 I have, like, I don't know, hundreds of videos, so. But anyways, let's start with this one. What's the volume of gas occupied by five grams of methane, CH4, at 25 degrees Celsius, and 1.018? So the one thing that you're going to be given with all of these problems is the ideal gas constant, which is all, OK? So let's write that down right now. So R equals 0.0821 liter ATM mole kelp, like that. So that's the way you'll be given. But that's like the text message way of writing, because you can write it in one line, or like the typewriter way. But it doesn't really help you in general chemistry, OK? So I want you to kind of be able to look at this and then change it to a form that'll help you a little bit more. And it's not very convenient. So you just take this, and where it says, per mole kelp, just erase that part, and divide the whole thing by one mole times kelp. Just makes your life a lot easier, I promise. And remember, what's the only equation we're going to remember for all these gas loss? What is it? PV equals nRT, OK? And if your problem says that your gas changes in some way, then you're going to do PV equals nRT divided by PV equals nRT. But if it doesn't show a change, you're just going to use PV equals nRT, OK? So let's read this problem now, OK? So what is the volume of gas occupied by 5 grams of methane at 25 degrees Celsius in 1 atm? Did anything change? No. So we're just going to use PV equals nRT. So let's write down what we know. PV, do we know that? What is it? 1 atm, 1.0 atm of V. Do we know that? No, that's what we're looking for. n, do we know that? Well, so we know the mass, right? We know the mass of it. So we don't know n, but we can get n from n, OK? So let's write down n for right now, 5.0 grams of methane. In fact, let's write down that we have methane, too. Of course, we know, given to us. And T, we know, too, 25 degrees Celsius. But in these gas laws, what kind of temperature you will have to have it in Kelvin? So what do you have to add to Celsius? 273. So yeah, exactly, 298. And remember your units, OK? Cool. Are moles from grams? You guys remember? So there's always a conversion factor to do something to something else, OK? So I'm looking for something where I can multiply and cancel out grams and get moles. Does everybody cool with that? So I want something that's like this. So it relates grams and moles. Does anybody know what does that? Anybody ever heard of anything that does that? The pot? No, that's Avogadro's number. All right. So it has something to do with this here. You guys remember? The molar mass. That makes sense, right? The molar mass. The molar mass would be the mass of one mole, right? The mass of one mole. That makes sense, right? The molar mass is the mass of one mole. That's why we call it the molar mass. So that should probably be able to relate grams and moles, right, because grams is the unit of mass, right? So how do we get the molar mass again? One carbine plus. Yeah, exactly. We add up the carbons and hydrogens, right? So we say 4 times 1.008. And we can say grams per mole, because that's the units we're looking for, plus 1, if you need to, 12.01 grams per mole. Where am I getting these numbers from? That's the periodic table. So we're finding the molar mass of methane here. OK, so let's take the calculator and add those up. So 12.01 plus 4 times 1.08. And remember your sig figs, 16.04 grams per mole. So that's going to relate, that's a conversion factor that relates grams and moles of methane. So how can we write that another way? We can say 16.04 grams of methane equals 1 mole of methane, right? So if we know that, and we've got 5 grams, right? We can just use that, right? So we say down here, 16.04, 1 mole, cancel, cancel. So 5 divided by that number. So how many moles of methane do we have? Very good. 3.12 moles. So we still, so now we're cool with the number of moles, right? But we now have to find the volume, OK? So do we have everything in the units that we wanted? Kelvin, yes. Moles, yes. ATM, yes, right? So volume we're going to have in liters, OK? So if it doesn't come out of liters, then we've done something incorrectly, OK? So can I erase all of this stuff down here? Well, it's recorded. You can watch it on YouTube, OK? Oh, thank you. So I'll give you the web address after. OK, so let's just write pv equals nrt. And we want to isolate the v variable, right? OK, so that's what we want. How do we do that? Divide both sides by p, cancel, cancel, like that. That's our new equation, OK? So let's just plug in. So do we have all these values? So this is now not m anymore. That's n, right? So do we have n, yes? Do we have r, yes? Do we have t, yes? p, yes, right? We're cool. So let's plug everything in. So n12 moles, r. So this is why I want you to write it like this. So I want you to make it big. It helps you. It helps out the introductory chemistry student. I promise you. t divided by p118, or 1.0. OK, is everybody cool with that? So now what are we going to do? Cancel our units, OK? Hopefully we get leaders up, OK? Moles on the numerator, moles on the denominator, cancel, cancel. Calving numerator, calving denominator, cancel, cancel. ATM numerator, ATM denominator, cancel, cancel. What are we left with in the numerator? Leaders. Is that cool, right? Because we're looking for volume, right? So let's just put this up here. Is that all right, OK? So let's just plug in. So 0.312 times the gas constant, 0.0821 times 298 divided by 1, of course. What did you guys get? So 7.63, if you go to too many sig figs, because we got one thing with two sig figs. So it's like kind of got to watch those things. So that's like, I would take half a point off for sig figs, and it's going to be, is everybody cool with that? OK, so if we have five grams of methane, that's a volume that's 7.6 liters big, if you can imagine that. So you want to think about things in the physical, too. It's not all from the chemical, OK? Any questions on this before I turn off the video? OK, cool.