 Okay, so we have a quiz on Friday, as you know. And it's just going to be exactly the way it was last Friday. The scantrons are going to be up there at the front. Grab one when you come in. Don't sit next to anybody. We had plenty of room last week, right? For some reason, they gave us this giant lecture hall, which is really nice. Well, I'm going to write the, I haven't written it yet, but I'll write it this afternoon. When I do that, I'm going to look at what I've been telling you for the last three lectures. I'm going to look at the discussion guide, too, which is posted now on the lectures page of our website. I'm going to look at the assigned homework, and then I'm going to brew up something with five questions on it. The first two questions are going to be pretty easy. The last three will be a little bit harder, okay? Just like it was last week, it's open book, open notes, open anything except we're not going to allow tablet computers and regular computers this time, okay? You can still bring your tablet computer and your laptop if you want, but, and you can use it after the quiz is over, but you can't use it during the quiz. Use any kind of calculator that you want, okay? And please do not use this as a reason to print out the lecture notes because that's very destructive, all right? Write anything down you want. Any formulas, put posted notes in your textbook, okay? Any questions on quiz two? Even though I try to include all this other stuff, I tend to mainly look at what's in the lectures. I try to make sure that I haven't asked you any questions on the quiz that I haven't answered in the lecture. That's what I tend to do. You might also want to look at quiz two from last year, which I posted on the announcements page because I'm lazy and I'm likely to steal problems off quiz two. It's just the way I am, okay? Today we're going to talk about a subject that's hardly discussed at all in your book, but I think it's really cool and it really helps us obtain an intuitive understanding of how the statistical mechanic stuff works, all right? In real molecules, the picture is considerably more complicated than this. This is where we started with the statistical mechanics and evenly spaced ladder of states such as that, which we would obtain if we had a harmonic oscillator, but real molecules don't even vibrate this way, even though we use the harmonic oscillator to approximate vibrations in real molecules, all right? There's anharmonicity in real molecules, all right? Translation comes pretty close to the harmonic oscillator picture, but even in this case, the states are not evenly spaced, even though I've drawn them evenly spaced here, in reality they're not. I'll show you that later on. And then these translational states are really close together. There's many of them between each rotational state of the molecule. Here's the rotational states, okay? And there's many of these translational states that are sandwiched in between each one of these rotational states, thousands in most cases. Hundreds of thousands. And then these rotational states, there are many of them in between each one of these vibrational states, aren't there? All right? So there's a lot of complexity in real molecules. Now we really haven't started to talk about it. It's been discussed in your book and you've been doing some homework problems that relate to this issue, but we really haven't talked about it in class. We're going to be lucky in that we can pick this problem apart. We can treat each of these energetic manifolds, all right? Each one of these things is what I'm calling a manifold. We can treat them separately and at the end of the day we can just take the partition function for translation and multiply it by the partition function for rotation and we can calculate each one of these partition functions separately multiplying together to get the total partition function for the whole molecule. And the reason this works is because these degrees of freedom are so-called weakly coupled. They don't talk to each other, okay? So we multiply them together. We can calculate them separately. We're going to get the total partition function that we're looking for for every different kind of molecule that we care about. Now we're going to start talking about that later on today, but in this lecture we're going to talk about a shortcut to calculating not this guy but something related to him, the heat capacity and the internal energy, a shortcut and it's called the equi-partition theorem. I think this is discussed right at the end of Chapter 13 but it's discussed very briefly, not in enough detail to really understand. And then at the end of this lecture we'll start to talk about the translational partition function. Okay, so when your book talks about the equi-partition theorem it concentrates on the internal energy of a particular molecule or the internal energy of a mole of molecules. These two variables are difficult to measure directly in the laboratory. In other words, if you're an experimental physical chemist and you go into the laboratory to actually measure the absolute value of the internal energy of a single molecule or a mole of molecules or any number of molecules, that turns out to be a hard thing to do. But it's a lot easier to measure the capacity of some volume of molecules to absorb heat, the heat capacity. That's an easier quantity to measure. You may recall