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Root-Mean-Square Speed of Ideal Gases

Patrick Ford
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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 gassed. Now this is something that your professor may not care about a whole lot, but just in case they don't want to break down very quickly, there's 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 the root mean, square is the type of average. So the root mean square speed is a type of average speed for ideal gas particles. Now, I'm gonna show you real quickly what the difference is between an average a real average and the RMS is. So let's check out this example here. We've got these three numbers 5, 11 and 32. We're going to calculate the average versus the R. M. S. So sorry, this is supposed to be part B. So the average here is going to be an average just in case you spend some time where you calculate an average is you just add up all the values five 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 three. So you just go ahead and work this out. You're gonna get is you're gonna get 16 because this works out to 48 divided by 3, 16. All right, So what's the RMS? So 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 each of the values or in this case for our MSP each of the particles speeds here. So the algorithm is a little bit different. We're gonna do is we're gonna take each of the values 5, 11 and 32. And first we're gonna square them. I like to sort of work backwards. So we're gonna square them first five squared plus 11 squared plus 32 squared. And now we're going to take the average or the mean of them. Again, there's three values here and now all you have to do, oops, sorry, this is gonna be 32 square. There's three values. So you just divide by three and then finally you're gonna take the square root of that average. So you're just gonna square root this entire thing here. Notice how these things look very similar but they're actually gonna work out the different things because we're squaring some of the numbers and then square rooting them. Right? So if you work this out, You're actually gonna get 19.7. So that's sort of the difference between the RMS and the average. They work out to almost kind of the same number. But in 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 gets squared and they get a little bit bigger. So they're kind of similar again, but they're a little bit different. All right, 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 gonna show you most just like with most of our problems or equations that we've seen so far, there's gonna be sort of two different forms, depending on which variables you have. So the V R. M. S is going to be equal to either three times bolts means constant times the temperature divided by mass. Or it's going to be equal to three R. T divided by big M. So, again, the difference is that this little M here is gonna be the mass in kilograms, but this big M here is gonna be the molar mass, that's kilograms per mole. And again, because we're working with teas and not delta teas are temperature has to be in kelvin's. 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 put the the molar masses. Alright, so basically we need V. R. M. S. So now, which one of these forms are we going to use? Well, let's see, we're given the molar mass which were given Big M. Over here, not little M. So we're probably gonna stick with that one. So that's gonna be the square roots of three times R. T, divided by big M. Now we have to work out some of the units, right? So we have to This is gonna be three times 8.314. But that's the universal gas constant. The temperature remember has to be in Kelvin's. So if t is equal to 27 degrees, then we have to add to 73 to it in order to put it into kelvin. So T k is gonna be 27 plus 2 73. And that's gonna be exactly 300 kelvin. So that's what goes inside of here. So this is gonna be 300. And now we're gonna have to divide by the molar mass, we're giving them all the 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 the M is equal to two g per mole. And in order to work this into our equations, we need to convert it to kilograms. So this is gonna be 0.2 kg per mole. Alright, So that's what we put in here at the bottom, 0.002. And if you go ahead and work the sandwich, you're getting, it is the RMS speed for hydrogen is about 1934 m per second. So, again, this is again like sort of a type of average, this is sort of like the average speed of the hydrogen particles in our in our atmosphere. Now, this is actually a pretty high speed. These things are flying around at thousands of MPH. 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 anyways, that's a little fun fact. That's for this one. Guys, let me know if you have any questions