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PROFESSOR: As you can probably
tell from the volume of my

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00:00:23,120 --> 00:00:25,110
voice, I'm not Moungi.

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00:00:25,110 --> 00:00:32,790
I'm Bob Field, and I will be
lecturing today as a special

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00:00:32,790 --> 00:00:37,640
booby prize, because Bawendi
is so wonderful.

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00:00:37,640 --> 00:00:44,460
Last time you heard about
work and heat.

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00:00:44,460 --> 00:00:52,760
So there's p v work and it's
given by the integral minus p

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00:00:52,760 --> 00:00:58,020
external dv or just
the integral from

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00:00:58,020 --> 00:01:01,560
one to two of dw.

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00:01:01,560 --> 00:01:05,040
And this little slash here
means an in inexact

16
00:01:05,040 --> 00:01:07,030
differential.

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00:01:07,030 --> 00:01:10,220
The reason for inexact doesn't
mean it's a crummy

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00:01:10,220 --> 00:01:14,150
measurement, it means that it's
path dependent, and so

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00:01:14,150 --> 00:01:16,730
the value of this integral
depends on how you get from

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00:01:16,730 --> 00:01:26,070
one to two. w greater than zero
means that work is done

21
00:01:26,070 --> 00:01:29,870
on the system, and it's really
important to keep track of the

22
00:01:29,870 --> 00:01:32,550
signs of these things.

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00:01:32,550 --> 00:01:40,870
For heat, heat is defined, or
the energy unit for heat,

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00:01:40,870 --> 00:01:45,710
calorie, is defined as the
amount of heat needed to raise

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00:01:45,710 --> 00:01:53,030
one gram of water from 14.5
degrees c to 15.5 degrees c.

26
00:01:53,030 --> 00:01:58,800
That great deal of specificity
implies that heat is also

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00:01:58,800 --> 00:02:05,070
path-dependent and again we have
the convention that if

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00:02:05,070 --> 00:02:08,580
heat is added to the system,
the quantity is

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00:02:08,580 --> 00:02:10,410
greater than zero.

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00:02:10,410 --> 00:02:14,470
OK, so that's a quick
statement of what

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00:02:14,470 --> 00:02:17,020
happened last time.

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00:02:17,020 --> 00:02:25,960
Today we're going to talk
about heat capacity.

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00:02:25,960 --> 00:02:37,490
The first law, a process that
takes a gas from p1, V1, T to

34
00:02:37,490 --> 00:02:47,790
p2, V2, T examined for
various paths.

35
00:02:47,790 --> 00:02:53,410
And what you'll find is that the
maximum work out is obtain

36
00:02:53,410 --> 00:02:57,130
for a reversible path.

37
00:02:57,130 --> 00:03:00,790
We'll then look at the quantity,
internal energy,

38
00:03:00,790 --> 00:03:05,990
which we define through the
first law, and we think of it

39
00:03:05,990 --> 00:03:10,460
as a function of two variables
T and V. And whenever we say

40
00:03:10,460 --> 00:03:15,450
something like that, we'll be
able to write a differential,

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00:03:15,450 --> 00:03:21,500
the partial of u, with respect
to T at constant V, dT, plus

42
00:03:21,500 --> 00:03:26,900
the partial of u with respect
to V at constant T dV.

43
00:03:26,900 --> 00:03:31,460
So, if we write something
like this, it implies

44
00:03:31,460 --> 00:03:33,950
this kind of equation.

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00:03:33,950 --> 00:03:38,380
And the main points of the
latter part of the lecture is

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00:03:38,380 --> 00:03:40,780
what are these quantities?

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00:03:40,780 --> 00:03:41,810
How do we measure them?

48
00:03:41,810 --> 00:03:44,430
How do we understand them?

49
00:03:44,430 --> 00:03:48,370
This one turns out to be the
heat capacity, and this one

50
00:03:48,370 --> 00:03:51,540
turns out to be something
that we measure in

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00:03:51,540 --> 00:03:58,490
the Joule-free expansion.

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00:03:58,490 --> 00:04:01,750
So, that's the menu for today,
at least that's what Moungi

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00:04:01,750 --> 00:04:10,540
assigned me to cover, and
I'll do the best I can.

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00:04:10,540 --> 00:04:18,420
So, let's talk about
heat capacity.

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00:04:18,420 --> 00:04:23,200
Heat capacity relates the amount
of heat that you add to

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00:04:23,200 --> 00:04:29,270
the system to the change in
temperature, and this is the

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00:04:29,270 --> 00:04:30,380
relationship.

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00:04:30,380 --> 00:04:33,370
The heat-added, temperature,
and this is a

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00:04:33,370 --> 00:04:36,350
proportionality constant.

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00:04:36,350 --> 00:04:43,570
Heat capacity depends on path.

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00:04:43,570 --> 00:04:51,140
So, here are two kinds
of experiments.

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00:04:51,140 --> 00:04:59,020
Here we have a fixed volume, and
we have a little candle,

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00:04:59,020 --> 00:05:05,520
and we're adding heat, and
when we add heat, the

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00:05:05,520 --> 00:05:06,520
pressure does what?

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00:05:06,520 --> 00:05:11,190
STUDENT:

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00:05:11,190 --> 00:05:15,650
PROFESSOR: It increases
and the temperature.

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00:05:15,650 --> 00:05:19,450
OK, we're you know,
this is MIT.

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00:05:19,450 --> 00:05:20,720
You guys know something.

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00:05:20,720 --> 00:05:23,430
You should yell it out,
not just whisper it.

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00:05:23,430 --> 00:05:24,495
You have the confidence --

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00:05:24,495 --> 00:05:29,910
I mean maybe up the street we
whisper, but here we know it.

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00:05:29,910 --> 00:05:37,250
And, so here is a different kind
of system where we have a

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00:05:37,250 --> 00:05:40,040
constant external pressure.

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00:05:40,040 --> 00:05:44,290
We add heat to the system.

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00:05:44,290 --> 00:05:49,590
And so here the volume
can change.

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00:05:49,590 --> 00:05:51,310
The temperature can change.

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00:05:51,310 --> 00:05:57,040
And so as we add heat here, the
temperature goes up and

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00:05:57,040 --> 00:06:03,240
the volume -- that was even
quieter than last time.

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00:06:03,240 --> 00:06:04,580
OK, but you know it.

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00:06:04,580 --> 00:06:05,430
You understand it.

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00:06:05,430 --> 00:06:11,720
Now, the coefficient that
relates the amount of heat in

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00:06:11,720 --> 00:06:14,240
to the temperature change
is obviously going to be

83
00:06:14,240 --> 00:06:17,130
different for these two cases.

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00:06:17,130 --> 00:06:21,140
in this case, where we have a
fixed volume, we say dq is

85
00:06:21,140 --> 00:06:26,230
equal that Cv, heat capacity
dT, where the v specifies

86
00:06:26,230 --> 00:06:29,340
what's constant for the path.

87
00:06:29,340 --> 00:06:32,400
And so this is one path,
constant volume.

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00:06:32,400 --> 00:06:35,350
This has a particular value.

89
00:06:35,350 --> 00:06:44,930
Over here, we have dq=Cp dT, the
heat, the proportionality

90
00:06:44,930 --> 00:06:49,160
between heat and temperature
rise is given by this, the

91
00:06:49,160 --> 00:06:51,820
constant pressure
heat capacity.

