Reconciling thermodynamic and state definitions of entropy

WEBVTT
Kind: captions
Language: en

00:00:00.650 --> 00:00:03.190
If you followed some of the
mathematics, and some of the

00:00:03.190 --> 00:00:06.590
thermodynamic principles in the
last several videos, what

00:00:06.590 --> 00:00:10.130
occurs in this video might
just blow your mind.

00:00:10.130 --> 00:00:12.570
So not to set expectations
too high, let's just

00:00:12.570 --> 00:00:14.506
start off with it.

00:00:14.506 --> 00:00:17.160
So let's say I have
a container.

00:00:17.160 --> 00:00:19.990
And in that container,
I have gas particles.

00:00:22.650 --> 00:00:25.480
Inside of that container,
they're bouncing around like

00:00:25.480 --> 00:00:29.000
gas particles tend to do,
creating some pressure on the

00:00:29.000 --> 00:00:30.760
container of a certain volume.

00:00:30.760 --> 00:00:32.784
And let's say I have
n particles.

00:00:36.520 --> 00:00:38.920
Now, each of these particles
could be

00:00:38.920 --> 00:00:40.860
in x different states.

00:00:40.860 --> 00:00:42.110
Let me write that down.

00:00:56.420 --> 00:00:57.800
What do I mean by state?

00:00:57.800 --> 00:00:59.640
Well, let's say I
take particle A.

00:00:59.640 --> 00:01:01.840
Let me make particle A
a different color.

00:01:01.840 --> 00:01:05.080
Particle A could be down here
in this corner, and it could

00:01:05.080 --> 00:01:06.810
have some velocity like that.

00:01:06.810 --> 00:01:08.190
It could also be in that
corner and have a

00:01:08.190 --> 00:01:08.940
velocity like that.

00:01:08.940 --> 00:01:10.080
Those would be two
different states.

00:01:10.080 --> 00:01:12.070
It could be up here, and have
a velocity like that.

00:01:12.070 --> 00:01:14.050
It could be there and have
a velocity like that.

00:01:14.050 --> 00:01:16.260
If you were to add up all the
different states, and there

00:01:16.260 --> 00:01:20.340
would be a gazillion, of
them, you would get x.

00:01:20.340 --> 00:01:22.550
That blue particle could have
x different states.

00:01:22.550 --> 00:01:23.470
You don't know.

00:01:23.470 --> 00:01:24.740
We're just saying, look.

00:01:24.740 --> 00:01:25.545
I have this container.

00:01:25.545 --> 00:01:26.470
It's got n particles.

00:01:26.470 --> 00:01:28.670
So we just know that each
of them could be

00:01:28.670 --> 00:01:30.330
in x different states.

00:01:30.330 --> 00:01:33.620
Now, if each of them can be in
x different states, how many

00:01:33.620 --> 00:01:37.950
total configurations are there
for the system as a whole?

00:01:37.950 --> 00:01:42.080
Well, particle A could be in x
different states, and then

00:01:42.080 --> 00:01:44.990
particle B could be in
x different places.

00:01:44.990 --> 00:01:46.340
So times x.

00:01:46.340 --> 00:01:48.880
If we just had two particles,
then you would multiply all

00:01:48.880 --> 00:01:51.130
the different places where
X could be times all the

00:01:51.130 --> 00:01:54.970
different places where the red
particle could be, then you'd

00:01:54.970 --> 00:01:57.290
get all the different
configurations for the system.

00:01:57.290 --> 00:01:58.570
But we don't have just
two particles.

00:01:58.570 --> 00:02:00.260
We have n particles.

00:02:00.260 --> 00:02:04.690
So for every particle, you'd
multiply it times the number

00:02:04.690 --> 00:02:06.840
of states it could have,
and you do that

00:02:06.840 --> 00:02:09.729
a total of n times.

00:02:09.729 --> 00:02:13.290
And this is really just
combinatorics here.

00:02:13.290 --> 00:02:14.540
You do it n times.

00:02:14.540 --> 00:02:16.840
This system would have
n configurations.

00:02:16.840 --> 00:02:19.940
For example, if I had two
particles, each particle had

00:02:19.940 --> 00:02:22.860
three different potential
states, how many different

00:02:22.860 --> 00:02:24.330
configurations could there be?

00:02:24.330 --> 00:02:28.210
Well, for every three that one
particle could have, the other

00:02:28.210 --> 00:02:30.540
one could have three different
states, so you'd have nine

00:02:30.540 --> 00:02:31.190
different states.

00:02:31.190 --> 00:02:31.970
You'd multiply them.

00:02:31.970 --> 00:02:33.690
If you had another particle with
three different states,

00:02:33.690 --> 00:02:35.050
you'd multiply that by
three, so you have

00:02:35.050 --> 00:02:36.530
27 different states.

00:02:36.530 --> 00:02:38.660
Here we have n particles.

00:02:38.660 --> 00:02:41.020
Each of them could be in
x different states.

00:02:41.020 --> 00:02:43.380
So the total number of
configurations we have for our

00:02:43.380 --> 00:02:47.580
system-- x times itself n times
is just x to the n.

00:02:47.580 --> 00:02:53.790
So we have x to the n states
in our system.

00:02:53.790 --> 00:02:57.530
Now, let's say that we like
thinking about how many states

00:02:57.530 --> 00:03:00.080
a system can have. Certain
states have less-- for

00:03:00.080 --> 00:03:02.340
example, if I had fewer
particles, I would have fewer

00:03:02.340 --> 00:03:03.510
potential states.

00:03:03.510 --> 00:03:07.140
Or maybe if I had a smaller
container, I would also have

00:03:07.140 --> 00:03:08.240
fewer potential states.

00:03:08.240 --> 00:03:10.510
There would be fewer potential
places for our little

00:03:10.510 --> 00:03:12.320
particles to exist.

00:03:12.320 --> 00:03:17.180
So I want to create some type
of state variable that tells

00:03:17.180 --> 00:03:21.230
me, well, how many states
can my system be in?

00:03:21.230 --> 00:03:24.030
So this is kind of a macrostate
variable.

00:03:24.030 --> 00:03:28.040
It tells me, how many states
can my system be in?

00:03:28.040 --> 00:03:31.730
And let's call it
s for states.

00:03:31.730 --> 00:03:33.640
For the first time in
thermodynamics, we're actually

00:03:33.640 --> 00:03:38.050
using a letter that in some way
is related to what we're

00:03:38.050 --> 00:03:39.930
actually trying to measure.

00:03:39.930 --> 00:03:41.650
s for states.

00:03:41.650 --> 00:03:44.750
And since the states, they can
grow really large, let's say I

00:03:44.750 --> 00:03:47.350
like to take the logarithm
of the number of states.

00:03:47.350 --> 00:03:49.860
Now this is just how I'm
defining my state variable.

00:03:49.860 --> 00:03:51.120
I get to define it.

00:03:51.120 --> 00:03:53.970
So I get to put a logarithm
out front.

00:03:53.970 --> 00:03:56.790
So let me just put
a logarithm.

00:03:56.790 --> 00:03:59.810
So in this case, it would be the
logarithm of my number of

00:03:59.810 --> 00:04:05.690
states-- so it would be x to the
n, where this is number of

00:04:05.690 --> 00:04:06.940
potential states.

00:04:10.250 --> 00:04:12.330
And you know, we need some
kind of scaling factor.

00:04:12.330 --> 00:04:14.710
Maybe I'll change the
units eventually.

00:04:14.710 --> 00:04:18.519
So let me put a little
constant out front.

00:04:18.519 --> 00:04:21.709
Every good formula needs
a constant to

00:04:21.709 --> 00:04:22.790
get our units right.

00:04:22.790 --> 00:04:24.500
I'll make that a lowercase k.

00:04:24.500 --> 00:04:25.580
So that's my definition.

00:04:25.580 --> 00:04:28.950
I call this my state variable.

00:04:28.950 --> 00:04:31.470
If you give me a system, I
should, in theory, be able to

00:04:31.470 --> 00:04:35.170
tell you how many states
the system can take on.

00:04:35.170 --> 00:04:35.760
Fair enough.

00:04:35.760 --> 00:04:38.760
So let me close that
box right there.

00:04:38.760 --> 00:04:46.160
Now let's say that I were to
take my box that I had-- let

00:04:46.160 --> 00:04:49.350
me copy and paste it.

00:04:49.350 --> 00:04:51.790
I take that box.

00:04:51.790 --> 00:04:53.860
And it just so happens
that there was an

00:04:53.860 --> 00:04:56.450
adjacent box next to it.

00:04:56.450 --> 00:05:00.120
They share this wall.

00:05:00.120 --> 00:05:02.505
They're identical in size,
although what I just drew

00:05:02.505 --> 00:05:03.560
isn't identical in size.

00:05:03.560 --> 00:05:04.290
But they're close enough.

00:05:04.290 --> 00:05:05.820
They're identical in size.

00:05:05.820 --> 00:05:09.210
And what I do, is I blow
away this wall.

00:05:09.210 --> 00:05:11.010
I just evaporate it,
all of a sudden.

00:05:11.010 --> 00:05:12.680
It just disappears.

00:05:12.680 --> 00:05:17.040
So this wall just disappears.

00:05:17.040 --> 00:05:18.790
Now, what's going to happen?

00:05:18.790 --> 00:05:22.350
Well, as soon as I blow away
this wall, this is very much

00:05:22.350 --> 00:05:25.210
not an isostatic process.

00:05:25.210 --> 00:05:25.700
Right?

00:05:25.700 --> 00:05:27.250
All hell's going
to break loose.

00:05:27.250 --> 00:05:29.350
I'm going to blow away this
wall, and you know, the

00:05:29.350 --> 00:05:33.020
particles that were about to
bounce off the wall are just

00:05:33.020 --> 00:05:34.550
going to keep going.

00:05:34.550 --> 00:05:34.890
Right?

00:05:34.890 --> 00:05:36.480
They're going to keep going
until they can maybe bounce

00:05:36.480 --> 00:05:37.280
off of that wall.

00:05:37.280 --> 00:05:40.110
So right when I blow away this
wall, there's no pressure

00:05:40.110 --> 00:05:42.580
here, because these guys have
nothing to bounce off to.

00:05:42.580 --> 00:05:44.180
While these guys don't
know anything.

00:05:44.180 --> 00:05:45.690
They don't know anything until
they come over here and say,

00:05:45.690 --> 00:05:46.680
oh, no wall.

00:05:46.680 --> 00:05:48.140
So the pressure is in flux.

00:05:48.140 --> 00:05:51.000
Even the volume is in flux, as
these guys make their way

00:05:51.000 --> 00:05:55.990
across the entire expanse
of the new volume.

00:05:55.990 --> 00:05:57.800
So everything is in flux.

00:05:57.800 --> 00:05:58.580
Right?

00:05:58.580 --> 00:05:59.820
And so what's our new volume?

00:05:59.820 --> 00:06:04.520
If we call this volume,
what's this?

00:06:04.520 --> 00:06:06.330
This is now 2 times
the volume.

00:06:09.120 --> 00:06:12.230
Let's think about some of the
other state variables we know.

00:06:12.230 --> 00:06:14.800
We know that the pressure
is going to go down.

00:06:14.800 --> 00:06:16.450
We can even relate it, because
we know that our

00:06:16.450 --> 00:06:17.670
volume is twice it.

00:06:17.670 --> 00:06:19.170
Is 2 times the volume.

00:06:19.170 --> 00:06:22.330
What about the temperature?

00:06:22.330 --> 00:06:23.890
Well, the temperature change.

00:06:23.890 --> 00:06:26.620
Temperature is average kinetic
energy, right?

00:06:26.620 --> 00:06:28.790
Or it's a measure of average
kinetic energy.

00:06:28.790 --> 00:06:31.910
So all of these molecules
here, each of them have

00:06:31.910 --> 00:06:32.700
kinetic energy.

00:06:32.700 --> 00:06:35.930
They could be different amounts
of kinetic energy, but

00:06:35.930 --> 00:06:38.410
temperature is a measure of
average kinetic energy.

00:06:38.410 --> 00:06:41.140
Now, if I blow away this wall,
does that change the kinetic

00:06:41.140 --> 00:06:42.680
energy of these molecules?

00:06:42.680 --> 00:06:43.310
No!

00:06:43.310 --> 00:06:44.670
It doesn't change it at all.

00:06:44.670 --> 00:06:46.700
So the temperature
is constant.

00:06:46.700 --> 00:06:51.480
So if this is T1, then the
temperature of this system

00:06:51.480 --> 00:06:53.280
here is T1.

00:06:53.280 --> 00:06:55.320
And you might say, hey, Sal,
wait, that doesn't make sense.

00:06:55.320 --> 00:06:59.780
In the past, when my cylinder
expanded, my

00:06:59.780 --> 00:07:01.450
temperature went down.

