Lecture 02. The Boltzmann Distribution Law.
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Transcript: English(auto-generated)
00:13
So let me remind you, we're going to take a quiz on Friday during the minutes of class. We'll be a stack of scantrons right at the front
00:22
of the lecture hall there. Pick them up. Pick up one, please, and find a seat. Turns out there's 116 of you, and there's 240 seats in this lecture hall. So what that means is we don't need two forms for this exam. Everyone, we can sit every other seat in this room,
00:41
and there shouldn't be a problem with that. Okay, so if you would, please make sure that you don't sit exactly next to somebody else. Leave an empty seat between you and the next person next to you. Okay, and we shouldn't have any problem. There's plenty of seats in here for us to do that.
01:03
Okay, it's going to be five short problems, key non-concepts covered in Lectures 1 and 2, Homework 1, and or Discussion Study Guide 1, which was posted to the website on Tuesday morning. And that's the Discussion Study Guide that you're going to be talking
01:20
about in discussion all week. I also posted three quizzes from last year. They're already posted to the announcements page so you can get an idea for what these problems are going to look like. One of the things you'll notice is that you have to be able to calculate a factorial using your calculator. This might be the first time you've had to do that in any class.
01:43
I don't know. But I know when I had to calculate factorial on my calculator, I had to pull out the owner's manual and figure out how to do it, because there's no key that says factorial on it. Okay, so Friday morning would be the wrong time to try and figure that out.
02:02
Look at it before that. Yes? I'm sorry? No, there's no calculator policy. Anything that you can carry in here is fine with me. All right, not only that, but this quiz is open book, open notes, but it would be a mistake to imagine
02:25
that you're going to have the luxury of time to sort through your notes and pause and look through the book and think about what you might want to, time is going to be an issue for you on this quiz. You're going to have to know what you're doing,
02:41
find the right equation if you don't remember it. Boom, knock it out. Right, there isn't going to be much time to do the problems on this quiz. You're going to have to know what you're doing. Okay, so write anything down that you want on a piece of paper if you want to have a list of constants or conversion factors or equations that have been involved
03:03
in the stuff that you've been studying this week, put that on a piece of paper. Anything you want. Yes? Since you have the slides not putting out the lecture notes, should we bring our laptops to have the lecture slides? You are not going to have time to click through the lecture slides.
03:22
What was on slide 40, Dave? No. You are not going to have time to do that. I, that would be a waste of time for you I think. All right, any questions on equation one? Yes?
03:40
Yes. Well, that's why I posted that discussion study guide on the website. Well, it turns out you can go to any discussion section. Right, you know, you might be assigned to the Friday discussion section, but you can go to the one on Tuesday, Wednesday, Thursday.
04:02
There are six of them. So if you want to go to an earlier one because you're concerned that, and you ought to be, okay, other questions? Okay, we're talking about the Boltzmann distribution law today.
04:21
It will be about microstates and configurations and W. We're going to derive the Boltzmann distribution law using a slightly different approach from the one that's taken in your book because you can read what's in your book already and you don't need to hear me regurgitate that to you. I'm going to introduce you to partition functions.
04:41
We're going to do a few examples. Boo, this is a molecule. Alright, it has evenly spaced vibrational states to zero order. We're going to assume it's a harmonic. Vibrations in this molecule are harmonic. Okay, now it's tedious for me to draw this monstrous thing.
05:03
I'm going to draw something smaller that looks like this. This is three molecules. We're going to label them A, B, and C. I've condensed these three molecules into this little notation that I'm showing you here now. These are the quantum numbers for the vibrational states.
05:22
B equals zero. That's the ground vibrational energy level, one, two, three. You remember all this from quantum mechanics. I'm calling this a three-dimensional array or a three-dimensional lattice because I want to emphasize that these three molecules are distinguishable
05:42
in terms of their position. In other words, I can tell where A is. I'm not going to confuse it with B because my lattice has A next to B next to C and I know where A, B, and C are and they're not going to move around. In other words, these molecules can be labeled and kept track of.