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00:06:51,820 --> 00:06:55,480
They're different quantities.

93
00:06:55,480 --> 00:06:59,890
If we want to know the total
heat added to the system, we

94
00:06:59,890 --> 00:07:02,670
can measure it, which is the
straightforward thing, but

95
00:07:02,670 --> 00:07:05,720
sometimes you want to calculate
in advance, or

96
00:07:05,720 --> 00:07:10,050
sometimes you want to calculate
it on an exam.

97
00:07:10,050 --> 00:07:15,000
So, we do an integral over a
path, for the heat capacity

98
00:07:15,000 --> 00:07:18,330
along that path, dT.

99
00:07:18,330 --> 00:07:23,490
So, for Cp and Cv, these are
often quantities that are

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00:07:23,490 --> 00:07:26,945
measured as a function of
temperature, and one could, in

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00:07:26,945 --> 00:07:28,720
fact, calculate this integral.

102
00:07:28,720 --> 00:07:33,130
For most problems on exams,
those quantities are constant,

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00:07:33,130 --> 00:07:35,590
independent of temperature.

104
00:07:35,590 --> 00:07:41,890
But there's no guarantee
that they will be.

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00:07:41,890 --> 00:07:46,250
Now I get to tell a fun story.

106
00:07:46,250 --> 00:07:50,400
The relationship between heat
and work was initially

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00:07:50,400 --> 00:07:54,380
proposed in the 1940's
by Joule.

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00:07:54,380 --> 00:07:58,780
Now, there are two stories about
Joule and how he came to

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00:07:58,780 --> 00:08:00,270
this insight.

110
00:08:00,270 --> 00:08:04,240
One is that he observed when
people were machining cannon

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00:08:04,240 --> 00:08:08,530
barrels, a lot of heat was
generated, and there was a lot

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00:08:08,530 --> 00:08:09,990
of work done.

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00:08:09,990 --> 00:08:14,840
And so maybe the work
generated the heat.

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00:08:14,840 --> 00:08:17,680
Or there was a relationship
between work and heat.

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00:08:17,680 --> 00:08:21,070
Well, this is ridiculous because
people were making

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00:08:21,070 --> 00:08:25,500
cannons long before 1840.

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00:08:25,500 --> 00:08:29,630
The other story, which is
probably just as untrue, is

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00:08:29,630 --> 00:08:37,580
that Joule, on his honeymoon,
took his wife to a mountain

119
00:08:37,580 --> 00:08:44,100
resort, and they were sitting at
the bottom of a waterfall,

120
00:08:44,100 --> 00:08:47,400
and he had this wonderful
idea.

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00:08:47,400 --> 00:08:51,405
And that was the water at the
bottom of the waterfall is

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00:08:51,405 --> 00:08:54,160
probably going to be hotter than
the water at the top of

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00:08:54,160 --> 00:08:55,510
the waterfall.

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00:08:55,510 --> 00:08:59,900
So he grabbed his thermometer,
and went and made a couple of

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00:08:59,900 --> 00:09:01,990
measurements and discovered
the first law of

126
00:09:01,990 --> 00:09:05,510
thermodynamics.

127
00:09:05,510 --> 00:09:08,360
In fact, the water at the bottom
of the waterfall is

128
00:09:08,360 --> 00:09:11,980
hotter than the water at the
top of the waterfall.

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00:09:11,980 --> 00:09:17,730
And the idea was that gravity
did work on the water and

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00:09:17,730 --> 00:09:23,780
falling, and that work led to
the generation of heat.

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00:09:23,780 --> 00:09:27,140
If you think about this, you
probably can come up with

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00:09:27,140 --> 00:09:30,190
several other ideas for how the
water at the bottom might

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00:09:30,190 --> 00:09:33,130
be warmer than at the top.

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00:09:33,130 --> 00:09:35,170
I mean it's flowing fast at
the top, and it's sort of

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00:09:35,170 --> 00:09:38,010
pooled at the bottom, and there
is sunlight and all

136
00:09:38,010 --> 00:09:43,170
sorts of things that could
make this coincidental as

137
00:09:43,170 --> 00:09:46,160
opposed to an insightful
observation.

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00:09:46,160 --> 00:09:49,120
But it's a good story,

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00:09:49,120 --> 00:09:54,600
Joule decided that there must
be a direct relationship

140
00:09:54,600 --> 00:09:55,830
between work and heat.

141
00:09:55,830 --> 00:09:57,340
They are the same quantity.

142
00:09:57,340 --> 00:10:00,270
They are both forms of energy.

143
00:10:00,270 --> 00:10:04,700
And so, we have this cartoon.

144
00:10:04,700 --> 00:10:07,900
Again, we have an open beaker
and a candle, and we're

145
00:10:07,900 --> 00:10:15,040
putting only heat into this
beaker, and the temperature

146
00:10:15,040 --> 00:10:18,900
goes from T1 to T2.

147
00:10:18,900 --> 00:10:25,180
And delta T is given by the
heat, which has to do with how

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00:10:25,180 --> 00:10:29,850
much of the candle burnt,
divided by the constant

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00:10:29,850 --> 00:10:32,650
pressure heat capacity.

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00:10:32,650 --> 00:10:35,480
Now you could do a similar sort
of thing, but instead of

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00:10:35,480 --> 00:10:39,380
having a candle, you have a
paddle wheel, and the paddle

152
00:10:39,380 --> 00:10:44,700
wheel is spun by a weight
that's dropping

153
00:10:44,700 --> 00:10:46,710
from here to here.

154
00:10:46,710 --> 00:10:50,820
So, this weight is being
spun-- now I'm really

155
00:10:50,820 --> 00:10:54,450
fantastic at drawing this
mechanical device, but you can

156
00:10:54,450 --> 00:10:57,120
imagine that dropping
a weight can cause a

157
00:10:57,120 --> 00:10:59,150
paddle wheel to turn.

158
00:10:59,150 --> 00:11:04,650
And we know about gravity,
and we know about work.

159
00:11:04,650 --> 00:11:14,400
So, if you're doing physics,
work is the integral of force,

160
00:11:14,400 --> 00:11:17,260
dx, x1 to x2.

161
00:11:17,260 --> 00:11:19,110
Right, that's work.

162
00:11:19,110 --> 00:11:24,630
Now, gravity, well force,
is equal to mass times

163
00:11:24,630 --> 00:11:26,810
acceleration, and the
acceleration due

164
00:11:26,810 --> 00:11:29,530
to gravity is g.

165
00:11:29,530 --> 00:11:34,590
The force, as this weight drops
is constant, and so the

166
00:11:34,590 --> 00:11:41,500
work is just going to be
m g h, where this is h.

167
00:11:41,500 --> 00:11:45,290
So, it's a trivial matter, by
looking at what is the weight,

168
00:11:45,290 --> 00:11:49,350
and how far does it drop, to say
OK, how much work is done

169
00:11:49,350 --> 00:11:51,370
by the paddle wheel.

170
00:11:51,370 --> 00:11:57,140
And this is in an isolated,
in an adiabatic container.

