 The rotational partition function. That's what this lecture is going to be about. Let me just point out that videos of these lectures are becoming available on the lecture page of our website. Here's the first four. Why are we posting these lectures? Why are we making videos of them? I mean, we're trying to disseminate this information as broadly as we can, even beyond this classroom. But for your purposes, this hopefully, if you miss a lecture for any reason, you can go back and listen to it if you want to. Or you can look at lectures before you study for the midterm exam. Maybe you think you might have missed something I said, and it's all going to be there. I will be an expendable component at the end of this quarter. All the information in the quarter will be on this website. That is a cause of some concern to me, actually. Okay. We talked on Wednesday about ideal gases a little bit. Mainly, we were talking about the translational partition function. Let's just review that quickly, and then we'll talk about the rotational partition function. As we said on Wednesday, if we neglect the electronic energy levels that are present in a molecule, we have three manifolds of states within molecules that determine their thermodynamic properties. Three manifolds of states, translation, rotation, and vibration. If we can neglect the electronic degrees of freedom. Why do we do that? Because electronic states are separated by huge amounts of energy, 20,000 wave numbers. It's a round number. Huge amounts of energy differences. Most of the time, we can think of these molecules as existing in a single electronic state. We can calculate the thermodynamic properties without considering the possibility that they might be able to access more than one electronic state. So we said also that we're fortunate we can decouple these various degrees of freedom, and we can express the total partition function for any molecule as a product between them, and what we're going to do is we're going to make the assumption that the partition function for the electronic degrees of freedom is just one. There's only one thermally accessible electronic state. Now we're going to focus attention on the translational partition function. This is what we did on Wednesday, and we said, look, the way statistical mechanics contends with translation is to imagine that it occurs within the confines of some volume. Let's say that that volume is well defined geometrically. Let's say it's a box. It can be a one-dimensional box or a two-dimensional box or a three-dimensional box. And if that's the case, we can use the simple particle in the box equations that we remember from quantum mechanics to calculate the available translational energies for the molecule. Okay? Here's our particle in the box expression for one dimension, x. And the zero point energy is not allowed, right? Here's the lowest allowed value of the energy, and here's the second, and you can see the energy spacing increases as a function of quantum number n. So the molecular partition function for q, all that involves is we only have to make a substitution for this energy, right? We use the particle in the box energy expression, and so that's our partition function for translation in one dimension. We just have the one-dimensional expression for the particle in a box. Very simple. One consequence of this expression though is that the energies that we're going to calculate here are very closely spaced. All right? And we did an example on Wednesday in which we considered a box that was one micron in size. We talked about the size of a micron. A micron is very small, about the size of a bacteria, which is about the smallest thing that you can see in an optical microscope. And so when we did this calculation, here's the difference in energy between the ground state and the first excited state. That difference in energy we can calculate, of course, is unbelievably small, right? One ten millionth of a wave number. We know wave number is a pretty small denomination of energy already. All right? So these translational states are quasi-continuous. All right? There's hardly any energy spacing between different translational states. Even in a really, really tiny box, and of course, as the box gets bigger, that state spacing gets smaller, doesn't it? So you can think of that as being sort of an upper limit that you would practically encounter. You're never going to see a box that's a micron in size, or very rarely. And the state spacing gets bigger, of course, as you go up in energy, the difference between states gets bigger, but it stays pretty small. I mean, it's 10 to the minus 28, roughly, for N equals 100. Between 100 and 101, that state spacing still on the order of 10 to the minus 28. Still a really, really tiny number. Okay? So if these states are quasi-continuous, we don't have to think about this summation. It's completely impractical. You know, there's going to be tens of thousands, or millions of states, millions of terms in this summation. We don't want to have to evaluate it explicitly. We don't have to, because it's quasi-continuous. We can think about it as an integration. This is the integral we have to evaluate. This is the trick that we talked about on Wednesday, and when we're done working the, evaluating that integral, here's the expression that we get for the one-dimensional translation. One-dimensional translational partition function. And so this is the calculation that we did for an argon atom in a one micron, one-dimensional box. We calculated 62,700 thermally accessible translational states. And if we did that in three dimensions, here's the three-dimensional expression for the translational partition function. It doesn't fit on the screen, but I can put it on two screens. And there, that's the number, all right? Ten to the 14 translational states at 300 degrees Kelvin. So we're never going to see in a laboratory experiment any evidence of this quantization of energy for translation. We're never going to see it, all right? We're never going to have the resolution to tell the difference between one state and another when they're this close together, all right? It wouldn't be practical to even look for that as an experimental outcome. It's going to be so hard to make that