Carnot cycle and Carnot engine

WEBVTT
Kind: captions
Language: en

00:00:00.690 --> 00:00:03.580
Let's start with this classic
system that I keep referring

00:00:03.580 --> 00:00:05.210
to in our thermodynamics
videos.

00:00:05.210 --> 00:00:06.910
I have a cylinder.

00:00:06.910 --> 00:00:10.240
It's got a little piston on the
top of it, or it's got a

00:00:10.240 --> 00:00:12.110
ceiling that's movable.

00:00:12.110 --> 00:00:16.890
The gas, and we're thinking of
monoatomic ideal gases in

00:00:16.890 --> 00:00:21.260
here, they're exerting pressure
onto this ceiling.

00:00:21.260 --> 00:00:23.220
And the reason why the ceiling
isn't moving all the way up,

00:00:23.220 --> 00:00:26.270
is because I've placed a bunch
of rocks on the top to offset

00:00:26.270 --> 00:00:29.520
the force per area of
the actual gas.

00:00:29.520 --> 00:00:32.229
And I start this gas when
it's in equilibrium.

00:00:32.229 --> 00:00:33.700
I can define its macrostates.

00:00:33.700 --> 00:00:35.030
It has some volume.

00:00:35.030 --> 00:00:37.415
It has some pressure that's
being offset by these rocks.

00:00:37.415 --> 00:00:40.140
And it has some well-defined
temperature.

00:00:40.140 --> 00:00:43.890
Now, what I'm going to do is,
I'm going to place this system

00:00:43.890 --> 00:00:46.830
here-- I'm going to place it
on top of a reservoir.

00:00:46.830 --> 00:00:49.280
And I talked about what a
reservoir was either in the

00:00:49.280 --> 00:00:51.040
last video or a couple
of videos ago.

00:00:51.040 --> 00:00:55.720
You can view it as an infinitely
large object, if

00:00:55.720 --> 00:00:57.460
you will, of a certain
temperature.

00:00:57.460 --> 00:01:00.850
So if I put it next to-- if I
put our system next to this

00:01:00.850 --> 00:01:03.930
reservoir-- and let's say I
start removing pebbles from

00:01:03.930 --> 00:01:06.290
our system.

00:01:06.290 --> 00:01:09.060
We learned a couple of videos
ago that if we did it

00:01:09.060 --> 00:01:11.270
adiabatically-- what does
adiabatically mean?

00:01:11.270 --> 00:01:14.230
If we removed these pebbles
in isolation, without any

00:01:14.230 --> 00:01:17.760
reservoir around, the volume
would increase, the pressure

00:01:17.760 --> 00:01:19.700
would go down, and actually
the temperature would

00:01:19.700 --> 00:01:20.420
decrease, as well.

00:01:20.420 --> 00:01:23.080
We showed that a couple
of videos ago.

00:01:23.080 --> 00:01:26.330
So by putting this big reservoir
there that's a lot

00:01:26.330 --> 00:01:29.480
larger than our actual canister,
this will keep the

00:01:29.480 --> 00:01:33.000
temperature in our
canister at T1.

00:01:33.000 --> 00:01:40.530
You can kind of view a reservoir
as-- say I had a cup

00:01:40.530 --> 00:01:42.230
of water in a stadium.

00:01:42.230 --> 00:01:46.710
And the air conditioner in the
stadium is at 60 degrees.

00:01:46.710 --> 00:01:49.880
Well, no matter what I do to
that water, I could put it in

00:01:49.880 --> 00:01:52.140
the microwave and warm it up,
but if I put it back in that

00:01:52.140 --> 00:01:53.850
stadium, that stadium
is going to keep

00:01:53.850 --> 00:01:55.480
that water at 60 degrees.

00:01:55.480 --> 00:01:58.610
And you might say, oh, won't
the reservoir's temperature

00:01:58.610 --> 00:02:00.650
decrease if it's throwing
off heat?

00:02:00.650 --> 00:02:03.980
Well, it would, but it's so much
larger that its impact

00:02:03.980 --> 00:02:04.680
isn't noticeable.

00:02:04.680 --> 00:02:09.690
For example, if I put a cup of
boiling water into a super

00:02:09.690 --> 00:02:14.360
large covered-dome stadium, the
water will get colder to

00:02:14.360 --> 00:02:16.140
the ambient temperature
of the stadium.

00:02:16.140 --> 00:02:18.770
The stadium will get warmer, but
it will be so marginally

00:02:18.770 --> 00:02:20.560
warmer than you won't
even notice it.

00:02:20.560 --> 00:02:22.200
So you can kind of view
that as a reservoir.

00:02:22.200 --> 00:02:24.450
And theoretically, this
is infinitely large.

00:02:24.450 --> 00:02:27.870
So the effect of this is, as we
remove these little rocks,

00:02:27.870 --> 00:02:30.390
we're going to keep the
temperature constant.

00:02:30.390 --> 00:02:32.030
And remember, if we're keeping
the temperature constant,

00:02:32.030 --> 00:02:34.090
we're also keeping the internal
energy constant,

00:02:34.090 --> 00:02:35.130
because we're not changing
the kinetic

00:02:35.130 --> 00:02:36.550
energy of the particles.

00:02:36.550 --> 00:02:39.130
So let me see what happens.

00:02:39.130 --> 00:02:41.320
So I keep doing that.

00:02:41.320 --> 00:02:47.220
And so I get to a point-- let
me see-- where my volume has

00:02:47.220 --> 00:02:52.660
increased-- so let me delete
some of my rocks here.

00:02:52.660 --> 00:02:54.600
Delete some of the rocks.

00:02:54.600 --> 00:02:57.370
So some of the rocks are gone.

00:02:57.370 --> 00:03:02.230
And now my overall volume
is going to be larger.

00:03:02.230 --> 00:03:04.980
Let me move this up
a little bit.

00:03:04.980 --> 00:03:08.190
And then let me color
this in black.

00:03:08.190 --> 00:03:09.670
Oh, whoops.

00:03:09.670 --> 00:03:14.670
Let me color this in, just
to give an idea.

00:03:14.670 --> 00:03:17.632
So our volume has gotten
a bit larger.

00:03:17.632 --> 00:03:21.320
And let me get my
pen correctly.

00:03:21.320 --> 00:03:25.620
So our volume has gotten larger
by roughly this amount.

00:03:25.620 --> 00:03:27.550
We have the same number
of particles.

00:03:27.550 --> 00:03:29.930
They're going to bump into
the ceiling a little less

00:03:29.930 --> 00:03:32.200
frequently, so my pressure
would have gone down.

00:03:32.200 --> 00:03:35.390
But because I kept this
reservoir here, because this

00:03:35.390 --> 00:03:40.160
reservoir was here the whole
time during this process, the

00:03:40.160 --> 00:03:42.210
temperature stayed at T1.

00:03:42.210 --> 00:03:45.860
And that was only because
of this reservoir.

00:03:45.860 --> 00:03:47.000
And I want to make that clear.

00:03:47.000 --> 00:03:50.410
And also, just as review, this
is a quasi-static process,

00:03:50.410 --> 00:03:51.680
because I'm doing
it very slowly.

00:03:51.680 --> 00:03:54.340
The system is in equilibrium
the whole time.

00:03:54.340 --> 00:04:02.270
So let's draw what we have so
far on our famous PV diagram.

00:04:02.270 --> 00:04:08.570
So this is the P-axis.

00:04:08.570 --> 00:04:09.820
That's the V-axis.

00:04:12.260 --> 00:04:14.600
You label them.

00:04:14.600 --> 00:04:15.840
This is P.

00:04:15.840 --> 00:04:17.040
This is V.

00:04:17.040 --> 00:04:19.200
Let me call this-- I'm going
to do it in a good color.

00:04:19.200 --> 00:04:21.980
This is state A of the system.

00:04:21.980 --> 00:04:23.940
This is state B of the system.

00:04:23.940 --> 00:04:27.530
So state A starts at some
pressure and volume-- I'll do

00:04:27.530 --> 00:04:29.960
it like that.

00:04:29.960 --> 00:04:31.450
That's state A.

00:04:31.450 --> 00:04:33.150
And it moves to state B.

00:04:33.150 --> 00:04:35.030
And notice, I kept the
temperature constant.

00:04:35.030 --> 00:04:36.910
And what did we learn in,
I think it was one

00:04:36.910 --> 00:04:38.250
or two videos ago?

00:04:38.250 --> 00:04:40.020
Well, we're at a constant
temperature, so we're going to

00:04:40.020 --> 00:04:42.170
move along an isotherm,
which is just

00:04:42.170 --> 00:04:45.360
a rectangular hyperbola.

00:04:45.360 --> 00:04:48.310
Because when your temperature
is constant, your pressure

00:04:48.310 --> 00:04:50.600
times your volume is going to
equal a constant number.

00:04:50.600 --> 00:04:51.530
And I went over that before.

00:04:51.530 --> 00:04:53.980
So we're going to move over--
our path is going to look

00:04:53.980 --> 00:04:59.380
something like this, and I'll
move here, to state B.

00:04:59.380 --> 00:05:01.590
I'll move over here
to state B.

00:05:01.590 --> 00:05:04.800
And the whole time, this was at
a constant temperature T1.

00:05:09.060 --> 00:05:10.660
Now, we've done a bunch
of videos now.

00:05:10.660 --> 00:05:13.470
We said, OK, how much work
was done on this system?

00:05:13.470 --> 00:05:15.160
Well, the work done on
the system is the

00:05:15.160 --> 00:05:16.160
area under this curve.

00:05:16.160 --> 00:05:19.680
So some positive work was-- not
done on the system, sorry.

00:05:19.680 --> 00:05:21.340
How much work was done
by the system?

00:05:21.340 --> 00:05:22.890
We're moving in this
direction.

00:05:22.890 --> 00:05:24.960
I should put the direction
there.

00:05:24.960 --> 00:05:26.980
We're moving from
left to right.

00:05:26.980 --> 00:05:28.900
The amount of work done
by the system is

00:05:28.900 --> 00:05:30.500
pressure times volume.

00:05:30.500 --> 00:05:32.000
We've seen that multiple
times.

00:05:32.000 --> 00:05:34.180
So you take this area of the
curve, and you have the work

00:05:34.180 --> 00:05:36.680
done by the system
from A to B.

00:05:36.680 --> 00:05:36.970
Right?

00:05:36.970 --> 00:05:39.590
So let's call that
work from A to B.

00:05:42.650 --> 00:05:45.790
Now, that's fair and everything,
but what I want to

00:05:45.790 --> 00:05:47.450
think more about, is
how much heat was

00:05:47.450 --> 00:05:48.910
transferred by my reservoir?

00:05:48.910 --> 00:05:51.550
Remember, we said, if this
reservoir wasn't there, the

00:05:51.550 --> 00:05:54.210
temperature of my canister
would have gone down as I

00:05:54.210 --> 00:05:56.910
expanded its volume, and as
the pressure went down.

00:05:56.910 --> 00:05:58.760
So how much heat came into it?

00:05:58.760 --> 00:06:01.970
Well, let's go back to our basic
internal energy formula.

00:06:01.970 --> 00:06:07.020
Change in internal energy is
equal to heat applied to the

00:06:07.020 --> 00:06:10.840
system minus the work
done by the system

00:06:10.840 --> 00:06:12.360
Now, what is the change
in internal

00:06:12.360 --> 00:06:14.120
energy in this scenario?

00:06:14.120 --> 00:06:15.500
Well, it was at a constant
temperature

00:06:15.500 --> 00:06:17.520
the whole time, right?

00:06:17.520 --> 00:06:21.330
And since we're dealing with a
very simple ideal gas, all of

00:06:21.330 --> 00:06:23.340
our internal energy is due
to kinetic energy, which

00:06:23.340 --> 00:06:24.640
temperature is a measure of.

00:06:24.640 --> 00:06:26.010
So, temperature didn't change.

00:06:26.010 --> 00:06:28.130
Our average kinetic energy
didn't change, which means our

00:06:28.130 --> 00:06:29.680
kinetic energy didn't change.

00:06:29.680 --> 00:06:36.440
So our internal energy did not
change while we moved from

00:06:36.440 --> 00:06:38.490
left to right along
this isotherm.

00:06:38.490 --> 00:06:40.970
So we could say our internal
energy is zero.

00:06:40.970 --> 00:06:45.180
And that is equal to the heat
added to the system minus the

00:06:45.180 --> 00:06:47.610
work done by the system.

00:06:47.610 --> 00:06:48.340
Right?