06:01
That's an important point because later on we're going to relax that constraint. But for the time being, we know where these molecules are. We can call them A, B, and C. We're not going to lose them. So now imagine you have a three-dimensional array of these molecules, each occupying identical position
06:20
in the lattice. To keep it simple, let's say that we've just got three molecules, A, B, and C. Now let's add three quantum energy to these three molecules. All right? There's three ways to do that. We can put all the energy into molecule A. We can put all the energy into molecule C. Or we can put all the energy into molecule B. That's one way
06:44
to configure the energy, put all the energy into one of the three molecules. We could also put two quanta into one of the molecules and one into the other, leaving the third with no energy in it at all. And there are six ways to do that.
07:04
Finally, we could just put one quantum of energy into each of the three molecules, and there's only one way to do that. I think you can see that intuitively. Now notice I'm not labeling these molecules anymore. Here they're labeled A, B, and C, but I'm going to get rid of that because I'm lazy.
07:24
Don't be confused. I'm always talking about three molecules here. Somebody sent me an email and asked me about that. It's a good question. A, B, and C are still here. Okay. We're going to call each one of these guys a microstate.
07:44
And we've got this short form notation that we can use to identify not microstates but families of microstates, configurations. These curly brackets are going
08:01
to help us identify different configurations that these microstates can be in. This first digit inside the curly brackets is going to be the number of molecules in the ground state. The next digit is going to be the number of molecules in the first excited state, and so on and so forth. That's what this notation will mean.
08:23
So you can see these guys are all in the same configuration. How do I know that? There's two that have the ground state occupied and one that has the excited state, the third, B plus three, occupied. Right? That's also true here, two and one. Also true here, two and one.
08:41
So they all have this configuration here, 2, 0, 0, 1. And these guys analogously 1, 1, 1, 0, boom. And of course there could be any number of zeros.
09:03
Okay? This guy is 0, 3, 0, 0. So we have a notation that we can use to identify these different configurations. That's all this is. There's no interesting science here. Okay. Now, short of figuring out
09:23
and drawing every single configuration, we need a formula that allows us to calculate how many microstates there are for each configuration. Let's see if we can derive one from the simple model that we've looked at already. Because instead of having three molecules, we're going
09:41
to have 10 to the 23. Right? So it will not be practical for us to think about these molecules and the terms that we've been using so far where we're just thinking about three. Okay? So as I build up, let's think about configuration two. You'll recall configuration two, that's one molecule
10:02
that has two quanta, one molecule has one quanta, and one molecule has zero quanta. Alright? As I build up that configuration, I have to, I can put that first V equals 2 into any one of these three molecules, A, B, and C, which are no longer on here.
10:20
Right? It can go into any one of the three molecules. So I put it in A, but it could have gone into B or it could have gone into C. In terms of where I put the second one, once I put that one there, I've got two possibilities of where I can put the second one. And the third one then, once I put that into A
10:40
and that one into C, the third one can only go into B. The last parcel of zero goes to the remaining one molecule. Okay? And so if I want to know how many versions of this are there, it's three times two times one. Three times two times one, that's just three factorial.
11:01
That means there should be six ways to do this and there are. Okay? What about configuration one? Configuration one was three quanta and zero zero.
11:23
Starting with the first, I can put the first one in A, B, or C. This time I relabeled these guys. Then the next one can all go into B here. In this case, it could go into A because I already put the three quanta into B. In this case, it can go into A because I already put the
11:41
three quanta into C. Okay? And then final increment, it's going to go in the place that's left over. I mean here, it can only go into C. Here, it can only go into C. Here, it can only go into B. Now, in principle, the counting
12:01
in this case works exactly the same as it worked in this case. Here, we've got six possibilities, but here I think you can see that there's only the possibility of having three microstates that have this configuration. What changed? What's the difference between configuration two and configuration three, or configuration one rather?
12:24
Well, the difference is that this state is degenerate. V equals zero. It's got two molecules, and in other words, there's two verbally distinguishable ways to create that microstate.