measurement. This equation makes sense, this equation right here because with increasing mass, the translational states are closer together, right? Here's our equation for the energy of these states. M is in the denominator, right? So we expect the translational partition function to go up with M, and it does. In other words, if there's a certain amount of thermal energy in the system, all right? There's going to be more accessible translational states if the mass of the thing is bigger. As mass goes up, Q goes up, right? In proportion to the 3 halves power. And the same thing is true of volume. This makes sort of intuitive sense. As V gets bigger, V is also effectively in the denominator of this expression, and so we expect the partition function to go up with volume also. It's directly proportional to volume. Okay, so there's some substitutions that we can make. We can simplify this expression somewhat in terms of the thermal wavelength, and we talked about that. And then we can calculate the translational energy, the internal energy due to translation, right? By using this expression for the energy that we derived earlier, and basically all we have to do is substitute V over whatever that Greek symbol is cubed, all right? There it is. We make that substitution there. We make the substitution here. We work through the derivative, and we find that the internal energy due to translation is 3 halves NKT, which is just 3 halves NRT, all right? Where N is the number of moles. If we make N N sub A, Y, that turns into an R. And this is the internal energy that we had been assuming in the Equipartition Theorem calculations that we were doing on Wednesday anyway, wasn't it? It's just a proof that that's right. Now, we haven't talked about the enthalpy, and we're really not going to talk about it today. But in the future, we will show that this equation is true. The enthalpy H is equal to the internal energy plus PV. And if that's true, PV equals NRT. And so the enthalpy is just 5 halves NRT. So we've got an expression from statistical mechanics for the internal energy and for the enthalpy. In fact, we know all of these things because earlier we talked about the entropy. This is the Sacre-Tetro equation that we derived last week. And so we've been able, using statistical mechanics, to calculate all of these thermodynamic parameters, including the partition function, which is not strictly speaking a thermodynamic parameter. The connection here with translation is all in the mass of the molecule. That's the attribute of the molecule that we're connecting to a thermodynamic property. At the beginning of this class, we said, look, what we want to do with statistical mechanics is connect the properties of the molecule, bond distances, geometry, mass, vibrational frequencies with the thermodynamic properties of the molecule that we can measure. Always we're measuring these thermodynamic properties on huge ensembles of molecules. So in the case of translation, we've done that. The only connection is in mass. That's the only molecular property that couples with this translational partition function. Okay? The translational partition function really only depends on the mass, and with this guy we can calculate that and that and that. Okay? So we have established a connection to molecular properties and thermodynamic properties, a connection that was absent in traditional thermodynamics. This begins to fulfill what we set out to do at the beginning of the class, just barely. We're just barely starting to do that now. We're doing it with translation. So in other words, we can measure these properties for molecules that don't store energy any other way. They don't rotate, they don't vibrate. Okay? So atomic gases. Now, what if the molecule can rotate? Then it's got another way to store energy. We have a manifold then of rotational states that looks like this, all right? Where this manifold of rotation state nests in between two vibrational states of the molecule, all right? There's a manifold like that here, and there's another one here, and there's another one here, and there's another one here, and then here, there's another one. Okay? So we're adding a lot of complexity to this picture. Now, I know you've seen this, and I had a whole lecture written that reviews rotational spectroscopy, but we've thrown it out. You'll notice it's still on the web page. We're not going to take the time to review all of rotational spectroscopy. You don't really need to know it for what we're doing, but what you do need to know is you have to understand what the energies of these states are and what their degeneracies are. You have to remember that. The energies are given by this expression, h bar squared over 2i times j times j plus 1, where j is the quantum number that applies here, right? j can be 0, 1, 2, 3, 4, and so on. What's i? Moment of inertia. Okay? So we can calculate these energies, right? Just by substituting it in different values of j, and if we remember that b is h bar squared over 2i, why we can just denominate these energies in terms of b instead of using other units? It's just convenient for us. All right? So the ground state, 0, first excited state is going to be 2b, 6, 12, 20. Hopefully what you recall is that these unequally spaced states give rise to perfectly equally spaced rotational lines in a rotational spectrum. Does everybody remember that? You remember the reason why? All right? There's a selection rule that j has to be equal to j. j final can be j plus 1 or j minus 1. Okay? So let me just emphasize that when I write b, I write it like this, all right? Because I'm writing it in joules, right? Do I really get joules when I evaluate these units of these other things? Yes, all right? h is js, joule seconds, i is kilogram meters squared, okay? And so when I work through those units, yes, I get joules, all the rest of the stuff cancels, okay? And I'm comfortable with that, all right? I like being able to calculate stuff in joules because then I can use my two conversion factors and get wave numbers or ev, all right? So I'm always doing this, I am very rarely doing this where I write b with a squiggle on it. That means b is denominated in terms of wave numbers and then you've got to do a conversion here and I just