the constant volume heat capacity is just the amount of energy a material compound can store per unit temperature so the constant volume heat capacity is just the partial derivative of the total energy of the system with temperature at constant volume. We can equally well define it in terms of the average internal energy of a particular molecule. Okay, so hypothetically, if the average internal energy of a particular molecule turned out to be kT over 2, then the heat capacity is just the derivative of kT over 2 with respect to T, so it's just k over 2. All right, so if we know the internal energy, we can get the heat capacity and vice versa. Now, I haven't told you why this is cool yet, but bear with me. This is a plot of the heat capacity as a function of temperature for some generic molecule. What we want to appreciate is that as the temperature increases along this axis from left to right, the capacity of a molecule to store energy increases in a stepwise fashion like this. Why is that? The answer is that at really low temperatures, only the translational states that are available to a molecule are occupied. In other words, the only way a molecule can store energy is by changing its velocity. Right? It can store more energy by speeding up. Once it gets to a temperature where the thermal energy available at that temperature equals the energy between rotational states of the molecule, now the molecule can start to rotate as well as translate. All right, so it's got a new manifold that it can access for storing energy. It can store energy as translation, but also it can store energy in terms of its rotational states because now it's reached the threshold here where the thermal energy available to it is high enough so that it can start to access. Now remember, these rotational states are much further apart in molecules than the translational states, much further apart. Finally, if the temperature is even higher, you get to the point where you're exciting many rotations, but suddenly you start to have just the threshold of energy you need to excite some vibrations of the molecule. Now you access the vibrational manifold and you've got three ways to store energy, translation, rotation, and now vibration. Okay, so let me point out a couple things about this diagram first. So it's not the slightest bit confusing. There's no units here, all right, but the units for this constant volume, he's capacitor are either k, in other words, 7 halves k, 5 halves k, 3 halves k. If we're talking about a single molecule, the units are in terms of k. If we're talking about a mole of molecules, they're in units of r because one mole, Avogadro's number times k is r, isn't it? Okay, so even though it doesn't say what the units are here, that's 3 halves r if we're talking about per mole. These guys, these temperatures here are the characteristic temperatures in your book called Thetis of v and Thetis of r. This is the characteristic rotational temperature and the characteristic vibrational temperature. The rotational temperature is just b written out in terms of joules, right, b, remember, is the rotational constant for the molecule. Write it out in terms of joules or write it in wave numbers and do a conversion to joules. That's what this is, always confusing to me. Divide by k and you get units of temperature, right, because k is joules per kelvin, right? If that's in joules, I'm going to get this in terms of kelvin, that's what that is right there, right? Likewise, that's the energy between vibrational states, h nu, if I write that in terms of joules and divide by k, I get that temperature right there. So that's the characteristic vibrational and the characteristic rotational temperature. We can calculate for that, these two temperatures for any molecule as long as we know b and h nu. Okay, so down here, we've got only translation going on. Up here, we've got translation and rotation because the thermal energy is high enough now so that the molecule can rotate as well as translate. And finally, up here, we've got all three things going on. Right, the capacity of the molecule to store heat increases as it has more channels in which to put the heat. Very intuitive idea, I think. Okay, so what the equi-partition does is it provides a shortcut method for estimating approximately now the internal energy and heat capacity of any molecule. What makes it interesting is it actually works. It's so simple and it actually works. It gives the right answer, approximately. How does it work? Consider the classical Hamiltonian for 1D harmonic oscillator, all right? Here's the way it works. You write the classical Hamiltonian for the molecule, the classical Hamiltonian, right? So if there's quantum mechanical stuff going on, we're going to miss it here. Then, we convert each one of these quadratic terms, well, let me show you. So consider the classical, so for a 1D harmonic oscillator, there's two terms in the Hamiltonian, a kinetic energy term that's not temperature and a potential energy term that's not volume. Kinetic energy, potential energy. The kinetic energy term is just p squared over 2m where p is the momentum, m is the mass. The potential energy is 1 1⁄2 kx squared, right? Just Hooke's law, all right? That's the force constant of the bond and that's the displacement of the bond from equilibrium, x, r minus r0, if you will, okay? So there's two terms in the classical Hamiltonian. The