00:06:48.340 --> 00:06:51.996
So if you just-- we put the work
done by the system on the

00:06:51.996 --> 00:06:55.250
other side, and then switch the
sides, you get heat added

00:06:55.250 --> 00:06:58.940
to the system is equal to the
work done by the system.

00:06:58.940 --> 00:06:59.740
And that makes sense.

00:06:59.740 --> 00:07:03.340
The system was doing some work
this entire time, so it was

00:07:03.340 --> 00:07:07.360
giving energy to-- well,
you know, it was giving

00:07:07.360 --> 00:07:08.530
essentially maybe
some potential

00:07:08.530 --> 00:07:10.370
energy to these rocks.

00:07:10.370 --> 00:07:11.910
So it was giving energy away.

00:07:11.910 --> 00:07:14.340
It was giving energy outside
of the system.

00:07:14.340 --> 00:07:17.650
So how did it maintain
its internal energy?

00:07:17.650 --> 00:07:19.530
Well, someone had to give
it some energy.

00:07:19.530 --> 00:07:24.610
And it was given that energy
by this reservoir.

00:07:24.610 --> 00:07:27.350
So let's say, and the convention
for doing this is

00:07:27.350 --> 00:07:30.400
to say, that it was given--
let me write this down.

00:07:30.400 --> 00:07:32.820
It was given some energy Q1.

00:07:32.820 --> 00:07:34.830
We just say, we just put this
downward arrow to say that

00:07:34.830 --> 00:07:37.610
some energy went into
the system here.

00:07:37.610 --> 00:07:38.820
Fair enough.

00:07:38.820 --> 00:07:46.050
Now let's take this state B and
remove the reservoir, and

00:07:46.050 --> 00:07:47.700
completely isolate ourselves.

00:07:47.700 --> 00:07:50.400
So there's no way that heat can
be transferred to and from

00:07:50.400 --> 00:07:51.150
our system.

00:07:51.150 --> 00:07:53.650
And let's keep removing
some rocks.

00:07:53.650 --> 00:07:56.375
So if we keep removing some
rocks, where do we get to?

00:07:56.375 --> 00:07:59.820
Let me go down here.

00:07:59.820 --> 00:08:03.170
So let's say we remove a
bunch of more rocks.

00:08:03.170 --> 00:08:07.300
So let me erase even more
rocks than we had in B.

00:08:07.300 --> 00:08:08.870
Maybe I only have
one rock left.

00:08:11.880 --> 00:08:15.690
And obviously, the overall
volume would have increased.

00:08:15.690 --> 00:08:20.810
So let me make our piston go up
like that, and I can make

00:08:20.810 --> 00:08:24.650
our piston is maybe
a lot higher now.

00:08:24.650 --> 00:08:28.720
And let me just fill in the rest
of our, just so that we

00:08:28.720 --> 00:08:32.100
don't have some empty
space there.

00:08:32.100 --> 00:08:37.520
So if I fill that in right
there-- OK Let

00:08:37.520 --> 00:08:39.510
me fill that in.

00:08:39.510 --> 00:08:43.510
And then I just use the blue--
I should be talking about

00:08:43.510 --> 00:08:44.870
thermodynamics, not drawing.

00:08:44.870 --> 00:08:46.520
But you get the idea.

00:08:46.520 --> 00:08:48.420
And then I have some
more-- you know, I

00:08:48.420 --> 00:08:49.670
shouldn't add particles.

00:08:49.670 --> 00:08:52.480
But my volume has increased
a good bit.

00:08:52.480 --> 00:08:56.200
My pressure will have gone down,
they're going to bump

00:08:56.200 --> 00:08:57.610
into the walls less.

00:08:57.610 --> 00:09:02.290
And because I removed the
reservoir, what's going to

00:09:02.290 --> 00:09:04.430
happen to the temperature?

00:09:04.430 --> 00:09:06.910
My temperature is going
to go down.

00:09:06.910 --> 00:09:09.000
This was an adiabatic process.

00:09:09.000 --> 00:09:12.050
So an adiabatic just means
we did it in isolation.

00:09:12.050 --> 00:09:16.020
There was no exchange of heat
from one system to another.

00:09:16.020 --> 00:09:19.150
So let me just-- this arrow
continues down here.

00:09:19.150 --> 00:09:20.400
I'll say adiabatic.

00:09:24.020 --> 00:09:26.040
Now, since I'm moving from
one temperature to

00:09:26.040 --> 00:09:27.960
another, this is at T2.

00:09:30.540 --> 00:09:32.920
So I will have moved to
another isotherm.

00:09:32.920 --> 00:09:34.550
This is the isotherm for T1.

00:09:34.550 --> 00:09:36.350
If I keep my temperature
constant, I

00:09:36.350 --> 00:09:38.380
move along this hyperbola.

00:09:38.380 --> 00:09:40.560
And I would have kept moving
along this hyperbola.

00:09:40.560 --> 00:09:43.820
But now that we didn't keep our
temperature constant, we

00:09:43.820 --> 00:09:45.250
now move like this.

00:09:45.250 --> 00:09:47.660
We move to another isotherm.

00:09:47.660 --> 00:09:49.660
So let's say I have another
isotherm at T2.

00:09:49.660 --> 00:09:51.990
It looks something like this.

00:09:51.990 --> 00:09:53.410
So let me draw like that.

00:09:53.410 --> 00:09:58.850
So let's say I have another-- it
should actually curve up a

00:09:58.850 --> 00:09:59.260
little bit.

00:09:59.260 --> 00:10:02.430
So let's say, everything at
temperature T2, depending on

00:10:02.430 --> 00:10:04.450
its pressure and volume, is
someplace along this curve

00:10:04.450 --> 00:10:06.530
that asymptotes up like
that, and then goes to

00:10:06.530 --> 00:10:07.760
the right like that.

00:10:07.760 --> 00:10:10.940
Now, I would have moved down
to this isotherm, and my

00:10:10.940 --> 00:10:13.400
pressure would have kept going
down, and my volume would have

00:10:13.400 --> 00:10:14.560
kept going down.

00:10:14.560 --> 00:10:20.290
So this move, from B to state
C, will look like this.

00:10:23.500 --> 00:10:24.760
Let me do to it in
another color.

00:10:24.760 --> 00:10:27.190
Let me do it in the orange
color of this arrow.

00:10:27.190 --> 00:10:30.000
So it will look like this.

00:10:30.000 --> 00:10:33.540
And now we're at state C.

00:10:33.540 --> 00:10:35.200
Now, this was adiabatic.

00:10:38.360 --> 00:10:40.070
Which means, there is
no exchange of heat.

00:10:40.070 --> 00:10:44.010
So I don't have to figure out
how much heat got transferred

00:10:44.010 --> 00:10:45.060
into the system.

00:10:45.060 --> 00:10:46.520
Now, there's something
interesting here.

00:10:46.520 --> 00:10:50.650
We still did do some work.

00:10:50.650 --> 00:10:52.670
We can take the area
under this curve.

00:10:52.670 --> 00:10:54.950
And we're going to leave it to
a future video to think about

00:10:54.950 --> 00:11:01.010
where that work energy-- well,
the main thing is, is what was

00:11:01.010 --> 00:11:02.880
reduced by that work energy.

00:11:02.880 --> 00:11:05.380
And, well, if you think
[UNINTELLIGIBLE]

00:11:05.380 --> 00:11:06.510
to leave it to future video.

00:11:06.510 --> 00:11:09.180
Our internal energy was
reduced, right?

00:11:09.180 --> 00:11:10.580
Because our temperature
went down.

00:11:10.580 --> 00:11:12.120
So our internal energy
went down.

00:11:12.120 --> 00:11:14.250
We'll talk more about that
in the future video.

00:11:14.250 --> 00:11:18.720
So now that we're at state C,
and we're at temperature T2.

00:11:18.720 --> 00:11:21.670
Let's put back another
sink here.

00:11:25.000 --> 00:11:31.230
But this sink, what it's going
to have is a reservoir.

00:11:31.230 --> 00:11:33.150
So let me put two things
right here.

00:11:36.700 --> 00:11:39.650
So I'm going to add--
let me erase some of

00:11:39.650 --> 00:11:43.190
these blocks in black.

00:11:45.890 --> 00:11:47.695
So now I'm going to
add blocks back.

00:11:50.210 --> 00:11:53.600
I'm going to add little
pebbles back into it.

00:11:53.600 --> 00:11:56.830
But I'm going to do it as
an isothermic process.

00:11:56.830 --> 00:12:00.220
I'm going to do it with
a reservoir here.

00:12:00.220 --> 00:12:02.520
But this reservoir here, it's
not going to be the same

00:12:02.520 --> 00:12:04.170
reservoir that I put up there.

00:12:04.170 --> 00:12:05.290
I swapped that one out.

00:12:05.290 --> 00:12:07.910
I got rid of any reservoir
when I went from B to C.

00:12:07.910 --> 00:12:09.730
And now I'm going to swap
in a new reservoir.

00:12:09.730 --> 00:12:13.780
Actually, let me make it blue.

00:12:13.780 --> 00:12:15.010
Because it's going to be--

00:12:15.010 --> 00:12:16.180
Because here's what's
happening.

00:12:16.180 --> 00:12:17.450
I'm now adding pebbles in.

00:12:17.450 --> 00:12:20.210
I'm compressing the gas.

00:12:20.210 --> 00:12:22.680
If this was an adiabatic
process, the gas would

00:12:22.680 --> 00:12:23.960
want to heat up.

00:12:23.960 --> 00:12:26.730
So what I'm doing is, I need to
put a reservoir to keep it

00:12:26.730 --> 00:12:29.750
at T2, to keep it along
this isotherm.

00:12:29.750 --> 00:12:31.840
So this is T2.

00:12:31.840 --> 00:12:34.320
Remember, this reservoir is
kind of a cold reservoir.

00:12:34.320 --> 00:12:36.200
It keeps the temperature down.

00:12:36.200 --> 00:12:37.520
As opposed to here.

00:12:37.520 --> 00:12:38.575
This was a hot reservoir.

00:12:38.575 --> 00:12:40.420
It kept the temperature up.

00:12:40.420 --> 00:12:42.370
So you can imagine.

00:12:42.370 --> 00:12:45.160
The heat generated in the
system, or internal energy

00:12:45.160 --> 00:12:47.490
being generated in the
system-- well, no, I

00:12:47.490 --> 00:12:48.590
shouldn't say that.

00:12:48.590 --> 00:12:52.140
The temperature of the system
will want to go up, but it's

00:12:52.140 --> 00:12:55.230
being released, because it's
able to transfer that heat

00:12:55.230 --> 00:12:57.550
into our new reservoir.

00:12:57.550 --> 00:12:58.910
And that amount of heat is Q2.

00:13:01.920 --> 00:13:03.970
So I move along this.

00:13:03.970 --> 00:13:04.700
This is right here.

00:13:04.700 --> 00:13:08.040
I'm moving along another
isotherm, I'm moving along

00:13:08.040 --> 00:13:09.250
this isotherm.

00:13:09.250 --> 00:13:14.580
Until I get to state D.

00:13:14.580 --> 00:13:17.420
We're almost there.

00:13:17.420 --> 00:13:21.040
This is state D.

00:13:21.040 --> 00:13:24.720
So state D will be someplace
here, along this isotherm

00:13:24.720 --> 00:13:25.670
right here.

00:13:25.670 --> 00:13:27.590
Maybe this is state D.

00:13:27.590 --> 00:13:30.720
And once again, you can make
the argument that we moved

00:13:30.720 --> 00:13:33.400
along an isotherm Our
temperature did not change

00:13:33.400 --> 00:13:35.040
from C to D.

00:13:35.040 --> 00:13:38.600
We know that our internal energy
went down from B to C,

00:13:38.600 --> 00:13:39.950
because we did some work.

00:13:39.950 --> 00:13:42.620
But from C to D, our temperature
stayed the same.

00:13:42.620 --> 00:13:47.290
It was at temperature-- let me
write it down-- T2, right?

00:13:47.290 --> 00:13:49.520
Because we had this
reservoir here.

00:13:49.520 --> 00:13:50.460
It stayed the same.

00:13:50.460 --> 00:13:52.440
If your temperature stays the
same, then your internal

00:13:52.440 --> 00:13:53.660
energy stays the same.

00:13:53.660 --> 00:13:56.100
At least for the system we're
dealing with, because it's a

00:13:56.100 --> 00:13:57.080
very simple gas.

00:13:57.080 --> 00:13:58.710
It's actually the system you'll
deal with most of the

00:13:58.710 --> 00:14:01.960
time, in an intro thermodynamics
course.