171
00:11:57,140 --> 00:12:01,780
All the energy that is inserted
into this, which

172
00:12:01,780 --> 00:12:08,770
might be turbulence initially,
becomes heat, or becomes -- it

173
00:12:08,770 --> 00:12:10,250
raises the temperature.

174
00:12:10,250 --> 00:12:16,620
And so, again, we see a
temperature increase, and we

175
00:12:16,620 --> 00:12:22,560
know the work, and the
temperature increase, it's a

176
00:12:22,560 --> 00:12:23,980
constant pressure thing.

177
00:12:23,980 --> 00:12:27,230
And so we now have
two observations.

178
00:12:27,230 --> 00:12:32,730
The same temperature increase,
work and heat, and we have a

179
00:12:32,730 --> 00:12:37,210
relationship between
heat and work.

180
00:12:37,210 --> 00:12:42,780
The first law of thermodynamics
is --

181
00:12:42,780 --> 00:12:43,310
I'm fine.

182
00:12:43,310 --> 00:12:45,070
I don't --

183
00:12:45,070 --> 00:12:47,230
I'm all set, I only
use this chalk.

184
00:12:47,230 --> 00:12:56,450
I brought it with me.

185
00:12:56,450 --> 00:12:58,470
The first law of thermodynamics,
written

186
00:12:58,470 --> 00:13:06,480
concisely is dw plus dq, two
inexact differentials,

187
00:13:06,480 --> 00:13:11,870
integrated over any closed
path, is zero.

188
00:13:11,870 --> 00:13:18,860
This is an abstract and
powerful mathematical

189
00:13:18,860 --> 00:13:21,310
statement of the first law
of thermodynamics.

190
00:13:21,310 --> 00:13:27,770
There are much better or more
appealing expressions.

191
00:13:27,770 --> 00:13:34,070
One is, du, u is called the
internal energy or just the

192
00:13:34,070 --> 00:13:42,210
energy, is equal
to dq plus dw.

193
00:13:42,210 --> 00:13:46,520
OK, notice we have two inexact
differentials and exact

194
00:13:46,520 --> 00:13:49,040
differentials.

195
00:13:49,040 --> 00:13:51,450
This is a condition.

196
00:13:51,450 --> 00:13:55,540
If you have a quantity which
is constant over any closed

197
00:13:55,540 --> 00:14:00,730
path, that quantity is a
thermodynamics state function.

198
00:14:00,730 --> 00:14:03,830
So, this observation is
equivalent to saying that

199
00:14:03,830 --> 00:14:09,150
there must be something that
is path independent.

200
00:14:09,150 --> 00:14:11,880
Therefore, we don't have
a cross through the d.

201
00:14:11,880 --> 00:14:16,490
And the first law says, well
heat and work are different

202
00:14:16,490 --> 00:14:19,890
forms of energy, and we can
add them, and the path

203
00:14:19,890 --> 00:14:23,720
dependence of these two things
is somehow cancelled in the

204
00:14:23,720 --> 00:14:28,100
fact that we have this
internal energy.

205
00:14:28,100 --> 00:14:38,040
So, we can also write delta u as
integral from 1 to 2 of du.

206
00:14:38,040 --> 00:14:46,390
That's u2 minus u1,
and it's q plus w.

207
00:14:46,390 --> 00:14:50,440
So, these two quantities, again,
are path dependent.

208
00:14:50,440 --> 00:14:52,050
This is not.

209
00:14:52,050 --> 00:14:54,000
That's the first law
of thermodynamics.

210
00:14:54,000 --> 00:14:59,130
It seems trivial, but it's
really important.

211
00:14:59,130 --> 00:15:01,850
It's almost as important
as the second law of

212
00:15:01,850 --> 00:15:07,980
thermodynamics, which you'll
see in a week or so.

213
00:15:07,980 --> 00:15:10,430
There is a corollary to this.

214
00:15:10,430 --> 00:15:15,860
If we say we have a system, and
it's in its surroundings.

215
00:15:15,860 --> 00:15:23,090
We can talk about du for the
system well, that's q plus w.

216
00:15:23,090 --> 00:15:27,650
And we can say du for
the surroundings.

217
00:15:27,650 --> 00:15:31,730
Well the system got its work
from the surroundings.

218
00:15:31,730 --> 00:15:35,990
It either had work, got its heat
from the surroundings, or

219
00:15:35,990 --> 00:15:39,210
it got worked on by
the surroundings.

220
00:15:39,210 --> 00:15:45,910
And so, we can immediately write
this and then we can

221
00:15:45,910 --> 00:15:51,790
write du for the universe,
which is system plus

222
00:15:51,790 --> 00:15:53,820
surroundings is equal to zero.

223
00:15:53,820 --> 00:15:55,000
This is the corollary.

224
00:15:55,000 --> 00:15:58,150
It's due to Clausius and it
says the energy in the

225
00:15:58,150 --> 00:16:01,030
universe is conserved.

226
00:16:01,030 --> 00:16:05,820
Fairly powerful statement, and
that's another form of the

227
00:16:05,820 --> 00:16:07,880
first law of thermodynamics.

228
00:16:07,880 --> 00:16:11,180
OK, enough for really
great discoveries.

229
00:16:11,180 --> 00:16:18,340
Now let's talk about some
simple observations on

230
00:16:18,340 --> 00:16:30,860
isothermal gas expansions.

231
00:16:30,860 --> 00:16:36,140
The purpose here is to look at a
series of processes in which

232
00:16:36,140 --> 00:16:39,750
temperature is held constant,
and we're going to calculate

233
00:16:39,750 --> 00:16:45,030
how much work we get from
allowing a gas to expand under

234
00:16:45,030 --> 00:16:46,990
various conditions.

235
00:16:46,990 --> 00:16:52,390
And what we'll discover is that
when we allow the gas to

236
00:16:52,390 --> 00:16:58,010
expand reversibly, we
get the most work.

237
00:16:58,010 --> 00:16:59,110
This is neat.

238
00:16:59,110 --> 00:17:01,620
This is important.

239
00:17:01,620 --> 00:17:06,850
OK, so we have constant
temperature, because it's

240
00:17:06,850 --> 00:17:07,970
isothermal.

241
00:17:07,970 --> 00:17:11,320
And let's do a series
of experiments.

242
00:17:11,320 --> 00:17:18,960
First let's set the external
pressure equal to zero.

243
00:17:18,960 --> 00:17:22,000
So we have an experiment
that looks like this.

244
00:17:22,000 --> 00:17:24,830
We have a piston.

245
00:17:24,830 --> 00:17:27,200
We have a pair of stops.

246
00:17:27,200 --> 00:17:28,580
We have another pair of stops.

247
00:17:28,580 --> 00:17:40,500
We start at p1, V1. and p
external is equal to zero.

248
00:17:40,500 --> 00:17:43,600
In other words, this
is vacuum.

249
00:17:43,600 --> 00:17:50,990
And so what happens when we
remove these stops is that the

250
00:17:50,990 --> 00:17:54,900
piston slams up against the
next pair of stops.

251
00:17:54,900 --> 00:17:58,480
It goes very fast.

252
00:17:58,480 --> 00:18:00,300
You could calculate the time.

