 So, thinking about gases as being made of molecules that are just moving around with some kinetic energy, as we've done with the kinetic theory of gases, can tell us not just about the velocities and speeds of those molecules, but it's also interesting to think about the collisions that those molecules experience. So, let me draw a picture here, and let's imagine we've got a box of gas molecules, so here's a molecule, there's a bunch of molecules of gas in the box. And I can ask if one of these molecules is moving through the gas, how crowded or how not crowded is that gas, how far will it go on average before it bumps into another molecule? So that's this concept that we're talking about of a mean free path on average, what's the mean for the path that will move freely before it bumps into another molecule? Maybe there's another molecule directly on its path? If we insist on thinking of these particles as small point particles, then they don't take up any space at all and they'll never collide with another one. So if we acknowledge that each of these particles has some finite size, has some radius, so I'll say r is the radius of one of these gas molecules, then it doesn't have to be directly on the path, it can be just near the path, and then as this molecule moves forward it sweeps out a region of space and it can collide with another molecule that's in its path. And in fact, the furthest a molecule can be and still experience a collision is when the center of this molecule, its radius, one factor of r below matches up with the boundary of another molecule which is a distance of r above the center of it. So there's actually what's called a collision tube that gets swept out, so the distance from the center of this molecule to the center of this molecule can actually be twice the radius. One radius for the molecule that's moving and one radius for the molecule that's being hit. So there's a tube of radius 2r or the diameter of the molecule that's being swept out as this molecule moves through the box. So we'd like to know as the molecule moves for some length, let's call that length lambda, how many molecules is it likely to bump into? So the, let's think about the surface area or the cross sectional area of this collision tube, so that's a circle of radius d, so there's pi times d squared is the cross sectional area of this circular tube as my spherical molecule sweeps its path out through the gas. If it moves for a distance lambda, then the volume of the tube that gets swept out from its starting point after it's moved a distance of lambda is the length times the cross sectional area of that cylinder, so it's lambda times pi d squared. What I really want to talk about is not the volume inside that cylinder, but the number of other molecules we expect to be inside that cylinder. So now let's think about how many molecules, what's the density of molecules we can expect to find in this box? So this density that I'm talking about is not the mass density, I don't really care how heavy the molecules are, all I care about is how many molecules there are in a unit volume. So just the number of molecules, so this is called a number density rather than a mass density. So that quantity I'm interested in the number of molecules in any given volume, that ratio. We can, because we know the ideal gas law, pressure times volume is n times k times t, or n times r times t if you prefer to use the gas constant. There's a relationship between n and v of course, so I can rewrite, if I bring the v over here and the kt over there, the number density n divided by volume for a given gas is the same as its pressure divided by kt or divided by rt. So that's the number density. So if I go back to this expression and I say how many collisions can I expect for a molecule that sweeps out this collision tube, if it travels for a distance lambda, sweeps out a volume of lambda times pi d squared, the number of molecules in that volume that I can expect. If I multiply that volume by the ratio of how many molecules there are per unit volume, so lambda times pi d squared times the number per unit volume, that should give me the expected number, the average number of collisions I should experience. And if I write that n over v as pressure over kt, I get this expression. So that's the expected number of collisions. In particular, the quantity that we want to calculate that's called the mean free path is not how many collisions are experienced on average if I move a distance of a meter, but how far do I have to move before the first collision? So not n collisions in a distance lambda, but what distance do I have to move in order to experience my first collision, just one collision? So if I use one collision, I can calculate the distance I have to move before I experience that first collision on average. So let's rearrange this expression solving for lambda. So kt moves to the top. The pi d squared p are all on the bottom. And that looks like what we would expect to have as our mean free path. On average, the distance we have to travel before we run into our first collision. There's one slight problem with this expression, however, which is the way I've set this problem up. I have one molecule moving and the rest of the molecules sitting still. Of course, in a gas, the other molecules aren't actually sitting still. They are moving around as well. So to illustrate why that matters, let's suppose for a moment that all these other molecules are moving in the same direction in the same speed as the initial molecule. So if that's true, then we're never going to experience any collision. All the molecules are running away from each other at exactly the same speed. So if the relative velocity of these molecules is zero, if they all are moving in the same direction in the same speed, then I won't see this many collisions. I'll see zero collisions instead. On the other hand, suppose the opposite. Suppose that all the other molecules are moving head on towards the molecule that I focused on in the beginning. So then not only would I collide with all the molecules in this collision tube, but I'd collide with some others further back that have moved their way into the collision tube by the time the first molecule got there. And in fact, if all the molecules are moving with the same speed and opposite direction, the number of collisions is going to be exactly twice what I originally expected. So again, if the relative velocity is twice this molecule's speed, then I'd get twice as many collisions. So this quantity of the relative velocity is important to understand. So that relative velocity as a vector, let's say, let's label our molecules. This is molecule A, and I have some other molecule B that might be moving in some different direction. And I can say the relative velocity is the velocity of A