00:07:01.450 --> 00:07:03.450
And the reason why the
temperature went down in that

00:07:03.450 --> 00:07:05.810
case is because your molecules
were doing work.

00:07:05.810 --> 00:07:08.540
They expanded the container
itself.

00:07:08.540 --> 00:07:09.890
They pushed up the cylinder.

00:07:09.890 --> 00:07:14.300
So they expended some of their
kinetic energy to do the work.

00:07:14.300 --> 00:07:15.830
In this case, I just blew
away that wall.

00:07:15.830 --> 00:07:18.350
These guys did no work
whatsoever, so they didn't

00:07:18.350 --> 00:07:21.990
have to expend any of their
kinetic energy to do any work.

00:07:21.990 --> 00:07:23.710
So their temperature
did not change.

00:07:23.710 --> 00:07:24.620
So that's interesting.

00:07:24.620 --> 00:07:25.530
Fair enough.

00:07:25.530 --> 00:07:29.310
Well, in this new world,
what happens?

00:07:29.310 --> 00:07:31.380
Eventually I get to a situation
where my molecules

00:07:31.380 --> 00:07:33.290
fill the container.

00:07:33.290 --> 00:07:33.590
Right?

00:07:33.590 --> 00:07:34.850
We know that from
common sense.

00:07:34.850 --> 00:07:36.500
And if you think about
it on a microlevel,

00:07:36.500 --> 00:07:37.330
why does that happen?

00:07:37.330 --> 00:07:38.540
It's not a mystery.

00:07:38.540 --> 00:07:41.070
You know, on this direction,
things were bouncing and they

00:07:41.070 --> 00:07:41.690
keep bouncing.

00:07:41.690 --> 00:07:43.750
But when they go here, there
used to be a wall, and then

00:07:43.750 --> 00:07:44.920
they'll just keep going,
and then they'll

00:07:44.920 --> 00:07:46.210
start bouncing here.

00:07:46.210 --> 00:07:49.010
So when you have gazillion
particles doing a gazillion of

00:07:49.010 --> 00:07:51.800
these bounces, eventually,
they're just as likely to be

00:07:51.800 --> 00:07:53.830
here as they are over there.

00:07:53.830 --> 00:07:54.220
Now.

00:07:54.220 --> 00:07:56.400
Let's do our computation
again.

00:07:56.400 --> 00:07:59.950
In our old situation, when we
just looked at this, each

00:07:59.950 --> 00:08:04.370
particle could be in one of x
places, or in one of x states.

00:08:04.370 --> 00:08:09.330
Now it could be in twice
as many states, right?

00:08:09.330 --> 00:08:16.790
Now, each particle could be
in 2x different states.

00:08:16.790 --> 00:08:17.990
Why do I say 2x?

00:08:17.990 --> 00:08:20.030
Because I have twice
the area to be in.

00:08:20.030 --> 00:08:23.320
Now, the states aren't just, you
know, position in space.

00:08:23.320 --> 00:08:29.610
But everything else-- so, you
know, before here, maybe I had

00:08:29.610 --> 00:08:36.960
a positions in space times b
positions, or b momentums, you

00:08:36.960 --> 00:08:38.789
know, where those are all the
different momentums, and that

00:08:38.789 --> 00:08:40.620
was equal to x.

00:08:40.620 --> 00:08:44.490
Now I have 2a positions in
volume that I could be in.

00:08:44.490 --> 00:08:46.270
I have twice the volume
to deal with.

00:08:46.270 --> 00:08:50.860
So I have 2a positions in volume
I can be at, but my

00:08:50.860 --> 00:08:53.680
momentum states are going to
still be-- I just have b

00:08:53.680 --> 00:08:56.150
momentum states-- so this
is equal to 2x.

00:08:56.150 --> 00:08:59.680
I now can be in 2x different
states, just because I have 2

00:08:59.680 --> 00:09:03.450
times the volume to travel
around in, right?

00:09:03.450 --> 00:09:05.570
So how many states are
there for the system?

00:09:05.570 --> 00:09:08.100
Well, each particle can
be in 2x states.

00:09:08.100 --> 00:09:11.080
So this is 2x times
2x times 2x.

00:09:11.080 --> 00:09:15.680
And I'm going to do
that n times.

00:09:15.680 --> 00:09:19.500
So my new s-- so this is, you
know, let's call this s

00:09:19.500 --> 00:09:24.540
initial-- so my s final, my
new way of measuring my

00:09:24.540 --> 00:09:27.880
states, is going to be equal to
that little constant that I

00:09:27.880 --> 00:09:32.000
threw in there, times the
natural log of the

00:09:32.000 --> 00:09:33.190
new number of states.

00:09:33.190 --> 00:09:33.620
So what is it?

00:09:33.620 --> 00:09:36.030
It's 2x to the n power.

00:09:40.880 --> 00:09:47.520
So my question to you is, what
is my change in s when I blew

00:09:47.520 --> 00:09:48.770
away the wall?

00:09:48.770 --> 00:09:51.710
You know, there was this room
here the entire time, although

00:09:51.710 --> 00:09:54.580
these particles really
didn't care because

00:09:54.580 --> 00:09:55.710
this wall was there.

00:09:55.710 --> 00:09:58.120
So what is the change in s when
I blew away this wall?

00:09:58.120 --> 00:09:58.760
And this should be clear.

00:09:58.760 --> 00:10:00.050
The temperature didn't
change, because no

00:10:00.050 --> 00:10:01.570
kinetic energy was expended.

00:10:01.570 --> 00:10:02.760
And this is all in isolation.

00:10:02.760 --> 00:10:03.156
I should have said.

00:10:03.156 --> 00:10:04.330
It's adiabatic.

00:10:04.330 --> 00:10:05.790
There's no transfer of heat.

00:10:05.790 --> 00:10:07.940
So that's also why the
temperature didn't change.

00:10:07.940 --> 00:10:10.390
So what is our change in s?

00:10:10.390 --> 00:10:16.700
Our change in s is equal to
our s final minus our s

00:10:16.700 --> 00:10:19.400
initial, which is equal to--
what's our s final?

00:10:19.400 --> 00:10:22.410
It's this expression,
right here.

00:10:22.410 --> 00:10:27.020
It is k times the natural
log--and we can write this as

00:10:27.020 --> 00:10:28.980
2 to the n, x to the n.

00:10:28.980 --> 00:10:30.400
That's just exponent rules.

00:10:30.400 --> 00:10:35.470
And from that, we're going to
subtract out our initial s

00:10:35.470 --> 00:10:38.900
value, which was this. k natural
log of x to the n.

00:10:42.590 --> 00:10:45.300
Now we can use our logarithm
properties to say, well, you

00:10:45.300 --> 00:10:47.970
know, you take the logarithm of
a minus the logarithm of b,

00:10:47.970 --> 00:10:49.400
you can just divide them.

00:10:49.400 --> 00:10:52.770
So this is equal to k-- you
could factor that out-- times

00:10:52.770 --> 00:10:59.120
the logarithm of 2 to the
N-- it's uppercase N,

00:10:59.120 --> 00:10:59.890
so let me do that.

00:10:59.890 --> 00:11:00.970
This is uppercase N.

00:11:00.970 --> 00:11:02.430
I don't want to get confused
with Moles.

00:11:02.430 --> 00:11:06.163
Uppercase N is the number of
particles we actually have. So

00:11:06.163 --> 00:11:11.430
it's 2 to upper case N times x
to the uppercase N divided by

00:11:11.430 --> 00:11:13.630
x to the uppercase N.

00:11:13.630 --> 00:11:15.890
So these two cancel out.

00:11:15.890 --> 00:11:21.310
So our change in s is equal to
k times the natural log of 2

00:11:21.310 --> 00:11:24.680
to the N-- or, if we wanted to
use our logarithm properties,

00:11:24.680 --> 00:11:26.105
we could throw that
N out front.

00:11:26.105 --> 00:11:31.510
And we could say, our change in
the s, whatever this state

00:11:31.510 --> 00:11:34.130
variable I've defined-- and this
is a different definition

00:11:34.130 --> 00:11:43.270
than I did in the last video--
is equal to big N, the number

00:11:43.270 --> 00:11:47.070
of molecules we have, times my
little constant, times the

00:11:47.070 --> 00:11:49.430
natural log of 2.

00:11:49.430 --> 00:11:55.200
So by blowing away that wall and
giving my molecules twice

00:11:55.200 --> 00:11:58.760
as much volume to travel around
in, my change in this

00:11:58.760 --> 00:12:02.010
little state function
I defined is Nk the

00:12:02.010 --> 00:12:03.470
natural log of 2.

00:12:03.470 --> 00:12:05.400
And what really happened?

00:12:05.400 --> 00:12:07.990
I mean, it clearly
went up, right?

00:12:07.990 --> 00:12:10.730
I clearly have a positive
value here.

00:12:10.730 --> 00:12:12.890
Natural log of 2 is
a positive value.

00:12:12.890 --> 00:12:14.540
N is positive value.

00:12:14.540 --> 00:12:16.360
It's going to be very large
number than the number of

00:12:16.360 --> 00:12:18.040
molecules we had.

00:12:18.040 --> 00:12:20.060
And I'm assuming my constant
I threw on there

00:12:20.060 --> 00:12:21.740
is a positive value.

00:12:21.740 --> 00:12:23.000
But what am I really
describing?

00:12:23.000 --> 00:12:26.310
I'm saying that look, by blowing
away that wall, my

00:12:26.310 --> 00:12:28.930
system can take on
more states.

00:12:28.930 --> 00:12:32.700
There's more different
things it can do.

00:12:32.700 --> 00:12:34.820
And I'll throw a little
word out here.

00:12:34.820 --> 00:12:38.020
Its entropy has gone up.

00:12:38.020 --> 00:12:39.490
Well, actually, let's
just define s

00:12:39.490 --> 00:12:40.670
to be the word entropy.

00:12:40.670 --> 00:12:43.110
We'll talk more about the
word in the future.

00:12:43.110 --> 00:12:47.110
Its entropy has gone up, which
means the number of states we

00:12:47.110 --> 00:12:48.040
have has gone up.

00:12:48.040 --> 00:12:50.090
I shouldn't use the word entropy
without just saying,

00:12:50.090 --> 00:12:51.622
entropy I'm defining
as equal to S.

00:12:51.622 --> 00:12:54.530
But let's just keep it
with s. s for states.

00:12:54.530 --> 00:12:58.010
The number of states we're
dealing with has gone up, and

00:12:58.010 --> 00:12:59.580
it's gone up by this factor.

00:12:59.580 --> 00:13:02.940
Actually, it's gone up by
a factor of 2 to the n.

00:13:02.940 --> 00:13:06.100
And that's why it becomes
n natural log of 2.

00:13:06.100 --> 00:13:07.110
Fair enough.

00:13:07.110 --> 00:13:07.690
Now you're saying, OK.

00:13:07.690 --> 00:13:09.280
This is nice, Sal.

00:13:09.280 --> 00:13:14.600
You have this statistical way,
or I guess you could, this

00:13:14.600 --> 00:13:18.110
combinatoric way of measuring
how many states this system

00:13:18.110 --> 00:13:18.860
can take on.

00:13:18.860 --> 00:13:20.360
And you looked at the
actual molecules.

00:13:20.360 --> 00:13:22.450
You weren't looking at the
kind of macrostates.

00:13:22.450 --> 00:13:23.460
And you were able to do that.

00:13:23.460 --> 00:13:25.840
You came up with this macrostate
that says, that's

00:13:25.840 --> 00:13:27.840
essentially saying, how many
states can I have?

00:13:27.840 --> 00:13:30.740
But how does that relate to
that s that defined in the

00:13:30.740 --> 00:13:32.440
previous video?

00:13:32.440 --> 00:13:34.550
Remember, in the previous video,
I was looking for state

00:13:34.550 --> 00:13:36.370
function that dealt with heat.

00:13:36.370 --> 00:13:42.460
And I defined s, or change in
s-- I defined as change in s--

00:13:42.460 --> 00:13:45.770
to be equal to the heat added
to the system divided by the

00:13:45.770 --> 00:13:50.290
temperature that the
heat was added at.

00:13:50.290 --> 00:13:57.020
So let's see if we can see
whether these things are

00:13:57.020 --> 00:13:57.900
somehow related.

00:13:57.900 --> 00:14:01.130
So let's go back to our
system, and go to a PV

00:14:01.130 --> 00:14:03.400
diagram, and see if we can do
anything useful with that.

00:14:08.920 --> 00:14:10.950
Alright.

00:14:10.950 --> 00:14:11.940
OK.

00:14:11.940 --> 00:14:14.800
So this is pressure,
this is volume.

00:14:14.800 --> 00:14:15.300
Now.

00:14:15.300 --> 00:14:18.150
When we started off, before we
blew away the wall, we had

00:14:18.150 --> 00:14:19.860
some pressure and some volume.

00:14:19.860 --> 00:14:23.010
So this is V1.

