:15:51 --> 00:15:53
surroundings is equal to zero.
 

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00:15:53 --> 00:15:55
This is the corollary.
 

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00:15:55 --> 00:15:58
It's due to Clausius and
it says the energy in the

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00:15:58 --> 00:16:01
universe is conserved.
 

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00:16:01 --> 00:16:05
Fairly powerful statement, and
that's another form of the

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00:16:05 --> 00:16:07
first law of thermodynamics.
 

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00:16:07 --> 00:16:11
OK, enough for really
great discoveries.

224
00:16:11 --> 00:16:18
Now let's talk about some
simple observations on

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00:16:18 --> 00:16:30
isothermal gas expansions.
 

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00:16:30 --> 00:16:36
The purpose here is to look at
a series of processes in which

227
00:16:36 --> 00:16:39
temperature is held constant,
and we're going to calculate

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00:16:39 --> 00:16:44
how much work we get from
allowing a gas to expand

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00:16:44 --> 00:16:46
under various conditions.
 

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00:16:46 --> 00:16:53
And what we'll discover is that
when we allow the gas to expand

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00:16:53 --> 00:16:58
reversibly, we get
the most work.

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00:16:58 --> 00:16:59
This is neat.
 

233
00:16:59 --> 00:17:01
This is important.
 

234
00:17:01 --> 00:17:06
OK, so we have constant
temperature, because

235
00:17:06 --> 00:17:07
it's isothermal.
 

236
00:17:07 --> 00:17:11
And let's do a series
of experiments.

237
00:17:11 --> 00:17:18
First let's set the external
pressure equal to zero.

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00:17:18 --> 00:17:22
So we have an experiment
that looks like this.

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00:17:22 --> 00:17:24
We have a piston.
 

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00:17:24 --> 00:17:27
We have a pair of stops.
 

241
00:17:27 --> 00:17:28
We have another pair of stops.
 

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00:17:28 --> 00:17:40
We start at p1, V1. and p
external is equal to zero.

243
00:17:40 --> 00:17:43
In other words, this is vacuum.
 

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00:17:43 --> 00:17:50
And so what happens when we
remove these stops is that

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00:17:50 --> 00:17:54
the piston slams up against
the next pair of stops.

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00:17:54 --> 00:17:58
It goes very fast.
 

247
00:17:58 --> 00:18:00
You could calculate the time.
 

248
00:18:00 --> 00:18:03
You could do all sorts of
stuff, but what is important is

249
00:18:03 --> 00:18:11
that this piston in moving from
here to here did no work on the

250
00:18:11 --> 00:18:17
surroundings because work is
the integral of pressure

251
00:18:17 --> 00:18:21
dv, and the pressure
external is zero.

252
00:18:21 --> 00:18:23
It doesn't matter what the
pressure internal is.

253
00:18:23 --> 00:18:28
This is a point that is often
confusing, because you can

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00:18:28 --> 00:18:30
think, well maybe I could
calculate what the internal

255
00:18:30 --> 00:18:34
pressure is even for this
very rapid process.

256
00:18:34 --> 00:18:36
It doesn't matter.
 

257
00:18:36 --> 00:18:41
Thermodynamics is asking you,
what work does this thing do

258
00:18:41 --> 00:18:45
on the surroundings or the
surroundings do on the system?

259
00:18:45 --> 00:18:50
And there is no work done on
the surroundings because

260
00:18:50 --> 00:18:54
pressure is zero.
 

261
00:18:54 --> 00:18:59
So, the work for this process
is the integral, or minus

262
00:18:59 --> 00:19:08
the integral, V1, V2, p dV.
 

263
00:19:08 --> 00:19:12
And it's p external and
p external is zero,

264
00:19:12 --> 00:19:15
so there's no work.
 

265
00:19:15 --> 00:19:23
OK, the next example is, well,
when this piston slams up

266
00:19:23 --> 00:19:31
against these stops, the gas
has expanded -- we're doing

267
00:19:31 --> 00:19:36
this isothermally, so this is
in contact with the heat bath.

268
00:19:36 --> 00:19:38
The gas has expanded.
 

269
00:19:38 --> 00:19:43
It has a particular pressure
and a particular volume.

270
00:19:43 --> 00:19:48
Every time you do the
experiment in equilibrium with

271
00:19:48 --> 00:19:53
the heat bath at T, you'll
get the same p2 and V2.

272
00:19:53 --> 00:19:55
One can know them.
 