00:14:01.960 --> 00:14:02.550
So.

00:14:02.550 --> 00:14:04.640
Given our internal energy didn't
change, we can apply

00:14:04.640 --> 00:14:10.490
the same argument that the heat
added to the system is

00:14:10.490 --> 00:14:13.230
equal to the work done
by the system.

00:14:13.230 --> 00:14:13.650
Right?

00:14:13.650 --> 00:14:15.580
Same math as we did up here.

00:14:15.580 --> 00:14:19.390
Now, in this case, the work
wasn't done by the system.

00:14:19.390 --> 00:14:20.820
The work was done
to the system.

00:14:20.820 --> 00:14:23.370
We compressed this piston.

00:14:23.370 --> 00:14:25.780
The force times distance
went the other way.

00:14:25.780 --> 00:14:29.660
So given that work was done to
the system, the heat added to

00:14:29.660 --> 00:14:31.260
the system was negative,
right?

00:14:31.260 --> 00:14:34.190
We're just applying
the same thing.

00:14:34.190 --> 00:14:37.910
If our internal energy is 0, the
heat added to the system

00:14:37.910 --> 00:14:40.250
is equal to the work
done by the system.

00:14:40.250 --> 00:14:42.260
The work done by the
system is negative.

00:14:42.260 --> 00:14:44.110
Work was done to it.

00:14:44.110 --> 00:14:47.600
So the heat added to the system
would be negative.

00:14:47.600 --> 00:14:49.330
Or another way to think
about it is that the

00:14:49.330 --> 00:14:51.030
system gave away heat.

00:14:53.950 --> 00:14:57.430
We put that with Q2.

00:14:57.430 --> 00:14:58.515
And where did it
give that heat?

00:14:58.515 --> 00:15:01.490
It gave it to this reservoir
that we put here, this kind of

00:15:01.490 --> 00:15:02.510
cold reservoir.

00:15:02.510 --> 00:15:05.210
You could almost view
it as a-- well, it's

00:15:05.210 --> 00:15:06.750
accepting the heat.

00:15:06.750 --> 00:15:06.890
OK.

00:15:06.890 --> 00:15:08.360
We're almost there.

00:15:08.360 --> 00:15:11.390
Now, let's say we remove this
reservoir from under our

00:15:11.390 --> 00:15:14.720
system again, so it's completely
isolated from

00:15:14.720 --> 00:15:17.080
everything else, at least
in terms of heat.

00:15:17.080 --> 00:15:21.830
And what we do is, we start
adding-- so state D, we still

00:15:21.830 --> 00:15:23.275
had a few less pebbles.

00:15:23.275 --> 00:15:26.290
But we start adding more
pebbles again.

00:15:26.290 --> 00:15:29.985
We start adding more pebbles
to get it to state A.

00:15:29.985 --> 00:15:32.770
So let me change my
pebble color.

00:15:32.770 --> 00:15:34.710
So we start adding more
pebbles again to

00:15:34.710 --> 00:15:38.620
get it to state A.

00:15:38.620 --> 00:15:40.530
So that's this process
right here.

00:15:43.430 --> 00:15:46.480
Let me do a different color.

00:15:46.480 --> 00:15:49.280
Let's say this is green.

00:15:49.280 --> 00:15:52.570
So as we add pebbles, that's
this movement right here.

00:15:52.570 --> 00:15:55.840
We're moving from one isotherm
up to another isotherm at a

00:15:55.840 --> 00:15:56.800
higher temperature.

00:15:56.800 --> 00:15:58.520
And remember, this whole
time we went

00:15:58.520 --> 00:16:01.330
this clockwise direction.

00:16:01.330 --> 00:16:05.240
So a couple of interesting
things are going on here.

00:16:05.240 --> 00:16:07.990
Because we're assuming an ideal
scenario, nothing was

00:16:07.990 --> 00:16:09.070
lost of friction.

00:16:09.070 --> 00:16:11.610
This piston just moves
up and down.

00:16:11.610 --> 00:16:13.340
No heat loss due to that.

00:16:13.340 --> 00:16:17.510
What we can say is that we've
achieved-- we are back at our

00:16:17.510 --> 00:16:19.250
original internal energy.

00:16:19.250 --> 00:16:21.960
In fact, this is one of the
properties of a state

00:16:21.960 --> 00:16:24.520
variable, is that if we're at
the same point on the PV

00:16:24.520 --> 00:16:27.550
diagram, the same exact
point, we have

00:16:27.550 --> 00:16:28.860
the same state variable.

00:16:28.860 --> 00:16:31.610
So now we have the same
pressure, volume, temperature,

00:16:31.610 --> 00:16:33.780
and internal energy as
what we started with.

00:16:33.780 --> 00:16:35.820
So we've done here is
completed a cycle.

00:16:35.820 --> 00:16:38.990
And this particular cycle, it's
an important one, it's

00:16:38.990 --> 00:16:41.660
called the Carnot cycle.

00:16:41.660 --> 00:16:44.390
It's named after a French
engineer who was trying to

00:16:44.390 --> 00:16:46.780
just optimize engines
in the early 1800s.

00:16:46.780 --> 00:16:48.030
So Carnot cycle.

00:16:50.810 --> 00:16:53.330
And we're going to study this
a lot in the next few videos

00:16:53.330 --> 00:16:55.860
to really make sure we
understand entropy correctly.

00:16:55.860 --> 00:16:57.830
Because in a lot of chemistry
classes, they'll

00:16:57.830 --> 00:16:58.760
throw entropy at you.

00:16:58.760 --> 00:17:00.060
Oh, it's measure of disorder.

00:17:00.060 --> 00:17:02.190
But you really don't know what
they're talking about, or how

00:17:02.190 --> 00:17:04.810
can you quantify it, or
measure it anyway.

00:17:04.810 --> 00:17:07.910
And we really need to deal with
the Carnot cycle in order

00:17:07.910 --> 00:17:10.650
to understand where the first
concepts of entropy really

00:17:10.650 --> 00:17:12.390
came from, and then relate
it to kind of more

00:17:12.390 --> 00:17:14.069
modern notions of it.

00:17:14.069 --> 00:17:19.099
Now, a system that completes
a Carnot cycle is called a

00:17:19.099 --> 00:17:19.950
Carnot engine.

00:17:19.950 --> 00:17:22.869
So our little piston here that's
moving up and down, we

00:17:22.869 --> 00:17:24.560
can consider this
a Carnot engine.

00:17:24.560 --> 00:17:25.800
You might say, oh, Sal,
this doesn't seem

00:17:25.800 --> 00:17:26.560
like a great engine.

00:17:26.560 --> 00:17:29.040
I have to move pebbles
and all of that.

00:17:29.040 --> 00:17:30.670
And you're right.

00:17:30.670 --> 00:17:33.890
You wouldn't actually implement
an engine this way.

00:17:33.890 --> 00:17:37.280
But it's a useful engine, or
it's a useful theoretical

00:17:37.280 --> 00:17:40.670
construct, in order for
understanding how heat is

00:17:40.670 --> 00:17:42.110
transferred in an engine.

00:17:42.110 --> 00:17:44.520
I mean, if you think about
what's happening here, is this

00:17:44.520 --> 00:17:48.680
first heat sink transferred some
heat to the system, and

00:17:48.680 --> 00:17:51.690
then the system transferred a
smaller amount of heat back to

00:17:51.690 --> 00:17:56.740
the other reservoir.

00:17:56.740 --> 00:17:57.290
Right?

00:17:57.290 --> 00:18:00.940
So this system was transferring
heat from one

00:18:00.940 --> 00:18:02.630
reservoir to another
reservoir.

00:18:02.630 --> 00:18:05.180
From a hotter reservoir
to a colder reservoir.

00:18:05.180 --> 00:18:07.890
And in the process, it was
also doing some work.

00:18:07.890 --> 00:18:09.880
And what was the work
that it did?

00:18:09.880 --> 00:18:13.500
Well, it's the area under this
curve, or the area inside of

00:18:13.500 --> 00:18:14.020
this cycle.

00:18:14.020 --> 00:18:19.890
So this is the work done
by our Carnot engine.

00:18:19.890 --> 00:18:22.890
And the way you think about it
is, when you're going in the

00:18:22.890 --> 00:18:24.770
rightward direction with
increasing volume, it's the

00:18:24.770 --> 00:18:27.850
area under the curve is the
work done by the system.

00:18:27.850 --> 00:18:29.560
And then when you move in the
leftward direction with

00:18:29.560 --> 00:18:33.350
decreasing volume, you subtract
out the work done to

00:18:33.350 --> 00:18:35.520
the system, and then you're
left with just

00:18:35.520 --> 00:18:37.170
the area in the curve.

00:18:37.170 --> 00:18:39.230
So we can write this Carnot
engine like this.

00:18:41.950 --> 00:18:48.280
It's taking, it's starting-- so
you have a reservoir at T1.

00:18:51.240 --> 00:18:55.420
And then you have your
engine, right here.

00:18:55.420 --> 00:18:59.360
And then it's connected--
so this takes Q1

00:18:59.360 --> 00:19:01.350
in from this reservoir.

00:19:01.350 --> 00:19:03.230
It does some work, all right?

00:19:03.230 --> 00:19:07.130
The work is represented by the
amount of-- the work right

00:19:07.130 --> 00:19:11.010
here is the area inside
of our cycle.

00:19:11.010 --> 00:19:14.430
And then it transfers Q2, or
essentially the remainder from

00:19:14.430 --> 00:19:17.190
Q1, into our cold reservoir.

00:19:17.190 --> 00:19:18.580
So T2.

00:19:18.580 --> 00:19:20.660
So it transfers Q2 there.

00:19:20.660 --> 00:19:22.420
So the work we did is really
the difference

00:19:22.420 --> 00:19:24.820
between Q1 and Q2, right?

00:19:24.820 --> 00:19:25.810
You say, hey.

00:19:25.810 --> 00:19:28.360
If I have more heat coming in
than I'm letting out, where

00:19:28.360 --> 00:19:29.950
did the rest of that heat go?

00:19:29.950 --> 00:19:32.590
It went to work.

00:19:32.590 --> 00:19:33.510
Literally.

00:19:33.510 --> 00:19:39.150
So Q1 minus Q2 is equal to the
amount of work we did.

00:19:39.150 --> 00:19:42.200
And actually, this is a good
time to emphasize again that

00:19:42.200 --> 00:19:44.650
heat and work are not
a state variable.

00:19:44.650 --> 00:19:48.040
A state variable has to be the
exact same value when we

00:19:48.040 --> 00:19:49.370
complete a cycle.

00:19:49.370 --> 00:19:52.610
Now, we see here that we
completed a cycle, and we had

00:19:52.610 --> 00:19:56.650
a net amount of work done, or a
net amount of heat added to

00:19:56.650 --> 00:19:57.110
the system.

00:19:57.110 --> 00:19:58.980
So we could just keep going
around the cycle, and keep

00:19:58.980 --> 00:20:00.430
having heat added
to the system.

00:20:00.430 --> 00:20:04.440
So there is no inherent heat
state variable right here.

00:20:04.440 --> 00:20:07.660
You can't say what the
value of heat is at

00:20:07.660 --> 00:20:08.740
this point in time.

00:20:08.740 --> 00:20:11.420
All you could say is what amount
of heat was added or

00:20:11.420 --> 00:20:14.010
taken away from the system, or
you can only say the amount of

00:20:14.010 --> 00:20:17.440
work that was done to, or
done by, the system.

00:20:17.440 --> 00:20:19.900
Anyway, I want to leave
you there right now.

00:20:19.900 --> 00:20:21.280
We're going to study
this a lot more.

00:20:21.280 --> 00:20:24.040
But the real important thing is,
and if you never want to

00:20:24.040 --> 00:20:25.830
get confused in a thermodynamics
class, I

00:20:25.830 --> 00:20:27.800
encourage you to even
go off on your

00:20:27.800 --> 00:20:29.820
own, and do this yourself.

00:20:29.820 --> 00:20:34.120
Kind of-- you can almost take
a pencil and paper, and redo

00:20:34.120 --> 00:20:35.430
this video that I just did.

00:20:35.430 --> 00:20:37.550
Because it's essential that
you understand the Carnot

00:20:37.550 --> 00:20:40.530
engine, understand this
adiabatic process, understand

00:20:40.530 --> 00:20:41.520
what isotherms are.