00:14:23.010 --> 00:14:26.060
And then we blew away the
wall, and we got to--

00:14:26.060 --> 00:14:27.350
Actually, let me do it a
little bit differently.

00:14:31.030 --> 00:14:35.290
I want that to be just
right there.

00:14:35.290 --> 00:14:38.270
Let me make it right there.

00:14:38.270 --> 00:14:41.770
So that is our V1.

00:14:41.770 --> 00:14:44.680
This is our original state
that we're in.

00:14:44.680 --> 00:14:46.990
So state initial, or
however we want it.

00:14:46.990 --> 00:14:47.970
That's our initial pressure.

00:14:47.970 --> 00:14:49.260
And then we blew away
the wall, and our

00:14:49.260 --> 00:14:51.300
volume doubled, right?

00:14:51.300 --> 00:14:54.860
So we could call this 2V1.

00:14:54.860 --> 00:14:58.180
Our volume doubled, our pressure
would have gone down,

00:14:58.180 --> 00:15:00.010
and we're here.

00:15:00.010 --> 00:15:00.330
Right?

00:15:00.330 --> 00:15:02.310
That's our state 2.

00:15:02.310 --> 00:15:04.660
That's this scenario right
here, after we

00:15:04.660 --> 00:15:05.750
blew away the wall.

00:15:05.750 --> 00:15:10.030
Now, what we did was not
a quasistatic process.

00:15:10.030 --> 00:15:13.170
I can't draw the path here,
because right when I blew away

00:15:13.170 --> 00:15:15.260
the wall, all hell broke
loose, and things like

00:15:15.260 --> 00:15:17.380
pressure and volume weren't
well defined.

00:15:17.380 --> 00:15:20.680
Eventually it got back to an
equilibrium where this filled

00:15:20.680 --> 00:15:22.840
the container, and nothing
else was in flux.

00:15:22.840 --> 00:15:25.100
And we could go back to here,
and we could say, OK, now the

00:15:25.100 --> 00:15:26.270
pressure and the
volume is this.

00:15:26.270 --> 00:15:28.680
But we don't know what happened
in between that.

00:15:28.680 --> 00:15:33.200
So if we wanted to figure out
our Q/T, or the heat into the

00:15:33.200 --> 00:15:35.760
system, we learned in the last
video, the heat added to the

00:15:35.760 --> 00:15:38.590
system is equal to the work
done by the system.

00:15:38.590 --> 00:15:41.790
We'd be at a loss, because the
work done by the system is the

00:15:41.790 --> 00:15:44.450
area under some curve, but
there's no curve to speak of

00:15:44.450 --> 00:15:47.710
here, because our system wasn't
defined while all the

00:15:47.710 --> 00:15:49.270
hell had broke loose.

00:15:49.270 --> 00:15:50.610
So what can we do?

00:15:50.610 --> 00:15:55.350
Well, remember, this is
a state function.

00:15:55.350 --> 00:15:56.610
And this is a state function.

00:15:56.610 --> 00:15:58.040
And I showed that in
the last video.

00:15:58.040 --> 00:16:00.530
So it shouldn't be dependent
on how we got

00:16:00.530 --> 00:16:02.090
from there to there.

00:16:02.090 --> 00:16:02.840
Right?

00:16:02.840 --> 00:16:06.375
So this change in entropy--
actually, let me be careful

00:16:06.375 --> 00:16:07.130
with my words.

00:16:07.130 --> 00:16:16.030
This change in s, so s2 minus
s1, should be independent of

00:16:16.030 --> 00:16:20.090
the process that got
me from s1 to s2.

00:16:20.090 --> 00:16:24.410
So this is independent of
whatever crazy path-- I mean,

00:16:24.410 --> 00:16:27.160
I could have taken some
crazy, quasistatic

00:16:27.160 --> 00:16:29.340
path like that, right?

00:16:29.340 --> 00:16:33.910
So any path that goes from this
s1 to this s2 will have

00:16:33.910 --> 00:16:37.370
the same heat going into the
system, or should have the

00:16:37.370 --> 00:16:39.180
same-- let me take that--

00:16:39.180 --> 00:16:43.480
Any system that goes from s1 to
s2, regardless of its path,

00:16:43.480 --> 00:16:46.550
will have the same change
in entropy, or their

00:16:46.550 --> 00:16:47.680
same change in s.

00:16:47.680 --> 00:16:50.250
Because their s was something
here, and it's something

00:16:50.250 --> 00:16:51.480
different over here.

00:16:51.480 --> 00:16:53.500
And you just take the difference
between the two.

00:16:53.500 --> 00:16:56.670
So what's a system that we
know that can do that?

00:16:56.670 --> 00:17:00.790
Well, let's say that we
did an isothermal.

00:17:00.790 --> 00:17:03.440
And we know that these are all
the same isotherm, right?

00:17:03.440 --> 00:17:06.400
We know that the temperature
didn't change.

00:17:06.400 --> 00:17:07.109
I told you that.

00:17:07.109 --> 00:17:09.760
Because no kinetic energy was
expended, and none of the

00:17:09.760 --> 00:17:11.280
particles did any work.

00:17:11.280 --> 00:17:15.390
So we can say, we can think of a
theoretical process in which

00:17:15.390 --> 00:17:20.290
instead of doing something like
that, we could have had a

00:17:20.290 --> 00:17:24.050
situation where we started off
with our original container

00:17:24.050 --> 00:17:26.260
with our molecules in it.

00:17:26.260 --> 00:17:29.890
We could have put a reservoir
here that's equivalent to the

00:17:29.890 --> 00:17:31.370
temperature that we're at.

00:17:31.370 --> 00:17:34.360
And then this could have been
a piston that was maybe, we

00:17:34.360 --> 00:17:37.640
were pushing on it with some
rocks that are pushing in the

00:17:37.640 --> 00:17:38.850
left-wards direction.

00:17:38.850 --> 00:17:42.840
And we slowly and slowly remove
the rocks, so that

00:17:42.840 --> 00:17:46.140
these gases could push the
piston and do some work, and

00:17:46.140 --> 00:17:49.760
fill this entire volume,
or twice the volume.

00:17:49.760 --> 00:17:52.650
And then the temperature would
have been kept constant by

00:17:52.650 --> 00:17:54.240
this heat reservoir.

00:17:54.240 --> 00:17:57.470
So this type of a process is
kind of a sideways version of

00:17:57.470 --> 00:18:00.250
what I've done in the Carnot
diagrams. That would be

00:18:00.250 --> 00:18:02.660
described like this.

00:18:02.660 --> 00:18:05.820
You'd go from this state to that
state, and it would be a

00:18:05.820 --> 00:18:09.106
quasistatic static process
along an isotherm.

00:18:09.106 --> 00:18:10.400
So it would look like that.

00:18:10.400 --> 00:18:13.280
So you could have
a curve there.

00:18:13.280 --> 00:18:17.430
Now, for that process,
what is the area

00:18:17.430 --> 00:18:22.330
under the curve there?

00:18:22.330 --> 00:18:27.240
Well, the area under the curve
is just the integral-- and

00:18:27.240 --> 00:18:29.950
we've done this multiple times--
from our initial

00:18:29.950 --> 00:18:35.280
volume to our second volume,
which is twice it, of p times

00:18:35.280 --> 00:18:36.840
our change in volume, right?

00:18:36.840 --> 00:18:40.100
p is our height, times our
little changes in volume, give

00:18:40.100 --> 00:18:41.030
us each rectangle.

00:18:41.030 --> 00:18:43.750
And then the integral is just
the sum along all of these.

00:18:43.750 --> 00:18:48.580
So that's essentially the work
that this system does.

00:18:48.580 --> 00:18:49.110
Right?

00:18:49.110 --> 00:18:51.690
And the work that this system
does, since we are on an

00:18:51.690 --> 00:18:55.570
isotherm, it is equal to the
heat added to the system.

00:18:55.570 --> 00:18:57.610
Because our internal energy
didn't change.

00:18:57.610 --> 00:18:58.490
So what is this?

00:18:58.490 --> 00:19:00.860
We've done this multiple times,
but I'll redo it.

00:19:00.860 --> 00:19:05.140
So this is equal to the
integral of V1 to 2V1.

00:19:05.140 --> 00:19:07.820
PV equals NRT, right?

00:19:07.820 --> 00:19:09.010
NRT.

00:19:09.010 --> 00:19:11.970
So P is equal to NRT/V.

00:19:11.970 --> 00:19:16.150
NRT over V dv.

00:19:16.150 --> 00:19:18.690
And the t is T1.

00:19:18.690 --> 00:19:21.340
Now, all of this is happening
along an isotherm, so all of

00:19:21.340 --> 00:19:23.770
these terms are constant.

00:19:23.770 --> 00:19:27.920
So this is equal to the integral
from V1 to 2V1 of

00:19:27.920 --> 00:19:31.610
NRT1 times 1 over V dv.

00:19:31.610 --> 00:19:34.130
I've done this integral
multiple times.

00:19:34.130 --> 00:19:37.870
And so this is equal to-- I'll
skip a couple of steps here,

00:19:37.870 --> 00:19:40.140
because I've done it in several
videos already-- the

00:19:40.140 --> 00:19:45.180
natural log of 2V1
over V1, right?

00:19:45.180 --> 00:19:47.100
The antiderivative of this
is the natural log.

00:19:47.100 --> 00:19:49.620
Take the natural log of that
minus the natural log of that,

00:19:49.620 --> 00:19:53.100
which is equal to the natural
log of 2V1 over V1.

00:19:53.100 --> 00:19:57.630
Which is just the same thing
as NRT1 times the

00:19:57.630 --> 00:20:00.190
natural log of 2.

00:20:00.190 --> 00:20:01.470
Interesting.

00:20:01.470 --> 00:20:06.530
Now, let's add one little one
little interesting thing to

00:20:06.530 --> 00:20:08.280
this to this equation.

00:20:08.280 --> 00:20:11.460
So this is NRT, but if I wanted
to write in terms of

00:20:11.460 --> 00:20:14.930
the number of molecules, N
is the number of moles.

00:20:14.930 --> 00:20:21.470
So I could rewrite N as the
number of molecules we have

00:20:21.470 --> 00:20:25.870
divided by, 6 times 10
to the 23 power.

00:20:25.870 --> 00:20:26.660
Right?

00:20:26.660 --> 00:20:28.710
That's what n could
be written as.

00:20:28.710 --> 00:20:33.260
So if we do it that way, then
what is our-- remember, all of

00:20:33.260 --> 00:20:36.040
this, we were trying to
find the amount of

00:20:36.040 --> 00:20:38.470
work done by our system.

00:20:38.470 --> 00:20:38.840
Right?

00:20:38.840 --> 00:20:41.290
But if we do it this way,
this equation will turn

00:20:41.290 --> 00:20:43.050
into-- so let's see.

00:20:43.050 --> 00:20:46.810
The work done by our system--
this is our quasistatic

00:20:46.810 --> 00:20:51.500
processes that's going from this
state to that state, but

00:20:51.500 --> 00:20:54.800
it's doing it in a quasistatic
way, so that we can get an

00:20:54.800 --> 00:20:57.020
area under the curve.

00:20:57.020 --> 00:20:59.810
So the work done by this
system is equal to--

00:20:59.810 --> 00:21:00.560
I'll just write it.

00:21:00.560 --> 00:21:08.190
N times R over 6 times 10
to the 23, times T1

00:21:08.190 --> 00:21:10.460
natural log of 2.

00:21:10.460 --> 00:21:11.560
Fair enough.

00:21:11.560 --> 00:21:14.140
Let's make this into
some new constant.

00:21:14.140 --> 00:21:18.000
For convenience, let me
call it a lowercase k.

00:21:18.000 --> 00:21:20.830
So the work we did is equal to
the number of particles we

00:21:20.830 --> 00:21:24.170
had, times some new constant--
we'll call that Boltzmann

00:21:24.170 --> 00:21:27.920
constant, so it's really
just 8 divided by that.

00:21:27.920 --> 00:21:32.270
Times T1, times the
natural log of 2.

00:21:32.270 --> 00:21:33.990
Fair enough.

00:21:33.990 --> 00:21:35.740
Now, that's only in
this situation.

00:21:35.740 --> 00:21:38.230
The other situation did
no work, right?

00:21:38.230 --> 00:21:41.930
So I can't talk about this
system doing any work.

00:21:41.930 --> 00:21:43.860
But this system did
do some work.

00:21:43.860 --> 00:21:48.270
And since it did it along an
isotherm, delta-- the change

00:21:48.270 --> 00:21:52.410
in internal energy is equal to
0, so the change in internal

00:21:52.410 --> 00:21:56.580
energy, which is equal to the
heat applied to the system

00:21:56.580 --> 00:21:59.390
minus the work done by the
system-- this is going to be

00:21:59.390 --> 00:22:02.760
equal to 0, since our
temperature didn't change.