12:41
I can put this guy here and this guy here or this guy here and this guy here. It makes no difference. I'm going to get the same microstate when I do that. So when I count microstates, even though there's in principle six ways, just like there was in the first configuration, three times two times one,
13:02
we're going to need to reduce the total number of configurations by the degeneracy factor. Okay, and in this case, there are two molecules that have the same energy, and so it will turn out to be the case that we need to reduce by dividing by two factorials. So instead of three factorial or six, we're going
13:23
to have three factorial divided by two factorial. That's six divided by two. That's three. So for me, there should only be three ways to make this state, and there are. Now, it may not be intuitively obvious that it has to work this way, and it's rarely the case for me
13:42
that anything is intuitively obvious, and so I work upon the examples, it's like, well, this turns out to be true, but it always works this way. In other words, you always need to divide by the number of states that have the same energy, and if there's multiple states that have the same energy,
14:02
you take the product, two factorial times three factorial times four factorial, for example, we'll do an example like that in just one second. In this case, there's three guys, right, in configuration three, and so that's three factorial, and three factorial divided by three factorial is one.
14:30
Okay, now you can do more examples like this and convince yourself that this is true as I have to when I first looked at this formula.
14:41
So, in other words, we do have a general formula that we can use, this is it. That's the number, total number of microstates, that big W there, that is the number of molecules, that's the number of molecules that have no energy, one quantum of energy, two quatoms of energy, and so on.
15:08
So, consider a system of eight molecules considering four energy quantum. Eight molecules, yep, one, two, three, four, five, seven, eight.
15:30
There's eight molecules there, four energy quantum, really, one, two, three, four, and these are all zeros. Alright, what's the shorthand notation for that configuration?
15:44
Well, there's five guys that have zero, two guys that have one, one guy that has three, and all the rest of these are going to be zeros. Problem one on quiz one.
16:07
What's W? A factorial, one, two, three, four, five.
16:22
Here's another one that's got two and one. What's zero factorial? Okay? Now, if you can't, if your calculator will not, do that, one times two, times three, times four, times five,
16:46
you know, so eight is sort of doable, but when you get to like 27, you know, you really need to figure out how your calculator does a factorial. Right? You will need to calculate factorials
17:00
on Friday during the quiz. Okay? Boom, 168 weights to generate that configuration. Pretty useful formula for us, it turns out. Now, let's look at this a little bit more closely
17:21
because there's something hidden here that is very, very important. Right? That we haven't noticed yet. Right? This is a fundamental fact of nature that you have never seen before. Almost certainly. Right? Really? Related to this?
17:41
I think so. Here it is. Right? Ten molecules now. One, two, three, four, five, six, seven, eight, nine, ten. Put all five quanta of energy. Now, let's make it five in this case. Put all five into that guy. All right? How many microstates are there? Ten molecules.
18:01
Nine of them are all in this energy state down here, so I'm going to divide by nine factorial there. There's ten ways to generate this configuration, and you can see immediately that that has to be true. Right? Ten ways to do this. Ten microstates. What about if I put four in one and one in another?
18:22
All right? What's the W going to be for that? Still ten. I'm just calculating factorial. Ten times nine times eight factorial is ten factorial. Right? There's eight guys that have this ground state energy,
18:41
so I'm going to have eight factorial there, one factorial, one factorial, a bunch of zero factorial. So, that's 90. All right? How do I know that? Eight factorial cancels with eight factorial, and I'm left with nine times ten is 90. So there's 90 ways to generate that microstate right there. Or 90 microstates that have that configuration.
19:04
You with me so far? Now, let's work out all of the other possibilities, which we can readily do. We understand how to do that now. All right? And here are the number of microstates. That's the one we calculated first. Then we calculated 90.
19:21
There's another 90, and this guy right here is 360. And this guy right here, he's 360. This guy is 840. 840 has two quanta in one, one quanta in three, and zero in six.
19:43
All right? And then there's a 252. That's this guy right here. Right? Those are the only possibilities that exist. What's profound about this? Well, that number is bigger than all the rest of these numbers.
