Hey, everybody. So let's get started with our problem here. We have a container whose volume is 1.6 liters. There's lots of numbers here, so I'm going to start writing out my numbers. So I've got \( v = 1.6 \, \text{l} \) (liters). It's filled with 200 grams of an ideal gas. Note that this is grams, not moles. This tells us the mass of this ideal gas. So this is \( m = 200 \, \text{g} \) (grams). This ideal gas has a molecular mass of 28 grams per mole. Molecular mass also means the same thing as molar mass. So remember, molar mass is not little m, it's big M. This is \( M = 28 \, \text{g/mol} \). And now we have the RMS speed of the molecules which is 600 meters per second. So we have \( v_{\text{RMS}} = 600 \, \text{m/s} \). Basically, what we want to calculate in this problem is the pressure of the gas. So that's ultimately my target variable. What is the pressure? Which equation do we use? Well, when it comes to pressures or volumes or something like that, almost always you're going to use either \( PV = nRT \) or \( nK_BT \). If you're ever unsure which one to use, always just stick with \( nRT \). That's probably going to be the easier one. All we have to do is isolate for \( P \). So this \\ P \) here, once you move the volume to the other side becomes \( \frac{nRT}{V} \). To calculate the pressure, I need 3 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 \( R \) is just a constant.
Do I have moles? Well, if you look at the problem, we actually don't because we're told that we have 200 grams as an 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 Avogadro's law. Remember that moles can be related to mass and molar mass by this equation here, \( \frac{m}{M} \). So that's what we're going to do here. So this little \( n \) here is equal to mass over molar mass, And we have both of those things. We have mass and molar mass. So what I'm going to do here is, I'm going to divide these two numbers, 200 grams and 28 grams per mole. You could convert them both to kilograms and then do the calculation. Or what you can do is if you just do 200 grams divided by 28 grams per mole, then what happens is the grams just cancel out anyway, regardless of you having to convert it back to kilograms. What you're going to end up with here is 7.14 moles. So you could convert it, and you're going to get the same answer, but it's 7.14 moles. Now that's for the moles.
What about the temperature here? How do we figure out \( T \)? Well, all we're told here is the volume and the mass, so we actually need to find out what \( T \) is. So let's move on to their second variable. So we have \( T \). How do we figure that out here? We have a bunch of equations that involve \( T \). Remember, we have kinetic energy average, we have \( E_{\text{internal}} \), but we also have the RMS velocity that is basically just a function of temperature and either mass or molar mass. Right? We have \( T \) 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. \( v_{\text{RMS}} = \sqrt{\frac{3RT}{M}} \). If I want to figure out the temperature, what I've got to do here, that's what I came over here for, is I've got to rearrange this equation. So I've got \( v_{\text{RMS}}^2 = \frac{3RT}{M} \). So now I'm just going to move this stuff to the other side like this to isolate the \( T \). So what I get here is that \( v_{\text{RMS}}^2 \), and actually I'm just going to go ahead and start plugging in some variables. Right? So I've got \( 600^2 \times M \), the molar mass. Now be careful because when I do this, I actually do have to convert this. This molar mass here is 28 grams per mole, but this is equal to 0.028 kilograms per mole. So I have to actually do this \( 0.028 \div (3 \times R) \), 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 going to get is the temperature of, let's see. This is going to be 404.1 Kelvin.
So now that we have the temperature, the last thing we need is the volume. And we have that. The volume is 1.6 liters, 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 sorted out. I've got a conversion table here to help us out with anything. Remember that 1 liter is 0.001 meters cubed. So if I have 1.6 liters, then if I convert it, it's going to be 0.0016. So that's just what goes inside of this equation here, and we're ready to go-ahead to plug everything in. So this is the number of moles, which is 7.14, that's what I got over here, times \( R \) 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 \( 1.5 \times 10^7 \) which is going to be in Pascals. And that's your final answer.
Hopefully, that made sense. Let me know if you have any questions.