00:22:02.760 --> 00:22:05.440
So the work is going to
be equal to the heat

00:22:05.440 --> 00:22:06.970
added to the system.

00:22:06.970 --> 00:22:12.550
So the heat added to the system
by our little reservoir

00:22:12.550 --> 00:22:16.050
there is going to be-- so the
heat is going to be the number

00:22:16.050 --> 00:22:19.310
of particles we had times
Boltzmann constant, times our

00:22:19.310 --> 00:22:21.200
temperature that we're on
the isotherm, times the

00:22:21.200 --> 00:22:23.090
natural log of 2.

00:22:23.090 --> 00:22:25.060
And all this is a byproduct
of the fact that

00:22:25.060 --> 00:22:26.900
we doubled our volume.

00:22:26.900 --> 00:22:32.560
Now, in the last video, I
defined change in s as equal

00:22:32.560 --> 00:22:35.550
to Q divided by the heat
added, divided by the

00:22:35.550 --> 00:22:37.570
temperature at which
I'm adding it.

00:22:37.570 --> 00:22:42.790
So for this system, this
quasistatic system, what was

00:22:42.790 --> 00:22:44.160
the change in s?

00:22:44.160 --> 00:22:47.750
How much did our s-term,
our s-state, change by?

00:22:47.750 --> 00:22:52.060
So change in s is equal to
heat added divided by our

00:22:52.060 --> 00:22:52.340
temperature.

00:22:52.340 --> 00:22:55.590
Our temperature is T1, so that's
equal to this thing.

00:22:55.590 --> 00:23:02.920
Nk T1 times the natural log
of 2, all of that over T1.

00:23:02.920 --> 00:23:05.050
So our delta-- these
cancel out.

00:23:05.050 --> 00:23:12.170
And our change in our s-quantity
is equal to Nk

00:23:12.170 --> 00:23:14.360
times the natural log of 2.

00:23:14.360 --> 00:23:18.720
Now, you should be starting to
experience an a-ha moment.

00:23:18.720 --> 00:23:21.500
When we defined in the previous
video, we were just

00:23:21.500 --> 00:23:23.720
playing with thermodynamics, and
we said, gee, we'd really

00:23:23.720 --> 00:23:26.380
like to have a state variable
that deals with heat, and we

00:23:26.380 --> 00:23:30.040
just made up this thing right
here that said, change in that

00:23:30.040 --> 00:23:33.350
state variable is equal to the
heat applied to the system

00:23:33.350 --> 00:23:36.320
divided by the temperature at
which the heat was applied.

00:23:36.320 --> 00:23:40.550
And when we use that definition
the change in our

00:23:40.550 --> 00:23:43.910
s-value from this position
to this position, for a

00:23:43.910 --> 00:23:48.290
quasi-static process,
ended up being this.

00:23:48.290 --> 00:23:50.870
Nk natural log of 2.

00:23:50.870 --> 00:23:53.100
Now, this is a state function.

00:23:53.100 --> 00:23:54.050
State variable.

00:23:54.050 --> 00:23:56.000
It's not dependent
on the path.

00:23:56.000 --> 00:24:00.720
So any process that gets from
here, that gets from this

00:24:00.720 --> 00:24:05.320
point to that point, has to
have the same change in s.

00:24:05.320 --> 00:24:09.190
So the delta s for any process
is going to be equal to that

00:24:09.190 --> 00:24:13.720
same value, which was N, in
this case, k, times the

00:24:13.720 --> 00:24:15.090
natural log of 2.

00:24:15.090 --> 00:24:17.610
Any system, by our definition,
right?

00:24:17.610 --> 00:24:18.380
It's the state variable.

00:24:18.380 --> 00:24:21.120
I don't care whether it
disappeared, or the path was

00:24:21.120 --> 00:24:22.390
some crazy path.

00:24:22.390 --> 00:24:22.715
It's a state.

00:24:22.715 --> 00:24:26.260
It's only a function of that and
of that, our change in s.

00:24:26.260 --> 00:24:29.410
So given that, even this
system-- we said that this

00:24:29.410 --> 00:24:32.520
system that we started the video
out with, it started off

00:24:32.520 --> 00:24:35.920
at this same V1, and it
got to the same V2.

00:24:35.920 --> 00:24:39.670
So by the definition of the
previous video, by this

00:24:39.670 --> 00:24:45.550
definition, its change in s is
going to be the number of

00:24:45.550 --> 00:24:48.880
molecules times some constant
times the natural log of 2.

00:24:48.880 --> 00:24:53.590
Now, that's the same exact
result we got when we thought

00:24:53.590 --> 00:24:56.050
about it from a statistical
point of view, when we were

00:24:56.050 --> 00:24:59.170
saying, how many more different
states can the

00:24:59.170 --> 00:25:01.330
system take on?

00:25:01.330 --> 00:25:04.980
And what's mind-blowing here
is that what we started off

00:25:04.980 --> 00:25:08.560
with was just kind of a nice,
you know, macrostate in our

00:25:08.560 --> 00:25:10.850
little Carnot engine world,
that we didn't really know

00:25:10.850 --> 00:25:11.760
what it meant.

00:25:11.760 --> 00:25:15.000
But we got the same exact result
that when we try to do

00:25:15.000 --> 00:25:17.110
it from a measuring the
number of states the

00:25:17.110 --> 00:25:18.670
system could take on.

00:25:18.670 --> 00:25:23.590
So all of this has been a long,
two-video-winded version

00:25:23.590 --> 00:25:25.210
of an introduction to entropy.

00:25:29.530 --> 00:25:33.660
And in thermodynamics, a change
in entropy-- entropy is

00:25:33.660 --> 00:25:38.180
s, or I think of it, s for
states-- the thermodynamic, or

00:25:38.180 --> 00:25:42.840
Carnot cycle, or Carnot engine
world, is defined as the

00:25:42.840 --> 00:25:44.740
change in entropy is defined
as the heat added to the

00:25:44.740 --> 00:25:46.230
system divided by the
temperature at

00:25:46.230 --> 00:25:47.510
which it was added.

00:25:47.510 --> 00:25:52.410
Now, in our statistical
mechanics world, we can define

00:25:52.410 --> 00:25:59.350
entropy as some constant-- and
it's especially convenient,

00:25:59.350 --> 00:26:04.540
this is Boltzmann's constant--
some constant times the

00:26:04.540 --> 00:26:09.080
natural log of the number of
states we have. Sometimes it's

00:26:09.080 --> 00:26:12.110
written as omega, sometimes
other things.

00:26:12.110 --> 00:26:14.450
But this time, it's the number
of states we have.

00:26:14.450 --> 00:26:17.150
And what just showed in this
video is, these are equivalent

00:26:17.150 --> 00:26:17.910
definitions.

00:26:17.910 --> 00:26:18.030
or.

00:26:18.030 --> 00:26:21.020
At least for that one case
I just showed you.

00:26:21.020 --> 00:26:22.490
These are equivalent
definitions.

00:26:22.490 --> 00:26:25.580
When we used the number of
states for this, how much did

00:26:25.580 --> 00:26:29.630
it increase, we got
this result.

00:26:29.630 --> 00:26:32.400
And then when we used the
thermodynamic definition of

00:26:32.400 --> 00:26:34.290
it, we got that same result.

00:26:34.290 --> 00:26:36.940
And if we assume that this
constant is the same as that

00:26:36.940 --> 00:26:40.610
constant, if they're both
Boltzmann's constant, both 1.3

00:26:40.610 --> 00:26:43.610
times 10 to the minus 23,
then our definitions are

00:26:43.610 --> 00:26:44.540
equivalent.

00:26:44.540 --> 00:26:47.990
And so the intuition of
entropy-- in the last one, we

00:26:47.990 --> 00:26:48.940
were kind of struggling
with that.

00:26:48.940 --> 00:26:50.730
We just defined it this way,
but we were like, what does

00:26:50.730 --> 00:26:51.990
that really mean?

00:26:51.990 --> 00:26:55.630
What change in entropy means,
is just how many more states

00:26:55.630 --> 00:26:57.640
can the system take on?

00:26:57.640 --> 00:27:00.310
You know, sometimes when you
learn it in your high school

00:27:00.310 --> 00:27:02.910
chemistry class, they'll
call it disorder.

00:27:02.910 --> 00:27:04.400
And it is disorder.

00:27:04.400 --> 00:27:08.600
But I don't want you to think
that somehow, you know, a

00:27:08.600 --> 00:27:11.950
messy room has higher entropy
than a clean room, which some

00:27:11.950 --> 00:27:13.800
people sometimes use
as an example.

00:27:13.800 --> 00:27:15.070
That's not the case.

00:27:15.070 --> 00:27:19.100
What you should say is, is that
a stadium full of people

00:27:19.100 --> 00:27:23.010
has more states than a stadium
without people in it.

00:27:23.010 --> 00:27:24.690
That has more entropy.

00:27:24.690 --> 00:27:26.640
Or actually, I should even
be careful there.

00:27:26.640 --> 00:27:32.590
Let me say, a stadium at a
high temperature has more

00:27:32.590 --> 00:27:36.180
entropy than the inside
of my refrigerator.

00:27:36.180 --> 00:27:38.630
That the particles in that
stadium have more potential

00:27:38.630 --> 00:27:41.910
states than the particles
in my refrigerator.

00:27:41.910 --> 00:27:44.610
Now I'm going to leave you
there, and we're going to take

00:27:44.610 --> 00:27:47.200
our definitions here, which I
think are pretty profound--

00:27:47.200 --> 00:27:50.650
this and this is the same
definition-- and we're going

00:27:50.650 --> 00:27:54.360
to apply that to talk about
the second law of

00:27:54.360 --> 00:27:54.960
thermodynamics.

00:27:54.960 --> 00:27:56.100
And actually, just
a little aside.

00:27:56.100 --> 00:28:02.940
I wrote omega here, but in our
example, this was 2 to the N.

00:28:02.940 --> 00:28:04.200
And so that's why
it's simplified.

00:28:04.200 --> 00:28:13.450
This was x the first time, and
then the second time, when we

00:28:13.450 --> 00:28:17.340
double the size of our room, or
our volume, it was x to the

00:28:17.340 --> 00:28:18.350
N times 2 to N.

00:28:18.350 --> 00:28:22.420
I just want to make sure you
realize what omega relates to,

00:28:22.420 --> 00:28:24.350
relative to what I just
went through.

00:28:24.350 --> 00:28:26.500
Anyway, see you in
the next video.


Proof: S (or entropy) is a valid state variable

WEBVTT
Kind: captions
Language: en

00:00:00.680 --> 00:00:03.090
I've talked a lot about the
general idea that in order to

00:00:03.090 --> 00:00:07.010
have a state variable like,
say, U, which is internal

00:00:07.010 --> 00:00:10.440
energy, at any point in this
PV diagram, that state

00:00:10.440 --> 00:00:12.710
variable should be that value.

00:00:12.710 --> 00:00:15.660
So for example, if at this
point, U is equal to 5, and I

00:00:15.660 --> 00:00:19.630
go do this whole Carnot cycle,
when I come back to state A, U

00:00:19.630 --> 00:00:21.630
should still be equal to 5.

00:00:21.630 --> 00:00:22.990
It should not have changed.

00:00:22.990 --> 00:00:25.770
It's not dependent upon what
we did to get there.

00:00:25.770 --> 00:00:30.530
So if we did some kind of crazy
path on our PV diagram,

00:00:30.530 --> 00:00:32.990
we got back there, U should
always be the same.

00:00:32.990 --> 00:00:34.760
That's what it means to
be a state variable.

00:00:34.760 --> 00:00:39.000
It's only dependent upon its
position in this PV diagram.

00:00:39.000 --> 00:00:41.960
It's only dependent on its
state, not how you got there.

00:00:41.960 --> 00:00:44.850
And because of that, heat is
something that we can't really

00:00:44.850 --> 00:00:46.110
use as a state variable.

00:00:46.110 --> 00:00:50.100
For example, if I tried to
define some heat-related state

00:00:50.100 --> 00:00:53.980
variable, let's say I call it
heat content, and I defined

00:00:53.980 --> 00:00:58.230
change in heat content as equal
to the amount of heat

00:00:58.230 --> 00:01:00.160
added to the system.

00:01:00.160 --> 00:01:02.470
Well, if we go back to our
Carnot cycle here, let's say

00:01:02.470 --> 00:01:05.850
that my heat content
here was 10.

00:01:05.850 --> 00:01:10.510
Well, I added some heat here,
in this process here.

00:01:10.510 --> 00:01:13.470
Nothing happened, because this
was adiabatic from B to C.

00:01:13.470 --> 00:01:16.030
Then from C to D, I took
out some heat.

00:01:16.030 --> 00:01:18.990
But I took out less heated
than was added here.

00:01:18.990 --> 00:01:21.140
And then here, nothing was
done with regard to heat.

00:01:21.140 --> 00:01:23.780
So I did add some heat
to the system.

