WEBVTT
00:00:00.310 --> 00:00:04.510
A system contains four particles evenly distributed between two boxes.
00:00:05.350 --> 00:00:12.890
Before heat transfer, two units of energy are distributed between the particles in one box, and the particles in the other box have zero energy.
00:00:13.790 --> 00:00:18.600
After heat transfer, the two units of energy are evenly distributed between the two boxes.
00:00:19.260 --> 00:00:24.380
Calculate, to four significant figures, the entropy change Δ𝑆 for this process.
00:00:25.310 --> 00:00:32.630
Okay, so in this question, we’ve been told that a system contains four particles that are evenly distributed between two boxes.
00:00:33.510 --> 00:00:39.480
In other words, each box has two particles in it because the four particles are evenly distributed between the two boxes.
00:00:40.270 --> 00:00:58.440
Now we’re also told that before the heat transfer process occurs between the boxes, two units of energy are distributed between the particles in one box — so let’s say that the orange box has two units of energy distributed between the particles — and the particles in the other box have zero energy.
00:00:59.130 --> 00:01:00.810
So that’s a big old zero here.
00:01:01.540 --> 00:01:12.740
Now we’re also told what happens after the heat transfer process, so let’s draw the two boxes again this time showing what would happen after the heat transfer process.
00:01:13.410 --> 00:01:21.270
So we’re told that after heat transfer, the two units of energy are evenly distributed between the two boxes.
00:01:22.180 --> 00:01:29.020
In other words, each box now has one unit of energy in it because the two units of energy are evenly distributed between the two boxes.
00:01:29.790 --> 00:01:37.320
So what we’re asked to do is to calculate, to four significant figures, the entropy change Δ𝑆 for this process.
00:01:38.010 --> 00:01:42.000
In other words, we don’t know what Δ𝑆 is and we’re trying to find this out to four significant figures.
00:01:42.900 --> 00:01:49.960
Alright so if we’re trying to work out the change in entropy, then the first thing that we need to know is the entropy of the system before and after heat exchange.
00:01:50.350 --> 00:01:55.380
So we need to know the values of 𝑆 sub 𝑏, the entropy before heat exchange, and 𝑆 sub 𝑎, the entropy after.
00:01:56.120 --> 00:02:01.350
And then the change in entropy Δ𝑆 is simply equal to the entropy after minus the entropy before.
00:02:01.770 --> 00:02:05.110
So this is how we will calculate the change in entropy that we’ve been asked to find.
00:02:05.720 --> 00:02:07.000
So let’s go about doing that.
00:02:07.540 --> 00:02:10.350
Alright, first things first, how do we find the entropy of a system?
00:02:10.930 --> 00:02:27.050
Well we can recall the Boltzmann formula for entropy, which tells us that the entropy of a system 𝑆 is equal to the Boltzmann constant 𝑘 𝐵 multiplied by the natural log of 𝛺 where 𝛺 is the total number of possible microstates that the system can occupy.
00:02:27.890 --> 00:02:31.400
And we’re using the Boltzmann entropy here because we’re looking at a microscopic system.
00:02:31.850 --> 00:02:42.310
In other words, we’re looking at the particle level rather than looking at the system at a macro scale, which is when we’d be considering the volume of the system or the pressure of the system or other set state variables.
00:02:42.830 --> 00:02:52.190
In that case, we’d be using the definition of change in entropy as the integral of the d𝑄 over 𝑇 where 𝑄 is the heat transfer of the system and 𝑇 is the temperature at which this occurs.
00:02:52.720 --> 00:02:59.530
However, like we said earlier this definition is for when we’re studying the system at a macroscopic scale, which is not what we’re doing here.
00:02:59.800 --> 00:03:01.940
Here, we’re studying it at as a microscopic scale.
00:03:02.370 --> 00:03:05.040
We’re looking at the individual particles that make up the system.
00:03:05.550 --> 00:03:07.970
So we need to use the Boltzmann formula for entropy.
00:03:08.480 --> 00:03:12.710
Anyway, so we know that 𝑆 is equal to 𝑘 𝐵 multiplied by the natural log of 𝛺.
00:03:13.250 --> 00:03:16.400
Now 𝑘 𝐵 is just a constant; that’s the Boltzmann constant.
