 So we've seen that the law of Dulong and Petit, which is the same thing as the predictions made by the echopartition theorem, tell us that for monatomic solids, for this list of solids that I've included a few examples of here, monatomic solids, the constant volume heat capacity should be 3R, which as Dulong and Petit would tell us, that's 24.9 joules per mole Kelvin as an actual number with units on it. So that works pretty well. We've seen that works pretty well for a lot of solids. Many of these numbers are pretty close to 24.9, some are very close to 24.9, but there's a few exceptions. Beryllium, for example, carbon, those both have constant volume heat capacities that are quite a bit lower than the echopartition theorem or Dulong and Petit would predict. So we need to be able to understand why that's happening. So the reason that happens, if you look at this list of solids, one thing might jump out at you, in particular about carbon. If you ask yourself what's different about solid carbon than the rest of these solids, the rest of these, chromium, copper, gold, iron, and so on, those are all metals, metallic solids. Carbon is not a metallic solid, right? That's a covalently bonded solid. And now that we've noticed that, beryllium, same idea. Beryllium is, if you look at a periodic table, beryllium is off on the left side relatively far away from the line that divides the metals and the nonmetals. So beryllium is also a relatively covalent solid. So it's suspicious that the two monatomic solids that are not obeying the echopartition theorem are the non-metallic solids. And we can also think back to what the echopartition theorem requires. If the echopartition theorem is going to hold, we need to be in the classical limit. That was one of the conditions of the echopartition theorem. So really, the statement of the echopartition theorem said the heat capacity should be 3R as long as we're hot enough to be in the classical limit, as long as we're hot enough to be able to treat the metal classically, the vibrations of the atoms in the metal classically. And one thing we know about covalent materials compared to metallic materials is metallic bonding is relatively weak compared to covalent bonding. That means the force constant, the bond energy of the force constant for a covalent bond, that's much stiffer, much stronger than the force constant and the bond energy for metallic bonding. So for covalent materials compared to metallic, the vibrational frequency is larger for carbon and beryllium than it would be for these other metals. That means if we think about the vibrational frequency, another way we have of talking about vibrational frequencies is via that vibrational temperature, which is equal to h nu over k. So when the vibrational frequency is larger, that means the vibrational temperature is also larger. So that's the sign that these materials will behave differently if they're covalent than they're metallic. The covalent materials have a higher vibrational temperature. And in fact, if I pull up in the fourth column of this table, a measurement of the vibrational temperature of these materials, this is not quite the same as the diatomic vibrational temperature that we've talked about previously, but it's related. We'll talk a little bit more about what that temperature this is actually called, something called the Debye temperature, D-E-B-Y-E temperature. But that's essentially a measurement of the vibrational frequency or the vibrational temperature of these materials. And we can see that those not only are the vibrational frequencies larger for carbon and beryllium, but this measurement of the vibrational temperature is also larger for those materials. If we remind ourselves what that means, let me just draw an axis here of temperatures. So I could take each of these metals and I could plot what its vibrational or its Debye temperature is on this axis. So let's put room temperature around here, 300 Kelvin. Many of these metals have temperatures, these Debye temperatures that are below room temperature, maybe 200 Kelvin, 215 Kelvin, maybe in the vicinity of room temperature like 318 Kelvin, maybe a little bit higher than room temperature like 400 Kelvin. So if I plot the various, so maybe 200 Kelvin, so many of these vibrational temperatures, Debye temperatures for these solids are a little below room temperature, near room temperature, a little above room temperature. So if that's the case, if the vibrational temperature, the Debye temperature is less than or in the vicinity of the actual temperature, room temperature, that means the actual temperature is hot compared to this value or is in the same vicinity as that value. And these systems behave relatively classically. On the other hand, if you're a beryllium, your Debye temperature is 1,000 Kelvin. If you're a carbon, your Debye temperature is 2,200 Kelvin. That's way up here for beryllium, way up here for carbon. So in these circumstances, the Debye temperature is quite a bit larger than the temperature. And that system is going to behave very quantum mechanically. So now perhaps it's not so much of a surprise that the Ecopartition Theorem answer doesn't work for carbon and beryllium, which are not in the classical limit. They're behaving quantum mechanically. We would have to heat them up to temperatures closer to 1,000 degrees or 2,200 degrees before we could trust that they would begin to behave classically. So what does that mean about the actual numerical values for their heat capacities? I'll remind you that let's use the extensive energy first. If we go back and think about diatomic molecules that has some vibrational frequency and some vibrational temperature, we can write the energy due to that vibration is remember the 0 point energy, 1 1⁄2 H nu. And then depending on how many average excitation that molecule has, the result we obtained when we talked about diatomic molecules was the internal energy was the 0 point energy, 1 1⁄2 H nu, plus some number of multiples of that excitation energy between different steps on the energy ladder. To talk about heat capacity, we need to take the derivative