 So the Maxwell-Boltzmann probability distribution tells us the probability that a molecule has a particular speed under certain circumstances, at a particular temperature, and that tells us this full distribution of having various different speeds. We also have found that the root mean square and the average speeds can be calculated directly from the temperature and the mass of the molecule, and those are not the same value. The average speed is a little larger than the peak of this distribution. The root mean square speed is a little bit larger still, but it's an interesting question to ask ourselves. What is this peak value? What is the value that has the highest probability? In other words, what's the most probable velocity? The single velocity that has the highest chance of representing one of the molecules in a gas. So that's what we'll call that VMP for most probable, and the thing we know about the peak of a distribution is that the slope of the distribution is flat at that point. So we know the derivative of probability with respect to the derivative of P with respect to V is equal to zero when we're at the peak. So when we evaluate that at the most probable velocity, then the slope of this distribution is equal to zero. So that's how we can find the most probable velocity. So let's go ahead and take the derivative of the Maxwell-Boltzmann distribution. So if we want dP dV, we want that to be equal to zero. So that derivative of this expression, I've got some constants out front, 4 pi times m over 2 pi kT to the 3 halves, and then the derivative has V showing up twice in it, V squared in the pre-factor and then a V squared in the exponent. So I need to use the product rule. So the derivative is these constants times the derivative of V squared is 2V multiplied by the exponential, and I also have a V squared leaving the V squared alone but taking the derivative of the exponential. So the derivative of this exponential is the exponential times the derivative of the exponent. So the derivative of the V squared up here gives me 2V. So if I include the constants minus m over 2 kT, I've got altogether minus 2 mV over 2 kT, and that's all multiplied by these constants out front. We can clean that up quite a bit by pulling some things out of the brackets, the constants out front. Inside the brackets, it looks like I have this exponential and this exponential are the same. This is out of the brackets. Likewise, I have a vector of V here and there's at least one vector of V in the second term that I can pull out of the brackets. So what's left after I pull those things out of the brackets is I have a 2. I have a negative sign from the second term. The 2's cancelled but I have an m and a kT and a V squared that doesn't get pulled out. And that's all multiplied by one vector of V e to the minus mV squared over 2 kT. And I want that all to be equal to zero. The only way to make that equal to zero, I can't make the constants equal to zero. I could make V equal to zero. So that means the slope of this curve is zero when I get to V equal to zero. That's not the interesting point. The interesting point is when this term in brackets is equal to zero and that's going to be the peak of this curve. So we're really looking for the point where zero is equal to the quantity in brackets, twice or two minus m over kT V squared. And when that's true is when we're at the most probable velocity. So rearranging that equation, two is equal to m over kT, most probable velocity squared. Bringing the kT and the m to the other side. My most probable velocity squared is twice kT over m. And then if I take the square root, V sub mp is equal to the square root of 2 kT over m. So that is yet a third different way of measuring a typical velocity or typical speed of a molecule in a gas. What's the average speed of a molecule in a gas? Roughly two and a half kT over m square rooted. The root mean square speed is a little bit larger, square root of 3 kT over m. The most probable velocity is smaller than either the average of the root mean square speed and that's the square root of 2 kT over m. So what we have is three different ways, three all different ways of conceptualizing what a typical velocity or speed is for a molecule in a gas. Are we interested in the speed that we see most often or the speed that we'd obtain by taking an average or the speed that we'd obtain by taking a root mean square? So we have lots of different ways of thinking about the speeds of molecules in a gas. What we can do next is stop thinking quite so much about velocities and go back to thinking about other things we can learn about gas molecules from the kinetic theory of gases.