Video: Determining the Difference in Molar Entropy between Systems of Distinguishable and Indistinguishable Particles

Consider two systems that are identical in all respects, except that in one the molecules are distinguishable, whereas in the other the molecules are indistinguishable. What would be the difference between the molar entropies of these two systems?

09:20

Video Transcript

Consider two systems that are identical in all respects, except that in one the molecules are distinguishable, whereas in the other the molecules are indistinguishable. What would be the difference between the molar entropies of these two systems?

Okay, so in this question, we’ve got two systems that we’ve been told are absolutely identical, except that in one system the molecules are distinguishable and at the other they’re indistinguishable. In other words, in the first system, we can tell the molecules apart from each other; we can distinguish one particle from another. And in the second system, we cannot do this; we cannot discern one particle from another. That’s why this system is made up of indistinguishable particles. Now in this case of course, the particles that we’re talking about are the molecules that make up the system.

Anyway, what we’ve been asked to do is to find the difference between the molar entropies of the two systems. In other words, we need to find the difference between the entropies of one mole of each one of these systems. So we’ve got two systems, and we need to find the entropy of each one and then find the difference between these entropies. In other words, we need to find, we need to find the difference between the entropy of the distinguishable system and the entropy of the indistinguishable system. Of course 𝑆 represents entropy and the subscript d and i tell us if it’s the entropy of the distinguishable system or the indistinguishable system.

And in each case, we’re calculating the molar entropies. So we know that we’ve got one mole of molecules in each system. So here are our two systems: the distinguishable system d and the indistinguishable system i. Both of these have one mole of molecules in them. There we go! The pink dots represent molecules. Now in the distinguishable system, the one on the left, we can tell molecules apart from each other. In other words, if we wanted to, we could label each one of these particles as particle one, particle two, particle three, particle four, and so on and so forth. And we could associate our particle number to each one of these particles.

Now of course it doesn’t matter how we number them. We could call any of these particles particle one, or particle two, or particle three, or so on and so forth. But the important thing is that we can distinguish them from each other, so we can assign a label to them. And even after some time has passed and the system has done its thing, so all the particles have moved around, we’d still be able to tell which one was particle one, which one is particle two, which one is particle three, and so on and so forth. And that’s what makes the system a system of distinguishable molecules.

However, with the indistinguishable system, we cannot do this. We can’t label them particle one, two, three, four and so on, until 6.02 times 10 to 23 which is Avogadro’s number cause that’s how many particles there are in one mole. We cannot do this, because if we were to label this particle, particle one, this one particle two, and so on and then leave the system to do its thing and let the particles move around, after some time we will not be able to tell which one is particle one, which one is two, so on and so forth because it’s made up of indistinguishable particles. We cannot tell them apart from each other and that is the key difference between the two systems.

So why is this relevant to calculating each one of their molar entropies? Well, we can recall that the Boltzmann formula for entropy is 𝑆 is equal to 𝑘 𝐵 multiplied by the natural log of 𝛺, where 𝑆 is the entropy of the system, 𝑘 𝐵 is the Boltzmann constant, and 𝛺 is the total number of possible microstates that the system can occupy. Now a microstate is simply one possible way of rearranging all of the particles in a system. So what we’re going to do here is to take a snapshot in time. We’re going to freeze these particles in time. And now let’s look at system d first: the distinguishable system.

Now all of the particles are found in the particular places that we’ve drawn in the diagram, but what we’ve drawn is simply one way of arranging all of those particles at those positions in space. More specifically, we’ve drawn one way of arranging all the particles with their specific velocities. So let’s say, for example, this particle is moving in this direction really quickly, this one is moving in this direction rather slowly, so on and so forth. And this diagram that we’ve drawn is simply one way of rearranging all of the 6.02 times 10 to the power of 23 molecules. But there are more than one ways of arranging this.

For example, we could if we wanted to swap particle one with particle four. So what we have now instead is particle four on the left and particle one on the right. And now this system is different to what we had earlier, because earlier we had particle one in the position where particle four is now and vice versa. And because the particles are distinguishable from each other, this is a different microstate. So we’ve just found a second possible microstate. But as you can imagine there are lots of different ways of arranging 6.02 times 10 to the power of 23 particles. In fact, the total number of ways of rearranging 6.02 times 10 to the power of 23 particles is given by 6.02 times 10 to the power of 23 factorial. So this is the total number of ways of rearranging our system. And hence it’s also the total number of microstates, 𝛺.