equi-partition theorem says that any quadratic term in this Hamiltonian having the form, for example, AP squared or Bx squared, the internal energy of the molecule is kT over 2 for each such term. How could it be that simple? I mean, any term at all, I just take the term and I multiply by kT over 2 and that's the internal energy. Yes, all right? And that's going to work. Well, we'll see. Okay, so the problem of applying the equi-partition theorem comes down to writing this classical Hamiltonian correctly, figuring out how many modes there are, right, that are actually participating in the energy storage and then assigning each one this magic number, kT over 2, or if you've got a mole, rT over 2, right? It's pretty easy. Now you recall that heat capacity is just the amount of energy, yes, we just said that. Okay, so if there's one term, if there's one quadratic term in the classical Hamiltonian, then this internal energy is kT over 2 and the heat capacity is k over 2 for a single molecule or r over 2 for a mole of molecules. Yes, just said that. Okay, so let's calculate something. Right now, all molecules translate and their classical Hamiltonian in three dimensions for translation is just this, p of x squared, the momentum in x squared, momentum in y squared, momentum in z squared divided by the two times the mass. Same for every molecule. How many quadratic terms are there? Three, one, two, three. Equipartition tells us that the translation, that the translation, that translation contributes three kT over 2 to the internal energy of a single molecule. Three because there's three quadratic terms. Okay, so the internal energy of a single molecule is going to be this, the internal energy of a mole molecule is going to be this and the heat capacity is just going to be derivative of that with respect to t. It's going to be 3r over 2. I just leave the t out. The contribution of molecular translation to the heat capacity is 3r over 2 for every molecule. Well, yeah, look at that, all right, 3r over 2. That's why that's 3r over 2. Now, is that only approximately correct? No, that's exactly correct. I'll show you in a second. Well, I'll show you at the end of the lecture. Okay, so this is also the total heat capacity for all monoatomic gases obviously because a monoatomic gas can't store energy any other way. It can't rotate, it can't vibrate and so this is the whole story for a monoatomic gas like neon or argon, right, it can't do anything else. All right, so this is the heat capacity of a noble gas for example. Full stop, there are no bumps, that doesn't happen and that doesn't happen, it's just boom. Okay, for molecules with more than one atom, vibration and rotation can also contribute to the heat capacity but vibration doesn't turn on until the temperature approaches the characteristic vibration temperature, same thing's true for rotation. Rotation doesn't turn on, when I say turn on, I mean it doesn't contribute to the heat capacity. So for a linear molecule, let's say that we're at a temperature that's higher than the rotational characteristic temperature but lower than the vibrational characteristic temperature. In other words, rotation is turned on but vibration isn't. All right, at moderately dull temperature, this turns out to be the case. Let's say below about 100 wave numbers in thermal energy. All right, so the Hamiltonian for rotation of a linear molecule now has got two terms, it can rotate an X and it can rotate an Y, these are the moments of inertia, these I's, so that should be I sub X and I sub Y. Okay, how many quadratic terms are there here? Two, could it be that simple? Rotation about the X axis, rotation about the Y axis, that's the whole story. Okay, so the internal energy of rotation now is going to be 2 times KT over 2 or for a mole of molecules, 2 times RT over 2, that's amazingly simple, isn't it? So the heat capacity then, the total heat capacity for the molecule in this temperature range has got two contributions to it. Here's the translational contribution, 3R over 2, that's always the same. That's always going to be 3R over 2. All right, here's the rotational contribution, boom. All right, it's 2R2 over 2 because the molecule's got two ways it can rotate. It can rotate an X, like it can tumble or it can rotate an Y. There's actually two coordinates that can rotate orthogonally to one another. All right, can it rotate along its axis like this? No, but it can rotate, if it's oriented like this it can rotate like that or it can rotate like that, right, that's the X and Y rotation that we're talking about. So the total heat capacity is just that some of these two things, 5R over 2, there's no vibrational contribution because we're way below the characteristic vibrational temperature, we said we're in this range here, we're way below the vibrational, the temperature where vibration would turn on. Okay, we're up here, right? That's 5 over 2, yes, that's just what we calculated. All right, so it looks like this plot probably applies to a linear molecule because if it wasn't a linear molecule, this wouldn't be 5R over 2, it would be that 2 would be a what? If it wasn't a linear molecule, it could rotate in all three dimensions, X, Y, and Z, right? So that 2 would be 3, now, for a nonlinear molecule, yes, this is just what I said, X, Y, and Z, 3KT over 2 or 3RT over 2, boom, all right, that would be the total heat capacity, 3R, if