 So the thermodynamic connection formulas have succeeded in showing us what the energy of an ideal gas is. The energy of an ideal gas is 3-halves NKT, we've discovered, but knowing something about the energy of the gas can also tell us something about the velocity of the molecules of the gas. So that's the next thing we'll explore. So first of all I'll point out that this energy, internal energy of 3-halves NKT, that's an extensive energy, it's proportional to the amount of gas that we have. The intensive energy, the energy per molecule per mole, if I just divide each side of that equation by N, then I find out that the intensive energy is equal to 3-halves KT. So that gives us an expression. We can use that to calculate the energy of a gas, so let's go ahead and do one quick example. I want to know, what is the energy per mole of a gas like nitrogen or oxygen, let's say in this room at a temperature of 298 Kelvin? The equation is quite easy. Plug into this expression, 3-halves times Boltzmann's constant, in fact let's not use Boltzmann's constant, I'd rather have this energy come out in units of Joules per mole, so let's go ahead and use the fact that Boltzmann's constant and the gas constant are the same thing. So I'll use Boltzmann's constant in the form of the gas constant, I'll multiply by 8.314 Joules per mole Kelvin, also multiply by the temperature 298 Kelvin. So one and a half times 298 times 8.314, if we use a calculator to calculate that we find that the result is 3,270 Joules per mole, the units of Kelvin have canceled or since we're in the thousands of Joules per mole, we typically write that as 3.27 kilojoules per mole. So that is the energy of a gas at 298 Kelvin. Notice one relatively important thing is that I didn't have to tell you what gas we were calculating that energy for, that could be argon gas or carbon dioxide or carbon monoxide or any gas that behaves like an ideal gas, in other words any gas that has the same partition function, the same energy as any other 3D particle in a box does, all those gases have the same energy as each other at the same temperature, so the energy is independent of what type of gas we're talking about, which is perhaps a little bit surprising. There's one more important thing we can use this expression for, now that we know the energy of these gas molecules, think about what that energy involves. If I have a gas in a box, the bunch of, let's go ahead and draw a picture of a box with some gas molecules bouncing around in it, in particular, since we solved this problem quantum mechanically using the 3 dimensional particle in a box model, one of our key assumptions in that model was that the potential energy was zero, in other words these molecules are not interacting with each other at all, that's one of the criteria for an ideal gas is the molecules don't interact with each other, so with non-interacting molecules, no potential energy at all, the only type of energy these molecules have is kinetic energy, so the energy that we've calculated here, 3 halves kT per molecule of this gas, that's purely kinetic energy, so we can say u equals 3 halves kT, intensive energy is 3 halves kT, that's the energy of each individual molecule, that's a microscopic, I'm sorry that's not microscopic, I'll come back to that statement in a minute, that's a macro state property, on the other hand the kinetic energy, all right the kinetic energy is e sub k, we know something about kinetic energy, kinetic energy is one half mv squared, so each of these molecules bouncing around has an energy of one half mv squared for the particular mass and particular velocity of that molecule, so this is a microscopic property, internal energy describing the energy of the entire box of gas, that's a macroscopic property, so there's a slight difference between these two variables, in particular every molecule of this gas may not have the same velocity as all the others, some might be moving faster, some might be moving slower, they each have their own individual velocity, so they each have their own individual kinetic energy, but this expression doesn't have any ability for the gas to be different for one molecule than the other, at a given temperature it just has this much energy, and the reason for that is because this macroscopic property describes the average behavior of the whole gas, so this macroscopic energy is like the average energy averaged over all the individual molecules in the box, so if we combine those two statements what we found is that the internal energy of three halves kt is the average of all the individual microscopic energy, so it's the average of the kinetic energy is one half mv squared for the individual molecules, one half and m, if all the molecules of gas in the box are the same type, if they're all nitrogen molecules they all have the same mass, so neither the one half nor the m varies from one molecule to the other, and the only thing that needs to be averaged is the v squareds, so there's a relationship between the temperature of the gas and this average of the squares of the velocities, so let's rearrange that and isolate the v squared average on one side, the twos cancel, and if I bring the m over to the denominator underneath the 3 kt I find that v squared averaged is equal to 3 kt over m, that's the relationship again between the temperature of the gas, its mass, and this average squared velocity of the gas, this quantity, the average of the squares of the velocity, we call that quantity a root mean square, so that just says the same thing in a word, sorry that's not a root mean square, that's a mean square, that's a mean square velocity, root