get all messed up when I have to do that. So your book will constantly use this equation. I will constantly use this equation. Don't be confused by that. Your book wants to do everything in terms of wave numbers. I'm going to do everything in joules like I always do and then convert later if I need to. Okay? So let's work out the expression for the rotational partition function. We've got this energy, we know what j is equal to and so let me remind you that there can be more than one moment of inertia in this expression depending on how symmetrical the molecule is, it may have a different moment of inertia in x than in y and in z. Okay? There can be three different moments of inertia and if that's the case, why the overall partition function for rotation is just going to be the product of x, y and z. We'll come back to that but for now q applies to each one of these three orthogonal axes, x, y and z and we can write an expression for one of them. This is just our standard expression for the partition function. Now we're just substituting in the energy for e and the degeneracy of each one of these states which is always equal to 2j plus 1. So you have to know that. Okay? So if we write out this series, let's write out the first free few terms, j equals 0, that's going to be a 1, right? If j equals 0, this is just going to be equal to 1. If j is equal to 2, this is going to be 3 and so on and so forth. Okay? And so we could write out the individual terms of this partition function if we want to. However, at room temperature, how many of these states are there going to be? How many thermally accessible rotational states are there going to be at room temperature? What's the state spacing order of magnitude in rotation? One to three wave numbers, let's just say one. How much thermal energy is there at room temperature in units of wave numbers? 207. So this is going to be first order hundreds, a hundred rotational states that are accessible at room temperature. Do we want to write a hundred terms in this expression? No. All right? And room temperature is a temperature that we care a lot about. Okay? So we can treat this expansion here much like we did in the case of translation. These states are way further apart than the translational states, but they're still quasi-continuous. Okay? And so we can, instead of writing a summation, we can do an integration in much the same way that we did with translation. And I'm not going to walk you through this integral, but you can see it if you want to go back later on and look at these slides. At the end of the day, this is what you calculate if you evaluate that integral, all right? The partition function for one dimension, right? 8 pi squared i, the moment of inertia times kT over h squared. We have to customize this expression for molecules of different shapes, three versions. Linear molecule lacking a center of symmetry. Boom. Any linear molecule. Boom. Nonlinear molecules, boom. Now you're never going to have to memorize equations like this. They'll always be available to you either in your book or on the exam. All right? But we have to be able to use them to calculate partition functions. Okay? So we're going to focus on that. First thing to understand, sigma. We've got to be able to figure out what sigma is for any molecule, all right? It's the symmetry factor. This a, b, and c here, those are a little bit cryptic. We haven't seen those before. Those are just the rotational constants about the three major axes of rotation x, y, and z. We won't need to say any more about them. But the symmetry factor, what is it? We do have to understand that and be able to figure out what it is for any molecule. Now in principle you can use group theory to do this and I understand Professor Martin beat you over the head with group theory. And I'm happy that she did that, but we want to just have sort of an intuitive feel. We want to be able to look at a molecule without doing the group theory on it. We want to be able to figure out what its symmetry factor is, okay? What's the symmetry number? HCl, that's the Cl, that's the H. There are no indistinguishable orientations of this molecule about any axis. What does that mean? Well, if I rotate the molecule about this axis right here, the chlorine will move from this side to this side, the hydrogen from this side to this side, and I can tell the difference. If I rotate it by 180 degrees I can see that it's rotated. If I put an axis through the molecule like this and I rotate it by 180 degrees I can tell that it's rotated. If I rotate this molecule I can always tell that it's rotated. There are no indistinguishable orientations of this molecule about any of its axes. All right, that means its symmetry number is one. Take HCl2 minus, this is a perfectly symmetrical molecule. It's got two indistinguishable orientations about its axis. Where are they? Well, if I label this chlorine 1 and this chlorine 2 because otherwise I can't tell them apart. If I label them and I do 180 degree rotation the two goes over here, the one goes over here, and if I rotate again the two goes over here and the one goes over here, those are indistinguishable from one another. And no matter how I put this axis through the molecule there's only two of them. The symmetry number is two. What is the consequences does that have? Well, the partition function is only going to be half as big for this guy, all right? It has direct consequences for the partition function. Ammonia, three indistinguishable orientations. What are they? Well, the axis that matters is the one that runs through the center of the nitrogen, the threefold axis of the molecule and if I label the hydrogens I can see that there's three indistinguishable orientations that exist around that axis, okay? And so the symmetry number here is three. Okay, what if I had a tetrahedral molecule? There's only one symmetry axis for this molecule that could, where rotation about that axis gives rise to an indistinguishable molecule about each symmetry axis. And this molecule is only one but we're going to talk about cases where there's more than one, okay? I never learned anything by actually writing down what the rules are. I mean, if I wrote down what the rules are for determining symmetry number I wouldn't