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00:19:55 --> 00:19:58
And so we can then say,
OK, let's do this second

274
00:19:58 --> 00:20:10
expansion. and let's set
p external equal to p2.

275
00:20:10 --> 00:20:14
And we do the same
sort of thing.

276
00:20:14 --> 00:20:16
We have stops here.
 

277
00:20:16 --> 00:20:21
We have p1, V1.
 

278
00:20:21 --> 00:20:23
And we don't really
need stops up here.

279
00:20:23 --> 00:20:32
We have a weight on the piston,
and that weight is chosen so

280
00:20:32 --> 00:20:40
that it acts as though p
external is equal to p2.

281
00:20:40 --> 00:20:47
We know the pressure is
equal to force per area.

282
00:20:47 --> 00:20:51
And so we know the force for a
particular mass, and we know

283
00:20:51 --> 00:20:55
the area of the piston. so we
know how to do this, and we

284
00:20:55 --> 00:21:00
know this thing when it hit
these stops, the pressure with

285
00:21:00 --> 00:21:02
p2, and the volume was V2.
 

286
00:21:02 --> 00:21:06
Now we could put stops here
at V2, but this piston

287
00:21:06 --> 00:21:07
would just kiss the stops.
 

288
00:21:07 --> 00:21:13
It would stop there
because we've chosen p2.

289
00:21:13 --> 00:21:23
The work for that process is
going to be minus V1, V2, p2,

290
00:21:23 --> 00:21:27
because that's what we chose p
external to be, dV, and that's

291
00:21:27 --> 00:21:34
going to be minus
p2, V2 minus V1.

292
00:21:34 --> 00:21:35
That's the work.
 

293
00:21:35 --> 00:21:40
V2 is larger than V1, and so
this quantity here is positive,

294
00:21:40 --> 00:22:14
and we have a negative sign
because the system did work.

295
00:22:14 --> 00:22:20
OK, so now we can take the
result from this and put it

296
00:22:20 --> 00:22:32
onto a p v diagram. p1, p2, V1,
V2 -- so here is the initial

297
00:22:32 --> 00:22:36
situation, and here is
the final situation.

298
00:22:36 --> 00:22:42
And the equation of state,
pressure versus volume at

299
00:22:42 --> 00:22:45
constant temperature, is going
to have some form, let's just

300
00:22:45 --> 00:22:47
draw it in there like that.
 

301
00:22:47 --> 00:22:48
So that's an equation of state.
 

302
00:22:48 --> 00:22:51
It's like the ideal gas
law, and one could know

303
00:22:51 --> 00:22:53
that in principle.
 

304
00:22:53 --> 00:23:03
OK, so we did work -- oh, I
should mention p times V

305
00:23:03 --> 00:23:11
has units of energy
or units of work.

306
00:23:11 --> 00:23:14
Remember that, because you're
going to find lots of

307
00:23:14 --> 00:23:17
thermodynamics quantities, and
you're often going to be

308
00:23:17 --> 00:23:21
writing on exams, deriving
things on exams, and you're

309
00:23:21 --> 00:23:24
going to almost always want the
combinations of quantities to

310
00:23:24 --> 00:23:28
have units of energy.
 

311
00:23:28 --> 00:23:31
So, dimensional analysis
is extremely valuable in

312
00:23:31 --> 00:23:34
thermodynamics, and here
is an example of it.

313
00:23:34 --> 00:23:40
Anyway, the work associated
with process number two

314
00:23:40 --> 00:23:44
is described by this box.
 

315
00:23:44 --> 00:23:49
Because we did work at constant
pressure, and so it's just

316
00:23:49 --> 00:23:53
volume difference
times pressure.

317
00:23:53 --> 00:24:00
OK, now we can do the
process in two steps.

318
00:24:00 --> 00:24:05
So we start out
with our piston.

319
00:24:05 --> 00:24:09
We have two weights on it.
 

320
00:24:09 --> 00:24:15
So we have p1, V1.
 

321
00:24:15 --> 00:24:17
We have stops.
 

322
00:24:17 --> 00:24:21
We've chosen two weights,
so that this is p3.

323
00:24:21 --> 00:24:23
External pressure is p3.
 

324
00:24:23 --> 00:24:31
It's higher than p2.
 

325
00:24:31 --> 00:24:42
So we remove the stops, and the
piston moves up, and now we

326
00:24:42 --> 00:24:46
remove one of these weights.
and when we remove one of the

327
00:24:46 --> 00:24:51
weights, this gives an
external pressure p2.

