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Finding Pressure from RMS Speed

Patrick Ford
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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.