20:01
Configuration six, it's just randomly labeled six. Six doesn't mean anything. It's favored by more than two to one compared with any of these other configurations. More than two to one, it's special.
20:20
It's not super special. I mean, it's only special by a factor of two or so. All right? Now I'm just plotting W versus the configuration number, which remember is just arbitrary. Don't worry about configuration number. It doesn't mean anything. But I'm just pointing out, here's that configuration, 840.
20:41
All right? It's special. Now, how special is it? Well, we've only got ten molecules here. What if we had ten to the 23rd? We've only got ten here.
21:01
What if we have, what we're dealing with in chemistry is not ten molecules very often. Once in a while. Most of the time we're dealing with ten to the 20th, ten to the 23rd, ten to the 24th, right? Ten to the 23rd is a mole, right? We're dealing with fractions of a mole or a mole.
21:22
All right? So it's that level that we want to be able to understand here. We want to be able to understand the statistics of large numbers of molecules. So let's make the system a little simpler. Let's make it a two-state system. Heads and tails.
21:40
Right? Only two states for the molecule now. Excited or not excited, let's say. Now our W, here's the number of coin flips. That's just a way to think about molecules.
22:01
Okay? Nothing has changed. I'm just going to evaluate the system multiple times. That's the number of heads, that's the number of tails, that's the number of microstates. For example, if I flip four times, I could get heads, heads, tails, tails.
22:21
I could get heads, tails, heads, tails. I could get tails, heads, heads, tails. I think this notation is now obvious to you. All right? If I want to know, I'm going to get two of each, all right? That's a configuration. How many microstates are there for that configuration?
22:41
Boom. Six. How do I know that? Four factorial, four flips, two of each. Six possible ways to do it. Boom. There they are. Here are all the different ways that you could do that. All heads, all tails, what's the probability of that?
23:02
One, two, three, four, five, six, seven, eight, nine, ten, eleven, twelve, thirteen, fourteen, fifteen, sixteen, one part in sixteen. You got one chance in sixteen of getting all heads flipped four times. Stay with me on this. All right, here's the profound thing.
23:24
On this axis, I'm plotting the number of microstates divided by the total maximum number of microstates, all right? In other words, the maximum number of microstates for different numbers of coin flips or different numbers of molecules, all right?
23:44
Here's six coin flips, ten, twenty, one hundred, one thousand. Okay? Of course, the number of microstates goes way up as the number of molecules goes up, but let's normalize by the maximum number of microstates that can exist
24:01
for each one of these systems as it gets larger. Because we want to see what happens to the shape of this distribution. All right? What happens is very interesting. Here's six coin flips. Here are the data points that I just showed you. Plotted on here, all right? And now, if you increase that to ten, twenty, one hundred,
24:23
one thousand, look what happens to this distribution. All right? It becomes needle sharp, all right? It becomes almost infinitely narrow. It's not infinitely narrow, but it is darn narrow.
24:40
That's only a thousand molecules. We're not dealing with a thousand molecules very often. We're dealing with ten to the twenty-third. What is that distribution going to look like for ten to the twenty-third? It's just going to be a needle. It's just going to be a delta function at this preferred configuration. What is on this axis here?
25:01
These are just different configurations for the system. Right? This axis is just different configurations. This axis is the normalized number of microstates. What am I saying here? I'm saying there is one configuration
25:21
or a small ensemble of configurations that dominate as the system gets bigger. That is a very surprising conclusion. Right? Are you telling me that I can't have four heads? No. You can't have four heads when you've got ten
25:40
to the twenty-third coin flips. In other words, you can't have ten to the twenty-third heads. No. That's not going to happen. Well, it's sort of intuitively obvious, isn't it? All right, but it's less obvious that you can't have any other possibility either.
26:01
In other words, there's really only one possibility that's going to be exactly probable it turns out to be half heads and half tails in the case of coin flips, all right? But more subtle if there's multiple states available to the molecule, more subtle.