328
00:24:51 --> 00:24:56
So, the piston moves
up again, one weight.

329
00:24:56 --> 00:24:58
So here is p2, p external.
 

330
00:24:58 --> 00:25:00
Here is p3.
 

331
00:25:00 --> 00:25:05
And so we have the work
coming from two steps.

332
00:25:05 --> 00:25:11
And it's trivial to calculate
what that will be, and when you

333
00:25:11 --> 00:25:22
do that, if we put p3 in on
this diagram, what you end up

334
00:25:22 --> 00:25:27
finding is that you get an
initial little bit of work

335
00:25:27 --> 00:25:32
corresponding to V3.
 

336
00:25:32 --> 00:25:38
So we break up our work
into three pieces, and

337
00:25:38 --> 00:25:52
we get more work out.
 

338
00:25:52 --> 00:26:06
So the final process involves
do it reversibly, or

339
00:26:06 --> 00:26:08
almost reversibly.
 

340
00:26:08 --> 00:26:14
We want p equal to p external
for the entire expansion, and

341
00:26:14 --> 00:26:19
p external is decreased
steadily from p1 to p2.

342
00:26:19 --> 00:26:21
How do we do this?
 

343
00:26:21 --> 00:26:29
Well, we start out with our
usual system, and we have a

344
00:26:29 --> 00:26:33
bunch of little pebbles on it.
 

345
00:26:33 --> 00:26:38
Each weighing, say, one 100th
of the weight needed to

346
00:26:38 --> 00:26:43
establish p external
is equal to p1.

347
00:26:43 --> 00:26:46
We remove a pebble,
system expands.

348
00:26:46 --> 00:26:49
We remove another
one, system expands.

349
00:26:49 --> 00:26:53
That's equivalent to doing the
integral, and so, what we

350
00:26:53 --> 00:26:59
end up getting is that the
reversible work is equal to

351
00:26:59 --> 00:27:06
minus integral V1, V2, p dV.
 

352
00:27:06 --> 00:27:11
Because what we've done is we
forced p, pressure here, to be

353
00:27:11 --> 00:27:13
equal to the external pressure.
 

354
00:27:13 --> 00:27:19
We've changed the external
pressure slowly, and again

355
00:27:19 --> 00:27:20
this is isothermal.
 

356
00:27:20 --> 00:27:25
There is a heat bath here that
keeps the temperature constant.

357
00:27:25 --> 00:27:29
So, the system does work
on the surroundings,

358
00:27:29 --> 00:27:32
hence the minus sign.
 

359
00:27:32 --> 00:27:36
Now, if this is an ideal gas,
we know that pressure is

360
00:27:36 --> 00:27:41
equal to nRT over volume.
 

361
00:27:41 --> 00:27:45
So we can put that in here
and do this integral.

362
00:27:45 --> 00:27:52
We have minus V1,
V2, nRT over V dV.

363
00:27:52 --> 00:27:54
These are all constant.
 

364
00:27:54 --> 00:27:55
It's isothermal.
 

365
00:27:55 --> 00:27:56
We can take them out.
 

366
00:27:56 --> 00:27:59
It's a closed system, so the
number moles doesn't change.

367
00:27:59 --> 00:28:02
The ideal gas constant doesn't
change, temperature doesn't

368
00:28:02 --> 00:28:08
change, and so we just have
minus nRT integral

369
00:28:08 --> 00:28:10
V1, V2, dV over V.
 

370
00:28:10 --> 00:28:19
Now, we are very gentle in this
course with respect to knowing

371
00:28:19 --> 00:28:23
integrals, but this is
one you have to know.

372
00:28:23 --> 00:28:31
The integral of one over a
quantity is the natural log.

373
00:28:31 --> 00:28:41
And so we can write this,
minus nRT log V2 over V1.

374
00:28:41 --> 00:28:43
That should not take a breath.
 

375
00:28:43 --> 00:28:47
You know that much
about integrals.

376
00:28:47 --> 00:28:56
OK, now we actually would like
to simplify this or to write

377
00:28:56 --> 00:28:59
this in terms of not
the volume change, but

378
00:28:59 --> 00:29:00
the pressure change.
 

379
00:29:00 --> 00:29:04
So, we have V2 over V1.
 

380
00:29:04 --> 00:29:10
Well, what we can write that
using the ideal gas law twice,

381
00:29:10 --> 00:29:14
V2 is equal -- pV = nRT.
 

