1

concept

## Root-Mean-Square Speed of Ideal Gases

5m

Play a video:

Was this helpful?

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

2

Problem

What is the temperature of a sample of CO_{2} molecules whose rms speed is 300 m/s? The molecular mass for Carbon and Oxygen is 12.01 g/mol and 16 g/mol, respectively.

A

158.7 K

B

529.2 K

C

1.7 × 10

^{5}KD

1391 K

3

example

## Finding Pressure from RMS Speed

6m

Play a video:

Was this helpful?

Hey everybody. So let's get started with our problem here. So we have a container. Its volume is 1.6 liters. There's lots of numbers here. So I'm gonna start running out my numbers. So I've got V. Equals 1.6 liters. Um I have that. It's filled with 200 g of an ideal gas. Now notice this is Graham's not moles. So what this is actually telling us here is this is the mass of this ideal gas. So this is M equals 200 g. Now this ideal gas has a molecular mass of 28 g per mol. Molecular mass also means the same thing as molar mass. So remember molar masses not little M it's big. M this is big M which is equal to 28 g per mole. And now we have the RMS speed of the molecules is 600 m per second. So we have V. R. M. S. So that's 600. And then basically what we want to calculate in this problem is what's the pressure of the gas? So that's ultimately my target variable. What is the pressure. So which equation do we use? Well, when it comes to pressures or volumes or something like that? Almost always are going to use either PV equals NRT or N. K. B. T. If you're ever unsure of which one to use, always just stick with N. R. T. That's probably gonna be the easier one. Um So this is gonna be PV equals and R. T. All right. So that's the equation we're gonna use and all we have to do is isolate for P. So this P. Here. Once you move the volume to the other side becomes N. R. T. Divided by V. So to calculate the pressure, I need three things. I need the moles. I need the temperature and I need the volume. So let's get started with the moles here because remember this are is just a constant. So do I have moles? Well, if you look at the problem, we actually don't because we're told that we have 200 g of ideal gas. But that's not the moles. That's the mass. So how do we figure out what this number of moles here is how do we figure out N well, let's see if we have the mass and we have the molar mass. Then we can use an old equation from when we talked about moles and avocados law. Remember that moles can be related to mass and molar mass by this equation here. Little M over big M. So that's what we're gonna do here. So this little end here is equal to mass over moller mass. And we have both of those things. We have mass and molar mass. So what I'm gonna do here is I'm gonna divide this uh these two numbers 200 g and 28 g per mole. Then there's two ways to do this. You could convert them both two kg. Uh and then do the calculation or what you can do is if you just do g divided by 28 g per mole, then what happens is the grams is just gonna cancel out anyways, regardless of you having to convert it back to kilograms. And what you're gonna end up with here is 7.14 moles. All right. So you couldn't convert it and you're gonna get the same answer, but it's 7.14 moles. So, that's done now. That's the that's the that's the end. What about the temperature here? How do we figure out t Well, we're told here is the volume and the mass. So that we actually need to find out what T. Is. So let's move on to their second variable. So, we have tea. So, how do we figure that out here? We have a bunch of equations that involve? T remember, we have kinetic energy average, we have E. Internal but we also have the RMS velocity. That is basically just a function of temperature and either mass or molar mass, right? We have tea inside of this equation. So, because we know what the RMS velocity is the 600 Here, we're going to start off with this equation RMS is equal to And then which of the forms are we going to use? We use this one or this one? Well, actually in this case it really doesn't matter because remember in this problem we have both the M or and we also have the molar mass. So it actually doesn't matter which one you use, you're still going to get the right answer. So it really just comes down to your preference. Um But let's go ahead and let's see. I I used I use just use the the three R. T. Divided by the molar mass. Right? So again you could have used the other one, you'll still get the same exact answer. All right. So if I want to figure out the temperature, what I've gotta do here, that's what I came over here for is I've got to rearrange this equation. So I've got the R. M. S. Squared equals three R. T. Divided by big M. So now I'm just gonna move this stuff to the other side like this to isolate the T. So what I get here is that V. R. M. S squared. And actually I'm just gonna go ahead and start plugging in some variables. Right? So I've got 600 squared times the M. The molar mass. Now be careful because when I do this I actually do have to convert this. This molar mass here is 28 g per mole, but this is equal to 0.028 kg per mole. Alright, so I have to actually convert this 0.028 Divided by and then the three times are which is 8.314. So that's going to give you the temperature. Remember that's what we came over here for and when you work everything out, what you're gonna get is the temperature of um Let's see this is going to be 404.1 Kelvin. Alright, so again, you know, a lot of this problem here is just figuring out the variable, looking for going off and using a separate equation and then bringing it back so we can plug everything into this equation here. So now that we have the temperature, the last thing we need is the volume and we have that The volume is 1. L but be careful because you have to convert that into the right unit. So a lot of these problems will try to trip you up in terms of the units, just make sure that you have all your units sort of squared away. I've got a conversion table here to help us out with anything, Remember that 1.1 leader is 0.1 m cubed. So if I have 1.6 liters then if I convert it, it's going to be 0.16. So that's just what goes inside of this equation here and we're ready to go ahead and plug everything in. So this is the number of moles which is 7.14, that's what I got over here times are which is 8.314 times the T, which I just figured out was 404.1 and then we divide that by the volume, which is 0.0016. And what you should get guys is a pressure of 1.5 times to the 7th. And that's going to be in pascal's and that's your final answer. Alright? So hopefully that made sense. Let me know if you have any questions.

Additional resources for Root-Mean-Square Velocity of Gases

PRACTICE PROBLEMS AND ACTIVITIES (8)

- Oxygen (O2) has a molar mass of 32.0 g/mol. What is (g) How many oxygen molecules traveling at this speed are ...
- Oxygen (O2) has a molar mass of 32.0 g>mol. What is (e) Suppose an oxygen molecule traveling at this speed ...
- For diatomic carbon dioxide gas (CO2, molar mass 44.0 g/mol) at T = 300 K, calculate (c) the root-mean-square ...
- Smoke particles in the air typically have masses of the order of 10-16 kg. The Brownian motion (rapid, irregul...
- At what temperature is the root-mean-square speed of nitrogen molecules equal to the root-mean-square speed of...
- Oxygen (O2) has a molar mass of 32.0 g>mol. What is (d) the momentum of an oxygen molecule traveling at thi...
- Martian Climate. The atmosphere of Mars is mostly CO2 (molar mass 44.0 g/mol) under a pressure of 650 Pa, whic...
- A flask contains a mixture of neon (Ne), krypton (Kr), and radon (Rn) gases. Compare (b) the root-mean-square ...