00:20:41.520 --> 00:20:43.790
Because if you understand that,
then a lot of what we're

00:20:43.790 --> 00:20:47.760
about to do in the next few
videos with regard to entropy

00:20:47.760 --> 00:20:49.320
will be a little bit more

00:20:49.320 --> 00:20:52.140
intuitive, and not too confusing.


Work done by isothermic process

WEBVTT
Kind: captions
Language: en

00:00:00.530 --> 00:00:03.570
Let's start with our classic
system that I keep using over

00:00:03.570 --> 00:00:04.340
and over again.

00:00:04.340 --> 00:00:07.270
And that's because it tends
to be very useful for

00:00:07.270 --> 00:00:07.645
instruction.

00:00:07.645 --> 00:00:11.400
It also tends to be the system
that is most covered in

00:00:11.400 --> 00:00:14.070
classrooms. So hopefully it'll
be productive for you and your

00:00:14.070 --> 00:00:14.890
school work.

00:00:14.890 --> 00:00:16.190
So I have this container.

00:00:16.190 --> 00:00:19.020
It has a movable piston
on top, or kind

00:00:19.020 --> 00:00:21.730
of a movable ceiling.

00:00:21.730 --> 00:00:24.630
Well, of course, inside of my
system I have a bunch of

00:00:24.630 --> 00:00:30.370
molecules or atoms bouncing
around, creating some type of

00:00:30.370 --> 00:00:31.850
pressure on the system.

00:00:31.850 --> 00:00:34.550
So let's say it has
some pressure, P1.

00:00:34.550 --> 00:00:37.255
This volume right here,
let's call that V1.

00:00:37.255 --> 00:00:39.265
And let's say it also has
some temperature that

00:00:39.265 --> 00:00:40.180
it starts off with.

00:00:40.180 --> 00:00:41.120
Everything is in equilibrium.

00:00:41.120 --> 00:00:42.600
Remember these are
macro states.

00:00:42.600 --> 00:00:44.660
The only way I can even tell
you what the volume, or the

00:00:44.660 --> 00:00:48.830
pressure, or the temperature
is, is if the system is in

00:00:48.830 --> 00:00:51.160
equilibrium, if everything
in it is uniform.

00:00:51.160 --> 00:00:53.230
The temperature is consistent
throughout.

00:00:53.230 --> 00:00:53.880
Fair enough.

00:00:53.880 --> 00:00:56.930
And in order to keep it placed
down, I have to put

00:00:56.930 --> 00:00:59.880
some rocks on top.

00:00:59.880 --> 00:01:03.430
And I've done this in multiple
processes so far.

00:01:03.430 --> 00:01:05.510
And, of course, I'm doing these
little pebbles, because

00:01:05.510 --> 00:01:07.090
I'm going to remove
them slowly.

00:01:07.090 --> 00:01:10.710
Because I want to approximate
a quasi-static process.

00:01:10.710 --> 00:01:13.790
Or I want to approximate a
system that's always close

00:01:13.790 --> 00:01:17.680
enough to equilibrium that I'm
cool with defining our macro

00:01:17.680 --> 00:01:21.005
states, our pressure, our
temperature, or our volume.

00:01:21.005 --> 00:01:22.550
Let me write V for volume.

00:01:22.550 --> 00:01:25.640
Now, in this video, I'm going
to study what's called an

00:01:25.640 --> 00:01:27.180
isothermic process.

00:01:32.210 --> 00:01:34.600
And really what it just means
is, I'm going to keep the

00:01:34.600 --> 00:01:36.110
temperature the same.

00:01:36.110 --> 00:01:37.680
Iso- just means the same.

00:01:37.680 --> 00:01:40.340
You probably remember when we
studied the periodic table.

00:01:40.340 --> 00:01:44.190
Isotopes, those are the same
element just with different

00:01:44.190 --> 00:01:44.960
mass numbers.

00:01:44.960 --> 00:01:47.740
So this is the same temperature
we're going to run

00:01:47.740 --> 00:01:48.900
our process.

00:01:48.900 --> 00:01:50.670
So my question is, how
can we do that?

00:01:50.670 --> 00:01:58.140
Because as I remove pebbles,
what's going to happen?

00:01:58.140 --> 00:02:02.010
If I just did this, without
any-- if it was completely

00:02:02.010 --> 00:02:03.620
isolated from the world.

00:02:03.620 --> 00:02:05.040
And actually I'll add in
a word right here.

00:02:05.040 --> 00:02:08.320
If it was an adiabatic
process-- If it was

00:02:08.320 --> 00:02:13.060
adiabatic-- fancy word.

00:02:13.060 --> 00:02:16.100
All that means is, completely
isolated from the world.

00:02:16.100 --> 00:02:20.010
So no heat is going into
or out of this system.

00:02:20.010 --> 00:02:24.540
If this was the case, what would
happen as I released or

00:02:24.540 --> 00:02:27.380
I took away some of these
little particles?

00:02:27.380 --> 00:02:31.110
Let me copy and paste it.

00:02:31.110 --> 00:02:33.500
Well, let me just redraw
it actually.

00:02:33.500 --> 00:02:37.660
So I have my one wall.

00:02:37.660 --> 00:02:38.910
I have another wall.

00:02:38.910 --> 00:02:40.710
I have another wall.

00:02:40.710 --> 00:02:44.600
As I release a couple of
pebbles, one at a time, my

00:02:44.600 --> 00:02:45.965
volume is going to increase.

00:02:45.965 --> 00:02:49.430
So I'm going to have a slightly
higher volume.

00:02:49.430 --> 00:02:51.370
My volume's going to go up.

00:02:51.370 --> 00:02:53.300
I have fewer pebbles here now.

00:02:55.980 --> 00:02:59.330
And since I have the same number
of molecules, they're

00:02:59.330 --> 00:03:02.000
going to bump into this less.

00:03:02.000 --> 00:03:04.430
So pressure is going
to go down.

00:03:04.430 --> 00:03:06.070
Volume is going to go up.

00:03:06.070 --> 00:03:10.920
And if I was adiabatic, if I had
no extra heat being added

00:03:10.920 --> 00:03:13.370
to the system, what do I know
is going to happen to the

00:03:13.370 --> 00:03:14.520
temperature?

00:03:14.520 --> 00:03:15.920
Well, think about it this way.

00:03:15.920 --> 00:03:17.200
Some work was done, right?

00:03:17.200 --> 00:03:21.290
Our old ceiling was maybe
some place around here.

00:03:21.290 --> 00:03:24.670
We pushed it up with some
force for some distance.

00:03:24.670 --> 00:03:26.070
So we did work.

00:03:26.070 --> 00:03:29.930
And so we changed some kinetic
energy, or we transferred some

00:03:29.930 --> 00:03:31.450
kinetic energy, out
of the system.

00:03:31.450 --> 00:03:33.870
That's essentially what
the work did.

00:03:33.870 --> 00:03:36.640
That kinetic energy was
turned into work.

00:03:36.640 --> 00:03:40.900
And temperature is just a
macro measure of average

00:03:40.900 --> 00:03:42.150
kinetic energy.

00:03:45.630 --> 00:03:48.170
In fact, we just-- well,
I won't go into it.

00:03:48.170 --> 00:03:51.780
But in the last video, the kind
of proof you want-- if

00:03:51.780 --> 00:03:53.340
you didn't watch it because you
didn't want to go through

00:03:53.340 --> 00:03:55.360
the math, which is completely
fair enough because it

00:03:55.360 --> 00:03:57.730
normally wouldn't be done in an
intro chemistry class-- I

00:03:57.730 --> 00:04:02.690
showed that the internal energy
is equal to the total

00:04:02.690 --> 00:04:10.970
kinetic energy, which was equal
to 3/2 times the number

00:04:10.970 --> 00:04:13.270
of moles, times R, times
temperature.

00:04:13.270 --> 00:04:16.899
So temperature is just, by some
scaling factor, a measure

00:04:16.899 --> 00:04:17.660
of kinetic energy.

00:04:17.660 --> 00:04:21.790
Now, when I do some work, it's
essentially a transfer of

00:04:21.790 --> 00:04:23.380
kinetic energy.

00:04:23.380 --> 00:04:25.920
And I can't replace that energy
with some heat because

00:04:25.920 --> 00:04:26.900
it's adiabatic.

00:04:26.900 --> 00:04:29.500
There's no heat going into
or out of the system.

00:04:29.500 --> 00:04:31.500
So in that situation, the
kinetic energy of

00:04:31.500 --> 00:04:32.790
the system went down.

00:04:32.790 --> 00:04:34.690
The average kinetic energy
of the system went down.

00:04:34.690 --> 00:04:36.390
So the temperature would've
also gone down.

00:04:39.020 --> 00:04:40.790
And actually, just as a bonus
point, what happened to the

00:04:40.790 --> 00:04:42.990
internal energy?

00:04:42.990 --> 00:04:45.210
Well, the internal energy is
the total kinetic energy of

00:04:45.210 --> 00:04:46.000
the system.

00:04:46.000 --> 00:04:48.540
And I could even right down
the original formula.

00:04:48.540 --> 00:04:54.140
Change in internal energy is
equal to change-- let me not

00:04:54.140 --> 00:04:57.350
do that, because I said I
shouldn't-- is equal to heat

00:04:57.350 --> 00:05:02.190
added to the system, minus
work done by the system.

00:05:02.190 --> 00:05:03.720
This is work done
by the system.

00:05:03.720 --> 00:05:05.250
That's why we're
subtracting it.

00:05:05.250 --> 00:05:07.830
Now, it's adiabatic,
so there's no heat

00:05:07.830 --> 00:05:09.120
added to the system.

00:05:09.120 --> 00:05:12.830
So the change in internal energy
is equal to the minus

00:05:12.830 --> 00:05:14.580
the work done by the system.

00:05:14.580 --> 00:05:16.630
Well, in this situation,
the system did do work.

00:05:16.630 --> 00:05:20.300
It pushed this piston up by some
distance with some force.

00:05:20.300 --> 00:05:22.880
So, your delta U is negative.

00:05:22.880 --> 00:05:24.680
It's less than 0.

00:05:24.680 --> 00:05:27.690
So U went down, and
that makes sense.

00:05:27.690 --> 00:05:30.090
If temperature changed,
then the internal

00:05:30.090 --> 00:05:31.060
energy is going to change.

00:05:31.060 --> 00:05:33.520
And for our simple system,
where internal energy is

00:05:33.520 --> 00:05:36.090
represented by the kinetic
energy of these molecules,

00:05:36.090 --> 00:05:38.330
that's always going
to be the case.

00:05:38.330 --> 00:05:40.690
If temperature doesn't change,
internal energy won't change.

00:05:40.690 --> 00:05:42.680
If temperature goes up, internal
energy goes up.

00:05:42.680 --> 00:05:45.320
If temperature goes down,
internal energy goes down.

00:05:45.320 --> 00:05:47.360
And, of course, they're not
the same thing, though.

00:05:47.360 --> 00:05:49.550
The difference between
internal energy and

00:05:49.550 --> 00:05:53.030
temperature is the scaling
factor, 3/2 times the number

00:05:53.030 --> 00:05:58.430
of molecules, times our
ideal gas constant.

00:05:58.430 --> 00:06:00.510
So fair enough.

00:06:00.510 --> 00:06:02.950
I went through this whole
exercise just to show you that

00:06:02.950 --> 00:06:06.280
if I was completely isolated,
and if I removed a couple of

00:06:06.280 --> 00:06:08.720
these pebbles, that
my temperature

00:06:08.720 --> 00:06:10.350
is going to go down.

00:06:10.350 --> 00:06:12.300
Now, I told you already
that I want to do

00:06:12.300 --> 00:06:13.640
an isothermic process.

00:06:17.090 --> 00:06:19.010
So I want to do this process
while keeping the

00:06:19.010 --> 00:06:20.440
temperature the same.

00:06:20.440 --> 00:06:21.600
So how can I do that?

00:06:21.600 --> 00:06:26.300
Well, what I'm going to do is
I'm going to place my system

00:06:26.300 --> 00:06:28.180
on top of what we'll
call, a reservoir.

00:06:31.680 --> 00:06:35.450
So a reservoir, you can kind of
view as an infinitely large

00:06:35.450 --> 00:06:38.220
amount of something that is
the temperature that we

00:06:38.220 --> 00:06:38.980
started off with.

00:06:38.980 --> 00:06:41.960
So this reservoir is T1.

00:06:41.960 --> 00:06:47.810
So even though if, I took
two relative things

00:06:47.810 --> 00:06:49.060
of comparable size.

00:06:51.470 --> 00:06:52.430
That says, temperature A.

