Hey, guys. So occasionally you might run into a problem that asks you to calculate something called the root mean square speed of an ideal gas. Now this is something that your professor may not care about a whole lot, but just in case they do, I want to break it down for you very quickly. There are a couple of very simple equations here. But first, I want to talk about what the root mean square actually is, what it means. Basically, all you need to know is that the root mean square is a type of average. So the root mean square speed is a type of average speed for ideal gas particles. Now, I'm going to show you real quickly what the difference is between an average, a real average, and the RMS. So let's check out this example here. We've got these three numbers 5, 11, and 32. We want to calculate the average versus the RMS.
So the average here is going to be: n a v e r a g e Just in case you spend some time where you calculate an average is you just add up all the values, 5 plus 11 plus 32, and you divide by the number of values that there are. So for instance, there are three numbers here, so you just divide by 3. So you just go ahead and work this out, and what you're going to get is 16 because this works out to 48 divided by 3 = 16.
So what's the RMS? Now we're going to calculate something different, the root mean square. Now as the name suggests, the root mean square is the square root of the mean, which is really just an average, of the squares of the values, or in this case for RMS, each of the particle's speeds here. So the process is a little bit different. We're going to take each of the values 5, 11, and 32, and first we're going to square them. I like to sort of work backwards. So we're going to square them first: 5 2 + 11 2 + 32 2 And now we're going to take the average or the mean of them. Again, there are 3 values here, so you just divide by 3. And then finally, you're going to take the square roots of that average: ( 5 2 + 11 2 + 32 2 3 ) So if you work this out, you're actually going to get approximately 19.7. So that's sort of the difference between the RMS and the average. They work out to almost the same number, but generally, what happens is that the root mean square of whatever value is going to be a little bit higher than the average. And that's because it's sort of skewed towards the higher numbers, the larger numbers that get squared, and they get a little bit bigger. They are similar, again, but they're a little bit different.
Alright. So let's move on to the RMS speed. The RMS speed really just depends for an ideal gas on the temperature and the mass of the gas particles. It's a very straightforward equation I'm going to show you. Most just like with most of our problems or equations that we've seen so far, there are going to be sort of 2 different forms depending on which variables you have. So the v RMS is going to be equal to either 3 × k T / m where k is Boltzmann's constant, T is the temperature, and m is the mass in kilograms; or it's going to be equal to 3 R T M . Again, the difference is that this little m here is going to be the mass in kilograms, but this big M here is going to be the molar mass, that's kilograms per mole. And again, because we're working with T and not delta T, our temperature has to be in Kelvins. That's really all there is to it. There's a, you know, sometimes your textbooks will sort of give the proof for this, but we don't really care about that, because you'll never actually be asked to prove it.
So let's take a look at our example here. We're going to calculate the RMS speed of hydrogen gas, that's for instance in the atmosphere, right, at 27 degrees Celsius. And we can go to the molar mass is.
Alright. So, basically, we need v RMS . So now, which one of these forms are we going to use? Well, let's see. We're given the molar mass, which we're given big M over here not little m, so we're probably going to stick with that one. So that's going to be 3 × 8.314 × 300 0.002 . Now we have to work out some of the units. Right? So, this is going to be 3 times 8.314, but that's the universal gas constant. The temperature, remember, has to be in Kelvins. So if \( T \) is equal to 27 degrees, then we have to add 273 to it in order to put it into Kelvin. So \( T_K \) is going to be 27 + 273 and that's going to be exactly 300 Kelvin. So that's what goes inside of here. So this is going to be 300, and now we're going to have to divide by the molar mass. We're given the molar mass in terms of grams per mole, but remember, this molar mass needs to be in kilograms per mole. Right? So basically, we're given that \( M \) is equal to 2 grams per mole, and in order to work this into our equations, we need to convert it to kilograms. So this is going to be 0.002 kilograms per mole. Alright? So that's what we put here on the bottom, 0.002. And if you go ahead and work this out, what you're getting is the RMS speed for hydrogen is about 1934 meters per second.
So again, this is again like a sort of type of average. This is sort of like the average speed of the hydrogen particles in our inner atmosphere. Now this is actually a pretty high speed. These things are flying around at thousands of miles per hour. And what you should know here is that this \( v RMS \) is actually sort of the average speed of the particles and many are actually going above and below that speed. This really high velocity for hydrogen gas is one of the reasons that we actually don't have a lot of hydrogen in our atmosphere. Anyway, that's a little fun fact. That's it for this one, guys. Let me know if you have any questions.
