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Worked example: Using the ideal gas law to calculate number of moles

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The ideal gas law relates four macroscopic properties of ideal gases (pressure, volume, number of moles, and temperature). If we know the values of three of these properties, we can use the ideal gas law to solve for the fourth. In this video, we'll use the ideal gas law to solve for the number of moles (and ultimately molecules) in a sample of gas. Created by Sal Khan.

Video transcript

- [Instructor] We're told an athlete takes a deep breath, inhaling 1.85 liters of air at 21 degrees Celsius and 754 millimeters of mercury. How many moles of air are in the breath? How many molecules? So pause this video, and see if you can figure this out on your own. All right, now let's work through this together. So let's think about what they are giving us and what we need to figure out. So, they are giving us a volume, right over here. They are also giving us a temperature, right over here. They're also giving us, I'm trying to use all of my colors here, they're giving us a pressure. And they want us to figure out the number of moles. I'm gonna use a green color here. So they want to know, so we often use the lowercase letter, n, to represent the number of moles. And so, do we know something that connects pressure, temperature, volume, and the number of moles? Well, you might be thinking of the Ideal Gas Law, which tells us that pressure times volume is equal to the number of moles, n, times the ideal gas constant, R, times temperature, T. And so we know everything here except for n, so we can solve for n. I know what some of you are saying, "Wait, do we know R?" Well, R is a constant. And it's going to be dependent on which units we use, and we'll figure out which version of R we use. But that's why I gave you this little table here, that you might see on a formula sheet, if you were taking something like an AP exam. So we actually do know what R is. So, we just need to solve for n. So, to solve for n, you just divide both sides by RT, and so you are going to get that n is equal to pressure times the volume over R times T, R times T. And so this is going to be equal to what? Well, our pressure is 754 millimeters of mercury. Now, over here, where they give us the ideal gas or the different versions of the ideal gas constants, you don't see any of them that deal with millimeters of mercury. But they do tell us that each millimeter of mercury is equal to a Torr. If you get very, very, very precise, they are slightly different. But for the purposes of a first-year chemistry class, you can view a millimeter of mercury as being a Torr. So, you can view the pressure here as 754 Torr. So, let me write that down. So, this is 754 Torr. And then we're going to multiply that times the volume. And here, they give the volume in liters in several of these, and we're probably going to be using this one, this version of the ideal gas constant, that has liters, Torr, moles, and Kelvin. And so let's multiply times the volume, so times 1.85 liters. And then that is going to be divided by the ideal gas constant. I'll use this version because it's using all of the units that I already have. I know what you're thinking, "Wait, the temperature's given in degrees Celsius." But it's easy to convert from degrees Celsius to Kelvin. You just have to add 273 to whatever you have in degrees Celsius to get to Kelvin, because none of these are given in degrees Celsius. And so, I will use this ideal gas constant. So this is going to be 62.36 liter Torr liter Torr, per mole Kelvin. Mole to the negative one is just one over mole, so I could write it like this. Kelvin to the negative one is just one over Kelvin. And then, I'm gonna multiply that times the temperature. So times, what is 21 degrees Celsius in terms of Kelvin? Well, I add 273 to that, so that's going to be 294 Kelvin. And we can validate that the units all work out. This liter cancels out with this liter. This Torr cancels out with that Torr. This Kelvin cancels out with this Kelvin. And so, we're going to be left with some calculation. And, it's going to be one over one over moles, or it's essentially going to simplify to just being a certain number of moles. And so, let's get our calculator out to figure out the number of moles in that breath. So n, I keep using slightly different colors, so n is going to be equal to 754 times 1.85 divided by 62.36 and then, also divided by, divided by 294, is equal to this thing. And let's see how many significant digits we have. We have three here, three here, three here, four here. So, when we're multiplying and dividing, we just want to use the fewest amount that I'm dealing with. So I wanna go to three significant figures. So 0.0, one, two, three significant figures, so 0.0761. This is going to be 0.0761. And I could say approximately 'cause I am rounding. But that's three significant figures there. So, that's the number of moles of air in the breath. Now, the next question is how many molecules is that? Well, we know that each mole has roughly 6.022 times 10 to the 23rd molecules in it, so we just have to multiply this times 6.022 times 10 to the 23rd. So, we could write it this way. We could write 0.0761 moles, I'll write mole, times 6.022 times 10 to the 23rd molecules, molecules per mole. Now these are going to cancel out, and I'm just going to be left with molecules. And I can just take the number that I had before 'cause it's nice to be able to retain precision until you have to think about your significant figures. And so, but once again, because we did this whole calculation, we're going to wanna round everything to three significant figures. So, let's just multiply this times 6.022. EE means times 10 to the, times 10 to the 23rd, is equal to that. And, if I round to three significant figures, because my whole calculation, that was my limiting significant figures, I have 4.58 times 10 to the 22nd. So, this is 4.58 times 10 to the 22nd molecules. Squeeze that in there, and we're done.