253
00:18:00,300 --> 00:18:03,390
You could do all sorts of stuff,
but what is important

254
00:18:03,390 --> 00:18:11,140
is that this piston in moving
from here to here did no work

255
00:18:11,140 --> 00:18:17,095
on the surroundings because
work is the integral of

256
00:18:17,095 --> 00:18:21,000
pressure dv, and the pressure
external is zero.

257
00:18:21,000 --> 00:18:23,560
It doesn't matter what the
pressure internal is.

258
00:18:23,560 --> 00:18:28,120
This is a point that is often
confusing, because you can

259
00:18:28,120 --> 00:18:30,960
think, well maybe I could
calculate what the internal

260
00:18:30,960 --> 00:18:34,390
pressure is even for this
very rapid process.

261
00:18:34,390 --> 00:18:36,510
It doesn't matter.

262
00:18:36,510 --> 00:18:41,850
Thermodynamics is asking you,
what work does this thing do

263
00:18:41,850 --> 00:18:45,840
on the surroundings or the
surroundings do on the system?

264
00:18:45,840 --> 00:18:50,530
And there is no work done on
the surroundings because

265
00:18:50,530 --> 00:18:54,340
pressure is zero.

266
00:18:54,340 --> 00:18:59,760
So, the work for this process is
the integral, or minus the

267
00:18:59,760 --> 00:19:08,390
integral, V1, V2, p dV.

268
00:19:08,390 --> 00:19:12,470
And it's p external and
p external is zero,

269
00:19:12,470 --> 00:19:15,550
so there's no work.

270
00:19:15,550 --> 00:19:23,510
OK, the next example is, well,
when this piston slams up

271
00:19:23,510 --> 00:19:31,030
against these stops, the gas
has expanded -- we're doing

272
00:19:31,030 --> 00:19:36,830
this isothermally, so this is in
contact with the heat bath.

273
00:19:36,830 --> 00:19:38,430
The gas has expanded.

274
00:19:38,430 --> 00:19:43,900
It has a particular pressure
and a particular volume.

275
00:19:43,900 --> 00:19:48,450
Every time you do the experiment
in equilibrium with

276
00:19:48,450 --> 00:19:53,300
the heat bath at T, you'll
get the same p2 and V2.

277
00:19:53,300 --> 00:19:55,470
One can know them.

278
00:19:55,470 --> 00:19:58,980
And so we can then say, OK,
let's do this second

279
00:19:58,980 --> 00:20:10,860
expansion. and let's set
p external equal to p2.

280
00:20:10,860 --> 00:20:14,730
And we do the same
sort of thing.

281
00:20:14,730 --> 00:20:16,970
We have stops here.

282
00:20:16,970 --> 00:20:21,430
We have p1, V1.

283
00:20:21,430 --> 00:20:23,790
And we don't really need
stops up here.

284
00:20:23,790 --> 00:20:31,620
We have a weight on the piston,
and that weight is

285
00:20:31,620 --> 00:20:35,150
chosen so that it
acts as though p

286
00:20:35,150 --> 00:20:40,000
external is equal to p2.

287
00:20:40,000 --> 00:20:47,840
We know the pressure is equal
to force per area.

288
00:20:47,840 --> 00:20:51,520
And so we know the force for a
particular mass, and we know

289
00:20:51,520 --> 00:20:55,750
the area of the piston. so we
know how to do this, and we

290
00:20:55,750 --> 00:21:00,290
know this thing when it hit
these stops, the pressure with

291
00:21:00,290 --> 00:21:02,720
p2, and the volume was V2.

292
00:21:02,720 --> 00:21:07,110
Now we could put stops here at
V2, but this piston would just

293
00:21:07,110 --> 00:21:07,890
kiss the stops.

294
00:21:07,890 --> 00:21:13,690
It would stop there because
we've chosen p2.

295
00:21:13,690 --> 00:21:23,370
The work for that process is
going to be minus V1, V2, p2,

296
00:21:23,370 --> 00:21:27,930
because that's what we chose p
external to be, dV, and that's

297
00:21:27,930 --> 00:21:34,630
going to be minus
p2, V2 minus V1.

298
00:21:34,630 --> 00:21:35,740
That's the work.

299
00:21:35,740 --> 00:21:39,710
V2 is larger than V1, and
so this quantity here is

300
00:21:39,710 --> 00:21:42,820
positive, and we have a negative
sign because the

301
00:21:42,820 --> 00:22:14,290
system did work.

302
00:22:14,290 --> 00:22:20,230
OK, so now we can take the
result from this and put it

303
00:22:20,230 --> 00:22:31,940
onto a p v diagram. p1, p2,
V1, V2 -- so here is the

304
00:22:31,940 --> 00:22:36,830
initial situation, and here
is the final situation.

305
00:22:36,830 --> 00:22:42,040
And the equation of state,
pressure versus volume at

306
00:22:42,040 --> 00:22:45,080
constant temperature, is going
to have some form, let's just

307
00:22:45,080 --> 00:22:47,200
draw it in there like that.

308
00:22:47,200 --> 00:22:48,930
So that's an equation
of state.

309
00:22:48,930 --> 00:22:51,740
It's like the ideal gas
law, and one could

310
00:22:51,740 --> 00:22:53,610
know that in principle.

311
00:22:53,610 --> 00:23:04,390
OK, so we did work -- oh, I
should mention p times V has

312
00:23:04,390 --> 00:23:11,110
units of energy or
units of work.

313
00:23:11,110 --> 00:23:14,750
Remember that, because you're
going to find lots of

314
00:23:14,750 --> 00:23:17,630
thermodynamics quantities, and
you're often going to be

315
00:23:17,630 --> 00:23:21,150
writing on exams, deriving
things on exams, and you're

316
00:23:21,150 --> 00:23:24,850
going to almost always want the
combinations of quantities

317
00:23:24,850 --> 00:23:28,470
to have units of energy.

318
00:23:28,470 --> 00:23:31,660
So, dimensional analysis is
extremely valuable in

319
00:23:31,660 --> 00:23:34,010
thermodynamics, and here
is an example of it.

320
00:23:34,010 --> 00:23:40,400
Anyway, the work associated
with process number two is

321
00:23:40,400 --> 00:23:44,320
described by this box.

322
00:23:44,320 --> 00:23:48,980
Because we did work at constant
pressure, and so it's

323
00:23:48,980 --> 00:23:53,530
just volume difference
times pressure.

324
00:23:53,530 --> 00:24:00,860
OK, now we can do the process
in two steps.

325
00:24:00,860 --> 00:24:05,780
So we start out with
our piston.

326
00:24:05,780 --> 00:24:09,980
We have two weights on it.

327
00:24:09,980 --> 00:24:15,100
So we have p1, V1.

328
00:24:15,100 --> 00:24:17,020
We have stops.

329
00:24:17,020 --> 00:24:21,800
We've chosen two weights,
so that this is p3.

330
00:24:21,800 --> 00:24:23,930
External pressure is p3.

331
00:24:23,930 --> 00:24:31,200
It's higher than p2.

332
00:24:31,200 --> 00:24:41,920
So we remove the stops, and the
piston moves up, and now

333
00:24:41,920 --> 00:24:46,490
we remove one of these weights.
and when we remove

334
00:24:46,490 --> 00:24:51,130
one of the weights, this gives
an external pressure p2.

335
00:24:51,130 --> 00:24:56,110
So, the piston moves up
again, one weight.

