# Video: Calculating the Entropy Change due to Heating

A system contains four particles evenly distributed between two boxes. 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. After heat transfer, the two units of energy are evenly distributed between the two boxes. Calculate to 4 significant figures the entropy change Δ𝑆 for this process.

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### Video Transcript

A system contains four particles evenly distributed between two boxes. 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. After heat transfer, the two units of energy are evenly distributed between the two boxes. Calculate, to four significant figures, the entropy change Δ𝑆 for this process.

Okay, so in this question, we’ve been told that a system contains four particles that are evenly distributed between two boxes. In other words, each box has two particles in it because the four particles are evenly distributed between the two boxes. 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. So that’s a big old zero here.

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. So we’re told that after heat transfer, the two units of energy are evenly distributed between the two boxes. 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. So what we’re asked to do is to calculate, to four significant figures, the entropy change Δ𝑆 for this process. In other words, we don’t know what Δ𝑆 is and we’re trying to find this out to four significant figures.

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. So we need to know the values of 𝑆 sub 𝑏, the entropy before heat exchange, and 𝑆 sub 𝑎, the entropy after. And then the change in entropy Δ𝑆 is simply equal to the entropy after minus the entropy before. So this is how we will calculate the change in entropy that we’ve been asked to find. So let’s go about doing that. Alright, first things first, how do we find the entropy of a system?

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. And we’re using the Boltzmann entropy here because we’re looking at a microscopic system. 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. 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.

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. Here, we’re studying it at as a microscopic scale. We’re looking at the individual particles that make up the system. So we need to use the Boltzmann formula for entropy. Anyway, so we know that 𝑆 is equal to 𝑘 𝐵 multiplied by the natural log of 𝛺. Now 𝑘 𝐵 is just a constant; that’s the Boltzmann constant. 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. That’s the number of possible microstates that the system can occupy.

So let’s first consider the number of possible microstates before heat transfer. So here are our two boxes, and now we can start drawing in how the particles will be behaving. 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. 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. So basically what we’re saying is that initially the two particles started out in the ground state. 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.

So this is one possible way of arranging the orange box. Now let’s quickly look at the pink box. Now before heat transfer in the pink box, there are zero units of energy. 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. And at this point, we’ve discovered one possible microstate for the system to be in. 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. However, this is not the only way of arranging the particles in the orange box.

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. Still the particles here are in the grand state because again they have no energy. They can’t be in an excited state. So we found yet another possible microstate of the system. However, there is a third way of arranging the particles in the orange box. Let’s look at that now. 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. And once again for the pink box, both of them are in the ground state. That’s the only possibility!

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. And therefore, altogether, the system can be arranged in three different ways. And at this point we’ve exhausted all the possibilities. 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. So let’s replace 𝛺 𝑏 over here with the number three; there we go! Now let’s find out the value of 𝛺 after heat transfer or 𝛺 𝑎.

Now after the heat transfer process has occurred, we know that each box has one unit of energy. 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. And the same logic can now be applied to the pink box because now the pink box also has a unit of energy. 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. And so this is the first possible microstate for the system after the heat transfer process has occurred. But there are other ways of arranging the particles in both boxes this time. Let’s look at them now.

This time let’s say that the particles in the orange box stay the same. So the first particle is in the first excited state and the second one is in the ground state. But we switch the particles in the second box. This time the first particle is in the ground state and the second one is in the first excited state. The pink box still only has one unit of energy. What’s changed is which particle has that unit for energy. And because we’ve changed one of the two boxes, the system is now different. So we found a second possible microstate. But you may have realized by now that there are still more ways of arranging the particles in the boxes.

A third way of arranging the particles is by switching around the order of the particles in the orange box this time. The first one is in the ground state and the second is in the first excited state. And for the pink box, the first particle is in the excited state and the second is in the ground state. So this is a third way of arranging all the particles, but there is one more. 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. So these are the four different ways of arranging the system so that each box has one unit of energy. And there are no other ways of arranging the system in this way.

So at this point we found out the value of 𝛺 sub 𝑎, which is equal to four because there are four different possibilities. And hence, we can replace 𝛺 sub 𝑎 with four. So we’ve got an expression for 𝑆 sub 𝑎 now as well. 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. So let’s plug that into our equation now. 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.

Now both of the terms have a factor of 𝑘 𝐵, so we can factorize this. This gives us Δ𝑆 is equal to 𝑘 𝐵 multiplied by log four minus log three. The value of 𝑘 𝐵 is 1.38064852 times 10 to the power of negative 23 meters squared kilograms per second squared per Kelvin. We can plug this value into our calculator for 𝑘 𝐵 and work out what log four minus log three is. And this should give us a value for Δ𝑆. 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.

Now this is what the question asked us to do: we were told to find the answer to four significant figures. So we’ve reached our final answer. Once again, Δ𝑆 is equal to 3.972 times 10 to the power of negative 24 joules per Kelvin to four significant figures.