00:06:52.430 --> 00:06:53.360
This is temperature B.

00:06:53.360 --> 00:06:54.840
And I've put them next
to each other.

00:06:54.840 --> 00:06:58.360
They're going to average out to
A plus B over 2, whatever

00:06:58.360 --> 00:06:59.090
their temperatures are.

00:06:59.090 --> 00:07:08.770
But if B is massive-- if A is
just a speck of particle--

00:07:08.770 --> 00:07:12.800
let's say it's iron dust-- while
B is the Eiffel Tower,

00:07:12.800 --> 00:07:14.610
then essentially B's
temperature will

00:07:14.610 --> 00:07:15.520
not change a lot.

00:07:15.520 --> 00:07:17.630
A will just become
B's temperature.

00:07:17.630 --> 00:07:18.690
Now, a reservoir is

00:07:18.690 --> 00:07:21.050
theoretically infinitely large.

00:07:21.050 --> 00:07:22.950
It's an infinitely
large object.

00:07:22.950 --> 00:07:25.600
So if something is next to a
reservoir and is given enough

00:07:25.600 --> 00:07:29.060
time, it'll always assume the
heat of the reservoir, or the

00:07:29.060 --> 00:07:30.650
temperature of the reservoir.

00:07:30.650 --> 00:07:31.960
So what's going to happen?

00:07:31.960 --> 00:07:35.130
So this is adiabatic, but now
I'm actually putting it next

00:07:35.130 --> 00:07:36.010
to a reservoir.

00:07:36.010 --> 00:07:38.090
So this isn't going to happen.

00:07:38.090 --> 00:07:40.630
The adiabatic situation
isn't going to happen.

00:07:40.630 --> 00:07:42.130
Now, I'm going to have a
situation where I'm going to

00:07:42.130 --> 00:07:44.470
stay the same temperature.

00:07:44.470 --> 00:07:48.770
So what's that going to look
like on the PV diagram?

00:07:48.770 --> 00:07:54.060
So let me draw the PV diagram.

00:07:56.960 --> 00:07:58.470
This is my pressure.

00:07:58.470 --> 00:08:00.050
This is my volume.

00:08:00.050 --> 00:08:03.770
So this is my starting
point right here.

00:08:03.770 --> 00:08:06.580
And what I'm saying is, if I'm
doing an isothermic process,

00:08:06.580 --> 00:08:09.970
so I just keep removing
these pebbles.

00:08:09.970 --> 00:08:12.950
So I start at this
state right here.

00:08:12.950 --> 00:08:14.800
Let me copy and paste it
since I've done so

00:08:14.800 --> 00:08:17.380
much art already here.

00:08:17.380 --> 00:08:24.430
So I'm going from there to
here where I'm removing a

00:08:24.430 --> 00:08:25.680
couple of the particles.

00:08:28.376 --> 00:08:30.980
So let's say I've removed a
couple of them over here.

00:08:30.980 --> 00:08:33.760
And because of that, I've
increased the volume.

00:08:33.760 --> 00:08:35.870
So let's say the volume,
it's not there anymore.

00:08:35.870 --> 00:08:37.020
Let's say it's a little
bit higher.

00:08:37.020 --> 00:08:39.720
Let's say the volume is--
just for the sake of our

00:08:39.720 --> 00:08:41.530
discussion-- let's say the
volume has expanded a little

00:08:41.530 --> 00:08:44.159
bit, because I've remove some
particles, the little pebbles

00:08:44.159 --> 00:08:45.870
on the top keeping it down.

00:08:45.870 --> 00:08:48.880
So it's like the adiabatic
process, but instead of the

00:08:48.880 --> 00:08:51.990
temperature going down, my
temperature stays at T1.

00:08:51.990 --> 00:08:55.040
My temperature's at T1 the
entire time, because I'm next

00:08:55.040 --> 00:08:59.550
to this theoretical thing
called a reservoir.

00:08:59.550 --> 00:09:03.090
So because of that, I will
travel along what we'll call

00:09:03.090 --> 00:09:04.420
an isotherm.

00:09:04.420 --> 00:09:06.060
So this is my first state.

00:09:06.060 --> 00:09:10.120
When I'm done, I might end
up some place over here.

00:09:10.120 --> 00:09:11.380
And so this is state 2.

00:09:11.380 --> 00:09:14.880
So this is state 2,
this is state 1.

00:09:14.880 --> 00:09:18.530
What I'm claiming is that my
path along this is going to be

00:09:18.530 --> 00:09:21.550
on some type of a rectangular
hyperbola, or at

00:09:21.550 --> 00:09:22.550
least part of it.

00:09:22.550 --> 00:09:27.620
If I were to add rocks to it and
compress it, I claim that

00:09:27.620 --> 00:09:29.920
my PV diagram would
go like this.

00:09:29.920 --> 00:09:33.030
If I were to keep removing rocks
from this diagram, I

00:09:33.030 --> 00:09:35.425
claim that my PV diagram would
keep going like that.

00:09:35.425 --> 00:09:36.810
And so what's the intuition?

00:09:36.810 --> 00:09:40.280
That if I keep the temperature
constant, that I'm essentially

00:09:40.280 --> 00:09:42.840
moving along this hyperbola.

00:09:42.840 --> 00:09:45.000
Well, let's just take out
the ideal gas formula.

00:09:45.000 --> 00:09:48.020
Let me box off all this
stuff over here.

00:09:48.020 --> 00:09:54.180
If I just take the ideal gas
formula, PV is equal to nRT.

00:09:54.180 --> 00:09:57.530
If T is constant-- we know that
R is a constant, it's the

00:09:57.530 --> 00:09:58.950
ideal gas constant.

00:09:58.950 --> 00:10:00.706
We know that we're not changing
the number of moles

00:10:00.706 --> 00:10:01.710
of particles.

00:10:01.710 --> 00:10:05.170
Then that means that PV is
equal to some constant.

00:10:05.170 --> 00:10:07.140
This whole thing is equal
to some constant.

00:10:07.140 --> 00:10:09.260
And then, if we wanted to write
P as a function of V, we

00:10:09.260 --> 00:10:13.550
would just write P is
equal to K over V.

00:10:13.550 --> 00:10:16.200
Now, this might not look 100%
familiar to you, but if I

00:10:16.200 --> 00:10:20.100
wrote it in algebraic terms,
if I told you to graph y is

00:10:20.100 --> 00:10:23.150
equal to 1 over x, what
does that look like?

00:10:23.150 --> 00:10:24.580
That's a rectangular
hyperbola.

00:10:24.580 --> 00:10:25.995
That looks like this.

00:10:25.995 --> 00:10:29.350
And this is the y-axis, that's
the x-axis, at least in this

00:10:29.350 --> 00:10:31.710
quadrant it looks like this.

00:10:31.710 --> 00:10:33.980
It also looks like that in the
third quadrant, but we won't

00:10:33.980 --> 00:10:35.550
worry about that too much.

00:10:35.550 --> 00:10:37.420
So whenever you hold temperature
constant, you're

00:10:37.420 --> 00:10:41.260
on some rectangular hyperbola
like this, like an isotherm.

00:10:41.260 --> 00:10:43.150
Now, if the temperature was a
different temperature, if it

00:10:43.150 --> 00:10:47.180
was a lower temperature, you'd
be on a different isotherm.

00:10:47.180 --> 00:10:48.910
So you would be on an isotherm
that looked like

00:10:48.910 --> 00:10:52.060
this, maybe over here.

00:10:52.060 --> 00:10:53.710
It would also be a rectangular
hyperbola

00:10:53.710 --> 00:10:55.130
but at a lower state.

00:10:55.130 --> 00:10:56.060
Why is that?

00:10:56.060 --> 00:11:04.020
Because, if you're at a lower
temperature, for any volume,

00:11:04.020 --> 00:11:05.910
you should have a lower pressure
and that works out.

00:11:05.910 --> 00:11:09.070
That's why this is some
temperature T2, that

00:11:09.070 --> 00:11:10.630
is lower than T1.

00:11:10.630 --> 00:11:12.480
So I want to do a couple of
things in this video.

00:11:12.480 --> 00:11:15.240
I inadvertently explained to you
what an adiabatic process

00:11:15.240 --> 00:11:17.940
is, and why the temperature
would naturally go down on its

00:11:17.940 --> 00:11:21.870
own if you didn't have
this reservoir here.

00:11:21.870 --> 00:11:23.920
But the whole reason why I even
thought about doing this

00:11:23.920 --> 00:11:26.240
video is because I wanted you
get comfortable with this idea

00:11:26.240 --> 00:11:30.720
of, one, that a reservoir will
keep you in kind of an

00:11:30.720 --> 00:11:31.880
isothermic state.

00:11:31.880 --> 00:11:33.610
It will keep the temperature
the same.

00:11:33.610 --> 00:11:35.920
And that if you keep the
temperature the same, that you

00:11:35.920 --> 00:11:39.940
will travel along this isotherm,
these rectangular

00:11:39.940 --> 00:11:40.600
hyperbolas.

00:11:40.600 --> 00:11:41.860
And that each temperature has

00:11:41.860 --> 00:11:44.340
associated with it an isotherm.

00:11:44.340 --> 00:11:46.740
So if you take that, let's
just do one more step.

00:11:46.740 --> 00:11:53.420
And let's think about the
actual work we did by

00:11:53.420 --> 00:11:55.470
traveling from this state
to this state.

00:11:55.470 --> 00:11:57.720
Or if you just want to think
of it in visual terms, from

00:11:57.720 --> 00:12:01.220
removing our pebbles slowly and
slowly with this reservoir

00:12:01.220 --> 00:12:04.420
down here the whole time, from
this state to this state.

00:12:04.420 --> 00:12:07.050
Where our volume has increased,
our pressure has--

00:12:07.050 --> 00:12:08.080
So our volume has increased.

00:12:08.080 --> 00:12:10.300
Our pressure has gone down. but
our temperature has stayed

00:12:10.300 --> 00:12:12.840
the same the entire time.

00:12:12.840 --> 00:12:17.120
So, several videos ago, we
learned that the work done is

00:12:17.120 --> 00:12:19.810
the area under this graph.

00:12:19.810 --> 00:12:23.410
It's the area under
that graph.

00:12:23.410 --> 00:12:26.480
Or, if we want to do in calculus
terms-- and I'm about

00:12:26.480 --> 00:12:29.290
to break into calculus, so if
you don't want to see calculus

00:12:29.290 --> 00:12:32.300
cover your eyes or ears.

00:12:32.300 --> 00:12:33.510
It would be the integral.

00:12:33.510 --> 00:12:35.730
And the rest of this video will
be a little bit mathy,

00:12:35.730 --> 00:12:39.430
and I guess I should make
that statement on the

00:12:39.430 --> 00:12:41.450
title of the video.

00:12:41.450 --> 00:12:44.130
But if I want to calculate
what this area is,

00:12:44.130 --> 00:12:45.230
I can now do this.

00:12:45.230 --> 00:12:49.130
The isotherm assumption makes
our math a little bit easier.

00:12:49.130 --> 00:12:55.410
Because we know that PV is equal
to nRT, ideal gas law.

00:12:55.410 --> 00:12:59.150
Or we could say P, if we divide
both sides by V, is

00:12:59.150 --> 00:13:02.740
equal to nRT, divided by V.

00:13:02.740 --> 00:13:03.580
So there we have it.

00:13:03.580 --> 00:13:05.800
We have P as a function of V.

00:13:05.800 --> 00:13:10.470
This function right here, this
graph right here, is this.

00:13:10.470 --> 00:13:14.450
We could write P as a function
of v is equal to nRT over V.

00:13:14.450 --> 00:13:16.990
So if we want to figure out the
area under the curve, we

00:13:16.990 --> 00:13:21.570
just integrate this function
from our starting, our V1 to

00:13:21.570 --> 00:13:24.340
our ending point, to our V2.

00:13:24.340 --> 00:13:26.190
So what is that going to be?

00:13:26.190 --> 00:13:29.695
Well, we're going to integrate
from V1 to V2.

00:13:29.695 --> 00:13:31.860
Actually, that shouldn't
be an equal.

00:13:31.860 --> 00:13:36.510
The work is going to be the
integral from V1 to V2, times

00:13:36.510 --> 00:13:42.050
our function, P as a function
of V, times dV.

00:13:42.050 --> 00:13:45.560
We're summing up all of the
little rectangles here.

00:13:45.560 --> 00:13:47.550
We did that a couple
of videos ago.