26:25
So we can, the main point here is that we really need to understand this configuration right here. What is that? What are its properties? If we can understand that configuration, that is sufficient for us to understand how this chemical system is going to function,
26:42
what its properties are. We only have to understand that guy. We don't have to understand this one or this one or this one or this one or any of the other ten of the ten configurations that may exist here. We only have to understand one for a small ensemble up there.
27:02
The properties of these other systems are going to be so rare that they're not going to contribute to the behavior that we observe in chemistry from macroscopic amounts of stuff. Imagine if that's ten to the twenty-third and not a thousand, boo.
27:26
What does the N stand for in Cornhusker Stadium? Who knows the answer?
27:41
Knowledge is the correct answer. Football players. Knowledge of this highly preferred configuration equates to knowledge of the system as a whole. We only have to know about that guy.
28:03
We only have to know about him. All the rest of these guys we don't need to know about. Okay. It gets sharper. That's the whole thing.
28:23
Now, your book derives the Boltzmann distribution law by finding the W where the maximum W. Right? Now, I think you can see that at the maximum W
28:42
by golly that's the one we want. That's the preferred configuration right there. So, if you find the maximum of any one of these curves, you're going to locate that point and that's what you want to do. Right? But there's another approach that one can take and that is that the derivative here is zero,
29:01
isn't it? The derivative with respect to configuration of W over W mass at this point is zero. In other words, if I draw a tangent there, I'm just going to get a horizontal line. That's another way to derive the Boltzmann distribution law. That's what I'm about to do.
29:21
Today, find the configuration for which W does not change when the configuration of the system is infinitesimally altered. In other words, if I shift the configuration left or right of that peak point slightly, all right, and W doesn't change because the derivative there is zero,
29:40
why I've located the peak of that distribution. I've located that magic configuration. That's the whole point of the Boltzmann distribution law. Consider an, okay, so we're going to do this derivation. You ready? Consider an isolated, in other words, the number of molecules is constant.
30:00
Q, what's Q? Q is the total number of energy quanta in the system. So, when I say isolated, I mean we're not going to change the number of particles. We're not going to change the amount of energy. We're going to leave that constant. Matroscopic, large number of molecules, assembly of,
30:20
and large, and oscillators that share a large number of energy quanta. Large number of oscillators, large amount of energy. Neither one of those things is going to change. Which configuration is preferred? In other words, where is the peak in that curve that I just showed you? Where is it?
30:41
Let's consider just three states. There could be thousands. Let's consider just three. Let's label them. L, M, and N, randomly. If we want to know how many microstates there are, we use our standard formula. That's the total number of molecules, and if this is L, M, and N, then there must be an A, B, C, B, E,
31:02
and so on, all the way to L, M, and N, and then O, P, Q, R, S, C, and so forth. Okay? Now what we're going to do is, you know, this is the configuration that we're starting with. Let's shift it slightly, all right,
31:22
and see if we can find the configuration that doesn't change W. That's the strategy here in this derivation. All right, we're going to shift it infinitesimally, and we're going to maintain the energy constant while we do that. How are you going to possibly, well, here's what you do.
31:42
Energy on this axis here. Here's our states, N, M, L. All right, here's the energy difference between M and L. The energy of M minus the energy of L. That's the difference between M and L. Here's the energy difference between N and M. It's just that difference right there, right? Graphically, this is our situation.
32:01
There's a bunch of states below this and a bunch of states above this. Okay? So now what we're going to do is we're going to shift. All right, we've got some number in state N, some number in state M, some number in state L. Now we're going to shift. Boom. We're going to move two guys from M into N
32:23
and P guys from M into L. We're going to choose P and Q so that the total energy here doesn't change. We're going to choose P and Q so that we maintain the same total energy. All right, but we're going to move some guys from here
32:42
to here and some guys from here to here. And I think you can see because we're going, we're moving some to lower energy and we're moving some to higher energy, it's reasonable that you could maintain the same total energy, right? Yes. No. They're not supposed to be evenly spaced, but thank you
33:00
for asking that question. Okay, so what concerns energy? Well, Q times that difference, N minus L, N minus M rather. Q times that difference minus P times that difference has to be equal to zero.