336
00:24:56,110 --> 00:24:58,810
So here is p2, p external.

337
00:24:58,810 --> 00:25:00,760
Here is p3.

338
00:25:00,760 --> 00:25:05,710
And so we have the work
coming from two steps.

339
00:25:05,710 --> 00:25:11,400
And it's trivial to calculate
what that will be, and when

340
00:25:11,400 --> 00:25:21,940
you do that, if we put p3 in on
this diagram, what you end

341
00:25:21,940 --> 00:25:27,910
up finding is that you get an
initial little bit of work

342
00:25:27,910 --> 00:25:32,380
corresponding to V3.

343
00:25:32,380 --> 00:25:38,560
So we break up our work into
three pieces, and we

344
00:25:38,560 --> 00:25:52,450
get more work out.

345
00:25:52,450 --> 00:26:06,610
So the final process involves
do it reversibly, or almost

346
00:26:06,610 --> 00:26:08,000
reversibly.

347
00:26:08,000 --> 00:26:14,050
We want p equal to p external
for the entire expansion, and

348
00:26:14,050 --> 00:26:19,800
p external is decreased steadily
from p1 to p2.

349
00:26:19,800 --> 00:26:21,550
How do we do this?

350
00:26:21,550 --> 00:26:29,730
Well, we start out with our
usual system, and we have a

351
00:26:29,730 --> 00:26:33,170
bunch of little pebbles on it.

352
00:26:33,170 --> 00:26:38,350
Each weighing, say, one 100th
of the weight needed to

353
00:26:38,350 --> 00:26:43,210
establish p external
is equal to p1.

354
00:26:43,210 --> 00:26:46,120
We remove a pebble,
system expands.

355
00:26:46,120 --> 00:26:49,040
We remove another one,
system expands.

356
00:26:49,040 --> 00:26:53,990
That's equivalent to doing the
integral, and so, what we end

357
00:26:53,990 --> 00:26:59,520
up getting is that the
reversible work is equal to

358
00:26:59,520 --> 00:27:06,850
minus integral V1, V2, p dV.

359
00:27:06,850 --> 00:27:11,430
Because what we've done is we
forced p, pressure here, to be

360
00:27:11,430 --> 00:27:13,550
equal to the external
pressure.

361
00:27:13,550 --> 00:27:19,120
We've changed the external
pressure slowly, and again

362
00:27:19,120 --> 00:27:20,160
this is isothermal.

363
00:27:20,160 --> 00:27:22,260
There is a heat bath
here that keeps

364
00:27:22,260 --> 00:27:25,160
the temperature constant.

365
00:27:25,160 --> 00:27:29,830
So, the system does work on the
surroundings, hence the

366
00:27:29,830 --> 00:27:32,010
minus sign.

367
00:27:32,010 --> 00:27:37,130
Now, if this is an ideal gas, we
know that pressure is equal

368
00:27:37,130 --> 00:27:41,380
to nRT over volume.

369
00:27:41,380 --> 00:27:45,510
So we can put that in here
and do this integral.

370
00:27:45,510 --> 00:27:52,720
We have minus V1, V2,
nRT over V dV.

371
00:27:52,720 --> 00:27:54,080
These are all constant.

372
00:27:54,080 --> 00:27:55,100
It's isothermal.

373
00:27:55,100 --> 00:27:56,060
We can take them out.

374
00:27:56,060 --> 00:27:59,570
It's a closed system, so the
number moles doesn't change.

375
00:27:59,570 --> 00:28:02,360
The ideal gas constant doesn't
change, temperature doesn't

376
00:28:02,360 --> 00:28:09,910
change, and so we just have
minus nRT integral V1, V2, dV

377
00:28:09,910 --> 00:28:17,710
over V. Now, we are very gentle
in this course with

378
00:28:17,710 --> 00:28:20,830
respect to knowing integrals,
but this is

379
00:28:20,830 --> 00:28:23,720
one you have to know.

380
00:28:23,720 --> 00:28:31,600
The integral of one over a
quantity is the natural log.

381
00:28:31,600 --> 00:28:41,340
And so we can write this, minus
nRT log V2 over V1.

382
00:28:41,340 --> 00:28:43,490
That should not take a breath.

383
00:28:43,490 --> 00:28:47,330
You know that much
about integrals.

384
00:28:47,330 --> 00:28:56,110
OK, now we actually would like
to simplify this or to write

385
00:28:56,110 --> 00:28:59,190
this in terms of not the
volume change, but the

386
00:28:59,190 --> 00:29:00,840
pressure change.

387
00:29:00,840 --> 00:29:04,540
So, we have V2 over V1.

388
00:29:04,540 --> 00:29:10,450
Well, what we can write that
using the ideal gas law twice,

389
00:29:10,450 --> 00:29:14,580
V2 is equal -- pV = nRT.

390
00:29:14,580 --> 00:29:23,170
So V2 = (nRT)/p2, and
V1 = (nRT)/p1.

391
00:29:26,620 --> 00:29:34,060
So, the nRT's cancel, and
we have p1 over p2.

392
00:29:34,060 --> 00:29:42,690
And so, we can rewrite this as
the work is equal to minus nRT

393
00:29:42,690 --> 00:29:53,880
log p1 over p2, or nRT
log p2 over p1.

394
00:29:53,880 --> 00:30:00,990
Now, p2 is less than p1, so this
is a negative quantity.

395
00:30:00,990 --> 00:30:07,770
The system has done work.

396
00:30:07,770 --> 00:30:15,880
So the reversible process, we
had this curve, and for the

397
00:30:15,880 --> 00:30:23,000
irreversible processes, we got
this, and then this, whoops,

398
00:30:23,000 --> 00:30:25,620
and now we get the
whole thing.

399
00:30:25,620 --> 00:30:31,740
So for the reversible process,
the work done is the integral

400
00:30:31,740 --> 00:30:35,090
under the pressure volume
state function,

401
00:30:35,090 --> 00:30:39,630
the function of state.

402
00:30:39,630 --> 00:30:49,480
OK, what time is it?

403
00:30:49,480 --> 00:30:54,680
I'm going to actually get caught
up, make up for but

404
00:30:54,680 --> 00:30:57,350
Bawendi's slow lecturing.

405
00:30:57,350 --> 00:31:00,060
That doesn't mean it's better,
it just means that I'm making

406
00:31:00,060 --> 00:31:02,330
up for a problem that
he created.

407
00:31:02,330 --> 00:31:06,670
OK, so what do we know so far?

408
00:31:06,670 --> 00:31:20,680
Maximum work out is by
reversible path.

409
00:31:20,680 --> 00:31:27,110
Delta u is q plus w.

410
00:31:27,110 --> 00:31:35,960
So the maximum work out required
the maximum heat in.

411
00:31:35,960 --> 00:31:42,820
So a reversible process leads
to requiring certain

412
00:31:42,820 --> 00:31:49,140
quantities to be maximized.

413
00:31:49,140 --> 00:31:54,210
Now, we have u, q and w.

414
00:31:54,210 --> 00:31:57,710
We have a relationship
between them.

415
00:31:57,710 --> 00:32:05,550
Often, for a particular state
change, it is easy to

416
00:32:05,550 --> 00:32:10,010
calculate two of these,
but not the third.