382
00:29:14 --> 00:29:23
So V2 = (nRT)/p2,
and V1 = (nRT)/p1.

383
00:29:23 --> 00:29:26
 


384
00:29:26 --> 00:29:34
So, the nRT's cancel,
and we have p1 over p2.

385
00:29:34 --> 00:29:40
And so, we can rewrite this as
the work is equal to minus

386
00:29:40 --> 00:29:53
nRT log p1 over p2, or
nRT log p2 over p1.

387
00:29:53 --> 00:30:00
Now, p2 is less than p1, so
this is a negative quantity.

388
00:30:00 --> 00:30:07
The system has done work.
 

389
00:30:07 --> 00:30:15
So the reversible process, we
had this curve, and for the

390
00:30:15 --> 00:30:23
irreversible processes, we got
this, and then this, whoops,

391
00:30:23 --> 00:30:25
and now we get the whole thing.
 

392
00:30:25 --> 00:30:31
So for the reversible process,
the work done is the integral

393
00:30:31 --> 00:30:36
under the pressure volume
state function, the

394
00:30:36 --> 00:30:39
function of state.
 

395
00:30:39 --> 00:30:49
OK, what time is it?
 

396
00:30:49 --> 00:30:54
I'm going to actually get
caught up, make up for but

397
00:30:54 --> 00:30:57
Bawendi's slow lecturing.
 

398
00:30:57 --> 00:31:00
That doesn't mean it's better,
it just means that I'm making

399
00:31:00 --> 00:31:02
up for a problem
that he created.

400
00:31:02 --> 00:31:06
OK, so what do we know so far?
 

401
00:31:06 --> 00:31:20
Maximum work out is
by reversible path.

402
00:31:20 --> 00:31:27
Delta u is q plus w.
 

403
00:31:27 --> 00:31:35
So the maximum work out
required the maximum heat in.

404
00:31:35 --> 00:31:42
So a reversible process
leads to requiring certain

405
00:31:42 --> 00:31:49
quantities to be maximized.
 

406
00:31:49 --> 00:31:54
Now, we have u, q and w.
 

407
00:31:54 --> 00:31:57
We have a relationship
between them.

408
00:31:57 --> 00:32:04
Often, for a particular
state change, it is easy

409
00:32:04 --> 00:32:10
to calculate two of
these, but not the third.

410
00:32:10 --> 00:32:13
And because there is an
explicit relationship between

411
00:32:13 --> 00:32:21
u, delta u, q and w, you can
always find the easy way to

412
00:32:21 --> 00:32:28
derive the change in internal
energy or the heat or the work.

413
00:32:28 --> 00:32:32
Even if the thing that you
really want is an integral that

414
00:32:32 --> 00:32:40
would be difficult to evaluate.
 

415
00:32:40 --> 00:32:51
OK, now, we're going to look at
the internal energy, and we're

416
00:32:51 --> 00:32:56
going to pretend that it is
explicitly a function of

417
00:32:56 --> 00:33:00
temperature and volume.
 

418
00:33:00 --> 00:33:04
We could choose any two
quantities, and, in fact, it

419
00:33:04 --> 00:33:07
turns out that these are going
to prove, after we have

420
00:33:07 --> 00:33:11
the second law, not to
be the best choice.

421
00:33:11 --> 00:33:16
But it's allowed to say the
internal energy is a function

422
00:33:16 --> 00:33:18
of temperature and volume.
 

423
00:33:18 --> 00:33:24
When you say that, it implies
that the differential is

424
00:33:24 --> 00:33:28
given by this pair of
partial derivatives.

425
00:33:28 --> 00:33:35
V dT, partial of u with respect
to the volume holding

426
00:33:35 --> 00:33:37
temperature constant, dV.
 

427
00:33:37 --> 00:33:42
So if you say this, it
requires you to say this.

428
00:33:42 --> 00:33:45
Now, the purpose of this
exercise is to give you a

429
00:33:45 --> 00:33:47
little bit of practice
in figuring out what

430
00:33:47 --> 00:33:49
these quantities are.
 

431
00:33:49 --> 00:33:51
And do a little practice
in manipulating

432
00:33:51 --> 00:33:57
these differentials.
 

433
00:33:57 --> 00:34:03
So, we have, we're interested
in the change in internal

434
00:34:03 --> 00:34:10
energy for various
experimental constraints.