mean square is in the next video lecture, or no the next line alright, so that is the mean or the average of the squares of all the individual velocities, so that's the expression we have here, that quantity which has units of velocity squared is not terribly intuitive, if I told you the mean square velocity of these molecules was some number of meter squared per second squared, that's not going to feel very intuitive because we're not used to thinking about the squares of the velocities, so we do often define when I jump to too quickly the root mean square velocity of a gas, if I take the square root of both sides of that expression, so the mean square velocity square rooted is equal to 3kT over M square rooted, so let me go ahead and put that expression in a box because it's one that we can use fairly commonly, if I want to know the root mean square velocity, in other words the square root of the mean of the squares of the velocities of each individual molecule, I can use this expression to get it, and let's do a numerical example to see how that works, so let's stick with a gas at 298 Kelvin, but now notice that I am going to have to tell you what gas we're going to calculate the root mean square velocity of, and let's go ahead and do this calculation for nitrogen gas because we're going to need to know what the mass of the gas is, so if we want to know the root mean square velocity of nitrogen gas at 298 Kelvin, and I suppose I should say also that in addition to calling this V squared averaged in square rooted, oftentimes we just write V sub RMS for root mean square velocity, the root mean square velocity of nitrogen molecules at a temperature of 298 Kelvin, it's just given by this expression, so I need the square root of 3, let's think about the denominator first, nitrogen with a mass of 14 grams per mole for nitrogen atom or 28 grams per mole for a nitrogen molecule goes in the denominator. To work in SI units I'm certainly not going to want units of grams, I'm going to want units of kilograms, so we can convert grams to kilograms, and now we could get rid of moles by using Avogadro's number or a shortcut to avoid having to do that is if we also use Boltzmann's constant with units of joules over moles, so I'm going to use the gas constant 8.314 joules per mole Kelvin in the numerator, also use the temperature 298 Kelvin in the numerator, then all the units should cancel, just to double check, I've got Kelvin canceling, I've got grams canceling grams and leaving kilograms, I've got one over moles in the denominator and in the numerator, and what I'm left with is joules divided by kilograms, a joule is a kilogram meter squared per second squared, so after canceling the kilograms I've just got meter squared per second squared, when I take the square root, that's going to end up with some number of meters per second, so the units work out fine, that's the units of a velocity, and if I plug those numbers into the calculator what I find is that the result is about 515 meters per second, so this calculation shows that if I have a box of nitrogen molecules at a temperature of 298 Kelvin, their root mean square velocity is 515 meters per second, so let's think a little more about what that root mean square velocity means, at this point it's okay just to think of that as a typical velocity, as a representative velocity, certainly not true that every molecule of nitrogen gas in this room at 298 Kelvin is moving at exactly 515 meters per second, because every molecule is moving with a different speed from every other one, it's also not correct to say that the average velocity of the molecules is 515 meters per second, because what I've done is not to calculate an actual average, what I've done is calculate the average of V squared and then take the square root of it, that's not at all the same thing as calculating the average, so at this point we can just say that this is a somewhat representative, not a mean, not an average, but a fairly representative velocity or speed of molecules of nitrogen at 298 Kelvin, we'll be able to talk soon about how to calculate that average a little more precisely. The last thing I'll point out is that for this calculation, again, because we did need to use the mass of the nitrogen molecules, here I would get a different answer if I computed the root mean square velocity of an oxygen molecule or an argon molecule or a carbon dioxide molecule and so on, the heavier molecules with larger masses are going to have smaller root mean square velocities because the mass is in the denominator of this square root. So we've seen, we've gotten a fair amount of mileage out of this expression, we have, by making a connection between the energy of a gas and the kinetic energy of the gas, in fact, I can point out that this result that we've obtained is one example of what's called the kinetic theory of gases. There's a bunch of results like this one that result from the assumption that the only thing worth worrying about for gas is its kinetic energy. So if its energy is this large, it must all be kinetic energy and that led us to these conclusions like this expression for the root mean square velocity. So this general idea of the kinetic theory of gases is the idea that we can treat gases purely by their kinetic properties and there's a number of other properties that we can learn about gases, starting with the fact that in order to calculate a true average rather than the slightly strange root mean square velocity, we need to understand a little more about the distribution of the different velocities that a molecule has in a box. So that's what we'll talk about next.