417
00:32:10,010 --> 00:32:13,500
And because there is an explicit
relationship between

418
00:32:13,500 --> 00:32:21,650
u, delta u, q and w, you can
always find the easy way to

419
00:32:21,650 --> 00:32:25,960
derive the change in
internal energy or

420
00:32:25,960 --> 00:32:28,750
the heat or the work.

421
00:32:28,750 --> 00:32:32,420
Even if the thing that you
really want is an integral

422
00:32:32,420 --> 00:32:40,590
that would be difficult
to evaluate.

423
00:32:40,590 --> 00:32:51,870
OK, now, we're going to look
at the internal energy, and

424
00:32:51,870 --> 00:32:56,720
we're going to pretend that it
is explicitly a function of

425
00:32:56,720 --> 00:33:00,080
temperature and volume.

426
00:33:00,080 --> 00:33:04,650
We could choose any two
quantities, and, in fact, it

427
00:33:04,650 --> 00:33:07,920
turns out that these are going
to prove, after we have the

428
00:33:07,920 --> 00:33:11,090
second law, not to be
the best choice.

429
00:33:11,090 --> 00:33:16,120
But it's allowed to say the
internal energy is a function

430
00:33:16,120 --> 00:33:18,420
of temperature and volume.

431
00:33:18,420 --> 00:33:25,240
When you say that, it implies
that the differential is given

432
00:33:25,240 --> 00:33:28,990
by this pair of partial
derivatives.

433
00:33:28,990 --> 00:33:35,530
V dT, partial of u with respect
to the volume holding

434
00:33:35,530 --> 00:33:37,850
temperature constant, dV.

435
00:33:37,850 --> 00:33:42,190
So if you say this, it requires
you to say this.

436
00:33:42,190 --> 00:33:45,440
Now, the purpose of this
exercise is to give you a

437
00:33:45,440 --> 00:33:47,530
little bit of practice in
figuring out what these

438
00:33:47,530 --> 00:33:49,270
quantities are.

439
00:33:49,270 --> 00:33:52,360
And do a little practice
in manipulating these

440
00:33:52,360 --> 00:33:57,490
differentials.

441
00:33:57,490 --> 00:34:03,800
So, we have, we're interested
in the change in internal

442
00:34:03,800 --> 00:34:10,660
energy for various experimental
constraints.

443
00:34:10,660 --> 00:34:16,300
And so, one constraint is the
process be done reversibly.

444
00:34:16,300 --> 00:34:25,270
Well then, du is equal to dq
reversible plus dw reversible,

445
00:34:25,270 --> 00:34:37,230
which is minus p dV, because p
is equal to p external for a

446
00:34:37,230 --> 00:34:43,600
reversible process, and
we can write that.

447
00:34:43,600 --> 00:34:48,430
We could have isolated.

448
00:34:48,430 --> 00:34:52,290
Suppose we're looking at the
system isolated from the

449
00:34:52,290 --> 00:34:53,650
outside world.

450
00:34:53,650 --> 00:35:00,830
Well, dq is equal to zero and
dw is equal to zero because

451
00:35:00,830 --> 00:35:02,230
it's isolated.

452
00:35:02,230 --> 00:35:21,590
So that implies du
is equal to zero.

453
00:35:21,590 --> 00:35:29,650
We could do an adiabatic
process.

454
00:35:29,650 --> 00:35:34,690
Adiabatic means there's no heat
transferred in or out of

455
00:35:34,690 --> 00:35:36,020
the system.

456
00:35:36,020 --> 00:35:41,780
We don't say anything about
whether the system does work

457
00:35:41,780 --> 00:35:47,760
or has worked on -- implicitly
here, we're talking about a

458
00:35:47,760 --> 00:35:52,900
closed system, so there's no
mass leaving the system.

459
00:35:52,900 --> 00:35:58,230
But if it's adiabatic, then dq
is equal zero, and for an

460
00:35:58,230 --> 00:36:11,580
adiabatic process, then
du is equal to dw.

461
00:36:11,580 --> 00:36:17,880
OK, now this is a point
we want to be careful.

462
00:36:17,880 --> 00:36:25,320
Be careful.

463
00:36:25,320 --> 00:36:29,510
Suppose we are doing an
adiabatic process.

464
00:36:29,510 --> 00:36:38,000
We can do it reversibly, or
we can do it irreversibly.

465
00:36:38,000 --> 00:36:42,140
So suppose we do it
irreversibly.

466
00:36:42,140 --> 00:36:46,580
Suppose we just remove stops,
and the system slams up

467
00:36:46,580 --> 00:36:50,080
against the other stops.

468
00:36:50,080 --> 00:36:52,310
Did no work.

469
00:36:52,310 --> 00:36:56,030
Does that mean du is zero?

470
00:36:56,030 --> 00:36:59,990
You bet it does not.

471
00:36:59,990 --> 00:37:06,450
So, it's important to write this
little thing on here, du

472
00:37:06,450 --> 00:37:10,750
is equal to the reversible
work, not just the work.

473
00:37:10,750 --> 00:37:14,760
You can have a process where
you can measure an

474
00:37:14,760 --> 00:37:16,710
irreversible process,
where you could

475
00:37:16,710 --> 00:37:17,890
calculate the work done.

476
00:37:17,890 --> 00:37:19,130
It could be zero.

477
00:37:19,130 --> 00:37:20,430
It could be something else.

478
00:37:20,430 --> 00:37:27,210
But if it's not reversible,
it's not du.

479
00:37:27,210 --> 00:37:29,840
And you will be invited to
make that mistake on

480
00:37:29,840 --> 00:37:33,280
an exam, I'm sure.

481
00:37:33,280 --> 00:37:41,140
OK, so, the thing about a state
function is that the

482
00:37:41,140 --> 00:37:46,200
function has a value for initial
conditions and at

483
00:37:46,200 --> 00:37:47,580
final conditions.

484
00:37:47,580 --> 00:37:49,550
And the difference
between those is

485
00:37:49,550 --> 00:37:52,980
what you could measure.

486
00:37:52,980 --> 00:37:59,020
If you measure something
else, you won't get du.

487
00:37:59,020 --> 00:38:03,340
Be careful, and this is going
to be especially complicated

488
00:38:03,340 --> 00:38:07,590
and confusing when we get to
quantities that have a more

489
00:38:07,590 --> 00:38:11,980
obscure meaning like entropy.

490
00:38:11,980 --> 00:38:15,820
I mean we can, we can sort of
understand why OK, the total

491
00:38:15,820 --> 00:38:20,030
energy, if we measure it, we
measure a process which is not

492
00:38:20,030 --> 00:38:20,940
reversible.

493
00:38:20,940 --> 00:38:26,430
Well it might not give the
energy change for that

494
00:38:26,430 --> 00:38:31,070
process, but when we have
a quantity which is more

495
00:38:31,070 --> 00:38:36,620
obscure, which the definition
of that quantity requires a

496
00:38:36,620 --> 00:38:40,700
very specific prescription for
calculating, you're going to

497
00:38:40,700 --> 00:38:41,680
get into trouble.

498
00:38:41,680 --> 00:38:47,460
So exam this and be sure
you understand that.

499
00:38:47,460 --> 00:38:53,410
OK, constant volume.