Now to save us the hassle of writing 6.02 times 10 to the power of 23 every time, we’ll just say that this is an 𝑁 sub 𝐴, also known as Avogadro’s number. And once again to reiterate the reason that we’re using Avogadro’s number is because we’ve got one mole of molecules in our system. That’s how we work out the molar entropies of these systems. So we found out the value of 𝛺 for system d. Hence, we’ll say 𝛺 sub d is equal to 𝑁 sub 𝐴 factorial. And at this point, we can plug this value back into our equation for the entropy to tell us that the entropy of the distinguishable system, 𝑆 sub d, is equal to the Boltzmann constant multiplied by the natural log of 𝑁 sub 𝐴 factorial.

So now that we’ve got an expression for 𝑆 sub d, let’s work out one for 𝑆 sub i: the entropy of the indistinguishable system. To do this, we need to work out what 𝛺 sub i is: the number of possible microstates of this system. Well this time for the indistinguishable particles as we said earlier, we cannot label them particles one, two, three, four, and so on, and so forth. So if we were to swap two particles, let’s say we swapped this particle here with this particle here, the resulting state would be exactly the same as before. Because to us, we cannot tell the difference; we don’t know the difference between two particles, so we don’t know the difference between the two states when the two particles have been swapped.

In other words, regardless of how we swap particles around, the system is identical to what it was before. And therefore, there is only one way of arranging the system. Because like we said earlier, if we swap particles around, the system is still exactly the same. And so 𝛺 sub i is equal to one. And at this point, we can work out the entropy of the indistinguishable system. 𝑆 sub i is equal to 𝑘 𝐵 multiplied by the natural log of 𝛺 sub i, and 𝛺 sub i was one. But the natural log of one is zero. And so we find that the entropy of the indistinguishable system is zero. Now at this point, we’ve got the value for 𝑆 sub d and we’ve got the value of 𝑆 sub i, which is zero. So the difference between the molar entropies of these two systems is simply 𝑆 sub d minus 𝑆 sub i.

But 𝑆 sub i is equal to zero, so the difference between the molar entropies simply as 𝑆 sub d. At this point, we can actually work out a numerical value for 𝑆 sub d. We know that 𝑆 sub d is equal to 𝑘𝐵 multiplied by the natural log of 𝑁 sub 𝐴 factorial. Now the natural log of 𝑁 sub 𝐴 factorial is something that your calculators will not be able to handle. In fact, many of them will fall at the first hurdle; they won’t even calculate 𝑁 sub 𝐴 factorial let alone the natural log of it. So to work out what the natural log of 𝑁 sub 𝐴 factorial is, we need to use something known as Stirling’s approximation. This approximation tells us that the natural log of some large value factorial is equal to that value 𝑛 multiplied by the natural log of 𝑛 minus 𝑛.

And we apply this approximation to our value of 𝑆 sub d. So we say that 𝑆 sub d is equal to 𝑘 𝐵 multiplied by 𝑁 sub 𝐴 log 𝑁 sub 𝐴 minus 𝑁 sub 𝐴, because we replaced the log of 𝑁 sub 𝐴 factorial with 𝑁 sub 𝐴 multiplied by the natural log of 𝑁 sub 𝐴 minus 𝑁 sub 𝐴, just like in Stirling’s approximation. Now at this point, this is something that calculators can generally compute. Using these values for the Boltzmann constant and Avogadro’s number, we can stick these into our calculator to give us a value of 𝑆 sub d. When we do this, we find that we get a value of 446.9 to four significant figures. But now what are the units? Well we’ve got the units of entropy, which is joules per Kelvin. But remember, this is a molar entropy. So we need to divide this by a further unit of moles.

In other words, this is the entropy per mole for the distinguishable system. And we can write the units more simply as joules per Kelvin multiplied by moles. Now finally we recall that 𝑆 sub d was the same as the value of 𝑆 sub d minus 𝑆 sub i, because 𝑆 sub i was equal to zero. So by finding the value of 𝑆 sub d, we have found the difference between the molar entropies of these two systems. And so just to be more clear, we’ll say that this is equal to 𝑆 sub d minus 𝑆 sub i. And at this point, we’ve arrived at our final answer: the difference between the molar entropies of these two systems is 446.9 joules per Kelvin per mole.

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