it was a nonlinear molecule. So this plot that I stole off Wikipedia obviously applies to a linear molecule, translation and rotation. Okay, now what about at higher temperatures where we start to excite not only rotation and translation but also vibration? As we said earlier, the classical Hamiltonian for vibration actually contains two terms, a little bit more complicated than for translation or rotation because even for a single mode, there's two terms in the classical Hamiltonian, the potential energy and the kinetic energy, all right, and we're going to sum these guys over either 3N minus 5 or 3N minus 6 vibrational modes per molecule, right, depending on whether molecule is linear or nonlinear. All right, if it's linear, it's going to be 3N minus 5. So following through with the predictions of the equi-partition theorem, we're going to get for each molecule 2KT over 2 per mode or for a molar molecule, 2RT over 2 per mode, 2KT over 2 per mode, all right, because the classical Hamiltonian contains two quadratic terms for each mode, so following through with the predictions of the equi-partition theorem, we've got translation for a nonlinear molecule, we've got rotation, right, 3 for X, Y and Z and we've got a contribution from vibration which is going to be either 3N minus 5 for linear molecules or 3N minus 6 for nonlinear molecules. That's translation, rotation and vibration and so the total is going to be that, shouldn't it be what? So 3N minus 5 is nonlinear molecules, 3N minus 5 is nonlinear molecules, yeah, sorry, right, sorry, yes, got to fix that, sorry, that should be nonlinear, okay, so for example, for a diatomic molecule, that should be 3N minus 6, is that right, yeah, so let's, yeah, okay, so if in fact that was a 6, this would be 3R over 2 for the translation of the diatomic molecule, it could rotate an X and Y, so it'd be 2R over 2 for its rotation and then it would be 3 times 2 for the number of atoms minus 6 times R, all right, so this would be 5R plus, 6 minus 0, plus 0, that's what we would have gotten for in nonlinear molecule, 7 halves are, let's do some examples, use the equi-partition theorem to estimate the constant volume molar heat capacity of I2 methane and benzene at 25 degrees C, okay, so the first thing that you want to figure out here is where you are on this plot in other words, how many terms are there going to be in your heat capacity expansion? Is translation only contributing? Translation and rotation or translation, rotation and vibration? The way that you figure that out is first of all, if you're at a temperature, near room temperature, what do we know about whether the rotations of the molecule are going to be excited or not? How much thermal energy is there at room temperature? What's KT at room temperature? In any units that you want to use, bless you. How much thermal energy is there at room temperature and wave numbers? Right, 207 or 200 roughly, right, 200 wave numbers. What is the energy spacing for rotation of a moderately sized molecule? Round numbers, energy spacing for rotation, what's B? Is it a thousand wave numbers? Anybody want to guess? 400, no that's too high, a handful, right, one, three, okay, one or a small number, right, for rotational states. How many wave numbers are there for vibration? Round number, OH stretching frequency. Remember that from organic chemistry, big blob over on the left-hand side of your spectrum, what were those energies, anybody remember? 3,000 wave numbers, that's an OH stretch, okay, so order of magnitude, 1,000 wave numbers, all right, so one or two for rotation, 1,000 for vibration, okay, so qualitatively this is going to help us figure out where we are on this diagram, all right, in case of I2, both of the vibrators are heavy, aren't they? Iodine is a big molecule, 126 grams per mole and so that's a pretty low frequency, right, that's a pretty low energy rather, 200 wave numbers, all right, but in the range where we sort of expected it to be, all right, 1,000 wave numbers is to one sig fig, that's how much energy there is in a vibration, okay, so do we have to think about whether this thing is going to be vibrating at 25 degrees C, well, at 25 degrees C we got 200 wave numbers of thermal energy, we know one or two is enough to excite rotation so this baby is rotating, all right, we don't have to worry about that, so we're definitely here, we're just not sure where we are here, are we up here, in other words, is that vibration excited or are we down here, is that vibration not excited? Well, we've already concluded that at 25 degrees C we got 200 wave numbers so we're close, will the vibration of I2 contribute to the heat capacity, well, if we're not sure, we can calculate the characteristic temperature from this 214 wave numbers, we just have to convert 214 to joules, divide by K and we get 308. Kelvin, all right, that's a little bit higher because that's not 200 wave numbers, it's 214, so that's why that's 308, okay, and so we're right here, all right, this line turns out to be 308, all right, and we're just below that, okay, and so this state, this vibrational mode of the I2 is significantly turned on, all right, it's starting to contribute to the heat capacity. We can either assume we're down here or assume we're out here and we always make the high temperature assumption, all right, if it's starting to contribute, we're going to use this to calculate our