435
00:34:10 --> 00:34:16
And so, one constraint is the
process be done reversibly.

436
00:34:16 --> 00:34:25
Well then, du is equal to dq
reversible plus dw reversible,

437
00:34:25 --> 00:34:37
which is minus p dV, because p
is equal to p external for a

438
00:34:37 --> 00:34:43
reversible process, and
we can write that.

439
00:34:43 --> 00:34:48
We could have isolated.
 

440
00:34:48 --> 00:34:51
Suppose we're looking
at the system isolated

441
00:34:51 --> 00:34:53
from the outside world.
 

442
00:34:53 --> 00:35:00
Well, dq is equal to zero
and dw is equal to zero

443
00:35:00 --> 00:35:02
because it's isolated.
 

444
00:35:02 --> 00:35:21
So that implies du
is equal to zero.

445
00:35:21 --> 00:35:29
We could do an
adiabatic process.

446
00:35:29 --> 00:35:34
Adiabatic means there's
no heat transferred in

447
00:35:34 --> 00:35:36
or out of the system.
 

448
00:35:36 --> 00:35:42
We don't say anything about
whether the system does work or

449
00:35:42 --> 00:35:47
has worked on -- implicitly
here, we're talking about a

450
00:35:47 --> 00:35:52
closed system, so there's no
mass leaving the system.

451
00:35:52 --> 00:35:58
But if it's adiabatic, then dq
is equal zero, and for an

452
00:35:58 --> 00:36:11
adiabatic process, then
du is equal to dw.

453
00:36:11 --> 00:36:17
OK, now this is a point
we want to be careful.

454
00:36:17 --> 00:36:25
Be careful.
 

455
00:36:25 --> 00:36:29
Suppose we are doing
an adiabatic process.

456
00:36:29 --> 00:36:38
We can do it reversibly, or
we can do it irreversibly.

457
00:36:38 --> 00:36:42
So suppose we do
it irreversibly.

458
00:36:42 --> 00:36:46
Suppose we just remove stops,
and the system slams up

459
00:36:46 --> 00:36:50
against the other stops.
 

460
00:36:50 --> 00:36:52
Did no work.
 

461
00:36:52 --> 00:36:56
Does that mean du is zero?
 

462
00:36:56 --> 00:36:59
You bet it does not.
 

463
00:36:59 --> 00:37:06
So, it's important to write
this little thing on here, du

464
00:37:06 --> 00:37:10
is equal to the reversible
work, not just the work.

465
00:37:10 --> 00:37:15
You can have a process where
you can measure an irreversible

466
00:37:15 --> 00:37:17
process, where you could
calculate the work done.

467
00:37:17 --> 00:37:19
It could be zero.
 

468
00:37:19 --> 00:37:20
It could be something else.
 

469
00:37:20 --> 00:37:27
But if it's not
reversible, it's not du.

470
00:37:27 --> 00:37:29
And you will be invited
to make that mistake

471
00:37:29 --> 00:37:33
on an exam, I'm sure.
 

472
00:37:33 --> 00:37:41
OK, so, the thing about a state
function is that the function

473
00:37:41 --> 00:37:46
has a value for initial
conditions and at

474
00:37:46 --> 00:37:47
final conditions.
 

475
00:37:47 --> 00:37:50
And the difference between
those is what you

476
00:37:50 --> 00:37:52
could measure.
 

477
00:37:52 --> 00:37:59
If you measure something
else, you won't get du.

478
00:37:59 --> 00:38:03
Be careful, and this is going
to be especially complicated

479
00:38:03 --> 00:38:07
and confusing when we get to
quantities that have a more

480
00:38:07 --> 00:38:11
obscure meaning like entropy.
 

481
00:38:11 --> 00:38:15
I mean we can, we can sort of
understand why OK, the total

482
00:38:15 --> 00:38:19
energy, if we measure it, we
measure a process which

483
00:38:19 --> 00:38:20
is not reversible.
 

484
00:38:20 --> 00:38:28
Well it might not give the
energy change for that process,

485
00:38:28 --> 00:38:32
but when we have a quantity
which is more obscure, which

486
00:38:32 --> 00:38:37
the definition of that quantity
requires a very specific

487
00:38:37 --> 00:38:40
prescription for calculating,
you're going to

488
00:38:40 --> 00:38:41
get into trouble.
 

489
00:38:41 --> 00:38:47
So exam this and be sure
you understand that.