00:13:47.550 --> 00:13:49.420
So what's P as a
function of V?

00:13:49.420 --> 00:13:56.870
So work done is equal to,
from V1 to V2, nRT

00:13:56.870 --> 00:14:01.700
over V, times dV.

00:14:01.700 --> 00:14:03.620
Now, this is our simplifying
assumption.

00:14:03.620 --> 00:14:07.070
We said we're sitting on
top of a reservoir.

00:14:07.070 --> 00:14:09.150
That this reservoir keeps
our temperature the

00:14:09.150 --> 00:14:10.160
same the whole time.

00:14:10.160 --> 00:14:12.360
And we're going to learn in a
second, it's doing that by

00:14:12.360 --> 00:14:14.130
transferring heat
into the system.

00:14:14.130 --> 00:14:16.270
And we're going to calculate how
much heat is transferred

00:14:16.270 --> 00:14:17.770
into the system.

00:14:17.770 --> 00:14:20.970
So, if we look at this right
here, temperature-- since

00:14:20.970 --> 00:14:23.310
we're assuming we're on an
isotherm, is a constant. n and

00:14:23.310 --> 00:14:24.940
R are definitely constants.

00:14:24.940 --> 00:14:29.330
So we can rewrite this integral
as the integral from

00:14:29.330 --> 00:14:34.380
V1 to V2 of 1 over V, dV.

00:14:34.380 --> 00:14:36.140
And then we could put
the nRT out here.

00:14:36.140 --> 00:14:37.950
I should have done that
first. nRT, it's

00:14:37.950 --> 00:14:39.910
just a constant term.

00:14:39.910 --> 00:14:42.920
Now, what's the antiderivative
of 1 over V?

00:14:42.920 --> 00:14:44.780
It's the natural log of V.

00:14:44.780 --> 00:14:53.970
So our work is equal to nRT
times the natural log-- this

00:14:53.970 --> 00:14:59.180
is the antiderivative-- of V,
evaluated at V2 minus it

00:14:59.180 --> 00:15:00.890
evaluated at V1.

00:15:00.890 --> 00:15:08.730
So that is equal to nRT times,
evaluated at V2, so the

00:15:08.730 --> 00:15:10.370
natural log of V2.

00:15:10.370 --> 00:15:12.890
Minus the natural log of V1.

00:15:12.890 --> 00:15:15.210
Now we know from logarithm
properties is the same thing

00:15:15.210 --> 00:15:26.010
as nRT times the natural
log of V2 over V1.

00:15:26.010 --> 00:15:27.310
So there you have
it, we actually

00:15:27.310 --> 00:15:28.670
calculated the real value.

00:15:28.670 --> 00:15:32.760
If we know our starting volume
and our finishing volume, we

00:15:32.760 --> 00:15:34.840
can actually figure out
the work done in

00:15:34.840 --> 00:15:36.960
this isothermic process.

00:15:36.960 --> 00:15:41.760
The work done in this isothermic
process is the area

00:15:41.760 --> 00:15:42.730
under this.

00:15:42.730 --> 00:15:44.560
And we figured out
what it was.

00:15:44.560 --> 00:15:46.630
By pushing up that piston,
it was nRT.

00:15:46.630 --> 00:15:49.800
These are the number of moles
we have, ideal gas constant.

00:15:49.800 --> 00:15:51.460
Our temperature that
we're sitting on.

00:15:51.460 --> 00:15:53.140
It would be T1 in this case.

00:15:53.140 --> 00:15:55.730
And the natural log of our
finishing volume divided by

00:15:55.730 --> 00:15:57.570
our starting volume.

00:15:57.570 --> 00:15:59.920
Now, let me ask a follow-up
question.

00:15:59.920 --> 00:16:06.350
How much heat was put into the
system by this isotherm here?

00:16:10.210 --> 00:16:12.540
It put in heat to keep the
temperature up, otherwise the

00:16:12.540 --> 00:16:13.980
temperature would have
gone down, right?

00:16:13.980 --> 00:16:15.660
Heat was going into the system
the entire time.

00:16:15.660 --> 00:16:17.190
How much was it?

00:16:17.190 --> 00:16:20.600
Well, since it's an isotherm,
since the temperature did not

00:16:20.600 --> 00:16:22.370
change, what do we know about
the internal energy?

00:16:22.370 --> 00:16:23.910
Did the internal
energy change?

00:16:23.910 --> 00:16:26.060
The temperature not changing
told us that the kinetic

00:16:26.060 --> 00:16:26.875
energy didn't change.

00:16:26.875 --> 00:16:29.590
If the kinetic energy didn't
change, then the internal

00:16:29.590 --> 00:16:31.820
energy did not change.

00:16:31.820 --> 00:16:37.950
And we know that the change in
internal energy is equal to

00:16:37.950 --> 00:16:41.570
the heat put into the
system minus the

00:16:41.570 --> 00:16:44.100
work done by the system.

00:16:44.100 --> 00:16:48.250
Now, if this is 0-- so we know
that this didn't change,

00:16:48.250 --> 00:16:50.180
because the temperature
didn't change.

00:16:50.180 --> 00:16:55.700
So that means 0 is equal
to Q minus W, or that

00:16:55.700 --> 00:16:59.500
Q is equal to W.

00:16:59.500 --> 00:17:01.520
So this is the work done
by the system.

00:17:01.520 --> 00:17:03.920
You'll end up getting
something in joules.

00:17:03.920 --> 00:17:06.839
And this is also equal to the
heat put into the system.

00:17:06.839 --> 00:17:09.240
It's also equal to Q.

00:17:09.240 --> 00:17:12.450
So when you look at this, if we
were to just draw this part

00:17:12.450 --> 00:17:17.119
of the curve-- let me redraw it
just to make things neat.

00:17:17.119 --> 00:17:19.390
I want to give you a little bit
of the convention of what

00:17:19.390 --> 00:17:23.724
people in the thermodynamics
world tend to do.

00:17:23.724 --> 00:17:26.160
I'll make a neat drawing here.

00:17:26.160 --> 00:17:28.470
We started here, at state 1.

00:17:28.470 --> 00:17:31.210
And we moved along this
rectangular hyperbola, which

00:17:31.210 --> 00:17:34.550
is an isotherm, to state 2.

00:17:37.380 --> 00:17:41.770
And now we calculated the area
under this, which is the work

00:17:41.770 --> 00:17:44.000
done, which was this
value right here.

00:17:44.000 --> 00:17:45.750
Let me write it there.

00:17:45.750 --> 00:17:50.750
It's nRT natural log
of V2 over V1.

00:17:50.750 --> 00:17:52.430
This is V2.

00:17:52.430 --> 00:17:52.990
This is V1.

00:17:52.990 --> 00:17:56.040
This whole axis, remember, was
the V-axis, volume axis.

00:17:56.040 --> 00:17:58.840
This axis here was the
pressure axis.

00:17:58.840 --> 00:18:02.400
And the convention is that
because we did work, but we

00:18:02.400 --> 00:18:05.340
were constant temperature, so
our internal energy didn't

00:18:05.340 --> 00:18:08.510
change, we had to add energy to
the system to make up for

00:18:08.510 --> 00:18:09.700
the work we did.

00:18:09.700 --> 00:18:12.170
So some heat must have been
added to the system.

00:18:12.170 --> 00:18:13.760
And then what they do is they
just put this little downward

00:18:13.760 --> 00:18:15.720
arrow and they write
a Q right there.

00:18:15.720 --> 00:18:21.450
So some heat was put into the
system during this isothermic

00:18:21.450 --> 00:18:22.450
process right there.

00:18:22.450 --> 00:18:26.840
And the value of that Q is
equivalent to the work we did.

00:18:26.840 --> 00:18:29.740
We put the exact amount of heat
into the system as the

00:18:29.740 --> 00:18:31.190
work that was performed.

00:18:31.190 --> 00:18:34.590
And because of that, our
internal energy didn't change.

00:18:34.590 --> 00:18:36.660
Or you could say our temperature
didn't change.

00:18:36.660 --> 00:18:37.380
Or you could go the other way.

00:18:37.380 --> 00:18:40.350
Because our temperature didn't
change, these two things have

00:18:40.350 --> 00:18:41.530
to be equal.

00:18:41.530 --> 00:18:42.860
Anyway, I want to
leave you there.

00:18:42.860 --> 00:18:45.140
Hopefully I gave you a little
bit more of an intuition of

00:18:45.140 --> 00:18:48.410
how PV diagrams work, a little
bit more intuition behind what

00:18:48.410 --> 00:18:50.290
isotherms and adiabatic mean.

00:18:50.290 --> 00:18:52.680
And the most important thing is,
once we get a little bit

00:18:52.680 --> 00:18:55.630
mathy, this result can be useful
for coming up with

00:18:55.630 --> 00:18:59.720
other interesting things about a
lot of these thermal systems

00:18:59.720 --> 00:19:01.000
that we're dealing with.

00:19:01.000 --> 00:19:02.860
See you in the next video.


Proof: U = (3/2)PV or U = (3/2)nRT

WEBVTT
Kind: captions
Language: en

00:00:00.710 --> 00:00:05.000
I've already told you multiple
times that big, uppercase U is

00:00:05.000 --> 00:00:07.754
the internal energy
of a system.

00:00:07.754 --> 00:00:10.220
And it's really everything
thrown in there.

00:00:10.220 --> 00:00:12.490
It's the kinetic energy
of the molecules.

00:00:12.490 --> 00:00:15.700
It has the potential energy if
the molecules are vibrating.

00:00:15.700 --> 00:00:19.660
It has the chemical energy
of the bonds.

00:00:19.660 --> 00:00:22.060
It has the potential energy of
electrons that want to get

00:00:22.060 --> 00:00:22.590
some place.

00:00:22.590 --> 00:00:25.200
But, for our sake, and
especially if we're kind of in

00:00:25.200 --> 00:00:29.130
an introductory chemistry,
physics, or thermodynamics

00:00:29.130 --> 00:00:32.229
course, let's just assume that
we're talking about a system

00:00:32.229 --> 00:00:33.570
that's an ideal gas.

00:00:33.570 --> 00:00:38.500
And even better, it's a kind
of a monoatomic ideal gas.

00:00:38.500 --> 00:00:43.390
So everything in on my system
are just individual atoms. So

00:00:43.390 --> 00:00:46.200
in that case, the only energy in
the system is all going to

00:00:46.200 --> 00:00:49.310
be the kinetic energy of each
of these particles.

00:00:49.310 --> 00:00:51.080
So what I want to do in this
video-- it's going to get a

00:00:51.080 --> 00:00:53.780
little bit mathy, but I think
it'll be satisfying for those

00:00:53.780 --> 00:00:57.880
of you who stick with it-- is
to relate how much internal

00:00:57.880 --> 00:01:01.010
energy there really is in a
system of a certain pressure,

00:01:01.010 --> 00:01:02.850
volume, or temperature.

00:01:02.850 --> 00:01:06.060
So we want to relate pressure,
volume, or temperature to

00:01:06.060 --> 00:01:07.190
internal energy.

00:01:07.190 --> 00:01:10.730
Notice all the videos we've done
up until now, I just said

00:01:10.730 --> 00:01:12.420
what's the change in
internal energy.

00:01:12.420 --> 00:01:15.810
And we related that to the heat
put into or taken out of

00:01:15.810 --> 00:01:18.600
a system, or the work
done, or done to,

00:01:18.600 --> 00:01:19.760
or done by the system.

00:01:19.760 --> 00:01:22.490
But now, let's just say before
we do any work or any heat,

00:01:22.490 --> 00:01:24.445
how do we know how much internal
energy we even have

00:01:24.445 --> 00:01:25.680
in a system?

00:01:25.680 --> 00:01:28.770
And to do this, let's do a
little bit of a thought

00:01:28.770 --> 00:01:29.800
experiment.

00:01:29.800 --> 00:01:33.740
There is a bit of a
simplification I'll make here.

00:01:33.740 --> 00:01:38.130
But I think you'll find it OK,
or reasonably satisfying.

00:01:38.130 --> 00:01:42.210
So let's say-- let me just
draw it-- I have a cube.

00:01:42.210 --> 00:01:44.810
And something tells me that I
might have already done this

00:01:44.810 --> 00:01:47.230
pseudo-proof in the physics play
list. Although, I don't

00:01:47.230 --> 00:01:50.090
think I related exactly
to internal energy.

00:01:50.090 --> 00:01:51.930
So I'll do that here.

00:01:51.930 --> 00:01:56.160
Let's say my system
is this cube.