equipartition theorem heat capacity, confident in the knowledge that we're going to overestimate it a little bit, right, because this isn't a perfect science, we're not going to get the heat capacity exactly to three sig figs, we're shooting for one sig fig here, okay, so we're going to say yes, that mode is turned on because we're somewhere on this rising portion of this curve, the molecule starting to store energy in its vibrational modes as well as rotation, and so we've got translational contribution to the heat capacity, we've got two rotational degrees of freedom because it's a linear molecule and we've got one vibrational mode or one R, we're going to include the whole mode, we don't split it up, so that's 5R over 2 plus R is 7R over 2, actually this is 3.5R, right, and the actual heat capacity of iodine is 3.4, so we overestimated the heat capacity slightly but not by that much, we did a really good job of guessing what it would be, right, just using equi-partition theorem, all right, I like simple, intuitive things like this that allow you to get the right answer, all right, we're always looking for better chemical intuition so that we can, at an order of magnitude level, figure out what's going on, all right, that's the real challenge, all right, later on we can calculate this to 3 sig figs if we want but we want to get the, we want to have an intuitive understanding of how big that number is and we can get that with this equi-partition theorem, note if we left out this mode, if we left out the vibrational mode, we'd be way too low, 2.5R, okay, so that's a justification for including it even if we're not all the way turned on here, we're here, we're not all the way up here, all right, but we're going to include it anyway and most of the time, that's going to get us closer to the right answer. Let's look at this guy, obviously more complicated, here are all the vibrational frequencies that apply to methane, 1367, 1582 and so on, all right, do these vibrations contribute to the heat capacity at room temperature, what do you think? 1367 is going to be storing energy for you at 25 degrees Kelvin, where the thermal energy is how many wave numbers? 200, you've got 200 wave numbers of thermal energy, all right, and the lowest vibrational energy level of the methane is 1367, all right, is that 1367 going to be storing energy for you? No, 200 wave numbers, 1300 wave numbers, you'd need 1300 wave numbers of thermal energy before this thing gets turned on, right? Not sure, plug in, plug in, take the lowest one of these numbers, take the 1367, plug it in, convert it to joules, divide by K, 1367 is the lowest one, 1966 degrees Kelvin, that's how hot it would have to be for this lowest mode to get turned on, all right, so at 298 K, it's not on, you see why? It's just the energy is too high for this vibration, it's not getting excited at 298, okay, so we're not up here, all right, we're going to assume none of these guys is turned on, we are right here, in other words, rotations are turned on, but no vibrations are turned on, all right, that's the hard part of this little calculation, it's figuring out what do you include? You include the vibrations or not, you're almost always including translation and rotation, unless you're at a really low temperature, you're including translation and rotation, all right, the question is usually concerns vibrations, which ones are turned on, which ones aren't, some might be turned on, others are not, okay, so the heat capacity is going to be translation plus rotation, this is not a linear molecule, so that's a 3, and then we're going to include no vibrational modes, because they're all at energies that are too high, so we're going to predict the heat capacity is 3R and the reality is, for methane, it's 3.2, okay, so the reality is these vibrational modes are contributing a little bit, we neglected them, but in reality they're contributing a little bit, we missed that, but we get awful close, we get the right answer to one sig fig, now a toughie, benzene, has got all of these different states, pay attention to this column, this is the actual frequency, 1100, 3,000, 1,300, okay, should we include any of these in our, is benzene going to be storing energy in any of these modes right here, they're all too high, how much thermal energy is there at room temperature? 200, never forget that number, all right, what about here? Here's another mode, the CH out of plain WAG, 600 wave numbers, huh, it's a little bit higher than 200, but not way higher, right, 3,000 forget it, 1,000 forget it, 900, very high, 684, another one that's a little bit lower than all the rest, 1,400, 1,100, all right, why don't we see what happens if we include two modes, all right, we'll include this guy, 684, and we'll include this guy, 651, all the rest of them are much higher, let's see how close we get, so let's go with 651 and 684 wave numbers only, translation, rotation, it's a nonlinear molecule, so that's a 3, if it was linear it would be a 2, right, 2 times 2, because, and I would have to tell you this, there's no way that you could know each one of these modes is doubly degenerate, all right, so there isn't one 651 wave number mode, there's two, all right, you have to include that, each degeneracy corresponds to another way the molecule can store the energy, all right, so there's really four here, so it's 4R, so 7R would be the total heat capacity