490
00:38:47 --> 00:38:53
OK, constant volume.
 

491
00:38:53 --> 00:38:57
Well for constant volume,
dw is equal to zero.

492
00:38:57 --> 00:38:58
Why?
 

493
00:38:58 --> 00:39:01
How does it do, what is work?
 

494
00:39:01 --> 00:39:04
Well it's p v work if the
volume doesn't change.

495
00:39:04 --> 00:39:05
There is no work.
 

496
00:39:05 --> 00:39:13
So in this case du is equal to
dq, and we put a little v on it

497
00:39:13 --> 00:39:23
to imply the work, the change
in internal energy is equal to

498
00:39:23 --> 00:39:28
the heat added at
constant volume.

499
00:39:28 --> 00:39:30
Nothing reversible about it.
 

500
00:39:30 --> 00:39:32
It's now, all we have to do
is say we're going to have

501
00:39:32 --> 00:39:36
heat at constant volume.
 

502
00:39:36 --> 00:39:41
OK, and now we return
to this differential.

503
00:39:41 --> 00:39:45
We want to ask the question,
what are these two quantities?

504
00:39:45 --> 00:39:49
How do we know what they are?
 

505
00:39:49 --> 00:39:54
This should be particularly
bothersome to you because, as

506
00:39:54 --> 00:39:58
you've already experienced in
5.60, there are a lot of

507
00:39:58 --> 00:40:00
partial derivatives. there
are a lot of variables.

508
00:40:00 --> 00:40:02
There are a lot of
things held constant.

509
00:40:02 --> 00:40:07
It's easy to get lost in this
sea of quantities, none of

510
00:40:07 --> 00:40:08
which have obvious meaning.
 

511
00:40:08 --> 00:40:12
So now what we're going to
do is start to extract

512
00:40:12 --> 00:40:15
what these things mean.
 

513
00:40:15 --> 00:40:22
OK, so for a constant volume
process, we can write du,

514
00:40:22 --> 00:40:24
partial derivative of u with
respect to T at constant V,

515
00:40:24 --> 00:40:34
dT, plus partial derivative
of u at constant V, dV.

516
00:40:34 --> 00:40:37
OK, for constant
volume, this is zero.

517
00:40:37 --> 00:40:47
So this term is gone, and we
rewrite this du, V is equal to

518
00:40:47 --> 00:40:53
du/dT, at constant V, dT v.
 

519
00:40:53 --> 00:40:57
So we, have a change in
temperature done at constant

520
00:40:57 --> 00:41:00
volume, and we have a change
in internal energy done at

521
00:41:00 --> 00:41:03
constant volume, and
we rearrange this.

522
00:41:03 --> 00:41:12
And we discover that du/dT
at constant V is equal

523
00:41:12 --> 00:41:20
to du/dT at constant V.
 

524
00:41:20 --> 00:41:24
What, have I done
something silly?

525
00:41:24 --> 00:41:37
Oh well, yes I have. dq v, so
du v is equal to dq v and so

526
00:41:37 --> 00:41:41
what I should have written
here, this is true, we can get

527
00:41:41 --> 00:41:46
a partial derivative by taking
two total derivatives at the

528
00:41:46 --> 00:41:50
same pressure, at the same
quantity, but what I really

529
00:41:50 --> 00:42:11
wanted to do is to write dq v
is equal to the partial

530
00:42:11 --> 00:42:19
derivative of T,
constant v dT v.

531
00:42:19 --> 00:42:27
OK, and now, we know the
relationship between heat and a

532
00:42:27 --> 00:42:33
temperature change is given by
a quantity, a heat capacity for

533
00:42:33 --> 00:42:36
a particular path,
and here it is.

534
00:42:36 --> 00:42:43
So what we've discovered from
this relationship that du at

535
00:42:43 --> 00:42:46
constant volume is
equal to dq v.

536
00:42:46 --> 00:42:52
We have discovered that this
partial derivative that appears

537
00:42:52 --> 00:42:56
in the definition, the abstract
definition of the differential

538
00:42:56 --> 00:42:59
for internal energy, is just
equal to the constant

539
00:42:59 --> 00:43:04
volume heat capacity.
 

540
00:43:04 --> 00:43:10
So, in this definition now, we
have one term which we know.

541
00:43:10 --> 00:43:11
It's something that
we can measure.