00:01:23.780 --> 00:01:26.500
The net heat that I added to
the system as I went around

00:01:26.500 --> 00:01:32.610
the cycle-- in this case, Q
would be equal to Q1 minus Q2.

00:01:32.610 --> 00:01:35.930
And we know that this number
is larger than this.

00:01:35.930 --> 00:01:38.730
The net amount of heat we added
to the system was the

00:01:38.730 --> 00:01:41.740
amount of work we did on the
system, because the internal

00:01:41.740 --> 00:01:43.260
energy didn't change.

00:01:43.260 --> 00:01:50.540
So if this is 0, then the amount
of heat we add to the

00:01:50.540 --> 00:01:52.480
system is the amount
of work we did.

00:01:52.480 --> 00:01:54.520
Our internal energy is
definitely 0 as we go all the

00:01:54.520 --> 00:01:55.680
way around.

00:01:55.680 --> 00:01:58.710
We did this shaded portion of
work-- I showed you that

00:01:58.710 --> 00:02:01.850
several videos ago, that the
area inside of our little

00:02:01.850 --> 00:02:03.570
cycle is the amount
of work we did.

00:02:03.570 --> 00:02:05.390
And so that's also the
net amount of heat

00:02:05.390 --> 00:02:06.960
added to the system.

00:02:06.960 --> 00:02:09.720
So if we added that amount of
heat to the system, if we

00:02:09.720 --> 00:02:12.680
started off at the heat of 10
here, or whatever my heat

00:02:12.680 --> 00:02:15.870
content mystical variable I made
up just now, when we go

00:02:15.870 --> 00:02:17.705
around, it would then
be 10 plus W.

00:02:17.705 --> 00:02:19.500
If we go around again,
it would be 10 plus

00:02:19.500 --> 00:02:21.990
2W and 10 plus 3W.

00:02:21.990 --> 00:02:25.370
So it can't be a legitimate
state variable, because it's

00:02:25.370 --> 00:02:27.330
completely dependent on what
we did to get there.

00:02:27.330 --> 00:02:29.350
And if we keep going around the
cycle, we can increase it,

00:02:29.350 --> 00:02:31.350
even though we get to
the same point.

00:02:31.350 --> 00:02:35.740
So this is an illegitimate state
variable, where I define

00:02:35.740 --> 00:02:39.310
the change in our little made up
heat content to be equal to

00:02:39.310 --> 00:02:40.460
the heat added to system.

00:02:40.460 --> 00:02:42.280
Not a valid state variable.

00:02:42.280 --> 00:02:43.750
Ignore it all.

00:02:43.750 --> 00:02:48.030
Now we know that Q1, we added
more heat here than we took

00:02:48.030 --> 00:02:49.680
away, so there was something
net heat added.

00:02:49.680 --> 00:02:51.930
But there's something
interesting here.

00:02:51.930 --> 00:02:54.980
We added it at a higher
temperature.

00:02:54.980 --> 00:02:58.660
And here we took less heat away
at a lower temperature.

00:02:58.660 --> 00:03:03.340
So maybe we can define another
state variable that can have

00:03:03.340 --> 00:03:06.230
the result that when we go
around the cycle, we do get

00:03:06.230 --> 00:03:08.085
back to our same value.

00:03:08.085 --> 00:03:10.620
Now let me just-- we're
just experimenting.

00:03:10.620 --> 00:03:12.410
Although I know where this
experiment will go.

00:03:12.410 --> 00:03:14.540
I wouldn't have been doing
it if I didn't.

00:03:14.540 --> 00:03:17.900
So let's say I define a
new state variable S.

00:03:17.900 --> 00:03:21.460
And I define a change in S.

00:03:21.460 --> 00:03:24.330
So I say, a change in S-- I'm
just making up a definition--

00:03:24.330 --> 00:03:29.140
is equal to the heat added to
the system divided by the

00:03:29.140 --> 00:03:32.310
temperature at which it was
added to the system.

00:03:32.310 --> 00:03:35.140
Now, I don't know what
this means just yet.

00:03:35.140 --> 00:03:38.160
In future videos, maybe we'll
get intuition about what this

00:03:38.160 --> 00:03:41.140
actually means in kind
of our minds.

00:03:41.140 --> 00:03:44.730
But let's see if at least this
is a valid state variable.

00:03:44.730 --> 00:03:47.980
If, as we go around the Carnot
cycle, whether our change in

00:03:47.980 --> 00:03:50.230
delta S is 0, right?

00:03:50.230 --> 00:03:52.430
To be a legitimate state
variable, we have some value

00:03:52.430 --> 00:03:53.130
for S here.

00:03:53.130 --> 00:03:54.160
Maybe it's 100.

00:03:54.160 --> 00:03:55.520
I don't know.

00:03:55.520 --> 00:03:58.920
Once we go back around the
Carnot cycle, it should be 100

00:03:58.920 --> 00:04:02.290
again, or our delta
S should be 0.

00:04:02.290 --> 00:04:05.000
So what's the delta S?

00:04:05.000 --> 00:04:08.580
So the delta S, as we go around
the whole cycle-- let

00:04:08.580 --> 00:04:11.860
me write delta S-- let
me do another color.

00:04:11.860 --> 00:04:13.400
Delta S.

00:04:13.400 --> 00:04:15.935
As we are go around, I'll say
c for the Carnot cycle.

00:04:15.935 --> 00:04:19.220
As we go around the Carnot
cycle, is going to be equal

00:04:19.220 --> 00:04:23.190
to-- well, when we went from A
to B, we were at a constant

00:04:23.190 --> 00:04:24.410
temperature, and we added Q1.

00:04:24.410 --> 00:04:29.230
So it's Q1, and we were
at temperature T1.

00:04:29.230 --> 00:04:30.330
Fair enough.

00:04:30.330 --> 00:04:34.150
Then when we went from B
to C, it was adiabatic.

00:04:34.150 --> 00:04:36.860
We added or took away no heat.

00:04:36.860 --> 00:04:39.580
So this value, Q over
T, would just be 0.

00:04:39.580 --> 00:04:41.780
So it's plus 0.

00:04:41.780 --> 00:04:43.420
Then we went from C to D.

00:04:43.420 --> 00:04:46.000
We were at a new temperature,
we were on a new isotherm.

00:04:46.000 --> 00:04:47.840
We were at T2.

00:04:47.840 --> 00:04:51.490
And we took away, or I won't put
the sign here, let's just

00:04:51.490 --> 00:04:53.150
say we added Q2 heat.

00:04:53.150 --> 00:04:55.240
We're going to actually
solve for it later.

00:04:55.240 --> 00:04:56.300
We added Q2 heat.

00:04:56.300 --> 00:04:58.260
We'lll see that it's actually
a negative value.

00:04:58.260 --> 00:05:03.000
And then finally, when we
went from D to A, it

00:05:03.000 --> 00:05:04.170
was adiabatic again.

00:05:04.170 --> 00:05:06.150
So no transfer of heat.

00:05:06.150 --> 00:05:07.730
So plus 0, right?

00:05:07.730 --> 00:05:10.730
The 0's are 0's over, you know,
changing temperature,

00:05:10.730 --> 00:05:12.270
but this is just 0.

00:05:12.270 --> 00:05:16.460
So this thing should be equal to
0 in order for this to be a

00:05:16.460 --> 00:05:18.260
valid state variable.

00:05:18.260 --> 00:05:20.510
So let's figure out what
this value is.

00:05:20.510 --> 00:05:27.230
What Q1-- so our change in our
mystical new candidate state

00:05:27.230 --> 00:05:31.810
variable, S, as we go around the
Carnot cycle, is equal to

00:05:31.810 --> 00:05:37.240
Q1 over T1, plus Q2 over T2.

00:05:37.240 --> 00:05:39.770
And we'll see, Q2 is negative.

00:05:39.770 --> 00:05:42.770
So what is Q1?

00:05:42.770 --> 00:05:44.890
Can we calculate Q1?

00:05:44.890 --> 00:05:48.200
Well, as we're on this top
isotherm, our temperature

00:05:48.200 --> 00:05:51.080
doesn't change, our internal
energy doesn't change.

00:05:51.080 --> 00:05:54.060
So if your internal energy
does not change, if your

00:05:54.060 --> 00:05:58.820
internal energy is 0, then the
heat added to the system is

00:05:58.820 --> 00:06:00.670
equal to the work done
by the system.

00:06:00.670 --> 00:06:02.300
So its the area under
this curve.

00:06:04.810 --> 00:06:06.240
Not just the area
in the cycle.

00:06:06.240 --> 00:06:08.570
It would be the whole area
under the curve.

00:06:08.570 --> 00:06:11.280
So what's the whole area
under the curve?

00:06:11.280 --> 00:06:13.930
Well, so let me do a
little aside here.

00:06:13.930 --> 00:06:20.240
So Q1 is equal to the work done
as we went from A to B.

00:06:20.240 --> 00:06:25.500
And work, remember, can just be
written as pressure times

00:06:25.500 --> 00:06:26.820
change in volume.

00:06:26.820 --> 00:06:28.710
We're going to do a little
calculus here, so I'll write

00:06:28.710 --> 00:06:30.605
dV for a small change
in volume.

00:06:33.295 --> 00:06:35.130
And we're going to integrate
it all over the

00:06:35.130 --> 00:06:36.230
little sums, right?

00:06:36.230 --> 00:06:39.220
This dV is this little
change in volume

00:06:39.220 --> 00:06:40.380
right there times pressure.

00:06:40.380 --> 00:06:41.710
That makes a little rectangle.

00:06:41.710 --> 00:06:49.110
And then we sum up all the
rectangles from our initial

00:06:49.110 --> 00:06:56.000
volume, which is VA, to our
final volume, which is VB.

00:06:56.000 --> 00:06:58.780
And then, what's Q2 going
to be equal to?

00:06:58.780 --> 00:07:03.210
Well, Q2 is going to be
essentially the same thing.

00:07:03.210 --> 00:07:07.540
It's going to be the sum of the
work done by our system,

00:07:07.540 --> 00:07:09.880
which in this case is going to
be negative, because work was

00:07:09.880 --> 00:07:14.090
done to our system as we
go from here to here.

00:07:14.090 --> 00:07:14.430
Right?

00:07:14.430 --> 00:07:18.640
That's when Q2 was operating,
the heat was being taken out

00:07:18.640 --> 00:07:19.560
of the system.

00:07:19.560 --> 00:07:23.770
So we're going to go from--
where was our starting point?

00:07:23.770 --> 00:07:29.250
VC and we go to VD.

00:07:29.250 --> 00:07:31.550
Now, how can we evaluate
these integrals?

00:07:31.550 --> 00:07:34.050
Well, we've done this before
in a previous video.

00:07:34.050 --> 00:07:37.250
We use both of these
circumstances-- When we go to

00:07:37.250 --> 00:07:41.140
A to B, and we go from C
to D-- both of these

00:07:41.140 --> 00:07:44.470
circumstances occur on
isotherms, right?

00:07:44.470 --> 00:07:46.000
So the only things that
are changing are

00:07:46.000 --> 00:07:47.060
pressure and volume.

00:07:47.060 --> 00:07:49.210
Temperature is not changing.

00:07:49.210 --> 00:07:55.510
And so if we go back to our
ideal gas equation-- PV is

00:07:55.510 --> 00:08:00.110
equal to nRT-- we can just
rewrite this by dividing both

00:08:00.110 --> 00:08:06.940
sides by V, as P is equal
to nRT over V.

00:08:06.940 --> 00:08:10.620
And substitute that back in
for P, in both cases.

00:08:10.620 --> 00:08:12.410
That is P as a function of V.

00:08:12.410 --> 00:08:14.740
We now have the equation
of the curve.

00:08:14.740 --> 00:08:17.630
And we're taking the area
under it in both cases.

00:08:17.630 --> 00:08:24.620
So Q1 is equal to the integral
from VA to VB

00:08:24.620 --> 00:08:30.120
of nRT over V, dv.

00:08:30.120 --> 00:08:40.210
And Q2 is equal to the integral
from VC to VD of nRT

00:08:40.210 --> 00:08:46.190
over V, dv I'm going to do two
integrals in parallel, just so

00:08:46.190 --> 00:08:48.280
that you kind of see that we're
essentially solving the

00:08:48.280 --> 00:08:49.640
same thing.

00:08:49.640 --> 00:08:50.040
OK.

00:08:50.040 --> 00:08:51.900
So how can we solve this?

00:08:51.900 --> 00:08:53.520
Well, we know in both of
these cases, we're

00:08:53.520 --> 00:08:54.910
moving along an isotherm.

00:08:54.910 --> 00:08:57.320
That our temperatures
are constant.

00:08:57.320 --> 00:08:59.470
And actually, we know the
temperatures are.

00:08:59.470 --> 00:09:02.820
When we're moving from VA to
VB, our temperature is T1.

00:09:02.820 --> 00:09:05.140
It was kept that way
by our reservoir.