relative to, or the velocity of A minus the velocity of molecule B. It turns out to be useful to think about what is that relative velocity squared if I take this velocity times itself? So VA minus VB squared, that's the same as saying VA minus VB times VA minus VB. And what multiplying means for vectors is I take the dot product of those two vectors. So that product, VA dotted to VA, gives me VA squared. I've got VA dot VB with a negative sign that shows up twice, once from this pair and once from this pair. And I've also got VB dotted into VB, the negative signs cancel, and I have VB squared. So that's one way of writing down what the relative velocity is. If I know the velocity of molecule B and molecule A, then I can calculate the relative velocities or the square of the relative velocity. Of course every molecule B is moving in some different direction with some different speed. So this is going to be different for every pair of molecules. On average, if I take the average of that quantity, let me go ahead and write it out fully, it's the average of this whole expression, VA squared minus twice VA dot VB plus VB squared. The average of those things added together, I can break that up into the sum of their various averages. So notice what we've got here, we've got a mean squared velocity. The mean squared relative velocity is equal to the mean squared velocity for molecule A plus the mean squared velocity for molecule B minus this term in the middle. But on average, if I take the average of VA dot VB, I don't know what direction this particular molecule is moving, but if I take the average over many molecules, some of them are moving towards me, some of them are moving away from me. In fact, there's exactly as many moving with a certain component of their velocity towards me as with the same component of their velocity moving away from me. So this term is equal to zero, not because the dot product is always equal to zero, but when I average over many of those dot products, there's as many positive ones as negative ones and that goes to zero. So this mean squared velocity, let me say mean squared velocity relative is equal to the mean squared velocity for molecule A plus the mean squared velocity for molecule B. That's not the root mean squared velocity, it's the mean squared velocity. But I can certainly convert that to a root mean squared relative velocity by taking the square root. So from here to here, I just take the square root of both sides of those equations. So I've got the square root of two, then the square root of the mean square velocity for any one of these molecules is just the root mean square velocity. So this is an interesting result. Turns out the root mean square relative velocity between any pair of molecules can be expected to be on average, square root of two, larger than the root mean square root of the velocities itself. Likewise, for the average velocities, I can say the average relative velocity is going to turn out to be root two bigger than the average velocity for one of the molecules. And what that means is instead of sweeping out, as this molecule moves in this direction and sweeps out the volume that it sweeps out, it's not going to collide with a number of molecules that we expected. It's going to collide with square root of two times as many molecules because the relative velocity is larger than the velocity of molecule A. So what that means is we need to stick an extra square root of two in here. So the number of collisions is square root of two times larger. So if I go back and retroactively throw in a square root of two into all of these equations, when I solve for lambda, the square root of two is going to end up in the denominator. And this is our final result for the mean free path of molecules in a gas. On average, the distance of molecule can be expected to move before it collides with another molecule is given by this expression, kT over root two pi times D, the diameter of the molecule squared divided by the pressure of the gas. So just to work one quick example, let's say we want to know about nitrogen gas at 298 Kelvin and one atmosphere pressure. The other thing we need to know is the diameter of the nitrogen molecule. So I'll give you what is experimentally observed to be the diameter, not angstroms, but angstrom, not angstrom squared, but angstroms. The, what's called the kinetic diameter of this molecule, the diameter as needed to do these calculations for collisions. Turns out to be about three and a half angstroms. So if we use that to calculate the mean free path, we just plug these terms in. So I need to know, let's see, I'll use k in units of joules per Kelvin. Multiply that by 298 Kelvin, as was given in the problem. If I divide by square root of two pi, rather than angstroms, if I convert to SI units, an angstrom is 10 to the minus 10 meters. I have to remember to square that. Also divide by pressure. In order to do this calculation in SI units, atmospheres is not an SI unit. So I'll convert one atmosphere to 101,325 Pascals, as perhaps the easiest way to keep us in SI units. And then if I plug all those numbers into a calculator just to get a size, an intuition for the size of these numbers, that turns out to be about 7 times 10 to the minus 8 meters. Double checking the units. Pascals, if we remember that the unit of a Pascale is the same as joule per cubic meter. Then in the denominator, I've got 1 over meters cubed times a meter squared, which leaves me with 1 over meters in the denominator. That converts to this meters. This joule cancels that joule. And these kelvins cancel those kelvins. So the units all work out and give me units of meters, which I'd expect for a distance like the mean free path. So if I convert this distance into nanometers, so 7 times 10 to the minus 8 meters or 70 nanometers, what that tells us is these molecules will move on average, typically 70 nanometers before they bump into another molecule in a gas of nitrogen at room temperature and room pressure. That sounds like a very short distance, of course, but keeping in mind that the diameter of one of these molecules is 3 and 1 half angstroms or 0.35 nanometers, that's roughly 200 molecular diameters. So a molecule will move roughly 200 times its own size before bumping into another molecule, which helps us get some idea, some intuition for how crowded or not crowded these gases are. A molecule will have to move 200 times its own size before colliding with another molecule. So we're not quite done with this type of calculation yet. It's often more useful to think about not the mean free path in terms of distance, but the mean free or the collision frequency, roughly how often the collisions will happen in units of time in a gas. So that's the next step is to convert the mean free path into a collision frequency.