 All right, so we have the Maxwell-Boltzmann distribution of molecular speeds. It tells us the probability that a molecule has a certain speed for a molecule of a particular mass at a particular temperature. And the valuable thing about this is it allows us to calculate average speeds of various different types. So for example, one that we've already seen, if we wanted to know the mean square speed of molecules at a particular temperature, we could. This is not how we did the calculation when we did it before, but we could evaluate that average of v squared as just the sum or the integral of v squareds multiplied by the probability of having that speed v integrated over all possible speeds from 0 up to infinity. So we could do that. That's an integral that we could calculate the answer to. We know what the answer would come out to be because we've done that problem already. We had a shorter way of calculating the mean squared speed, and that turned out to be 3 kT over m. But now we have an alternate way that we could have calculated it. The advantage is, in cases where we don't have a shortcut, if we want to know the average speed, not the average squared speed, but the average speed itself, we can now calculate that. So the average speed is going to be sum up all the speeds multiplied by the probability that a molecule has that speed, again, integrate from 0 to infinity, and that will tell us the average speed of a molecule. And that's something we don't yet know the answer to. So let's go ahead and compute what that is. So we know this long expression, the Maxwell-Boltzmann distribution, for the probability. So we can just insert that into this integral. So the integral we want to calculate has some constants out front. 4 pi m over 2 pi kT to 3 halves. And inside the integral, I need to keep the v. So I have a v here and a v squared from Maxwell-Boltzmann. Gives me v cubed altogether. And a Gaussian, e to the minus v squared, with some constants up in the exponent. Essentially, kinetic energy over kT in the exponent. And I'm integrating over speeds from 0 to infinity. So we need to know how to do this integral. Integral of variable cubed times a Gaussian, e to the minus variable squared up in the exponent. For now, we'll just make use of what we could look up in an integral table. I'll tell you that if we want to know the integral of x cubed, e to the minus sub-number of x squareds. Again, as a definite integral from 0 to infinity, what that works out to be, the integral tables we'll tell you is 1 over twice the exponent coefficient itself squared. So this type of integral shows up an awful lot in pKm. So it's also useful to know how to do that integral without looking it up. So there'll be a separate video on how to perform these Gaussian integrals. But for now, we can just use that result to say that if I keep the constants out front, performing this integral gives me this quantity 1 over 2 and a 1 over alpha squared. So in this case, by 1 over alpha squared, alpha is the coefficient in the Gaussian, the coefficient that multiplies the variable squared. So our variable squared is v squared. So the coefficient is the m over 2 kT. So I've got alpha is m over 2 kT in the denominator. And I'm squaring that whole thing. And so now we've just reduced this to a bunch of constants and thermodynamic variables. And we just have to clean it up and simplify it a bit. Let me go ahead and rewrite this last term without the fraction of a fraction. So let's see. I'll call that 4 pi m over 2 pi kT to the 3 halves. Another 1 over 2. And then m over 2 kT in the denominator is like 2 kT over m right side up. And that's all squared. So clearly, there's going to be a lot of simplification here. So let's take that step by step. In terms of the factors of 2, let's do those first. I have a 4, which is divided by 2. So that leaves me 2 multiplied by 2 squared. So that's 8 altogether. Divide by 2 to the 3 halves. So that gets rid of 1 factor of 2 down to 4. And another factor of 2 leaves me with 2 root 2. Let's see. That takes care of my factors of 2. I've dealt with all those. Let's do the m's and the kT's next. I've got a kT squared divided by a kT to the 3 halves. So that leaves me with a net amount of kT in the denominator that's been square rooted. So that's all of the kT's taken care of. m's are exactly the opposite. m in the denominator squared cancels all and more of this m to the 3 halves in the numerator. So all I'm left with, actually, I've done that wrong, the kT's. kT squared in the numerator is bigger than kT to the 3 halves in the denominator. So I end up with a kT under the square root in the numerator. m squared in the denominator is bigger than the m to the 3 halves in the numerator. And I end up with a m in the denominator. So that's dealt with the m's. All I have left is the pi's. So I have a pi to the 1 and a pi to the 3 halves in the denominator. So the net result is a pi underneath the square root in the denominator. And to combine all those together, 2 squared of 2, if I want to bring that underneath the square root sign, that would be like the square root of 8. Square root of 8 is 2 root 2. So after all of that cancellation, what I'm left with is that the average speed for molecules at a particular temperature is the square root of 8 kT over pi m. So let me go ahead and write that down over here. And sometimes we refer to that with this average notation, v in angle brackets is the average v average speed. We can also see, say, v sub av for the average. And that is sort of to make it comparable to the value for the root mean square speed that we've seen so far. Remember, the result for that was 3 kT over m. When we did that via a different approach, the square root of the mean square velocity, the root mean square speed, is squared of 3 kT over m. So we get a different result for the average speed and for the root mean square speed. I'll point out that this ratio, 8 divided by pi. If we calculate what 8 divided by pi is, that works out to be not exactly, but somewhat close to 2 and 1 half. I'll write that where you can see it. That one is 2.5 kT divided by mass. So the root mean square velocity, square root of 3 kT over m is larger than the average velocity 2 and 1 half kT over m inside a square root. So when we're looking at the velocity distribution, Maxwell-Boltzmann tells us the probability of having a certain speed has a shape like this. Both the root mean square and the average speed are larger than the peak in this distribution. The peak in this distribution would be the mean only if the distribution were symmetric. If the tail on the left side and the tail are on the right side, we're mirror images of each other. What the Maxwell-Boltzmann distribution tells us is the speeds at the high speed and the tails on the right side are much fatter to curve. And the root mean square speed, because we're averaging the squares rather than the velocities themselves, that shifts it even further because the squares of these large numbers in this tail are even larger than the speeds themselves. So the VRMS is over here. The average is smaller but still greater than the peak in the middle. Just to give you an idea of the numerical values, if we were to calculate for a molecule like nitrogen at a temperature of, let's say, 298 Kelvin. So computationally, this works very much like when we did the calculation of root mean square speed. After think about the same sort of units, what units use to use either K or R and being careful to convert the molecular weight of nitrogen from grams per mole into either kilograms or kilograms per mole to make sure all the units cancel. So I won't plug the numbers in to work that out, but if you use 28 grams per mole and 298 Kelvin and Boltzmann's constant and an 8 and a pi, what you find is that the average speed of a molecule of nitrogen at room temperature at 298 Kelvin, that works out to be 474 meters per second. And if you recall, the root mean square speed was 515 meters per second. So again, the root mean square speed because of this factor of 3 is larger by a little bit than the average speed because it has a factor of 8 over pi or roughly 2 and 1 half instead. So that tells us the formula we need to calculate the average speed, which is perhaps more useful. Certainly easier to think about the average speed than the root mean square speed. This diagram, however, points out that there's another important speed that we might be interested in, which is what's the most typical speed, the speed at the top of this distribution? So we can also calculate what that is, and that's the subject of the next video lecture.