542
00:43:11 --> 00:43:17
We can measure the heat
capacity at constant volume,

543
00:43:17 --> 00:43:22
and now we have another term,
and if we can figure out how to

544
00:43:22 --> 00:43:26
measure it, we'll have a
complete form for this

545
00:43:26 --> 00:43:29
differential which will enable
us to calculate du

546
00:43:29 --> 00:43:36
for any process.
 

547
00:43:36 --> 00:43:49
So let me write where I am now,
or we are now. du is equal to

548
00:43:49 --> 00:43:56
Cv dT, plus partial of u
with respect to volume at

549
00:43:56 --> 00:44:00
constant temperature dV.
 

550
00:44:00 --> 00:44:04
So, we've simplified the
expression by replacing one of

551
00:44:04 --> 00:44:08
the partial derivatives by a
quantity that we can measure.

552
00:44:08 --> 00:44:12
And we like to know
what about this.

553
00:44:12 --> 00:44:15
So we need an experiment
that will enable us to

554
00:44:15 --> 00:44:18
measure this quantity.
 

555
00:44:18 --> 00:44:23
And that's where we get Joule,
and I like to say it Joule's

556
00:44:23 --> 00:44:30
free expansion -- it's usually
referred to as the Joule free

557
00:44:30 --> 00:44:35
expansion, which sort of
implies that no energy flows,

558
00:44:35 --> 00:44:39
which actually is true, but
it's an experiment proposed by

559
00:44:39 --> 00:44:45
Joule, and the experiment
involves an adiabatic box.

560
00:44:45 --> 00:44:51
So we have system insulated
from the outside world,

561
00:44:51 --> 00:44:57
and we have two bulbs.
 

562
00:44:57 --> 00:44:58
There's a valve between them.
 

563
00:44:58 --> 00:45:05
And so we have gas
and we have vacuum.

564
00:45:05 --> 00:45:11
So the Joule free expansion
involves opening this valve and

565
00:45:11 --> 00:45:18
asking what happens when this
gas moves into the other bulb

566
00:45:18 --> 00:45:22
or distributes between the two.
well since the gas is expanding

567
00:45:22 --> 00:45:25
into vacuum no work is done.
 

568
00:45:25 --> 00:45:29
Since it's isolated,
no heat is added.

569
00:45:29 --> 00:45:39
So du is equal to zero because
dq and dw are both zero.

570
00:45:39 --> 00:45:44
So we can now take this
expression and rewrite it

571
00:45:44 --> 00:45:48
under the condition of
du is equal to zero.

572
00:45:48 --> 00:45:53
So we have du equal to
zero, just a zero here.

573
00:45:53 --> 00:46:00
Cv dT constant u plus the
derivative du/dV at constant

574
00:46:00 --> 00:46:06
T, dV at constant u.
 

575
00:46:06 --> 00:46:07
This is just a number.
 

576
00:46:07 --> 00:46:12
We don't have to specify that
it's measured at constant u.

577
00:46:12 --> 00:46:14
It's just a number.
 

578
00:46:14 --> 00:46:22
So now we rearrange this
expression, and so we get du/dV

579
00:46:22 --> 00:46:36
at constant T is equal to minus
Cv times dT u over dV u or

580
00:46:36 --> 00:46:42
minus Cv partial derivative of
temperature with respect for

581
00:46:42 --> 00:46:52
volume a constant, free energy,
a constant internal energy.

582
00:46:52 --> 00:46:57
So, this is the
quantity we want.

583
00:46:57 --> 00:47:03
It's related to the heat
capacity, the constant

584
00:47:03 --> 00:47:07
volume of heat capacity and
something you could measure.

585
00:47:07 --> 00:47:11
What happens as you
expand into volume?

586
00:47:11 --> 00:47:14
Does the temperature go up?
 

587
00:47:14 --> 00:47:16
Does it not change?
 

588
00:47:16 --> 00:47:21
Joule actually did this
experiment, and he observed

589
00:47:21 --> 00:47:27
that for the gas expansions
that he could do, that

590
00:47:27 --> 00:47:32
the temperature did not
increase measurably.

591
00:47:32 --> 00:47:41
So he made an
incorrect conclusion.

592
00:47:41 --> 00:47:45
Because something was small and
unmeasurable, he said, well the

593
00:47:45 --> 00:47:56
best of my knowledge dT/dV at
constant u is equal to zero.

594
00:47:56 --> 00:48:06
And that implies that since the
quantity we want is given by

595
00:48:06 --> 00:48:10
this quantity, which is zero
times a constant, the quantity

596
00:48:10 --> 00:48:13
we want is also zero.
 