33:21
So we're going to choose P and Q so that they satisfy this equality right here. Now, so before, here's our number of microstates. We've already said they're given by this equation. Before the shift, all right, this equation is true. After the shift, I've added some number of molecules to L.
33:44
I've removed molecules from M, moved them to P, moved P to L and Q to M, all right, and N got larger by N amount Q. And so this is the shift here described by these three equations right here.
34:01
All right, and what I'm saying is that we want to consider the case where the before equals the after. We want to find the microstate where after we do this infinitesimal shift, W doesn't change. That's the whole point.
34:20
All right, so if the rate of change of W with configuration is zero, that means that these guys have to be equal to one another. In other words, right at the peak of this curve, W before equals W after if, and only if,
34:43
we are right there. If we're here and we shift in either direction, it's not going to be true. W changes with configuration, right? It only doesn't change with configuration up there.
35:00
So that has to equal that. And if you work out the math, we're going to skip some steps. Canceling or rearranging terms. You don't have to do this on quiz one. Take my word for it, lots of cancellations occur here
35:23
and when you're done, this equality holds. But a more useful version of this is if you just take the natural logarithm of both sides here, you get this, all right, P times log of NL over NM equals Q times log NM over NN.
35:41
It doesn't look super useful. Okay, and substituting from the expression for the energy defined earlier, we have, all right, I plug this in for Q. I plug this in for P. And I get these guys right here after actually I cross multiply.
36:05
And then you've got this equation right here that looks completely useless, but what you have to realize is that M, M and L, M and N are just random letters. I could have chosen any three states and done this manipulation, okay?
36:25
And so it turns out to be the case that these two things can only be equal to one another. You know, if I permutate these states, right, if it's the case that I could have used A, B and C
36:46
or X, Y and Z and gotten the same result, that implies that this has to be a constant, right, or this has to be a constant and that may not be intuitively obvious because it's really not intuitively obvious to me, but it turns out to be the case.
37:05
This can only be true if the left and right sides of this equality are a constant. They do not depend on exactly which states are involved, which states we chose. We chose L, M and N. Let's call this constant beta.
37:20
In other words, I'm going to take one of these guys and just set them equal to a constant. We're going to call that constant beta. Okay, so now I can rearrange this. I can multiply by that energy difference. Boom. It moves over to the right-hand side. So I've got log of N of N over N of I. I chose just N and I now to label these states.
37:47
Completely general expression. It doesn't matter which two states you choose, right? Subject to the constraint that that's the energy difference between the two states right there. That's the only constraint.
38:02
Okay, now if we choose one of these guys to be the ground state, let's say that guy, I, right, so instead of N sub I, we'll have N sub zero. Let's assume that the energy of the ground state is zero. In other words, that there's no zero point energy. Okay? And so instead of having E sub I minus E sub N,
38:23
we're just going to have minus beta times E sub N because E sub I is equal to zero ground state, all right? So we get this expression here and if we just take E to the both sides, take the exponential of both sides rather, we get this expression right here.
38:44
This is the famous Boltzmann distribution law. What it says is the ratio of the populations of these two states, state N, which is elevated by energy E above the ground state, right? The energy of the state is E higher than this ground state, right?
39:04
We can calculate the population of that excited energy level using this equation right here where beta turns out to be equal to one over KT, where K is Boltzmann's constant. Is that on the next slide? No. Beta is one over Boltzmann's constant times temperature, one over KT.
39:32
Leonard Nash, who is my hero, says for any isolated microscopic assembly of species with any kind of spacing between the states, any kind of state spacing,
39:43
the predominant configuration has been fully defined by an explicit functional relation between the energy and the population of each quantum level. In other words, this equation holds true not for every configuration, right? For the only one that matters, right?
40:02
The one that dominates the number of configurations that are out there. Remember, when you have a large system, it's dominated by one configuration or a very small number of them, all right? What we're saying is that this equation here characterizes that preferred configuration. That preferred configuration is very important.
