 So we spent an awful lot of time talking about gases so far, monatomic gases, diatomic, polyatomic gases. We talked about ideal gases and Van der Waals gases and various other real gas equations of state. It's about time to start moving more towards condensed phases and talking about solids and liquids. So we'll start with talking about solids and begin by asking ourselves what physical chemistry and Boltzmann, the Boltzmann distribution can tell us about the properties of solids. So we'll start with something that we already know how to do which is to calculate heat capacities. We can use the Equipartition Theorem to calculate heat capacities. So we'll see how well we can do at predicting the heat capacities of a solid. And just to give you an idea of what it is we're trying to predict, I'll pull up a table here of the heat capacities of several monatomic solids. Solids that are made up of just a single element, various different elements here. You can see if we look at the heat capacities in units of joules per gram Kelvin, those numbers vary all over the map. That might be about what you'd expect to find. Heat capacity is a property that's different for different materials. If you look the heat capacities up in a table of heat capacities you'll find that can be as large as perhaps three and a half joules per gram Kelvin for lithium or as low as perhaps point one joules per gram Kelvin for uranium. But then something either stranger or more interesting or perhaps nicer happens if we look at the heat capacities in units of joule per mole Kelvin. And most, not all the values, but most of these values are the same. If we look on this side of the table most of these values are 24, 25, 26 joules per mole Kelvin. So nearly all those values have the same quantitative value, roughly 25 joules per mole Kelvin. So why is it true that the heat capacity of all these metals, all these very different summer metals, in fact some are not metals, the heat capacity of all these substances seems to be 25 joules per mole Kelvin? Well we can answer that question using the equipartition theorem which we know how to use to estimate heat capacities. So if we're talking about a solid, let's draw a picture of the solid that we're talking about. So just a cartoon really. So here's a block of metal. It might be chromium or copper or whatever. That metal has a bunch of atoms. They might be face centered cubic or body centered cubic. Those details are not terribly important to me right now. But the solid is a three-dimensional lattice of atoms packed into this solid. So let's say we've got a chunk of this metal or whatever the solid is that contains n atoms. So we've begun essentially thinking of this block of metal, block of our material, as a large molecule containing many many atoms. So if it's a macroscopic chunk of this metal, the number of atoms we're talking about is Avogadro's number or so. Many many many more than just a handful of atoms. So we've got a essentially a macroscopically large molecule that consists of many atoms of this particular element. So if we then to use the equipartition theorem start asking ourselves how many translational or rotational or vibrational degrees of freedom that weird molecule has, we know the total numbers of degrees of freedom is going to have to be three n. One for each dimension x, y, z for each one of the atoms. And this molecule like any molecule can only translate in three directions, can translate in the three x, y, z coordinates of our three dimensional world. It's not a linear molecule so it can rotate in three different ways. And that leaves three n minus six different vibrational modes. So if n is Avogadro's number there's no way we want to count and identify what are the nearly three times Avogadro's number of different vibrational modes in this molecule, but they're in there somewhere. There's three n minus six different vibrational modes that in principle we could identify. So the next question is what is the heat capacity of this molecule and how much of that contribution comes from the kinetic energy and how much of it comes from the potential energy. Remember the equipartition theorem just tells us that's one half R for each one of these degrees of freedom that is quadratic, that contributes quadratically to the energy. Kinetic energy is always contribute quadratically so three halves are from the translational energy, the kinetic energy of translation, three halves are from the kinetic energy of rotation. That would be the rotation of this entire block of metal. And then three n minus six factors of one half R from all these different vibrational modes of the atoms within this crystal lattice. For the potential energy now, we have to think a little more carefully because potential energy doesn't always contribute to the heat capacity. And again, thinking of this block as one molecule, if I translate the molecule, I don't necessarily change its potential energy, certainly not quadratically. So that does not contribute to the heat capacity. Likewise, the rotation does not contribute to the heat capacity. But if we treat these vibrations as harmonic oscillator degrees of freedom, then they do have a potential energy that is quadratic. So I get another three n minus six factors of one half R contributing to the heat capacity from the potential energy of these vibrational degrees of freedom. So if I add all these up, I've got a whole bunch of factors of one half R. I've got three n minus six and three n minus six that adds up to six n minus 12, six n minus 12 plus these two factors of three half that add up to three. So my six n minus 12 becomes six n minus nine. If I add up all those factors of one half R, so that's the same as three n minus nine halves factors of R. And now is the occasion when I'll look back and say for a macroscopically large chunk of this solid, Avogadro's number is or however many atoms I have is much larger than one. So if I subtract four and a half from three times Avogadro's number, that number is insignificant. So I can completely neglect that nine halves when subtracting from a n times a very large value of n times three. So this is equal essentially to three n times the gas constant or times Boltzmann's constant if we prefer. Remember that is the heat capacity, that would be the heat capacity of blocks of metal. So when we list the heat capacity or when we look it up in a table, what we expect to find is the heat capacity in joules not per mole of samples of the metal, but joules per mole of atoms. So we want to recalculate the heat capacity in joules per mole of atoms rather than joules per mole of molecules or blocks of metal. So this will be the heat capacity of this macroscopic sample, the heat capacity of an individual atom. If I just divide by the number of atoms in the sample, that's going to give me three R. So dividing by the total number of atoms. Three times R, of course, is three times the gas constant in units of joules per mole of Kelvin. That would be 8.3 joules per mole Kelvin. So that's going to work out to 24.9 joules per mole Kelvin. And that value we see is exactly the value predicted in this table for not exactly the same value, but roughly the right value for most of the entries in this table. So most of these values are 25 plus or minus one, so pretty close to 24.9. So that is in fact this title statement, the law of du long and petite. This fact has been known for a very long time that the heat capacity of metals, of in fact monatomic solids, whether they're metals or not, is typically very close to 24.9 joules per mole Kelvin. We can, with our additional information about the co-partition theorem, simplify that a bit and say that the heat capacity is three R for any monatomic solids. Like I said, that fact has been known for a long time. Du long and petite, as you might be able to guess from the names, they were French scientists that were, they were working in the early 1800s when they first observed this trend that the heat capacity is very predictable for monatomic solids. 25 joules per mole Kelvin, although they weren't working in those units. So we have this, the co-partition theorem gives us this nice simple rule. We can predict at least roughly the heat capacity of many different solids. Now is the time though to look a little more closely and say that, for example, carbon doesn't follow the law of du long and petite. The heat capacity of carbon is 6.1 joules per mole Kelvin. That's pretty far from 24.9. Likewise, some of the other elements, cadmium is probably the one that's actually tin or uranium. In fact, uranium is 27.7. That's starting to look a little too high, much larger than 24.9 joules per Kelvin. So some of these elements have heat capacities that are lower than the law of du long and petite would predict. A few of them have heat capacities that are higher than the law of du long and petite would predict. So we'll be able to do a little better than just the equi-partition theorem if we look a little more closely at how solids behave.