00:03:16.950 --> 00:03:24.030
So to work out the values of 𝑆 sub 𝑏 and 𝑆 sub 𝑎, what we need to first find out other values of 𝛺 𝑏 and 𝛺 𝑎.
00:03:24.450 --> 00:03:28.020
That’s the number of possible microstates that the system can occupy.
00:03:28.700 --> 00:03:32.560
So let’s first consider the number of possible microstates before heat transfer.
00:03:33.070 --> 00:03:37.620
So here are our two boxes, and now we can start drawing in how the particles will be behaving.
00:03:38.190 --> 00:03:44.560
Now in the orange box, we know that we’ve got two units of energy, and we also know that the orange box has two particles in it.
00:03:45.150 --> 00:03:56.250
So one way of arranging the particles in the orange box is if one particle has two units of energy, so is in the second excited state, and the other particle is in the ground state so it has zero energy.
00:03:56.690 --> 00:04:01.100
So basically what we’re saying is that initially the two particles started out in the ground state.
00:04:01.270 --> 00:04:09.550
And then when we gave this box two units of energy, this particle here went from the ground state up to the second excited state because it had two units of energy.
00:04:10.150 --> 00:04:12.680
So this is one possible way of arranging the orange box.
00:04:13.280 --> 00:04:15.150
Now let’s quickly look at the pink box.
00:04:15.380 --> 00:04:18.900
Now before heat transfer in the pink box, there are zero units of energy.
00:04:19.520 --> 00:04:26.010
So logically, both particles must be in the ground state because there’s no energy for them to be in any of the excited states.
00:04:26.670 --> 00:04:30.480
And at this point, we’ve discovered one possible microstate for the system to be in.
00:04:31.250 --> 00:04:41.820
That’s this microstate here where in the orange box one particle is in the second excited state and one of them is the ground state and in the pink box both of them are in the ground state.
00:04:42.600 --> 00:04:46.400
However, this is not the only way of arranging the particles in the orange box.
00:04:46.890 --> 00:05:00.660
So let’s look at another way to arrange the system, a second way to arrange particles in the orange box is if this time the other particle has two units of energy and therefore is in the second excited state and the first particle is in the ground state.
00:05:01.630 --> 00:05:06.500
Still the particles here are in the grand state because again they have no energy.
00:05:06.500 --> 00:05:08.100
They can’t be in an excited state.
00:05:08.590 --> 00:05:11.220
So we found yet another possible microstate of the system.
00:05:12.740 --> 00:05:16.330
However, there is a third way of arranging the particles in the orange box.
00:05:16.460 --> 00:05:17.360
Let’s look at that now.
00:05:17.890 --> 00:05:25.520
The third way is if the two particles in the orange box each have one unit of energy, therefore each one of them is in the first excited state.
00:05:26.130 --> 00:05:28.850
And once again for the pink box, both of them are in the ground state.
00:05:29.100 --> 00:05:30.150
That’s the only possibility!
00:05:30.720 --> 00:05:37.890
So in this system, it’s the Orange Box that can be arranged in multiple different ways whereas the pink box in this case can only be arranged in one way.
00:05:38.470 --> 00:05:42.530
And therefore, altogether, the system can be arranged in three different ways.
00:05:43.130 --> 00:05:45.410
And at this point we’ve exhausted all the possibilities.
00:05:45.690 --> 00:05:54.080
And since there are three ways of arranging the system, we found that 𝛺 𝑏 the total number of possible micro States before heat transfer is equal to three.
00:05:54.680 --> 00:05:58.820
So let’s replace 𝛺 𝑏 over here with the number three; there we go!
00:05:59.190 --> 00:06:03.120
Now let’s find out the value of 𝛺 after heat transfer or 𝛺 𝑎.
00:06:03.510 --> 00:06:07.930
Now after the heat transfer process has occurred, we know that each box has one unit of energy.
00:06:08.540 --> 00:06:18.900
Therefore, one possible way of arranging the orange box is if the first particle in the box has one unit of energy, so it’s in the first excited state, and the second particle is still in the ground state.
00:06:19.580 --> 00:06:24.950
And the same logic can now be applied to the pink box because now the pink box also has a unit of energy.
00:06:25.430 --> 00:06:33.830
So let’s say that one way of arranging the two particles in the pink box is that the first one is in the first excited state and the second is in the ground state.