that we estimate if we include two modes, and forget all the rest, and you can see we don't quite get one sig fig of accuracy, the actual heat capacity is 8.8, so that means benzene can use some of these modes at sort of a thousand wave numbers, they can contribute a little bit to the heat capacity, even though we're way lower temperature, all right, we're at a thermal energy of 200 wave numbers, we shouldn't be turning these things on until we get to 1,000, but they get tickled a little bit even at this low temperature, they contribute a little bit, that's the difference between the seven that we calculated and the 8.8 that benzene actually has, yes, we should have, but let's go back and look, all right, these guys, where's the 900 wave number mode? Yeah, you know, who's going to know, all right, you can't expect it, in reality if we include the 900 wave number mode, we're going to get closer to the right answer, but how would we know that? So this is a very approximate arc, all right, and in this particular case it wouldn't be exactly clear which modes to choose and which ones to not choose, all right, and it can even be more complicated than this, all right, but in many, for simpler molecules, all right, in most cases it's fairly obvious which modes to include and which ones not to include. Now the way most people treat equi-partition theorem is they just include all the modes. The equi-partition theorem gives you the high temperature limit for the heat capacity, okay, and you can see where it would, all right, if you include all the modes and the molecule has to be really hot before it can access all these modes, in the case of benzene, okay, definitely a quiz question on this coming at you for Friday. Okay, now with that intuition, we need to be able to calculate exactly how big each one of these guys is, and I know you've been doing that already by if you've been doing the homework, going to discussion, but now we're going to talk about it in lecture just briefly today. We're going to start by talking about translation. Here's a molecule, that's the translational energy for it moving in three dimensions. This is my little pay no attention. And the quantum mechanical gas energies are given by the particle in the box model, so here's the classical, right, if we know the velocity in x, y, and z we can calculate the kinetic energy, okay, but quantum mechanically, we use the particle in the box, all right, where these are the dimensions of the box, Lx, Ly that should be not x, Lz that should be not x, sorry, y, and these are the quantum numbers for each of those dimensions. Remember this, way back probably from fall quarter, okay? So these are what those wave functions look like for goodness sakes in three dimensions, all right, blasts from the past. So now we're going to concentrate attention on ideal monotomic gases just for the moment. Such gases have no internal energy in form of rotations of vibration, we'll assume that just the ground electronic state of the system used to be considered in our analysis, so one of the tacit assumptions we've been making for the last 20 minutes is that the electronic states of the molecule are not contributing anything to the heat capacity because we're only occupying a single electronic state. Now for some molecules that would be a bad assumption, but there's relatively few where you've got low-lying electronic states that contribute to the heat capacity. We talked about one, NL, right? But there's very few examples like that. Okay, because these various energy manifolds, rotation, vibration, translation can be separated, the solution to the monotomic gas, translational energy will also provide us with a general expression for the translational energy of any gas, no matter how many atoms it has. All we need to know is how big it is. Okay, so consider, first of all, a monotomic gas in one dimension, we've only got a one-dimensional term here, we've only got a quantum number for X and a dimension for X. The molecular partition function is just, we just have to plug this energy into our expression for the partition function, boom, right? That's all I did there. Now these energies are very closely spaced. Consider, for example, if I put this argon atom in a one micron box, one micron, how big is one micron? Well, it's 10 to the minus 6 meters, all right? How big is a red blood cell? What's the smallest thing that you can see in an optical microscope? Anybody know, you've got an optical microscope, let's say you buy the world's best Zeiss optical microscope, you pay $12,000 for it, it's got objectives like beer cans on it, all right? You look through it, what's the smallest thing that you can see? What's the smallest size of the thing that you can see? How many people have had microbiology class? Come on, you guys, microbiology people should know the answer to this, how big is the bacteria? About a micron, can you see a bacteria? Yes, just barely, yes, all right? One micron, all right, in an optical microscope, you can see a one micron object, all right? It doesn't matter how much you pay for it because you can't see more than a fraction of a wavelength of light, all right? What's the wavelength of green light? Half a micron, all right? Turns out that's the smallest dimension you're going to see, doesn't matter how much money you pay for your microscope, all right? If you don't pay enough, you won't see