00:01:56.160 --> 00:01:58.180
And let's say the dimensions
of the cube

00:01:58.180 --> 00:02:00.110
are x in every direction.

00:02:00.110 --> 00:02:04.010
So it's x high, x wide,
and x deep.

00:02:04.010 --> 00:02:07.510
So its volume is, of course,
x to the third.

00:02:07.510 --> 00:02:09.389
And let's say I have
n particles in my

00:02:09.389 --> 00:02:12.600
system, capital N.

00:02:12.600 --> 00:02:15.480
I could have written lowercase
n moles, but let's just keep

00:02:15.480 --> 00:02:16.260
it straightforward.

00:02:16.260 --> 00:02:17.510
I have N particles.

00:02:20.260 --> 00:02:23.810
So they're all doing
what they will.

00:02:23.810 --> 00:02:26.030
Now, this is where I'm going
to make the gross

00:02:26.030 --> 00:02:27.100
simplification.

00:02:27.100 --> 00:02:28.610
But I think it's reasonable.

00:02:28.610 --> 00:02:31.210
So in a normal system, every
particle, and we've done this

00:02:31.210 --> 00:02:34.410
before, is just bouncing off
in every which way, every

00:02:34.410 --> 00:02:36.330
possible random direction.

00:02:36.330 --> 00:02:38.950
And that's what, when they
ricochet off of each of the

00:02:38.950 --> 00:02:41.420
sides, that's what causes
the pressure.

00:02:41.420 --> 00:02:43.750
And they're always bumping into
each other, et cetera, et

00:02:43.750 --> 00:02:45.360
cetera, in all random
directions.

00:02:45.360 --> 00:02:48.170
Now, for the sake of simplicity
of our mathematics,

00:02:48.170 --> 00:02:51.000
and just to be able to do it
in a reasonable amount of

00:02:51.000 --> 00:02:53.120
time, I'm going to make
an assumption.

00:02:53.120 --> 00:02:55.340
I'm going to make an assumption
that 1/3 of the

00:02:55.340 --> 00:03:00.210
particles are going-- well, 1/3
of the particles are going

00:03:00.210 --> 00:03:01.730
parallel to each of the axes.

00:03:01.730 --> 00:03:07.660
So 1/3 of the particles are
going in this direction, I

00:03:07.660 --> 00:03:09.710
guess we could say,
left to right.

00:03:09.710 --> 00:03:13.440
1/3 of the particles are
going up and down.

00:03:17.300 --> 00:03:21.970
And then 1/3 of the particles
are going forward and back.

00:03:21.970 --> 00:03:24.520
Now, we know that this isn't
what's going in reality, but

00:03:24.520 --> 00:03:26.350
it makes our math
a lot simpler.

00:03:26.350 --> 00:03:29.640
And if you actually were to do
the statistical mechanics

00:03:29.640 --> 00:03:32.370
behind all of the particles
going in every which way, you

00:03:32.370 --> 00:03:34.350
would actually end up getting
the same result.

00:03:34.350 --> 00:03:35.820
Now, with that said, I'm
saying it's a gross

00:03:35.820 --> 00:03:36.590
oversimplification.

00:03:36.590 --> 00:03:39.840
There is some infinitesimally
small chance that we actually

00:03:39.840 --> 00:03:43.120
do fall onto a system where
this is already the case.

00:03:43.120 --> 00:03:44.860
And we'll talk a little bit
later about entropy and why

00:03:44.860 --> 00:03:46.400
it's such a small probability.

00:03:46.400 --> 00:03:48.200
But this could actually
be our system.

00:03:48.200 --> 00:03:49.720
And this system would
generate pressure.

00:03:49.720 --> 00:03:51.420
And it makes our math
a lot simpler.

00:03:51.420 --> 00:03:53.770
So with that said, let's
study this system.

00:03:53.770 --> 00:03:57.070
So let's take a sideways view.

00:03:57.070 --> 00:03:59.010
Let's take a sideways
view right here.

00:04:01.770 --> 00:04:03.800
And let's just study
one particle.

00:04:03.800 --> 00:04:05.280
Maybe I should have
done it in green.

00:04:05.280 --> 00:04:07.670
But let's say I have
one particle.

00:04:07.670 --> 00:04:12.660
It has some mass, m, and
some velocity, v.

00:04:17.500 --> 00:04:21.890
And this is one of the capital
N particles in my system.

00:04:21.890 --> 00:04:25.190
But what I'm curious is how
much pressure does this

00:04:25.190 --> 00:04:28.010
particle exert on this
wall right here?

00:04:30.530 --> 00:04:34.180
We know what the area of
this wall is, right?

00:04:34.180 --> 00:04:37.260
The area of this wall
is x times x.

00:04:37.260 --> 00:04:40.740
So it's x squared area.

00:04:40.740 --> 00:04:43.910
How much force is being exerted
by this particle?

00:04:43.910 --> 00:04:44.880
Well, let's think about
it this way.

00:04:44.880 --> 00:04:47.960
It's going forward, or left
to right just like this.

00:04:47.960 --> 00:04:51.240
And the force will be exerted
when it changes its momentum.

00:04:51.240 --> 00:04:53.420
I'll do a little bit of review
of kinetics right here.

00:04:53.420 --> 00:04:59.800
We know that force is equal to
mass times acceleration.

00:04:59.800 --> 00:05:02.700
We know acceleration can be
written as, which is equal to

00:05:02.700 --> 00:05:10.630
mass times, change in velocity
over change in time.

00:05:10.630 --> 00:05:12.940
And, of course, we know that
this could be rewritten as

00:05:12.940 --> 00:05:15.800
this is equal to-- mass is a
constant and shouldn't change

00:05:15.800 --> 00:05:18.500
for the physics we deal
with-- so it's delta.

00:05:18.500 --> 00:05:20.140
We could put that inside
of the change.

00:05:20.140 --> 00:05:23.820
So it's delta mv over
change in time.

00:05:23.820 --> 00:05:26.730
And this is just change
in momentum, right?

00:05:26.730 --> 00:05:30.890
So this is equal to change in
momentum over change in time.

00:05:30.890 --> 00:05:33.030
So that's another way
to write force.

00:05:33.030 --> 00:05:34.560
So what's the change
in momentum going

00:05:34.560 --> 00:05:35.980
to be for this particle?

00:05:35.980 --> 00:05:37.800
Well, it's going to bump
into this wall.

00:05:37.800 --> 00:05:41.180
In this direction, right now,
it has some momentum.

00:05:41.180 --> 00:05:43.250
Its momentum is equal to mv.

00:05:43.250 --> 00:05:45.420
And it's going to bump into this
wall, and then going to

00:05:45.420 --> 00:05:47.130
ricochet straight back.

00:05:47.130 --> 00:05:49.530
And what's its momentum
going to be?

00:05:49.530 --> 00:05:50.990
Well, it's going to
have the same mass

00:05:50.990 --> 00:05:52.220
and the same velocity.

00:05:52.220 --> 00:05:54.190
We'll assume it's a completely
elastic collision.

00:05:54.190 --> 00:05:56.270
Nothing is lost to heat
or whatever else.

00:05:56.270 --> 00:05:57.980
But the velocity is in
the other direction.

00:05:57.980 --> 00:06:01.640
So the new momentum is going
to be minus mv, because the

00:06:01.640 --> 00:06:04.380
velocity has switched
directions.

00:06:04.380 --> 00:06:08.590
Now, if I come in with a
momentum of mv, and I ricochet

00:06:08.590 --> 00:06:10.950
off with a momentum of
minus mv, what's

00:06:10.950 --> 00:06:12.890
my change in momentum?

00:06:12.890 --> 00:06:17.720
My change in momentum, off of
that ricochet, is equal to--

00:06:17.720 --> 00:06:19.220
well, it's the difference
between these two,

00:06:19.220 --> 00:06:20.470
which is just 2mv.

00:06:22.830 --> 00:06:24.230
Now, that doesn't give
me the force.

00:06:24.230 --> 00:06:26.860
I need to know the change in
momentum per unit of time.

00:06:32.390 --> 00:06:33.840
So how often does this happen?

00:06:33.840 --> 00:06:35.410
How frequently?

00:06:35.410 --> 00:06:39.040
Well, it's going to happen
every time we come here.

00:06:39.040 --> 00:06:40.020
We're going to hit this wall.

00:06:40.020 --> 00:06:41.670
Then the particle is going to
have to travel here, bounce

00:06:41.670 --> 00:06:43.080
off of that wall, and
then come back

00:06:43.080 --> 00:06:45.370
here and hit it again.

00:06:45.370 --> 00:06:47.600
So that's how frequently
it's going to happen.

00:06:47.600 --> 00:06:51.280
So how long of an interval do
we have to wait between the

00:06:51.280 --> 00:06:52.450
collisions?

00:06:52.450 --> 00:06:55.040
Well, the particle has to
travel x going back.

00:06:55.040 --> 00:06:56.300
It's going to collide.

00:06:56.300 --> 00:06:58.790
It's going to have to travel
x to the left.

00:06:58.790 --> 00:07:00.090
This distance is x.

00:07:00.090 --> 00:07:02.890
Let me do that in a
different color.

00:07:02.890 --> 00:07:04.720
This distance right here is x.

00:07:04.720 --> 00:07:07.310
It's going to have to
travel x to go back.

00:07:07.310 --> 00:07:08.800
Then it's going to have
to travel x back.

00:07:08.800 --> 00:07:11.740
So it's going to have to
travel 2x distance.

00:07:11.740 --> 00:07:15.060
And how long will it take it
to travel 2x distance?

00:07:15.060 --> 00:07:19.100
Well, the time, delta T, is
equal to, we know this.

00:07:19.100 --> 00:07:22.710
Distance is equal to
rate times time.

00:07:22.710 --> 00:07:29.390
Or if we do distance divided by
rate, we'll get the amount

00:07:29.390 --> 00:07:30.630
of time we took.

00:07:30.630 --> 00:07:33.900
This is just our basic
motion formula.

00:07:33.900 --> 00:07:35.500
Our delta T, the distance
we have to

00:07:35.500 --> 00:07:36.830
travel is back and forth.

00:07:36.830 --> 00:07:41.040
So it's 2 x's, divided
by-- what's our rate?

00:07:41.040 --> 00:07:42.720
Well, our rate is
our velocity.

00:07:42.720 --> 00:07:43.970
Divided by v.

00:07:47.810 --> 00:07:48.260
There you go.

00:07:48.260 --> 00:07:51.440
So this is our delta
T right here.

00:07:51.440 --> 00:08:03.680
So our change in momentum per
time is equal to 2 times our

00:08:03.680 --> 00:08:04.620
incident momentum.

00:08:04.620 --> 00:08:07.500
Because we ricocheted back with
the same magnitude, but

00:08:07.500 --> 00:08:08.430
negative momentum.

00:08:08.430 --> 00:08:09.475
So that's our change
in momentum.

00:08:09.475 --> 00:08:12.890
And then our change in time
is this value over here.

00:08:12.890 --> 00:08:15.380
It's the total distance we
have to travel between

00:08:15.380 --> 00:08:18.800
collisions of this wall, divided
by our velocity.

00:08:18.800 --> 00:08:27.920
So it is, 2x divided by v, which
is equal to 2mv times

00:08:27.920 --> 00:08:30.780
the reciprocal of this-- so
this is just fraction

00:08:30.780 --> 00:08:32.755
math-- v over 2x.

00:08:32.755 --> 00:08:34.010
And what is this equal to?

00:08:34.010 --> 00:08:35.710
The 2's cancel out.

00:08:35.710 --> 00:08:41.710
So that is equal to mv
squared, over x.

00:08:41.710 --> 00:08:42.260
Interesting.

00:08:42.260 --> 00:08:45.250
We're getting someplace
interesting already.

00:08:45.250 --> 00:08:47.590
And if it doesn't seem too
interesting, just hang on with

00:08:47.590 --> 00:08:48.260
me for a second.

00:08:48.260 --> 00:08:52.520
Now, this is the force being
applied by one particle, is

00:08:52.520 --> 00:08:56.730
this-- force from one particle
on this wall.

00:09:02.330 --> 00:09:03.840
Now, what was the area?

00:09:03.840 --> 00:09:05.090
We care about the pressure.

00:09:12.200 --> 00:09:14.610
We wrote it up here.

00:09:14.610 --> 00:09:16.890
The pressure is equal to
the force per area.

00:09:21.360 --> 00:09:24.160
So this is the force
of that particle.

00:09:24.160 --> 00:09:28.570
So that's mv squared
over x, divided by

00:09:28.570 --> 00:09:29.840
the area of the wall.