00:06:34.700 --> 00:06:39.270
And so this is the first possible microstate for the system after the heat transfer process has occurred.
00:06:40.060 --> 00:06:43.400
But there are other ways of arranging the particles in both boxes this time.
00:06:43.840 --> 00:06:44.670
Let’s look at them now.
00:06:45.110 --> 00:06:47.770
This time let’s say that the particles in the orange box stay the same.
00:06:47.770 --> 00:06:51.170
So the first particle is in the first excited state and the second one is in the ground state.
00:06:51.740 --> 00:06:54.350
But we switch the particles in the second box.
00:06:54.760 --> 00:06:58.430
This time the first particle is in the ground state and the second one is in the first excited state.
00:06:59.030 --> 00:07:01.300
The pink box still only has one unit of energy.
00:07:01.830 --> 00:07:04.680
What’s changed is which particle has that unit for energy.
00:07:05.110 --> 00:07:08.320
And because we’ve changed one of the two boxes, the system is now different.
00:07:08.830 --> 00:07:11.340
So we found a second possible microstate.
00:07:12.490 --> 00:07:16.330
But you may have realized by now that there are still more ways of arranging the particles in the boxes.
00:07:16.820 --> 00:07:21.880
A third way of arranging the particles is by switching around the order of the particles in the orange box this time.
00:07:22.320 --> 00:07:25.470
The first one is in the ground state and the second is in the first excited state.
00:07:26.070 --> 00:07:30.600
And for the pink box, the first particle is in the excited state and the second is in the ground state.
00:07:31.270 --> 00:07:35.580
So this is a third way of arranging all the particles, but there is one more.
00:07:36.080 --> 00:07:47.300
The final way is if particle one in the orange box is in the ground state and the second is in the first excited state and, in the pink box, the first particle is in the ground state and the second is in the first excited state as well.
00:07:48.250 --> 00:07:53.220
So these are the four different ways of arranging the system so that each box has one unit of energy.
00:07:53.780 --> 00:07:56.510
And there are no other ways of arranging the system in this way.
00:07:57.050 --> 00:08:03.270
So at this point we found out the value of 𝛺 sub 𝑎, which is equal to four because there are four different possibilities.
00:08:03.820 --> 00:08:06.390
And hence, we can replace 𝛺 sub 𝑎 with four.
00:08:06.810 --> 00:08:09.670
So we’ve got an expression for 𝑆 sub 𝑎 now as well.
00:08:10.560 --> 00:08:17.620
At this point, we can work out the change in entropy which is what we’ve been asked to find in the question because we’ve got expressions for 𝑆 sub 𝑎 and 𝑆 sub 𝑏.
00:08:18.100 --> 00:08:19.820
So let’s plug that into our equation now.
00:08:20.450 --> 00:08:27.460
Our expression for Δ𝑆 turns out to be 𝑘 𝐵 multiplied by the natural log of four minus 𝑘 𝐵 multiplied by the natural log of three.
00:08:28.180 --> 00:08:30.950
Now both of the terms have a factor of 𝑘 𝐵, so we can factorize this.
00:08:31.590 --> 00:08:36.170
This gives us Δ𝑆 is equal to 𝑘 𝐵 multiplied by log four minus log three.
00:08:36.810 --> 00:08:45.010
The value of 𝑘 𝐵 is 1.38064852 times 10 to the power of negative 23 meters squared kilograms per second squared per Kelvin.
00:08:45.560 --> 00:08:50.320
We can plug this value into our calculator for 𝑘 𝐵 and work out what log four minus log three is.
00:08:50.740 --> 00:08:52.600
And this should give us a value for Δ𝑆.
00:08:53.010 --> 00:09:00.950
What we find is that Δ𝑆 is equal to 3.972 times 10 to the power of negative 24 joules per Kelvin once we’ve rounded it to four significant figures.
00:09:01.400 --> 00:09:05.760
Now this is what the question asked us to do: we were told to find the answer to four significant figures.
00:09:05.930 --> 00:09:07.950
So we’ve reached our final answer.
00:09:08.500 --> 00:09:15.760
Once again, Δ𝑆 is equal to 3.972 times 10 to the power of negative 24 joules per Kelvin to four significant figures.