# Ideal gas equation example 3

## Video transcript

Let's do a few more examples with the ideal gas equation. So we have pressure times volume is equal to the number of molecules we have in moles times the ideal gas constant times temperature in Kelvin. And I just got a comment on one of the videos saying it made sense except for the fact that R seems kind of like this mysterious thing in here. And the way you think about R-- let me rearrange this a little bit. Let me write it as pressure times volume is equal to R times nT. So we've established, or hopefully we've established, the intuition that the pressure times the volume should be proportional to essentially the total energy we have in the system. Temperature is average energy, or energy per molecule, and this is the total number molecules we had. What R is, the ideal gas constant is, when you multiply moles times Kelvin, you get mole Kelvin. And with pressure times volume, it's force per area times the volume. You get force times distance, or force times one dimension of distance, which is joules, so this is in joules. So what the ideal gas constant essentially does is it converts-- we know that this is proportional to this, but it sets the exact constant of proportionality and it also makes sure we get the units right, so that's all it is. It just helps us translate from a world dealing with moles and Kelvins to a world of dealing with, well, in this case, atmospheres and liters, or bars and meters cubed, or kilopascals and meters cubed. But no matter what the units of the pressure and the volume are, the whole unit of pressure times volume is going to be joules. So it's just a translation between the two and we know that they're proportional. Now, with that said, let's do some more problems. So let's say we want to know how many grams of oxygen. So we want to know grams of O2. So how many grams of O2 are in a 300-milliliter container that has a pressure of 12 atmospheres and the temperature is 10 degrees Celsius? Well, we've already broken out our ideal gas equation. Let's see, pressure is 12 atmospheres. So we can say 12. I'll keep the units there. 12 atmospheres times the volume. The volume should be in liters, so this is 300 milliliters, or 300 thousandths or 3 tenths. So that's 0.3 liters. 300 one thousandths of a liter is 0.3 liters. And that is equal to the number of moles we have of this, and that's what we need to figure out. If we know the number of moles, we know the number of grams. So this is equal to n times R. Now which R should we use? We're dealing with liters and atmospheres. We'll deal with this one. Liters, atmospheres, mole-Kelvin, 0.082. Times 0.082. And what's our temperature? It's 10 degrees Celsius. We always have to do everything in Kelvin. So it's 283 degrees Kelvin. So all we have to solve for is n. We just have to divide both sides of this equation by this right here, and so we have n is equal to 12 atmospheres-- I'm just swapping the sides-- times 0.3 liters divided by 0.082 times 283 degrees. Our answer will be in moles. And if you want to verify that, you can plug in all the units and use units for the ideal gas constant. The number of moles we're dealing with, so it's 12 times 0.3 divided by 0.082 divided by 283 is equal to 0.155 moles. Now how many grams are in one mole of an oxygen molecule? So one mole of O2, well, we know oxygen's atomic mass. It's 16. One molecule of oxygen, of gaseous oxygen, has two atoms in it, so it has an atomic mass of 32. So it's molar mass, it's mass per mole, is going to be 32 grams. Now we don't have one mole. We have 0.155 moles. So to figure out how many grams we have, we multiply 32 grams per mole times 0.155 moles, and we'll get our answer. So let's do that. So we have 32 times 0.155 is equal to 4.96 grams, or roughly 5 grams. So we have approximately 5 grams of molecular oxygen.