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PROFESSOR: As you can probably
tell from the volume of
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00:00:22 --> 00:00:25
my voice, I'm not Moungi.
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00:00:25 --> 00:00:32
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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00:00:37 --> 00:00:44
Last time you heard
about work and heat.
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00:00:44 --> 00:00:52
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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00:01:07 --> 00:01:11
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
22
00:01:30 --> 00:01:32
of these things.
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00:01:32 --> 00:01:40
For heat, heat is defined, or
the energy unit for heat,
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00:01:40 --> 00:01:45
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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00:01:58 --> 00:02:05
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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00:02:08 --> 00:02:10
is greater than zero.
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00:02:10 --> 00:02:17
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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00:02:37 --> 00:02:47
p2, V2, T examined
for various paths.
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00:02:47 --> 00:02:53
And what you'll find is that
the maximum work out is obtain
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00:02:53 --> 00:02:57
for a reversible path.
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00:02:57 --> 00:03:00
We'll then look at the
quantity, internal energy,
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00:03:00 --> 00:03:05
which we define through the
first law, and we think
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00:03:05 --> 00:03:09
of it as a function of
two variables T and V.
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00:03:09 --> 00:03:12
And whenever we say something
like that, we'll be able to
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00:03:12 --> 00:03:18
write a differential, the
partial of u, with respect to T
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00:03:18 --> 00:03:23
at constant V, dT, plus the
partial of u with respect
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00:03:23 --> 00:03:26
to V at constant T dV.
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00:03:26 --> 00:03:31
So, if we write something
like this, it implies
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00:03:31 --> 00:03:33
this kind of equation.
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00:03:33 --> 00:03:38
And the main points of the
latter part of the lecture is
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what are these quantities?
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00:03:40 --> 00:03:41
How do we measure them?
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00:03:41 --> 00:03:44
How do we understand them?
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00:03:44 --> 00:03:48
This one turns out to be the
heat capacity, and this one
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00:03:48 --> 00:03:52
turns out to be something
that we measure in the
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00:03:52 --> 00:03:58
Joule-free expansion.
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00:03:58 --> 00:04:01
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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00:04:10 --> 00:04:18
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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00:06:05 --> 00:06:11
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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00:06:51 --> 00:06:55
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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00:06:59 --> 00:07:02
can measure it, which is the
straightforward thing, but
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00:07:02 --> 00:07:06
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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00:07:10 --> 00:07:15
So, we do an integral over a
path, for the heat capacity
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00:07:15 --> 00:07:18
along that path, dT.
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00:07:18 --> 00:07:23
So, for Cp and Cv, these are
often quantities that are
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00:07:23 --> 00:07:26
measured as a function of
temperature, and one could, in
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fact, calculate this integral.
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00:07:28 --> 00:07:33
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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00:07:41 --> 00:07:46
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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00:07:50 --> 00:07:54
in the 1940's by Joule.
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00:07:54 --> 00:07:58
Now, there are two stories
about Joule and how he
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00:07:58 --> 00:08:00
came to this insight.
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00:08:00 --> 00:08:04
One is that he observed when
people were machining cannon
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00:08:04 --> 00:08:08
barrels, a lot of heat was
generated, and there was
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00:08:08 --> 00:08:09
a lot of work done.
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00:08:09 --> 00:08:14
And so maybe the work
generated the heat.
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00:08:14 --> 00:08:17
Or there was a relationship
between work and heat.
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00:08:17 --> 00:08:21
Well, this is ridiculous
because people were making
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00:08:21 --> 00:08:25
cannons long before 1840.
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00:08:25 --> 00:08:29
The other story, which is
probably just as untrue, is
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00:08:29 --> 00:08:37
that Joule, on his honeymoon,
took his wife to a mountain
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00:08:37 --> 00:08:44
resort, and they were sitting
at the bottom of a waterfall,
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00:08:44 --> 00:08:47
and he had this wonderful idea.
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00:08:47 --> 00:08:51
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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00:08:53 --> 00:08:55
top of the waterfall.
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00:08:55 --> 00:08:59
So he grabbed his thermometer,
and went and made a couple of
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00:08:59 --> 00:09:05
measurements and discovered the
first law of thermodynamics.
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00:09:05 --> 00:09:08
In fact, the water at the
bottom of the waterfall is
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00:09:08 --> 00:09:11
hotter than the water at
the top of the waterfall.
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00:09:11 --> 00:09:17
And the idea was that gravity
did work on the water and
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00:09:17 --> 00:09:23
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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00:09:27 --> 00:09:30
several other ideas for how
the water at the bottom might
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00:09:30 --> 00:09:33
be warmer than at the top.
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00:09:33 --> 00:09:35
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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00:09:43 --> 00:09:46
an insightful observation.
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00:09:46 --> 00:09:49
But it's a good story,
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00:09:49 --> 00:09:54
Joule decided that there must
be a direct relationship
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00:09:54 --> 00:09:55
between work and heat.
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00:09:55 --> 00:09:57
They are the same quantity.
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00:09:57 --> 00:10:00
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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00:10:35 --> 00:10:40
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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00:10:59 --> 00:11:04
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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00:11:17 --> 00:11:19
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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00:11:29 --> 00:11:35
The force, as this weight drops
is constant, and so the work is
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00:11:35 --> 00:11:41
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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00:11:45 --> 00:11:49
and how far does it drop,
to say OK, how much work is
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00:11:49 --> 00:11:51
done by the paddle wheel.
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00:11:51 --> 00:11:57
And this is in an isolated,
in an adiabatic container.
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00:11:57 --> 00:12:02
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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00:12:10 --> 00:12:16
And so, again, we see a
temperature increase, and we
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00:12:16 --> 00:12:22
know the work, and the
temperature increase, it's a
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00:12:22 --> 00:12:23
constant pressure thing.
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00:12:23 --> 00:12:27
And so we now have
two observations.
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00:12:27 --> 00:12:32
The same temperature increase,
work and heat, and we have a
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00:12:32 --> 00:12:37
relationship between
heat and work.
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The first law of thermodynamics
is -- I'm fine.
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00:12:43 --> 00:12:47
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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00:12:58 --> 00:13:06
concisely is dw plus dq, two
inexact differentials,
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00:13:06 --> 00:13:11
integrated over any
closed path, is zero.
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00:13:11 --> 00:13:18
This is an abstract and
powerful mathematical
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00:13:18 --> 00:13:21
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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00:13:55 --> 00:14:00
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
194
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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00:14:24 --> 00:14:28
have this internal energy.
200
00:14:28 --> 00:14:38
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.
211
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.
215
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.
219
00:15:55 --> 00:15:58
It's due to Clausius and
it says the energy in the
220
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.
223
00:16:07 --> 00:16:11
OK, enough for really
great discoveries.
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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
228
00:16:39 --> 00:16:44
how much work we get from
allowing a gas to expand
229
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
231
00:16:53 --> 00:16:58
reversibly, we get
the most work.
232
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.
238
00:17:18 --> 00:17:22
So we have an experiment
that looks like this.
239
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.
242
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.
244
00:17:43 --> 00:17:50
And so what happens when we
remove these stops is that
245
00:17:50 --> 00:17:54
the piston slams up against
the next pair of stops.
246
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
254
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.
273
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