500
00:38:53,410 --> 00:38:57,500
Well for constant volume,
dw is equal to zero.

501
00:38:57,500 --> 00:38:58,320
Why?

502
00:38:58,320 --> 00:39:01,270
How does it do, what is work?

503
00:39:01,270 --> 00:39:04,640
Well it's p v work if the
volume doesn't change.

504
00:39:04,640 --> 00:39:05,610
There is no work.

505
00:39:05,610 --> 00:39:13,150
So in this case du is equal to
dq, and we put a little v on

506
00:39:13,150 --> 00:39:22,820
it to imply the work, the change
in internal energy is

507
00:39:22,820 --> 00:39:28,410
equal to the heat added
at constant volume.

508
00:39:28,410 --> 00:39:30,060
Nothing reversible about it.

509
00:39:30,060 --> 00:39:32,460
It's now, all we have to do is
say we're going to have heat

510
00:39:32,460 --> 00:39:36,280
at constant volume.

511
00:39:36,280 --> 00:39:41,330
OK, and now we return to
this differential.

512
00:39:41,330 --> 00:39:45,320
We want to ask the question,
what are these two quantities?

513
00:39:45,320 --> 00:39:49,310
How do we know what they are?

514
00:39:49,310 --> 00:39:54,090
This should be particularly
bothersome to you because, as

515
00:39:54,090 --> 00:39:58,070
you've already experienced in
5.60, there are a lot of

516
00:39:58,070 --> 00:40:00,750
partial derivatives. there
are a lot of variables.

517
00:40:00,750 --> 00:40:02,610
There are a lot of things
held constant.

518
00:40:02,610 --> 00:40:07,290
It's easy to get lost in this
sea of quantities, none of

519
00:40:07,290 --> 00:40:08,840
which have obvious meaning.

520
00:40:08,840 --> 00:40:13,240
So now what we're going to do is
start to extract what these

521
00:40:13,240 --> 00:40:15,840
things mean.

522
00:40:15,840 --> 00:40:22,570
OK, so for a constant volume
process, we can write du,

523
00:40:22,570 --> 00:40:24,990
partial derivative of u with
respect to T at constant V,

524
00:40:24,990 --> 00:40:34,210
dT, plus partial derivative
of u at constant V, dV.

525
00:40:34,210 --> 00:40:37,950
OK, for constant volume,
this is zero.

526
00:40:37,950 --> 00:40:47,300
So this term is gone, and we
rewrite this du, V is equal to

527
00:40:47,300 --> 00:40:56,000
du/dT, at constant V, dT v.
So we, have a change in

528
00:40:56,000 --> 00:40:59,350
temperature done at constant
volume, and we have a change

529
00:40:59,350 --> 00:41:01,930
in internal energy done at
constant volume, and we

530
00:41:01,930 --> 00:41:03,690
rearrange this.

531
00:41:03,690 --> 00:41:18,930
And we discover that du/dT at
constant V is equal to du/dT

532
00:41:18,930 --> 00:41:24,780
at constant V. What, have
I done something silly?

533
00:41:24,780 --> 00:41:37,390
Oh well, yes I have. dq v, so
du v is equal to dq v and so

534
00:41:37,390 --> 00:41:41,740
what I should have written here,
this is true, we can get

535
00:41:41,740 --> 00:41:46,910
a partial derivative by taking
two total derivatives at the

536
00:41:46,910 --> 00:41:50,070
same pressure, at the same
quantity, but what I really

537
00:41:50,070 --> 00:42:11,140
wanted to do is to write dq
v is equal to the partial

538
00:42:11,140 --> 00:42:21,670
derivative of T, constant v dT
v. OK, and now, we know the

539
00:42:21,670 --> 00:42:29,550
relationship between heat and a
temperature change is given

540
00:42:29,550 --> 00:42:35,700
by a quantity, a heat capacity
for a particular path, and

541
00:42:35,700 --> 00:42:36,370
here it is.

542
00:42:36,370 --> 00:42:43,020
So what we've discovered from
this relationship that du at

543
00:42:43,020 --> 00:42:49,920
constant volume is equal to dq
v. We have discovered that

544
00:42:49,920 --> 00:42:53,860
this partial derivative that
appears in the definition, the

545
00:42:53,860 --> 00:42:56,730
abstract definition of the
differential for internal

546
00:42:56,730 --> 00:43:04,140
energy, is just equal to the
constant volume heat capacity.

547
00:43:04,140 --> 00:43:10,200
So, in this definition now, we
have one term which we know.

548
00:43:10,200 --> 00:43:11,950
It's something that
we can measure.

549
00:43:11,950 --> 00:43:17,030
We can measure the heat capacity
at constant volume,

550
00:43:17,030 --> 00:43:22,160
and now we have another term,
and if we can figure out how

551
00:43:22,160 --> 00:43:26,000
to measure it, we'll have
a complete form for this

552
00:43:26,000 --> 00:43:28,770
differential which
will enable us to

553
00:43:28,770 --> 00:43:36,720
calculate du for any process.

554
00:43:36,720 --> 00:43:48,600
So let me write where I am
now, or we are now. du is

555
00:43:48,600 --> 00:43:56,620
equal to Cv dT, plus partial of
u with respect to volume at

556
00:43:56,620 --> 00:44:00,050
constant temperature dV.

557
00:44:00,050 --> 00:44:04,490
So, we've simplified the
expression by replacing one of

558
00:44:04,490 --> 00:44:08,550
the partial derivatives by a
quantity that we can measure.

559
00:44:08,550 --> 00:44:12,570
And we like to know
what about this.

560
00:44:12,570 --> 00:44:15,360
So we need an experiment
that will enable us to

561
00:44:15,360 --> 00:44:18,910
measure this quantity.

562
00:44:18,910 --> 00:44:23,940
And that's where we get Joule,
and I like to say it Joule's

563
00:44:23,940 --> 00:44:30,300
free expansion -- it's usually
referred to as the Joule free

564
00:44:30,300 --> 00:44:35,160
expansion, which sort of implies
that no energy flows,

565
00:44:35,160 --> 00:44:39,430
which actually is true, but it's
an experiment proposed by

566
00:44:39,430 --> 00:44:45,370
Joule, and the experiment
involves an adiabatic box.

567
00:44:45,370 --> 00:44:51,770
So we have system insulated from
the outside world, and we

568
00:44:51,770 --> 00:44:57,360
have two bulbs.

569
00:44:57,360 --> 00:44:58,840
There's a valve between them.

570
00:44:58,840 --> 00:45:05,700
And so we have gas and
we have vacuum.

571
00:45:05,700 --> 00:45:11,020
So the Joule free expansion
involves opening this valve

572
00:45:11,020 --> 00:45:17,510
and asking what happens when
this gas moves into the other

573
00:45:17,510 --> 00:45:21,830
bulb or distributes between the
two. well since the gas is

574
00:45:21,830 --> 00:45:25,680
expanding into vacuum
no work is done.

575
00:45:25,680 --> 00:45:29,120
Since it's isolated,
no heat is added.

576
00:45:29,120 --> 00:45:39,270
So du is equal to zero because
dq and dw are both zero.

577
00:45:39,270 --> 00:45:44,550
So we can now take this
expression and rewrite it

578
00:45:44,550 --> 00:45:48,100
under the condition of
du is equal to zero.