understand them. All right, you have to, the only way I ever learn anything is by doing examples. Okay, tetrahedron, what's its symmetry number? Twelve, all right, all right, why is it twelve? All right, there are four threefold axes, four. Where are they? Each axis starts at the, one of these points and goes through the center of the triangle across from it. One, two, three, four vertices in a tetrahedron. All right, so there's got to be four axes that go right through the center. So for this guy it's going to be right through there. For this guy it's right through here. For this guy it's right through there. And hopefully you can imagine that if you rotate about an axis that goes through this triangle here, there are three equivalent orientations around each one of those axes. The other way to think about it is that there are six twofold axes. Huh? Yeah, in other words, if I draw an axis through the center of, I'm having a hard time, yeah, so imagine an axis that goes through the center of this edge and the center of this edge. See an axis that goes like that? Or an axis that goes from the center of this edge to the center of this edge. Those edge centers are twofold axes for a tetrahedron. All right, and there are six of those guys, right, each one of those twofold axes has twofold symmetry. Six times two is 12, or three times four is 12. I probe the same symmetry elements independent of which one of those two sets of axes I choose. I can't choose them both. I get 24, that's the wrong number. I've counted the same symmetry elements twice if I do that. All right, so you choose one set of equivalent axes. You either choose to go from the vertex to the center of the triangle or from the center of this edge to the center of that edge over there, or the center of this edge to that edge over there. You either use the edges or you use the vertices, you can't use them both. But you should get the same answer. So it's a nice way to check whether your symmetry number is correct. You should get the same answer no matter which set of vertices, which set of axes you elect to use. For these other geometries, that guy's 24, that guy's 24, that guy's 60, that guy's 60. See if you can get those numbers. That's eight times three, it's also four times six. That's six times four, it's also three times eight. That's 12 times five, it's also 20 times three. In other words, there's different sets of equivalent axes for each one of these highly symmetrical geometries, probably more than two. Let's do an example. What is the symmetry number of CO2? Linear molecule, two, right? Ozone, also two. SL3, six, why? How come it's not like ammonia? So the first question to ask is SL3 flat or is it trigonal bipyramidal? It turns out it's flat, 24 electrons, S ends up with no extra electrons, the sulfur has no lone pairs on it, it's got to be flat. If it's flat, how does this work? While there's two ways to get this number, there's a symmetry axis for the molecule right through the oxygen and the sulfur and there's three of those. I think you'll agree that each one of those axes has two equivalent orientations. This is the one that is absolutely clear. So you've got two, two, and two or six total for the symmetry number. The other way to think about this is what if you had an axis that goes right through the sulfur perpendicular to the screen, such an axis would have three-fold symmetry, wouldn't it? I could rotate this oxygen down to there, this oxygen over to here, this oxygen up to here. Three different ways to look at that. But I have to consider that axis not only in this direction but also in this direction. There's two axes, they're coincident with one another, running in different directions. Why is that true for this molecule? Because there's a plane of symmetry in the screen for this molecule. Any symmetry axis that has a plane of symmetry in it, you use twice. With ammonia, that plane of symmetry doesn't exist along the symmetry axis, does it? The molecule's got this triangular shape. If I put a mere plane along that axis, I can't reflect the ammonia. It doesn't stay symmetrical. The ammonia's got this triangular shape. If I put the plane there, then these oxygens would be like coming out of the screen. So ammonia has a symmetry number of three, SO3 has a symmetry number of six because it's flat. It's more symmetrical. That's the reason. Okay, if you understood what I just said, you can tell me what the symmetry number is of this highly symmetrical octahedral molecule. Twenty-four, how did I get that? There's a, there are six axes. There's at least three ways to do this problem but let's do it the simplest way. One, two, three, but I've got to use each one of these axes twice because if I draw, there's mere symmetry of the molecule. If I put a mere right here, okay, if I put a mere right here, this looks exactly the same as this. Okay, so I'm going to use that axis twice. There's two axes there. Two, four, six axes and around each axis, there's four equivalent orientations and there's at least two other sets of symmetry axes that'll give you the same answer. Right, in other words, I could draw a symmetry axis through the bisecting, this fluorine-self-refloring bond. Right, molecule's got two-fold symmetry there but there are 12 of those guys if you look carefully. If you build the molecule and you look at it, how about this guy, aluminum chloride. These are chlorines. Okay, if I put an axis through here and I look at the center of that axis, is the molecule going to be symmetrical about it? It is. All right, or if I put an axis through here, I can probe the same symmetry elements two ways. All right, does this side of the molecule look the same as this side of the molecule along that axis? Yes. Okay, so it's two times two is four. There's, the symmetry number for aluminum chloride is four. All right, two in this direction, two in this direction, two in this direction, or if you'd rather, two in this direction and two in that direction. In this case, there's only two equivalent axes. Now, I've never seen it explained this way in any textbook but this little algorithm that we just described, I think always works. I haven't found an example where this doesn't work. All right, try it out yourself and see what you think. All right.