40:25
Any chemical system is dominated by it. But there are all these other ones that we're really going to ignore. And we ignore them in all the thermodynamics, at least all of them that I know about.
40:42
Okay. Now, we also need a different form of the Boltzmann distribution law. And you can look at these slides later, and there's a little derivation of that here. Total number of molecules, molecules in the ground state plus the number of molecules
41:05
in the first excited state plus total molecules in the second excited state. We can write a summation formula in terms of that. There it is. These are summation over all of the states that are occupied. And so we can write one of these Boltzmann distribution laws for each one of these terms.
41:22
Look at that. Isn't that nice? All right, N2, if that's N2, then that's got to be E2. Yeah, sure enough, there it is. All right, N2, E2, N0.
41:41
Yes. Let me just click through this. So solving for N0, we have this equation right here. And I can then plug this N0 into this N0 right here. Nope. Let me do that backwards. Let's put solve for N0 here, plug that into this guy, I get this equation right here.
42:03
This is your equation, 13.6A. You don't actually have to do this derivation yourself. I'm just showing you where this equation came from. It's not totally obvious where this thing comes from. All right, you can just derive it from the simplest version of the Boltzmann distribution law.
42:21
All right, what matters is how we use this equation. This thing in the denominator looks innocuous, but it turns out to be super-homogeneous. It's super-important, all right. It's something called the partition function. All right, we're going to keep coming back to this.
42:41
As innocuous as this seems, this is the central idea that we keep returning to in statistical mechanics. All right, what is that thing? It doesn't look like anything in particular, does it? It's the summation of this exponential where we're going to put a bunch of different energies in here and sum them up.
43:02
I mean, how important could it possibly be? Okay, so this is the molecular partition function for that particular case that we're talking about, where we've got evenly spaced states and a ladder partition function for that case.
43:26
The molecular partition function, we usually call it Q. There's two versions, it turns out. There's a version that contains the degeneracy, G. The degeneracy is the number of molecules that have the same energy in a particular state, all right.
43:43
So if the ground state happens to have four molecules in it, no, no. Let me say that a different way. If there are four ground state energies that all have the same energy of zero,
44:00
the degeneracy would be four. We'll do an example. Okay, this is your equation 13.9. All right, two versions of this equation. I never learn anything until we start to do examples. So let's do some examples. So if any of this starts to make sense.
44:23
What are the relative populations of the states of a two-level system when the temperature is infinite? We've got two states that are going to be very high. What's the relative population of those two states?
44:43
In a system that's got a lot of molecules in it, right? This happens to be a case that we encounter all the time. Here's our equation, all right, for the population of some state I. Two states, so let's call one state two and one state one.
45:05
Let's write one of these equations for state two and one of these equations for state one, all right. What's the difference between them? I've got E2 here for state two. I've got E1 here for state one. Otherwise, the denominator is the same summation.
45:22
It's identical, okay? So if I cancel terms, I'm asking about two states in the same system, so big N is the same. There's only one system here that has N molecules in it, right? This thing in the denominator, this partition function,
45:41
that's the same between these two terms, right? The only thing that's different is this exponential, and so I can just pull that out. Everything else cancels, so I've got this exponential for energy two corresponding to N2, and this exponential for energy one corresponding to state one, and so I can simplify that, all right,
46:02
and here's what the resulting exponential looks like. Okay, and remember, beta is one over KT where K is Boltzmann's constant, all right? And so if I want to calculate this, I just plug in one over KT, but I said T is infinite.
46:26
That symbol is supposed to represent, okay? And so if the denominator here becomes very large, I'm going to have E to the zero. E to the zero is one, so what we conclude is that the population in these two states will be equal.
46:43
If the thermal energy of the system is way higher than either the energy spacing between them, I think that's already counterintuitive, all right? Wouldn't you expect that the excited state to be preferentially occupied at high temperature? I mean, it's my simple way of thinking about it.
47:02
That's what makes sense to me, but that's wrong, all right? If the energy, the thermal energy of the system dictated by the temperature, if that is high compared to the energy difference between these two states, all right, they will be equally occupied.