597
00:48:13 --> 00:48:20
So it would imply that du
was equal to only the

598
00:48:20 --> 00:48:25
first term Cv dT.
 

599
00:48:25 --> 00:48:28
Now, this is an important
lesson in what you

600
00:48:28 --> 00:48:30
do in science.
 

601
00:48:30 --> 00:48:32
You make an observation.
 

602
00:48:32 --> 00:48:36
You make an observation doing
an experiment that is as

603
00:48:36 --> 00:48:38
accurate as you can do.
 

604
00:48:38 --> 00:48:42
And so an experiment said
the gas didn't increase

605
00:48:42 --> 00:48:46
its temperature when it
expanded the vacuum.

606
00:48:46 --> 00:48:49
And so the next thing you do
is you call up a journal

607
00:48:49 --> 00:48:55
and say I've discovered a
fundamental law of nature.

608
00:48:55 --> 00:49:03
And, so you propose that there
is no, that this derivative is

609
00:49:03 --> 00:49:07
zero, and that the internal
energy is given simply

610
00:49:07 --> 00:49:09
by this quantity.
 

611
00:49:09 --> 00:49:14
It turns out that this quantity
here, which is called eta of

612
00:49:14 --> 00:49:21
J the Joule free expansion
parameter, is not quite zero.

613
00:49:21 --> 00:49:25
When you expand a real
gas into vacuum, the

614
00:49:25 --> 00:49:29
temperature goes down.
 

615
00:49:29 --> 00:49:38
So this is a very small
number and for ideal gases,

616
00:49:38 --> 00:49:41
eta J is equal to zero.
 

617
00:49:41 --> 00:49:46
This quantity is exactly zero
for an ideal gas and we'll

618
00:49:46 --> 00:49:49
discover why eventually it has
to do with what we mean by

619
00:49:49 --> 00:49:51
an ideal gas it turns out.
 

620
00:49:51 --> 00:49:56
And so for many, many problems,
especially on exams, especially

621
00:49:56 --> 00:50:02
on this first exam, you will be
able to say that this is the

622
00:50:02 --> 00:50:07
relationship between internal
energy and temperature.

623
00:50:07 --> 00:50:12
That u is a function
of temperature only.

624
00:50:12 --> 00:50:15
It doesn't matter what
other things change.

625
00:50:15 --> 00:50:19
The value of the internal
energy is only determined

626
00:50:19 --> 00:50:20
by temperature.
 

627
00:50:20 --> 00:50:22
But this is only true for
an ideal gas and it's

628
00:50:22 --> 00:50:27
approximately true
for other things.

629
00:50:27 --> 00:50:41
So, delta u is equal to
zero for all isothermal

630
00:50:41 --> 00:50:47
ideal gas processes.
 

631
00:50:47 --> 00:50:49
And it's approximately
equal to zero for all

632
00:50:49 --> 00:50:51
real gas processes.
 

633
00:50:51 --> 00:50:58
And so that means that delta u
is always calculable from Cv(T)

634
00:50:58 --> 00:51:05
dT for any ideal gas change.
 

635
00:51:05 --> 00:51:08
So even if work is done,
it doesn't matter.

636
00:51:08 --> 00:51:12
All you care about is what
was the temperature change?

637
00:51:12 --> 00:51:19
And this is always the easy way
to calculate du, and it's often

638
00:51:19 --> 00:51:25
the easy way to calculate
either q or w, using

639
00:51:25 --> 00:51:26
the first law.
 

640
00:51:26 --> 00:51:35
So, since du is equal to q plus
w, and for an isothermal

641
00:51:35 --> 00:51:40
process this is equal
to zero, q = -w.

642
00:51:40 --> 00:51:45
For an ideal gas, for
an isothermal process.

643
00:51:45 --> 00:51:49
It simplifies your
life enormously.

644
00:51:49 --> 00:51:52
You don't have to
calculate much.

645
00:51:52 --> 00:51:56
OK, so it's time to stop, I
did, in fact, get caught up.

646
00:51:56 --> 00:52:00
I hope you didn't mind
the enhanced velocity.

647
00:52:00 --> 00:52:02
Now Bawendi can do the
beautiful stuff on

648
00:52:02 --> 00:52:04
lecture number four.
 

649
00:52:04 --> 00:52:05
Thank you.
 

650
00:52:05 --> 00:52:07