579
00:45:48,100 --> 00:45:53,030
So we have du equal to zero,
just a zero here.

580
00:45:53,030 --> 00:46:00,910
Cv dT constant u plus the
derivative du/dV at constant

581
00:46:00,910 --> 00:46:06,720
T, dV at constant u.

582
00:46:06,720 --> 00:46:07,730
This is just a number.

583
00:46:07,730 --> 00:46:12,880
We don't have to specify that
it's measured at constant u.

584
00:46:12,880 --> 00:46:14,110
It's just a number.

585
00:46:14,110 --> 00:46:18,880
So now we rearrange this
expression, and so we get

586
00:46:18,880 --> 00:46:34,700
du/dV at constant T is equal to
minus Cv times dT u over dV

587
00:46:34,700 --> 00:46:41,630
u or minus Cv partial derivative
of temperature with

588
00:46:41,630 --> 00:46:50,710
respect for volume a constant,
free energy, a constant

589
00:46:50,710 --> 00:46:52,630
internal energy.

590
00:46:52,630 --> 00:46:57,890
So, this is the quantity
we want.

591
00:46:57,890 --> 00:47:03,390
It's related to the heat
capacity, the constant volume

592
00:47:03,390 --> 00:47:07,300
of heat capacity and something
you could measure.

593
00:47:07,300 --> 00:47:11,780
What happens as you expand
into volume?

594
00:47:11,780 --> 00:47:14,160
Does the temperature go up?

595
00:47:14,160 --> 00:47:16,780
Does it not change?

596
00:47:16,780 --> 00:47:21,770
Joule actually did this
experiment, and he observed

597
00:47:21,770 --> 00:47:27,180
that for the gas expansions
that he could do, that the

598
00:47:27,180 --> 00:47:32,570
temperature did not increase
measurably.

599
00:47:32,570 --> 00:47:41,330
So he made an incorrect
conclusion.

600
00:47:41,330 --> 00:47:44,720
Because something was small
and unmeasurable, he said,

601
00:47:44,720 --> 00:47:53,280
well the best of my knowledge
dT/dV at constant

602
00:47:53,280 --> 00:47:56,560
u is equal to zero.

603
00:47:56,560 --> 00:48:06,380
And that implies that since the
quantity we want is given

604
00:48:06,380 --> 00:48:09,620
by this quantity, which is zero
times a constant, the

605
00:48:09,620 --> 00:48:13,290
quantity we want is also zero.

606
00:48:13,290 --> 00:48:20,800
So it would imply that du
was equal to only the

607
00:48:20,800 --> 00:48:25,250
first term Cv dT.

608
00:48:25,250 --> 00:48:28,200
Now, this is an important
lesson in

609
00:48:28,200 --> 00:48:30,380
what you do in science.

610
00:48:30,380 --> 00:48:32,560
You make an observation.

611
00:48:32,560 --> 00:48:36,080
You make an observation doing
an experiment that is as

612
00:48:36,080 --> 00:48:38,370
accurate as you can do.

613
00:48:38,370 --> 00:48:42,600
And so an experiment said the
gas didn't increase its

614
00:48:42,600 --> 00:48:46,910
temperature when it expanded
the vacuum.

615
00:48:46,910 --> 00:48:50,050
And so the next thing you do is
you call up a journal and

616
00:48:50,050 --> 00:48:55,270
say I've discovered a
fundamental law of nature.

617
00:48:55,270 --> 00:49:03,820
And, so you propose that there
is no, that this derivative is

618
00:49:03,820 --> 00:49:06,840
zero, and that the internal
energy is given

619
00:49:06,840 --> 00:49:09,860
simply by this quantity.

620
00:49:09,860 --> 00:49:14,490
It turns out that this quantity
here, which is called

621
00:49:14,490 --> 00:49:19,430
eta of J the Joule free
expansion parameter, is not

622
00:49:19,430 --> 00:49:21,170
quite zero.

623
00:49:21,170 --> 00:49:25,870
When you expand a real
gas into vacuum, the

624
00:49:25,870 --> 00:49:29,130
temperature goes down.

625
00:49:29,130 --> 00:49:39,715
So this is a very small number
and for ideal gases, eta J is

626
00:49:39,715 --> 00:49:41,050
equal to zero.

627
00:49:41,050 --> 00:49:46,220
This quantity is exactly zero
for an ideal gas and we'll

628
00:49:46,220 --> 00:49:49,820
discover why eventually it has
to do with what we mean by an

629
00:49:49,820 --> 00:49:51,690
ideal gas it turns out.

630
00:49:51,690 --> 00:49:56,290
And so for many, many problems,
especially on exams,

631
00:49:56,290 --> 00:50:02,420
especially on this first exam,
you will be able to say that

632
00:50:02,420 --> 00:50:05,780
this is the relationship between
internal energy and

633
00:50:05,780 --> 00:50:07,080
temperature.

634
00:50:07,080 --> 00:50:12,110
That u is a function of
temperature only.

635
00:50:12,110 --> 00:50:15,630
It doesn't matter what
other things change.

636
00:50:15,630 --> 00:50:19,210
The value of the internal energy
is only determined by

637
00:50:19,210 --> 00:50:20,020
temperature.

638
00:50:20,020 --> 00:50:22,880
But this is only true for
an ideal gas and it's

639
00:50:22,880 --> 00:50:27,110
approximately true
for other things.

640
00:50:27,110 --> 00:50:37,660
So, delta u is equal
to zero for all

641
00:50:37,660 --> 00:50:47,540
isothermal ideal gas processes.

642
00:50:47,540 --> 00:50:49,660
And it's approximately
equal to zero for

643
00:50:49,660 --> 00:50:51,720
all real gas processes.

644
00:50:51,720 --> 00:50:58,240
And so that means that delta
u is always calculable from

645
00:50:58,240 --> 00:51:05,180
Cv(T) dT for any ideal
gas change.

646
00:51:05,180 --> 00:51:08,470
So even if work is done,
it doesn't matter.

647
00:51:08,470 --> 00:51:12,710
All you care about is what was
the temperature change?

648
00:51:12,710 --> 00:51:19,410
And this is always the easy way
to calculate du, and it's

649
00:51:19,410 --> 00:51:25,230
often the easy way to calculate
either q or w, using

650
00:51:25,230 --> 00:51:26,410
the first law.

651
00:51:26,410 --> 00:51:35,730
So, since du is equal to q plus
w, and for an isothermal

652
00:51:35,730 --> 00:51:40,490
process this is equal
to zero, q = -w.

653
00:51:40,490 --> 00:51:45,500
For an ideal gas, for an
isothermal process.

654
00:51:45,500 --> 00:51:49,230
It simplifies your
life enormously.

655
00:51:49,230 --> 00:51:52,040
You don't have to
calculate much.

656
00:51:52,040 --> 00:51:56,500
OK, so it's time to stop, I did,
in fact, get caught up.

657
00:51:56,500 --> 00:52:00,200
I hope you didn't mind the
enhanced velocity.

658
00:52:00,200 --> 00:52:03,070
Now Bawendi can do the beautiful
stuff on lecture

659
00:52:03,070 --> 00:52:04,040
number four.

660
00:52:04,040 --> 00:52:05,520
Thank you.