00:09:05.140 --> 00:09:08.340
When we moved from VC to VD,
our temperature was T2.

00:09:08.340 --> 00:09:10.330
It was kept that way by
our reservoir, right?

00:09:10.330 --> 00:09:13.280
T2, when we move from C to D.

00:09:13.280 --> 00:09:16.830
And T1, we move from A to B
Those were our temperatures.

00:09:16.830 --> 00:09:18.570
And they're constant.

00:09:18.570 --> 00:09:19.340
Fair enough.

00:09:19.340 --> 00:09:21.320
So we can take-- so
n is constant.

00:09:21.320 --> 00:09:23.380
R is definitely a constant.

00:09:23.380 --> 00:09:25.810
n is just a number of molecules
we have. And then

00:09:25.810 --> 00:09:27.520
our temperature is also
constant, so we can take it

00:09:27.520 --> 00:09:28.890
out of the integral.

00:09:28.890 --> 00:09:37.640
So we can rewrite Q1 is equal to
nRT1 over the integral from

00:09:37.640 --> 00:09:49.430
VA to VB times 1/V dv, and Q2 we
can write as nRT2 times the

00:09:49.430 --> 00:09:56.000
integral from VC
to VD, 1/V dv.

00:09:56.000 --> 00:09:56.560
All right.

00:09:56.560 --> 00:09:58.900
Now this integral is fairly
straightforward to evaluate.

00:09:58.900 --> 00:10:02.660
The antiderivative of 1/V
is the natural log of V.

00:10:02.660 --> 00:10:11.470
So we get Q1 is equal to nRT1
times the natural log of V

00:10:11.470 --> 00:10:16.050
evaluated at VB, minus
it evaluated at VA.

00:10:16.050 --> 00:10:19.650
And Q2-- well, let me just
solve this whole equation

00:10:19.650 --> 00:10:20.580
right here.

00:10:20.580 --> 00:10:22.320
So this is equal to what?

00:10:22.320 --> 00:10:24.710
This is equal to a natural
log of VB minus the

00:10:24.710 --> 00:10:26.160
natural log of VA.

00:10:26.160 --> 00:10:33.390
Which is the same thing as the
natural log of VB over VA

00:10:33.390 --> 00:10:36.590
times nRT1.

00:10:36.590 --> 00:10:38.850
And all that is equal to Q2.

00:10:38.850 --> 00:10:42.690
Now, the same logic, Q2, is
going to be equal to what?

00:10:42.690 --> 00:10:48.220
Q2 is going to be
equal to nRT2.

00:10:48.220 --> 00:10:50.090
Now the only difference with
this integral is where I had

00:10:50.090 --> 00:10:52.440
VB, now I have VD.

00:10:52.440 --> 00:10:53.960
Sorry.

00:10:53.960 --> 00:10:56.790
So then it becomes natural
log of VD.

00:10:56.790 --> 00:11:01.300
And where I had VA, now
I have VC, So over VC.

00:11:01.300 --> 00:11:02.290
All right.

00:11:02.290 --> 00:11:04.870
Now what was our original
question that we

00:11:04.870 --> 00:11:05.730
were dealing with?

00:11:05.730 --> 00:11:09.020
We said, this is a legitimate
state variable.

00:11:09.020 --> 00:11:12.510
If the change in this-- whatever
this value S as we go

00:11:12.510 --> 00:11:14.350
around the cycle-- is
equal to 0, that

00:11:14.350 --> 00:11:15.690
means it didn't change.

00:11:15.690 --> 00:11:19.400
So these two things, when you
sum them, have to equal 0.

00:11:19.400 --> 00:11:21.800
Q1 over T1, plus Q2 over T2.

00:11:21.800 --> 00:11:23.860
So let's add them.

00:11:23.860 --> 00:11:29.820
So Q1 over T1 is equal
to that over T1.

00:11:29.820 --> 00:11:31.150
That cancels out.

00:11:31.150 --> 00:11:35.110
Q2 over T2 is equal
to that over T2.

00:11:35.110 --> 00:11:36.730
That cancels out.

00:11:36.730 --> 00:11:41.360
So our change in our mystical
state variable, as we go

00:11:41.360 --> 00:11:45.340
around the Carnot cycle,
is equal to Q1 over T1,

00:11:45.340 --> 00:11:47.730
plus Q2 over T2.

00:11:47.730 --> 00:11:56.770
Which is equal to nR times the
natural log of VB over VA.

00:11:56.770 --> 00:11:58.760
That's that, right there.

00:11:58.760 --> 00:12:06.070
And then plus Q2 over T2, which
is just nR times the

00:12:06.070 --> 00:12:15.120
natural log of VD over V.

00:12:15.120 --> 00:12:16.310
This is VC.

00:12:16.310 --> 00:12:18.460
This is a VA here.

00:12:18.460 --> 00:12:19.360
All right.

00:12:19.360 --> 00:12:20.140
Now let's see what we can do.

00:12:20.140 --> 00:12:22.570
This is equal to--
almost there.

00:12:22.570 --> 00:12:23.720
Home stretch.

00:12:23.720 --> 00:12:24.820
nR.

00:12:24.820 --> 00:12:26.930
We can factor out an nR.

00:12:26.930 --> 00:12:30.130
And then the natural log of A
plus the natural log of B is

00:12:30.130 --> 00:12:32.080
just the same thing as the
natural log of AB.

00:12:32.080 --> 00:12:40.590
So this is equal to times the
natural log of VB over VA,

00:12:40.590 --> 00:12:45.620
times VD over VC.

00:12:45.620 --> 00:12:47.440
All right.

00:12:47.440 --> 00:12:51.280
So this is our change in our
S, state variable that we're

00:12:51.280 --> 00:12:52.660
playing with right now.

00:12:52.660 --> 00:12:53.910
Now what is this equal to?

00:12:59.890 --> 00:13:02.030
Let me think of the best
way to say this.

00:13:02.030 --> 00:13:06.030
So let's divide the numerator
and the denominator by VC.

00:13:06.030 --> 00:13:10.420
So let me take this expression
here, and divide its numerator

00:13:10.420 --> 00:13:14.590
and its denominator,
essentially, by VC over VD.

00:13:14.590 --> 00:13:16.260
Or let me multiply
its numerator and

00:13:16.260 --> 00:13:17.720
denominator VC over VD.

00:13:17.720 --> 00:13:24.260
So I can rewrite this as the
natural log of VB over VA,

00:13:24.260 --> 00:13:26.000
divided by-- right?

00:13:26.000 --> 00:13:28.330
Instead of multiplying it times
this, I can divide it by

00:13:28.330 --> 00:13:31.500
this reciprocal, VC over VD.

00:13:34.960 --> 00:13:35.980
So I just rewrote it.

00:13:35.980 --> 00:13:38.030
I just did a little bit
of fraction math.

00:13:38.030 --> 00:13:38.590
That's all it is.

00:13:38.590 --> 00:13:40.690
Instead of multiplying it times
this, I divided by its

00:13:40.690 --> 00:13:42.030
reciprocal.

00:13:42.030 --> 00:13:47.530
Now you see why the previous
video I did was done.

00:13:47.530 --> 00:13:49.610
What is this equal to?

00:13:49.610 --> 00:13:54.230
On the previous video, I showed
you that VB over VA is

00:13:54.230 --> 00:13:56.340
equal to VC over VD.

00:13:56.340 --> 00:14:00.630
We did that big convoluted hairy
proof to prove this.

00:14:00.630 --> 00:14:04.190
And now that we proved it, we
can use this to know that this

00:14:04.190 --> 00:14:06.270
quantity is equal to
this quantity.

00:14:06.270 --> 00:14:08.920
So if you divide something by
itself, they're equal to each

00:14:08.920 --> 00:14:11.770
other, this is equal to 1.

00:14:11.770 --> 00:14:15.570
If that's equal to 1, then
what's the natural log of 1?

00:14:15.570 --> 00:14:20.780
So our change in our mystical
S state variable is nR times

00:14:20.780 --> 00:14:22.280
the natural log of 1.

00:14:22.280 --> 00:14:23.200
What's the natural log of 1?

00:14:23.200 --> 00:14:25.210
E to the what power
is equal to 1?

00:14:25.210 --> 00:14:27.690
e to the 0 is equal to 1.

00:14:27.690 --> 00:14:30.760
n times R times 0-- I don't
care how big or

00:14:30.760 --> 00:14:32.220
whatever-- this is 0.

00:14:32.220 --> 00:14:33.990
So it equals 0.

00:14:33.990 --> 00:14:34.960
So there we have it.

00:14:34.960 --> 00:14:38.260
We've stumbled upon a legitimate
state variable that

00:14:38.260 --> 00:14:40.770
deals with heat.

00:14:40.770 --> 00:14:49.600
If we define change in S is
equal to the heat added to the

00:14:49.600 --> 00:14:52.810
system, divided by the
temperature at which the heat

00:14:52.810 --> 00:14:54.890
was added to the system,
this is a

00:14:54.890 --> 00:14:57.100
legitimate state variable.

00:14:57.100 --> 00:14:59.350
Now, we don't have much
intuition about what it really

00:14:59.350 --> 00:15:02.860
means at kind of a micro
state level.

00:15:02.860 --> 00:15:04.230
But at least we've
stumbled upon

00:15:04.230 --> 00:15:06.210
some property of something.

00:15:06.210 --> 00:15:10.530
If S is 10 here, and we go
around here, our change in S

00:15:10.530 --> 00:15:12.750
will be 0, S is 10 again.

00:15:12.750 --> 00:15:15.270
If S is, I don't know, let's
say S is 15 here, and we go

00:15:15.270 --> 00:15:18.170
around some crazy cycle and we
come back here, our change in

00:15:18.170 --> 00:15:19.860
S is going to be 0 again.

00:15:19.860 --> 00:15:21.420
Or sorry, it's going
to be 15 again.

00:15:21.420 --> 00:15:24.750
So we didn't have-- our change
in S will be 0, so our s

00:15:24.750 --> 00:15:26.470
itself will be 15 again.

00:15:26.470 --> 00:15:29.410
So S is a legitimate state
variable, but we don't have a

00:15:29.410 --> 00:15:31.795
good sense of what it
actually means.

00:15:34.570 --> 00:15:37.290
We'll leave that to
a future video.


Proof: Volume ratios in a carnot cycle

WEBVTT
Kind: captions
Language: en

00:00:00.940 --> 00:00:04.750
The goal in this video is to
essentially prove a pretty

00:00:04.750 --> 00:00:05.920
simple result.

00:00:05.920 --> 00:00:08.710
And that's that the ratio
between the volumes-- let me

00:00:08.710 --> 00:00:11.850
write this down-- that the ratio
between the volume at

00:00:11.850 --> 00:00:17.580
state B and the volume at state
A-- so the ratio of that

00:00:17.580 --> 00:00:21.930
volume to that volume-- is equal
to, in our Carnot cycle,

00:00:21.930 --> 00:00:25.620
is equal to the ratio between
the volume at state C.

00:00:25.620 --> 00:00:28.860
So this volume and
that volume.

00:00:28.860 --> 00:00:33.920
So volume at C to
the volume at D.

00:00:33.920 --> 00:00:36.930
So this is what I'm about
to embark to prove.

00:00:36.930 --> 00:00:39.720
A fairly simple result, that
maybe is even, if you look at

00:00:39.720 --> 00:00:42.090
this, that looks about right.

00:00:42.090 --> 00:00:43.780
So if you're happy just knowing
that, you don't have

00:00:43.780 --> 00:00:44.850
to watch the rest
of the video.

00:00:44.850 --> 00:00:47.440
But if you are curious how we
get there, I encourage you to

00:00:47.440 --> 00:00:49.400
watch it, although it gets
a little bit math-y.

00:00:49.400 --> 00:00:51.460
But I think the fun part
about it, even, it will

00:00:51.460 --> 00:00:52.420
one, satisfy you.

00:00:52.420 --> 00:00:53.090
That this is true.

00:00:53.090 --> 00:00:55.150
But the other thing is, we'll
be able to study adiabatic

00:00:55.150 --> 00:00:56.890
processes a little bit more.

00:00:56.890 --> 00:00:59.820
So just kind of launching off
of that, the whole proof

00:00:59.820 --> 00:01:03.250
revolves around to this step
right here and this step right

00:01:03.250 --> 00:01:05.850
here. when we go from D to A.

00:01:05.850 --> 00:01:10.090
So by definition, an adiabatic
process is one where there's

00:01:10.090 --> 00:01:12.060
no transfer of heat.

00:01:12.060 --> 00:01:16.530
So our heat transfer in an
adiabatic process is 0.

00:01:16.530 --> 00:01:19.010
So if we go back to our original
definitions-- let me

00:01:19.010 --> 00:01:20.650
show you that here.

00:01:20.650 --> 00:01:23.150
Right here, at the step
and this step, we have

00:01:23.150 --> 00:01:25.470
no transfer of heat.