00:09:29.840 --> 00:09:31.730
Well, what's the area
of the wall?

00:09:31.730 --> 00:09:35.200
The area of the wall here,
each sideis x.

00:09:35.200 --> 00:09:37.610
And so if we draw the wall
there, it's x times x.

00:09:37.610 --> 00:09:39.360
It's x squared.

00:09:39.360 --> 00:09:43.390
So divided by the area of
the wall, is x squared.

00:09:43.390 --> 00:09:44.250
And what does this equal?

00:09:44.250 --> 00:09:52.370
This is equal to mv squared
over x cubed.

00:09:52.370 --> 00:09:54.790
You can just say, this is times
1 over x squared, when

00:09:54.790 --> 00:09:55.760
this all becomes x cubed.

00:09:55.760 --> 00:09:57.530
This is just fraction math.

00:09:57.530 --> 00:09:59.120
So now we have an interesting
thing.

00:09:59.120 --> 00:10:07.160
The pressure due to this one
particle-- let's just call

00:10:07.160 --> 00:10:14.220
this from this one particle--
is equal to m v

00:10:14.220 --> 00:10:16.750
squared over x cubed.

00:10:16.750 --> 00:10:18.650
Now, what's x cubed?

00:10:18.650 --> 00:10:20.960
That's the volume of
our container.

00:10:20.960 --> 00:10:22.210
Over the volume.

00:10:22.210 --> 00:10:27.120
I'll do that in a
big V, right?

00:10:27.120 --> 00:10:29.130
So let's see if we can relate
this to something else that's

00:10:29.130 --> 00:10:29.630
interesting.

00:10:29.630 --> 00:10:32.820
So that means that the pressure
being exerted by this

00:10:32.820 --> 00:10:36.220
one particle-- well, actually
let me just take another step.

00:10:36.220 --> 00:10:39.280
So this is one particle
on this wall, right?

00:10:39.280 --> 00:10:40.900
This is from one particle
on this wall.

00:10:40.900 --> 00:10:45.890
Now, of all the particles-- we
have N particles in our cube--

00:10:45.890 --> 00:10:47.860
what fraction of them
are going to be

00:10:47.860 --> 00:10:49.170
bouncing off of this wall?

00:10:49.170 --> 00:10:50.730
That are going to be doing
the exact same

00:10:50.730 --> 00:10:52.950
thing as this particle?

00:10:52.950 --> 00:10:53.820
Well, I just said.

00:10:53.820 --> 00:10:55.380
1/3 are going to be going
in this direction.

00:10:55.380 --> 00:10:56.820
1/3 are going to be
going up and down.

00:10:56.820 --> 00:10:58.930
And 1/3 are going to go
be going in and out.

00:10:58.930 --> 00:11:02.420
So if I have N total particles,
N over 3 are going

00:11:02.420 --> 00:11:06.195
to be doing exactly what this
particle is going to be doing.

00:11:08.740 --> 00:11:10.350
This is the pressure
from one particle.

00:11:10.350 --> 00:11:12.850
If I wanted the pressure from
all of the particles on that

00:11:12.850 --> 00:11:15.960
wall-- so the total pressure
on that wall is going to be

00:11:15.960 --> 00:11:18.000
from N over 3 of
the particles.

00:11:18.000 --> 00:11:19.990
The other particles aren't
bouncing off that wall.

00:11:19.990 --> 00:11:21.700
So we don't have to
worry about them.

00:11:21.700 --> 00:11:26.620
So if we want the total pressure
on that wall-- I'll

00:11:26.620 --> 00:11:29.150
just write, pressure
sub on the wall.

00:11:29.150 --> 00:11:31.060
Total pressure on the wall is
going to be the pressure from

00:11:31.060 --> 00:11:37.400
one particle, mv squared, over
our volume, times the total

00:11:37.400 --> 00:11:40.950
number of particles
hitting the wall.

00:11:40.950 --> 00:11:44.750
The total number of particles
is N divided by 3, because

00:11:44.750 --> 00:11:47.270
only 3 will be going
in that direction.

00:11:47.270 --> 00:11:50.800
So, the total pressure on that
wall is equal to mv squared,

00:11:50.800 --> 00:11:52.790
over our volume of our
container, times the total

00:11:52.790 --> 00:11:54.115
particles divided by 3.

00:11:54.115 --> 00:11:57.610
Let's see if we can manipulate
this thing a little bit.

00:11:57.610 --> 00:12:02.420
So if we multiply both sides
by-- let's see what we can do.

00:12:02.420 --> 00:12:14.390
If we multiply both sides by 3v,
we get pv times 3 is equal

00:12:14.390 --> 00:12:21.730
to mv squared, times N, where N
is the number of particles.

00:12:21.730 --> 00:12:24.580
Let's divide both sides by N.

00:12:24.580 --> 00:12:34.270
So we get 3pv over-- actually,
no, let me leave the N there.

00:12:34.270 --> 00:12:41.240
Let's divide both sides
of this equation by 2.

00:12:41.240 --> 00:12:43.520
So we get, what do we get?

00:12:43.520 --> 00:12:49.320
We get 3/2 pv is equal to--
now this is interesting.

00:12:49.320 --> 00:12:54.780
It's equal to N, the number of
particles we have, times mv

00:12:54.780 --> 00:12:57.920
squared over 2.

00:12:57.920 --> 00:12:59.610
Remember, I just divided
this equation right

00:12:59.610 --> 00:13:00.650
here by 2 to get this.

00:13:00.650 --> 00:13:02.280
And I did this for a very
particular reason.

00:13:02.280 --> 00:13:05.260
What is mv squared over 2?

00:13:05.260 --> 00:13:09.910
mv squared over 2 is the kinetic
energy of that little

00:13:09.910 --> 00:13:11.140
particle we started off with.

00:13:11.140 --> 00:13:12.900
That's the formula for
kinetic energy.

00:13:12.900 --> 00:13:20.070
Kinetic energy is equal
to mv squared over 2.

00:13:20.070 --> 00:13:21.950
So this is the kinetic energy
of one particle.

00:13:28.570 --> 00:13:30.890
Now, we're multiplying that
times the total number of

00:13:30.890 --> 00:13:33.280
particles we have, times N.

00:13:33.280 --> 00:13:35.890
So N times the kinetic energy of
one particle is going to be

00:13:35.890 --> 00:13:37.490
the kinetic energy of
all the particles.

00:13:37.490 --> 00:13:39.430
And, of course, we also made
another assumption.

00:13:39.430 --> 00:13:41.160
I should state that I assumed
that all the particles are

00:13:41.160 --> 00:13:43.510
moving with the same velocity
and have the same mass.

00:13:43.510 --> 00:13:45.880
In a real situation, the
particles might have very

00:13:45.880 --> 00:13:47.060
different velocities.

00:13:47.060 --> 00:13:48.980
But this was one of our
simplifying assumptions.

00:13:48.980 --> 00:13:50.850
So, we just assumed they
all have that.

00:13:50.850 --> 00:13:53.890
So, if I multiply N times that--
this statement right

00:13:53.890 --> 00:13:56.285
here-- is the kinetic energy
of the system.

00:14:01.700 --> 00:14:03.140
Now, we're almost there.

00:14:03.140 --> 00:14:04.480
In fact, we are there.

00:14:04.480 --> 00:14:08.610
We just established that the
kinetic energy of the system

00:14:08.610 --> 00:14:12.730
is equal to 3/2 times the
pressure, times the volume of

00:14:12.730 --> 00:14:13.650
the system.

00:14:13.650 --> 00:14:16.270
Now, what is the kinetic
energy of the system?

00:14:16.270 --> 00:14:17.320
It's the internal energy.

00:14:17.320 --> 00:14:19.290
Because we said all the energy
in the system, because it's a

00:14:19.290 --> 00:14:22.900
simple ideal monoatomic gas,
all of the energy in the

00:14:22.900 --> 00:14:25.770
system is in kinetic energy.

00:14:25.770 --> 00:14:30.980
So we could say the internal
energy of the system is equal

00:14:30.980 --> 00:14:33.370
to-- that's just the total
kinetic energy of the system--

00:14:33.370 --> 00:14:38.020
it's equal to 3/2 times our
total pressure, times our

00:14:38.020 --> 00:14:38.620
total volume.

00:14:38.620 --> 00:14:41.070
Now you might say, hey, Sal,
you just figured out the

00:14:41.070 --> 00:14:42.130
pressure on this side.

00:14:42.130 --> 00:14:43.960
What about the pressure on that
side, and that side, and

00:14:43.960 --> 00:14:45.730
that side, or on every
side of the cube?

00:14:45.730 --> 00:14:46.850
Well, the pressure on
every side of the

00:14:46.850 --> 00:14:48.250
cube is the same value.

00:14:48.250 --> 00:14:50.530
So all we have to do is find
in terms of the pressure on

00:14:50.530 --> 00:14:52.420
one side, and that's essentially
the pressure of

00:14:52.420 --> 00:14:54.480
the system.

00:14:54.480 --> 00:14:55.590
So what else can we
do with that?

00:14:55.590 --> 00:15:00.700
Well, we know that pv is equal
to nRT, our ideal gas formula.

00:15:00.700 --> 00:15:06.475
pv is equal to nRT, where this
is the number of moles of gas.

00:15:06.475 --> 00:15:08.410
And this is the ideal
gas constant.

00:15:08.410 --> 00:15:10.240
This is our temperature
in kelvin.

00:15:10.240 --> 00:15:12.880
So if we make that replacement,
we'll say that

00:15:12.880 --> 00:15:17.140
internal energy can also be
written as 3/2 times the

00:15:17.140 --> 00:15:20.500
number of moles we have, times
the ideal gas constant, times

00:15:20.500 --> 00:15:22.290
our temperature.

00:15:22.290 --> 00:15:24.560
Now, I did a lot of work, and
it's a little bit mathy.

00:15:24.560 --> 00:15:27.780
But these results are,
one, interesting.

00:15:27.780 --> 00:15:30.140
Because now you have a
direct relationship.

00:15:30.140 --> 00:15:32.860
If you know the pressure and the
volume, you know what the

00:15:32.860 --> 00:15:37.530
actual internal energy, or
the total kinetic energy,

00:15:37.530 --> 00:15:38.660
of the system is.

00:15:38.660 --> 00:15:41.350
Or, if you know what the
temperature and the number of

00:15:41.350 --> 00:15:44.320
molecules you have are, you also
know what the internal

00:15:44.320 --> 00:15:45.630
energy of the system is.

00:15:45.630 --> 00:15:48.520
And there's a couple of key
takeaways I want you to have.

00:15:48.520 --> 00:15:52.070
If the temperature does not
change in our ideal situation

00:15:52.070 --> 00:15:56.930
here-- if delta T is equal to
0-- if this doesn't change,

00:15:56.930 --> 00:15:58.680
the number particles aren't
going to change.

00:15:58.680 --> 00:16:04.750
Then our internal energy does
not change as well.

00:16:04.750 --> 00:16:07.030
So if we say that there is
some change in internal

00:16:07.030 --> 00:16:10.790
energy, and I'll use this in
future proofs, we could say

00:16:10.790 --> 00:16:16.360
that that's equal to 3/2 times
nR times-- well, the only

00:16:16.360 --> 00:16:18.850
thing that can change, not the
number molecules or the ideal

00:16:18.850 --> 00:16:21.350
gas constant-- times
the change in T.

00:16:21.350 --> 00:16:27.000
Or, it could also be written as
3/2 times the change in pv.

00:16:27.000 --> 00:16:28.450
We don't know if either
of these are constant.

00:16:28.450 --> 00:16:30.520
So we have to say the change
in the product.

00:16:30.520 --> 00:16:32.780
Anyway, this was a
little bit mathy.

00:16:32.780 --> 00:16:34.050
And I apologize for it.

00:16:34.050 --> 00:16:36.840
But hopefully, it gives you a
little bit more sense that

00:16:36.840 --> 00:16:39.300
this really is just the sum
of all the kinetic energy.

00:16:39.300 --> 00:16:42.420
We related it to some of these
macro state variables, like

00:16:42.420 --> 00:16:44.590
pressure, volume, and time.

00:16:44.590 --> 00:16:47.640
And now, since I've done the
video on it, we can actually

00:16:47.640 --> 00:16:50.990
use this result in
future proofs.

00:16:50.990 --> 00:16:53.220
Or at least you won't complain
too much if I do.

00:16:53.220 --> 00:16:54.470
Anyway, see you in
the next video.