that, okay? One micron is a tiny box, it's the smallest thing that you could possibly see in an optical microscope. So we're not giving the molecule very far to move, not only that, we're only considering its motion in one direction, all right, not Y and Z, okay? So these energies are very closely spaced, considering where we're going to want them, all right? So delta E, all right, what's the state spacing between the ground translational energy level and the first excited translational level? Let's just calculate that and find out what it is. H, M, in units of kilograms now, all right? So when you're doing this on the quiz on Friday, make sure that you use kilograms, okay? That's 10 to the minus 6 meters, all right? L is squared, okay, and we've got N equals 2, so 2 squared is 4 minus 1 is 3, all right, that's the energy we get, 2.48 times 10 to the minus 30 joules. Big energy is small, who knows, all right? It's joules, all right, it always looks small, all right? If it was big, it'd be 10 to the minus 20, 10 to the minus 18, it still seems like a small number, all right? We convert it to wave numbers, we know that's small, all right? 1.25 times 10 to the minus 7, wave numbers, tiny unit of energy, right? One wave number to get the molecule to rotate, 10 to the minus 7, okay? So the state spacing is really, really tiny, all right? Here's a log scale, here's 2.48 times 10 to the minus 30, it's right there, all right? As I increase the quantum number, I'm looking at the state spacing for higher and higher, look what happens, all right, the states can even get closer together, okay? So these states are quasi-continuous, there's a tremendous density of states. They're so close together that they're almost a continuous distribution, all right? And since that's the case, we can evaluate, we can turn this summation into an integral. We're going to integrate over all the states, 0 to infinity, all right? Just move that guy into the integral and we're going to integrate across all of the states, n sub x. And we're going to use a little trick to get the integral right. And when we plug everything in, we're just going to substitute alpha for everything here except for n, because n is the integration variable, okay? And when we do that and we evaluate with the integrate what the, what q is equal to, this is the expression that we get. After integration, we find out that the partition function for one-dimension is just the dimension divided by h root 2 pi m over beta, all right? Now we can calculate exactly what the partition function is in one-dimension, as many sig figs as we want. This is the translational function, partition function in one-dimension for any molecule, any molecule. All we need to know is its mass. Okay, calculate the partition function in one-dimension for an argon atom confined to one micron, one-dimensional box at 300 degrees Kelvin, boom, okay? Ten to the minus six, 1 over kT, right? We have to know T, it's 300. We have to know m in units of kilograms, kilograms, am I emphasizing this point enough? We want the mass of one atom, so we divide by, so if I look at the periodic table, it says 39.948 grams per mole, but we're going to write down kilograms. We're not going to forget that, or we're going to get none of the above when the right answer is actually right there. Okay? And so when we calculate this number, it's 62,000. There's 62,000 translational states in this one micron box? Amazing, all right? There are, in principle, 62,000 thermally accessible translational states at room temperature in this one micron box. That's a large number. Okay, what if it was a three-dimensional box? Well, we just have to cube the same expression we just arrived, we cube it so that's not a square root anymore, it's three-halves, and now we have to include Lx, Ly, Lz, got to cube the H as well. Okay? We cube everything because the translational partition function overall is just Qx times Qy times Qz, they're separable. Okay? And so this is just the volume, obviously. And so if I just move things around, that's the expression that I've got, and there's an even simpler expression. If I substitute something called the thermal wavelength and you've seen all this already, if you've been looking at the homework problems, we talked about the thermal wavelength already, all right? This is just a way to parameterize this equation a little more conveniently because if we define the thermal wavelength this way, then the translational partition function is just that, really, really easy to remember, okay? And so we can calculate the energy from this by using the equation that we derive, the least intuitive equation in chemistry, okay? Q is just this, so we can plug that in for Q. Where did we do? There we did it right there. There's Q, okay, so it's just thermal wavelength cubed over V, dd beta of V over the thermal wavelength cubed with the thermal wavelength cubed equal to that, all right? And if you trust me, all right, that's what we get for the internal energy for, it should be a mole. This should be the average internal energy because we've got K here unless we use N equal to Avogadro's number, okay? So the bottom line is this is 3 halves RT for one mole. Well, we knew that from Equipartition Theorem, didn't we? Okay, it's 3 halves RT for translation. Now we've proved that that's exactly right, okay? What is that, 100 slides, pretty good. So we'll see you on Friday.