00:01:25.470 --> 00:01:26.565
So if we go back to there.

00:01:26.565 --> 00:01:28.680
Adiabatic-- we're completely
isolated from

00:01:28.680 --> 00:01:29.550
the rest of the world.

00:01:29.550 --> 00:01:32.220
So there's nothing to transfer
heat to or from.

00:01:32.220 --> 00:01:36.660
So if we go to our definition,
almost, or our first law of

00:01:36.660 --> 00:01:42.540
thermodynamics, we know that the
change in internal energy

00:01:42.540 --> 00:01:48.500
is equal to the heat applied to
the system minus the work

00:01:48.500 --> 00:01:49.660
done by the system.

00:01:49.660 --> 00:01:51.530
And the work done by the
system is equal to the

00:01:51.530 --> 00:01:54.420
pressure of the system times
some change in volume.

00:01:54.420 --> 00:01:56.720
At least, maybe it's a very
small change in volume, while

00:01:56.720 --> 00:01:57.790
the pressure is constant.

00:01:57.790 --> 00:02:00.580
But if we're doing a
quasi-static process, we can

00:02:00.580 --> 00:02:01.670
write this.

00:02:01.670 --> 00:02:04.710
Pressure, you can view it as
kind of constant for that very

00:02:04.710 --> 00:02:06.920
small change in volume.

00:02:06.920 --> 00:02:08.300
So that's what we have there.

00:02:08.300 --> 00:02:11.165
Now, if it's adiabatic, we
know that this is 0.

00:02:14.960 --> 00:02:18.650
And if that's 0, we can add P
delta V to both sides of this

00:02:18.650 --> 00:02:22.645
equation, and we will get that--
this is only true if it

00:02:22.645 --> 00:02:27.000
were adiabatic-- that delta U,
our change in internal energy,

00:02:27.000 --> 00:02:30.530
plus our pressure times
our change in

00:02:30.530 --> 00:02:34.095
volume, is equal to 0.

00:02:34.095 --> 00:02:36.530
And let's see if we can do
this somehow, we can do

00:02:36.530 --> 00:02:39.020
something with this equation to
get to that result that I'm

00:02:39.020 --> 00:02:40.450
trying to get to.

00:02:40.450 --> 00:02:45.100
So a few videos ago, I proved
to you that U, the internal

00:02:45.100 --> 00:02:48.050
energy at any point in time--
let me write it here.

00:02:48.050 --> 00:02:54.890
The internal energy at any point
in time is equal to 3/2

00:02:54.890 --> 00:02:58.400
times n times R times T.

00:02:58.400 --> 00:03:01.440
Which is also equal
to 3/2 times PV.

00:03:01.440 --> 00:03:04.520
Now, if I have a change in
internal energy, what can

00:03:04.520 --> 00:03:05.460
change on this side?

00:03:05.460 --> 00:03:07.000
Something must have changed.

00:03:07.000 --> 00:03:08.980
Well, 3/2 can't change.

00:03:08.980 --> 00:03:09.850
n can't change.

00:03:09.850 --> 00:03:12.190
We're not going to change the
number of molecules we have.

00:03:12.190 --> 00:03:13.990
The universal gas constant
can't change.

00:03:13.990 --> 00:03:16.050
So the temperature
must change.

00:03:16.050 --> 00:03:16.930
So there you have it.

00:03:16.930 --> 00:03:21.636
You have delta U could be
rewritten as delta-- let me do

00:03:21.636 --> 00:03:26.930
it in a different color-- delta
U could be written as

00:03:26.930 --> 00:03:32.810
3/2 n times R times our
change in temperature.

00:03:32.810 --> 00:03:34.880
And that's why I keep saying
in this-- especially when

00:03:34.880 --> 00:03:36.770
we're dealing with the situation
where all of the

00:03:36.770 --> 00:03:39.440
internal energy is essentially
kinetic energy-- that if you

00:03:39.440 --> 00:03:40.925
don't have a change in
temperature, you're not going

00:03:40.925 --> 00:03:42.620
to have a change in
internal energy.

00:03:42.620 --> 00:03:44.520
Likewise, if you don't have a
change in internal energy,

00:03:44.520 --> 00:03:47.030
you're not going to have a
change in temperature.

00:03:47.030 --> 00:03:48.620
So let me put this
aside right here.

00:03:48.620 --> 00:03:50.850
I'm going to substitute
it back there.

00:03:50.850 --> 00:03:56.000
But let's see if we can do
something with this P here.

00:03:56.000 --> 00:04:00.920
Well, we'll just resort to
our ideal gas equation.

00:04:00.920 --> 00:04:03.470
Because we're dealing with an
ideal gas, we might as well.

00:04:03.470 --> 00:04:06.720
PV is equal to nRT.

00:04:06.720 --> 00:04:10.970
This should be emblazoned in
your mind, at this point.

00:04:10.970 --> 00:04:15.150
So if we want to solve
for P, we get O is

00:04:15.150 --> 00:04:19.700
equal to nRT over V.

00:04:19.700 --> 00:04:20.250
Fair enough.

00:04:20.250 --> 00:04:23.840
So let's put both of these
things aside, and substitute

00:04:23.840 --> 00:04:25.730
them into this formula.

00:04:25.730 --> 00:04:27.580
So delta U is equal
to this thing.

00:04:27.580 --> 00:04:38.910
So that means that 3/2 nR delta
T plus P-- P is this

00:04:38.910 --> 00:04:50.740
thing-- plus nRT over V times
delta V is equal to 0.

00:04:50.740 --> 00:04:51.810
Interesting.

00:04:51.810 --> 00:04:53.670
So what can we do
further here?

00:04:53.670 --> 00:04:56.080
And I'll kind of tell you where
I'm going with this.

00:04:56.080 --> 00:04:59.570
So that tells me, my change in
internal energy over a very

00:04:59.570 --> 00:05:03.660
small delta T-- this tells me
my work done by the system

00:05:03.660 --> 00:05:05.330
over a very small delta V.

00:05:05.330 --> 00:05:07.380
And we're saying that, you know,
over each little small

00:05:07.380 --> 00:05:11.160
increments, they're going
to add up to zero.

00:05:11.160 --> 00:05:13.740
So let me just go back to
the original graph.

00:05:13.740 --> 00:05:17.090
So this is over a very small
delta V right there.

00:05:17.090 --> 00:05:19.060
Let me do it in a more
vibrant color.

00:05:19.060 --> 00:05:21.560
A very small change,
as we go from there

00:05:21.560 --> 00:05:22.730
to, let's say, there.

00:05:22.730 --> 00:05:24.700
We're going to have some
change in our volume.

00:05:24.700 --> 00:05:26.590
And you don't see the
temperature here.

00:05:26.590 --> 00:05:29.030
So don't try to even imagine,
when we do the integral, that

00:05:29.030 --> 00:05:30.970
we should think of it in
some terms of area.

00:05:30.970 --> 00:05:32.410
But we're going to integrate
over the change in

00:05:32.410 --> 00:05:33.820
temperature, as well.

00:05:33.820 --> 00:05:35.180
The temperature changes
a little bit

00:05:35.180 --> 00:05:36.400
from there to there.

00:05:36.400 --> 00:05:39.540
So what I want to do--
this is, right here,

00:05:39.540 --> 00:05:40.360
over a small change.

00:05:40.360 --> 00:05:43.250
I want to integrate eventually
over all of the changes that

00:05:43.250 --> 00:05:46.470
occur during our adiabatic
process.

00:05:46.470 --> 00:05:48.650
So let's see if we can simplify
this before I break

00:05:48.650 --> 00:05:50.110
out the calculus.

00:05:50.110 --> 00:05:54.970
So if we divide both sides
by nRT, what do we get?

00:05:54.970 --> 00:06:01.130
So let's divide it by nRT,
let's divide it by nRT.

00:06:01.130 --> 00:06:04.770
And we have to do it to both
sides of the equation. nRT.

00:06:04.770 --> 00:06:08.850
Well, on this term, the n's
cancel out, the R cancels out.

00:06:08.850 --> 00:06:12.110
Over here, this nRT cancels
out with this nRT.

00:06:12.110 --> 00:06:13.950
And what are we left with?

00:06:13.950 --> 00:06:19.820
We're left with 3/2-- we have
this 1 over T left-- times 1

00:06:19.820 --> 00:06:30.130
over T delta T plus 1 over V
delta V is equal to-- well,

00:06:30.130 --> 00:06:34.650
zero divided by anything
is just equal to 0.

00:06:34.650 --> 00:06:36.910
Now we're going to integrate
over a bunch of really small

00:06:36.910 --> 00:06:40.470
delta T's and delta V's.

00:06:40.470 --> 00:06:43.250
So let me just change those to
our calculus terminology.

00:06:43.250 --> 00:06:46.740
We're going to do an infinite
sum over infinitesimally small

00:06:46.740 --> 00:06:49.410
changes in delta
T and delta V.

00:06:49.410 --> 00:07:01.030
So I'll rewrite this as 3/2 1
over T dt plus 1 over V dv is

00:07:01.030 --> 00:07:01.750
equal to 0.

00:07:01.750 --> 00:07:03.860
Remember, this just
means a very, very

00:07:03.860 --> 00:07:06.690
small change in volume.

00:07:06.690 --> 00:07:09.840
This is a very, very, very,
small change, an

00:07:09.840 --> 00:07:12.580
infinitesimally small change,
in temperature.

00:07:12.580 --> 00:07:15.540
And now I want to do the total
change in temperature.

00:07:15.540 --> 00:07:17.240
I want to integrate over the
total change in temperature

00:07:17.240 --> 00:07:18.960
and the total change
in volume.

00:07:18.960 --> 00:07:20.880
So let's do that.

00:07:20.880 --> 00:07:26.620
So I want to go from always
temperature start to

00:07:26.620 --> 00:07:27.870
temperature finish.

00:07:30.010 --> 00:07:33.060
And this will be going
from our volume

00:07:33.060 --> 00:07:39.350
start to volume finish.

00:07:39.350 --> 00:07:40.630
Fair enough.

00:07:40.630 --> 00:07:42.250
Let's do these integrals.

00:07:42.250 --> 00:07:45.790
This tends to show up a lot
in thermodynamics, these

00:07:45.790 --> 00:07:46.620
antiderivatives.

00:07:46.620 --> 00:07:49.800
The antiderivative of 1 over
T is natural log of T.

00:07:49.800 --> 00:07:57.190
So this is equal to 3/2 times
the natural log of T.

00:07:57.190 --> 00:08:00.870
We're going to evaluate it at
the final temperature and then

00:08:00.870 --> 00:08:06.370
the starting temperature, plus
the natural log-- the

00:08:06.370 --> 00:08:09.340
antiderivative of 1 over V is
just the natural log of V--

00:08:09.340 --> 00:08:17.120
plus the natural log of V,
evaluated from our final

00:08:17.120 --> 00:08:18.310
velocity, and we're going
to subtract out

00:08:18.310 --> 00:08:19.300
the starting velocity.

00:08:19.300 --> 00:08:21.400
This is just the
calculus here.

00:08:21.400 --> 00:08:23.540
And this is going to
be equal to 0.

00:08:23.540 --> 00:08:23.790
Right?

00:08:23.790 --> 00:08:27.610
I mean, we could integrate both
sides-- well, if every

00:08:27.610 --> 00:08:31.010
infinitesimal change is equal to
the sum is equal to 0, the

00:08:31.010 --> 00:08:33.039
sum of all of the infinitesimal
changes are also

00:08:33.039 --> 00:08:33.840
going to be equal to 0.

00:08:33.840 --> 00:08:37.460
So this is still equal to 0.

00:08:37.460 --> 00:08:38.640
See what we can do here.

00:08:38.640 --> 00:08:45.670
So we could rewrite this green
part as-- so it's 3/2 times

00:08:45.670 --> 00:08:50.170
the natural log of TF minus the
natural log of TS, which

00:08:50.170 --> 00:08:53.320
is just, using our log
properties, the natural log of

00:08:53.320 --> 00:08:56.900
TF over the natural log of TS.

00:08:56.900 --> 00:08:57.610
Right?

00:08:57.610 --> 00:09:00.030
When you evaluate, you get
natural log of TF minus the

00:09:00.030 --> 00:09:01.180
natural log of TS.

00:09:01.180 --> 00:09:03.160
That's the same thing as this.

00:09:03.160 --> 00:09:09.820
Plus, for the same reason, the
natural log of VF over the

00:09:09.820 --> 00:09:12.760
natural log of VS. When you
evaluate this, it's the

00:09:12.760 --> 00:09:15.890
natural log of VF minus the
natural log of VS, which can

00:09:15.890 --> 00:09:18.380
be simplified this to this,
just from our logarithmic

00:09:18.380 --> 00:09:19.560
properties.

00:09:19.560 --> 00:09:21.220
So this equals 0.