of this energy with respect to temperature to get the constant volume heat capacity. I'll add one thing to this expression first. This is the energy of a single vibration for a single diatomic molecule. Now that we're talking about a solid with many atoms in it, many different vibrational modes, the total energy due to vibration of the solid with n atoms, each of which can move in three different directions, is going to be 3n of these vibrational degrees of freedom. So if I take the derivative of this expression, that's actually a fairly tedious derivative to take with these temperatures showing up in the denominator of exponentials, perhaps themselves in the denominator of the fraction. So rather than watch me take that derivative, I will tell you what that derivative is equal to. You can certainly double check me. So that result ends up being similar to the original expression. First of all, the 0 point energy, there's no temperature dependence there. That temperature derivative just disappears. These h nu's, I've converted h nu is equal to k times theta. So that k times one of these theta's came from this h nu. The derivative brought down several more factors of these exponentials and some simplification happened. So the end result ends up looking like this. The vibrational temperature theta divided by t, that quantity squared, still an exponential e to the minus theta over t on top in the denominator. I've still got a 1 minus e to the minus theta over t, but now it's squared in the denominator. This 3n is still there. I can take the whole thing and multiply by 3n. And I get that result. And I guess I'll point out two things about this result. Number one, if I take the limit where the temperature becomes quite large compared to the vibrational temperature, and again, I won't run through the math of this in particular. But if I do that and let this exponential get very close to 1, for example, and make appropriate simplifications, what we'll find out is that everything from this theta over t squared on is equal to 1 in this approximation. And my heat capacity simplifies to just 3nk, or the molar heat capacity would after dividing by n, I get 3k or 3r. So this is, in fact, the echopartition limit. That's the result we've had all along for the law of du long and petite. This just tells us the temperature-dependent form of the heat capacity that in the classical limit, when the temperature gets large, becomes 3r. So if I were to actually plot a graph of what that temperature-dependent heat capacity as a function of temperature looks like, we've just figured out that if the temperature gets hot enough, the value is going to approach 3r. And what happens in between is it starts out small, starts out zero, increases, and then asymptotically approaches this limit of 3r as the temperature gets large. And when you get in the vicinity of the vibrational temperature, then you've gotten pretty close to the classical limit or the du long and petite limit of 3r. But if you're at temperatures below the vibrational temperature, then you might have some value of the heat capacity that's very far from this limit of 3r. So that's exactly what's happening at room temperature, for example, for carbon, where if the actual temperature is 300, but the vibrational temperature is 2,000, if this value is 2,000, then room temperature is down here somewhere, and the heat capacity of carbon is nowhere near the 24.9 echopartition limit yet. These other materials, the metallic solids, they have vibrational temperatures that are already fairly low. So room temperature might be here, might be here, might be here, but it's already close to the 3r echopartition limit. So last things I have to say about this particular approach for talking about the heat capacities of monatomic solids. Number one, this Dubai temperature that we're talking about, or the vibrational temperature for these metals, unlike for a diatomic molecule, that's not derived directly from the properties of the molecules, it's derived from some experimental data. If I measure the heat capacity of a material at lots of different temperatures, I can figure out what value of this vibrational temperature makes this curve fit the data well, and that's where these Dubai temperatures actually came from. And what that means is that temperature is different for every substance. Each one of these solids has its own different value of that Dubai temperature. The other thing that's worth noting is in this particular model, I haven't used this word yet, but this is called the Einstein model for the heat capacity of solids. In this model, developed of course originally by the guy that's named after, Albert Einstein. In this model, we assume that every one of the vibrations happening in this solid is happening with the same vibrational frequency, and therefore there's only one value of this vibrational temperature that's appropriate for every one of these vibrations. That turns out to be somewhat oversimplified approximation. A more advanced model developed by Dubai leads to a more sophisticated model. But for our purposes, the basic behavior is the same. In both the Einstein model and the Dubai model for solids, the heat capacity starts out at zero, increases with temperature, and then asymptotically approaches this limit of three R as the temperature gets large. And then the last thing to point out is that despite the fact that we have a much better understanding now of these heat capacities of solids, most of them are in the equipartition limit, predictive value near 24.9. The ones that are not yet hot enough to be close to their Dubai temperature, their vibrational temperature, are lower than they are expected to be by the equipartition theorem. But notice there's a few. I'll highlight 10, for example. That's quite a bit larger than the equipartition limit. 26.9 is larger than 24.9. Nothing I've said so far would allow us to understand how a metal could have a heat capacity that's larger than three R. So that seems to indicate that there's something we don't quite understand yet, and that's in fact correct. So that's the next thing we'll explore.