47:22
That's an important thing to understand, it turns out, in lots of different kinds of physical chemistry that we're going to be getting to later on. Let's do another example. A certain atom has a threefold degenerate ground level. The non-degenerate electronically excited level of 3500 wave numbers and the threefold degenerate level of 4700 wave numbers.
47:41
Calculate the partition function for these electronic states at 1900 degrees Kelvin. Here's the situation. What do these words mean? A threefold degenerate ground level, boom, energy zero.
48:09
An electronically excited energy level of 3500 wave numbers, boom, 3500 wave numbers. A threefold degenerate level at 4700 wave numbers, boom. All right, calculate the partition function
48:23
of these electronic states at 1900 degrees Kelvin. Partition function, what is the partition function? It's this guy, this is the denominator of that expression that we were looking at earlier. I'm using the version, there's two partition functions,
48:42
I'm using the version that has the degeneracy in it because we've been told about the degeneracy. Here's the partition function, here's the version with the degeneracy, here's the version without it, we need this one because we have degenerate states in this system.
49:01
Okay, so what does the summation mean? It means that we're going to have to add up, we've got three states, we've got three terms in our summation. Ground state, first excited state, second excited state. Let's figure out what those terms are. The degeneracy of the ground state is three, boom.
49:25
Okay, so that's three times the exponential minus beta times zero because the energy of the ground state is zero. Plus non-degenerate e to the minus beta times 3500 wave numbers plus triply degenerate e
49:41
to the minus beta 4700 wave numbers. Three terms because there's three states. Okay, now, one of the things that we're going to have
50:02
to do constantly is move between different energy units in this class. Wave numbers, joules, eV, all right. The way that I do this is clumsy but reliable. All right, I remember two constants, the conversion from wave numbers to eV, that's 8065.5 wave numbers per eV.
50:25
And the conversion from joules to eV, there's 1.602 times 10 to the minus 19 joules per eV. If you remember those two conversion factors, I submit to you that you can obtain any unit that happens to be a joule, eV, or a wave number.
50:45
For example, 3500 wave numbers, what is that in joules? Why do I care? Because Boltzmann's constant is almost always 1.381 times 10 to the minus 23 joules per Kelvin. Boltzmann's constant is always in joules. So other energies, if they're in joules, it's very helpful.
51:02
Because then you'd have to know Boltzmann's constant in different units. Always use 1.381 times 10 to the minus 23 for Boltzmann's constant and change the units of the other stuff. So 3500 wave numbers, what is that? 3500 wave numbers divided by 80, you know, go back to your Chem 1 training and make sure
51:21
that you cancel units and get the right ones, all right? Because we're going to be moving between energy units all the time. And this is the way I do it, all right? Divide by 80, 65.5, wave numbers cancel. So that product gives me energy in eV.
51:41
Then I multiply by 1.62 times 10 to the minus 19 joules per eV. Now I get energy in joules. And then I divide by kT and I get 2.649. 2.649 is that number right there. There will be a minus sign in front of that, right there.
52:03
All right, so these three terms then are going to be 3 because that's just e to the 0 times 3. So that's 3 times 1, that's 3. That guy turns out to be .07069. That guy turns out to be .0855. Add them up, I get 3.156.
52:25
Does that make sense? What is the partition function? It's the number of thermally accessible states. How much thermal energy is there in the system? What is kT? kT turns out to be 1300 wave numbers. If you convert that number, 1900 kelvin,
52:43
convert it into units of wave numbers, it's 1321 wave numbers, all right? And so what this calculation tells us is that there should be a little more than three thermally accessible states in this system. Well, there are. These three states are thermally accessible. They're the ground states for good mistakes.
53:03
All right, and there's a tiny probability that these other states will be occupied. That's what this number indicates. 3.1 thermally accessible states. Yeah, because there's not enough thermal energy to significantly populate these excited states. The thermal energy is too low.
53:25
Okay, so we'll review this on Friday. Thanks for watching!