00:09:21.220 --> 00:09:23.940
And now we can-- this
coefficient out front, we can

00:09:23.940 --> 00:09:25.320
use our logarithmic
properties.

00:09:25.320 --> 00:09:27.990
Instead of putting a 3/2 natural
log of this, we can

00:09:27.990 --> 00:09:34.150
rewrite this as the natural log
of TF over TS to the 3/2.

00:09:34.150 --> 00:09:36.110
Now we can keep doing our
logarithm properties.

00:09:36.110 --> 00:09:38.350
You take the log of something
plus the log of something.

00:09:38.350 --> 00:09:40.530
That's equal to the log
of their product.

00:09:40.530 --> 00:09:45.520
So this is equal to-- I'll
switch colors-- The natural

00:09:45.520 --> 00:09:56.870
log of TF over TS to the 3/2
power, times the natural log

00:09:56.870 --> 00:10:03.090
of VF over VS. And this
is a fatiguing proof.

00:10:03.090 --> 00:10:03.830
All right.

00:10:03.830 --> 00:10:07.730
And all that is going
to be equal to 0.

00:10:07.730 --> 00:10:09.010
Now what can we say?

00:10:09.010 --> 00:10:12.040
Well, we're saying that e to the
0 power-- the natural log

00:10:12.040 --> 00:10:16.470
is log base e-- e to the 0 power
is equal to this thing.

00:10:16.470 --> 00:10:18.410
So this thing must be 1.

00:10:18.410 --> 00:10:20.830
E to the 0 power is 1.

00:10:20.830 --> 00:10:28.020
So we can say-- we're almost
there-- that TF, our final

00:10:28.020 --> 00:10:33.900
temperature over our starting
temperature to the 3/2 power,

00:10:33.900 --> 00:10:38.860
times our final volume
over our starting

00:10:38.860 --> 00:10:43.400
volume is equal to 1.

00:10:43.400 --> 00:10:47.280
Now let's take this result that
we worked reasonably hard

00:10:47.280 --> 00:10:48.530
to produce.

00:10:50.430 --> 00:10:52.940
Remember all of this, we just
said, we're dealing with an

00:10:52.940 --> 00:10:56.450
adiabatic process, and we
started from the principle of

00:10:56.450 --> 00:10:58.890
just what the definition
of internal energy is.

00:10:58.890 --> 00:11:03.150
And then we substitute it with
our PV equals nRT formulas.

00:11:03.150 --> 00:11:06.090
Although this was kind of PV--
this is internal energy at any

00:11:06.090 --> 00:11:09.290
point is equal to
3/2 times PV.

00:11:09.290 --> 00:11:11.920
And then we integrated over all
the changes, and we said,

00:11:11.920 --> 00:11:13.240
look, this is adiabatic.

00:11:13.240 --> 00:11:17.410
So the total change-- the sum
of all of our change in

00:11:17.410 --> 00:11:19.780
internal energy and work done
by the system has to be 0,

00:11:19.780 --> 00:11:22.560
then we use the property of
log to get to this result.

00:11:22.560 --> 00:11:24.850
Now let's do these for both
of these adiabatic

00:11:24.850 --> 00:11:26.890
processes over here.

00:11:26.890 --> 00:11:30.110
So the first one we could do is
this one where we go from

00:11:30.110 --> 00:11:35.690
volume B at T1 to
volume C at T2.

00:11:35.690 --> 00:11:40.370
Watch the Carnot cycle video,
if you forgot that.

00:11:40.370 --> 00:11:42.860
This was the VB All of these
things up here were at

00:11:42.860 --> 00:11:43.690
temperature 1.

00:11:43.690 --> 00:11:46.206
All of the things down here were
at temperature 2 So we're

00:11:46.206 --> 00:11:49.480
at temperature 1 up here,
and temperature 2 down

00:11:49.480 --> 00:11:50.940
here, volume C.

00:11:50.940 --> 00:11:51.840
So let's look at that.

00:11:51.840 --> 00:11:57.850
So on that right part, that
right process, our final

00:11:57.850 --> 00:11:59.880
temperature was temperature 2.

00:11:59.880 --> 00:12:00.900
So let me write it down.

00:12:00.900 --> 00:12:02.710
Temperature 2.

00:12:02.710 --> 00:12:04.875
Our initial temperature was
temperature 1, where we

00:12:04.875 --> 00:12:09.110
started off at point
B to the 3/2.

00:12:09.110 --> 00:12:11.660
Times-- what was our
final volume?

00:12:11.660 --> 00:12:15.040
Our final volume was our volume
at C divided by the

00:12:15.040 --> 00:12:16.680
volume at B.

00:12:16.680 --> 00:12:19.460
And that's going to
be equal to 1.

00:12:19.460 --> 00:12:20.600
Neat.

00:12:20.600 --> 00:12:23.260
That's the result we got from
this adiabatic process.

00:12:23.260 --> 00:12:25.370
We got that formula saying, this
is adiabatic, we did a

00:12:25.370 --> 00:12:27.380
bunch of math, and then we
just substitute for our

00:12:27.380 --> 00:12:30.990
initial and final volumes
and temperatures.

00:12:30.990 --> 00:12:35.310
Let's do it the same way, but
let's go from D to A.

00:12:35.310 --> 00:12:42.760
So when you go from D to A,
what's your final temperature?

00:12:42.760 --> 00:12:44.330
Don't want to get you
dizzy going up.

00:12:44.330 --> 00:12:46.320
Well your final temperature,
we're going from D to A.

00:12:46.320 --> 00:12:48.150
So our final temperature
is T1.

00:12:48.150 --> 00:12:51.940
and our final volume
is the volume at A.

00:12:51.940 --> 00:12:53.480
Go back down.

00:12:53.480 --> 00:12:56.580
So our final temperature
is T1.

00:12:56.580 --> 00:12:58.690
Our initial temperature is T2.

00:12:58.690 --> 00:13:06.430
We're going from D to A to the
3/2 power is times-- let me

00:13:06.430 --> 00:13:07.520
write our form formula there.

00:13:07.520 --> 00:13:10.250
Our final volume is
the volume at A.

00:13:10.250 --> 00:13:12.630
That's where we moved to.

00:13:12.630 --> 00:13:14.380
And we moved from
our volume at D.

00:13:14.380 --> 00:13:16.580
And this is going to
be equal to 1.

00:13:16.580 --> 00:13:16.990
OK.

00:13:16.990 --> 00:13:20.110
We're almost there,
if your eyes are

00:13:20.110 --> 00:13:21.310
beginning to glaze over.

00:13:21.310 --> 00:13:22.280
But this is interesting.

00:13:22.280 --> 00:13:25.490
And if anything, it's a little
bit of fun mathematics to wake

00:13:25.490 --> 00:13:26.760
you up in the morning.

00:13:26.760 --> 00:13:27.910
So let's see.

00:13:27.910 --> 00:13:29.890
We can almost relate
these two things.

00:13:29.890 --> 00:13:31.880
We could set them equal to each
other, but it's not quite

00:13:31.880 --> 00:13:33.690
satisfying yet.

00:13:33.690 --> 00:13:38.670
Let's take the reciprocal of
both sides of this equation

00:13:38.670 --> 00:13:39.710
right here.

00:13:39.710 --> 00:13:44.890
So obviously if we take the
reciprocal of this-- and we

00:13:44.890 --> 00:13:47.990
could just say, this is T2
over T1 to the minus 3/2

00:13:47.990 --> 00:13:53.670
power, which is the same
thing as T1 over

00:13:53.670 --> 00:13:56.580
T2, to the 3/2 power.

00:13:56.580 --> 00:13:56.780
Right?

00:13:56.780 --> 00:13:58.000
That's just the reciprocal.

00:13:58.000 --> 00:14:00.540
And we're taking both sides to
the negative 1 power, so we're

00:14:00.540 --> 00:14:01.990
going to have to take this
to the negative 1 power.

00:14:01.990 --> 00:14:04.635
VB over VC.

00:14:04.635 --> 00:14:06.390
And when you take the reciprocal
of 1, that just

00:14:06.390 --> 00:14:09.530
equals 1, That still equals 1.

00:14:09.530 --> 00:14:12.230
Which this also equals, so we
could say, that equals this

00:14:12.230 --> 00:14:13.560
thing over here.

00:14:13.560 --> 00:14:19.550
So that is equal to T1 over
T2, to the 3/2 power,

00:14:19.550 --> 00:14:24.180
times VA over VD.

00:14:24.180 --> 00:14:25.600
Now, these things are
equal to each other.

00:14:25.600 --> 00:14:26.500
We can get rid of the 1.

00:14:26.500 --> 00:14:30.120
These two-- actually, let me
just erase some of this.

00:14:30.120 --> 00:14:31.580
I don't want to make it
say not equal to.

00:14:31.580 --> 00:14:32.540
They're equal to each other.

00:14:32.540 --> 00:14:34.980
They both equal 1.

00:14:34.980 --> 00:14:36.960
So they both equal each other.

00:14:40.070 --> 00:14:41.960
This thing and this thing
are the same thing.

00:14:41.960 --> 00:14:44.700
T1 over T2 to the 3/2, T1
over T2 to the 3/2.

00:14:44.700 --> 00:14:47.370
So let's just divide
both sides by that.

00:14:47.370 --> 00:14:48.750
Those cancel out.

00:14:48.750 --> 00:14:50.190
And what are we left with?

00:14:50.190 --> 00:14:52.380
I think you can see
the finish line.

00:14:52.380 --> 00:14:54.140
The finish line is near.

00:14:54.140 --> 00:15:02.730
We have VB over VC is
equal to VA over VD.

00:15:02.730 --> 00:15:04.740
Now that's not quite the result
we wanted, but it takes

00:15:04.740 --> 00:15:07.300
a little bit of simple
arithmetic to get there.

00:15:07.300 --> 00:15:09.240
Let's just cross-multiply.

00:15:09.240 --> 00:15:18.160
And you get VB times VD is
equal to VC times VA.

00:15:18.160 --> 00:15:30.710
Now if we divide both sides by
VBVC-- actually, let's do it

00:15:30.710 --> 00:15:31.070
the other way.

00:15:31.070 --> 00:15:32.350
Let's divide both
sides by VDVA.

00:15:40.010 --> 00:15:41.340
What do we get?

00:15:41.340 --> 00:15:44.780
These cancel out, and
these cancel out.

00:15:44.780 --> 00:15:53.870
And we are left with VB over
VA is equal to VC over VD.

00:15:53.870 --> 00:15:56.800
All that work for a nice and
simple result, but that's

00:15:56.800 --> 00:15:59.620
better than doing a lot
of work for a hairy

00:15:59.620 --> 00:16:01.140
and monstrous result.

00:16:01.140 --> 00:16:03.440
So that's what we set
out to prove.

00:16:03.440 --> 00:16:06.960
That VB over VA is equal
to VC over VD.

00:16:06.960 --> 00:16:09.190
And we got it all from the
notion that we're dealing with

00:16:09.190 --> 00:16:13.220
adiabatic process, that our
change in heat is 0.

00:16:13.220 --> 00:16:16.080
and we just went to our formula,
or our definition of

00:16:16.080 --> 00:16:18.260
our change in internal energy,
the first law of

00:16:18.260 --> 00:16:21.130
thermodynamics, that if we have
any change in internal

00:16:21.130 --> 00:16:23.450
energy, it must be equal
to the amount of

00:16:23.450 --> 00:16:25.470
work done by the system.

00:16:25.470 --> 00:16:27.880
Or at least a negative of the
work done by the system.

00:16:27.880 --> 00:16:29.720
When you add them up,
you get to 0.

00:16:29.720 --> 00:16:33.120
Then we use that result from a
few videos ago, where we said

00:16:33.120 --> 00:16:36.660
the internal energy at any
point is 3/2 times nRT.

00:16:36.660 --> 00:16:39.490
So the change in internal energy
is that times delta T,

00:16:39.490 --> 00:16:41.210
because that's the only
thing that can change.

00:16:41.210 --> 00:16:42.890
We used PV equal nRT.

00:16:42.890 --> 00:16:45.430
And then we just integrated
along all of the little

00:16:45.430 --> 00:16:47.290
changes in temperature
and volume, as we

00:16:47.290 --> 00:16:49.030
moved along this line.

00:16:49.030 --> 00:16:51.700
As we moved along the line,
we took the integral.

00:16:51.700 --> 00:16:54.110
We said that had to
be equal to 0.

00:16:54.110 --> 00:16:56.110
And we ended up with this
formula over here, and then we

00:16:56.110 --> 00:16:58.010
just applied it to our two
adiabatic processes.

00:16:58.010 --> 00:17:01.580
And we went from B to C, and
we went from D to A.

00:17:01.580 --> 00:17:02.695
And we got these results.

00:17:02.695 --> 00:17:05.849
And we got to our finish line.

00:17:05.849 --> 00:17:07.619
See you in the next video.