Ideal Gas Law Practice Problems with Density

In this video we are going to do two

practice

problems

that have to do with the use of

density

and molar mass with the

ideal

gas law. So far we have been using this equation here: PV equals NRT, which allows us to solve for pressure, volume, moles and temperature of a gas sample, if we know three of these variables, we can always calculate the fourth when we come across a problem when we try to ask a question like this that asks what is the

density

of a particular gas sample and that is a problem because Density is not one of these four variables that I have here, so how do I solve a problem like this?

It turns out that I can take PV equals NRT and transform it into another version of the

ideal

gas law that I wrote about here. allows us to solve for slightly different variables, in particular it allows us to solve for the capital density D and the molar mass of a capital gas M, as well as the pressure and temperature, so that when we ask a question about the ideal gas law that asks us about the density of a gas we have to use this alternative form of the ideal gas law equation instead of PV equals NRT Now, if you're interested in how I got from here to here, I'll show you how to do it at end Of the video you can watch it if you want, but let's start right now with how to solve this type of problem, so I've set some of the variables here, obviously density is what we're going to solve. and let's start with the temperature, it's 65 degrees Celsius, we know it won't work with gases.

Celsius temperatures are not good with gas, we are going to have to convert them to Kelvin temperatures by adding 273 and I end up getting 338 Kelvin now we're going to work with a gas equation, let's look at pressure, the pressure units don't match my units of pressure in our problem, we have millimeters of mercury here and I have ATMs there, that means I will have to take my pressure, multiply it by a conversion factor to convert from millimeters of mercury to ATM, so when I do that I will get a point two three atm, now the pressures match and I'm ready, the last thing I'll need What I do is solve for the molar mass of the gas I'm using, which here is sulfur dioxide so2.

To get the molar mass of so2, I will have to take the molar mass of sulfur and add it to twice the molar mass. molar mass of oxygen because I have two oxygens when I do that I will get sixty four point one grams per mole now I have my molar mass temperature and pressure and I'm ready to go ahead and solve the equation now when you use this form of the gas law ideals, the fractions you get are very ugly, you will see in just a minute that I am going to have to do a 4 story fraction, there is no good way to avoid this when we are doing PV equals NRT.

I showed you a trick to avoid getting more than a top and bottom in a fraction, but unfortunately there's no good way to avoid it here, so I'll walk you through it slowly and then the most important thing. I'll show you how to cancel units with one of these four-story fractions. It's complicated, but it's really not that difficult, so take this equation: Density equals pressure 1.23 ATM times molar mass 60 4.1 grams per mole, that's the top of the fraction now. we have our times T 338 Kelvin, as you can see, there are four different levels of this fraction. Now what I want to do is cancel out the units and this is how I do it when I'm dealing with something that looks like This the first thing is I just look at the top of the fraction.

Are there units I can cancel here? Now there aren't any. I look only at the bottom of the fraction. Are there units I can cancel here. There is one that I have. Calvin up here and I have Calvin down there so they can cancel right now. I look at the top above and the top below. I have an ATM here and an ATM there so they can cancel it and then I look. the bottom of the top and the bottom of the bottom moles here moles there I've canceled out everything I have left with grams and liters, so I plug these calculations into my calculator and the answer I get rounded to three significant digits is It's going to be two point eight four, what are the units in this well?

I have grams on top of the fraction and I have liters on the bottom of the fraction so my final units will be grams per liter which makes a lot of sense because that's how I report the density as grams per liter so that's how we solve this guy equation, let's continue, let's move on to the next example. Well, here is our second problem. A gas has a density of three point zero one grams per liter in STP. What is its density? molar mass, so first of all, since we're using density, we're going to want to use this form of the ideal gas law that has density and molar mass now instead of giving us a bunch of numbers that this guy here just gives us. ask. for STP, don't be scared by this, some people remember what STP is.

STP stands for standard temperature and pressure and is a particular set of zero degrees Celsius and 1 atm conditions, so we can go ahead and plug these directly into our pressure variables. The temperature of 1 atm is zero degrees Celsius, which we can't use because it's Celsius, we add 273 to that to get Kelvin, so 0 plus 270 three will be 273 Kelvin. Here are all our variables laid out and where we are ready to go to work now. I don't want to have to go through all the steps to rearrange the equation, so I'll just show you what I'm going to do here.

I have this and then I take our T. I move it over to this side. of the fraction I divide both sides by P and then I just turn it around so that I get M equals RT D divided by P and that's the form of the equation I'm going to use here, so M equals R 0.0821 liters ATM / Kelvin moles times temperature 273 Kelvin times density 3.01 grams per liter, all of that divided by pressure 1 atm, okay, now we have to cancel out units in this ugly three-story fraction, okay , let's just look up, what can we cancel here?, liters up here liters down there Kelvin up here Kelvin down there okay, that's it now we look at the top of the top and the bottom here ATM is there ATM is there lo The only thing we have left is grams per mole.

It's not that funny because it turns out that molar mass is always reported in grams per mole, so anyway I plug this into my calculator and the calculations I'm going to do are 0.0821 times 273 times 3.01, all of that divided by one and rounded to two three significant figures I will get sixty-seven point five grams per mole and this is how I solve for the molar mass using density and the ideal gas law. Now finally, if you are interested, I will show you how we can do it. from PV equals NRT to the transformed version of the ideal gas law that uses density and molar mass, so I'll start here with PV equals NRT.

I'm going to solve it for n, so I'm going to get N equals PV divided by RT the first thing I want to do is get the molar mass into the equation. Remember that moles to get moles we take the mass of a sample like 40 grams of a piece of carbon or something and divide it by the molar mass to find out how many moles there are, so what I have written is moles here. I can rewrite it as mass divided by molar mass, so I'm going to substitute the little M divided by the big m mass divided by the molar mass in n so the mass divided by the molar mass is equal to PV is equal to RT so there is molar mass , that's how we enter that into the equation.

How are we going to include density in the equation? Remember that density is mass divided by volume, so let's rearrange this to get mass over volume. What I'm going to do is move this M up here and move this video here, so I'm going to divide M by V equals P m RT, now mass over volume I can rewrite as density because It's the same thing, so now I have the density equal to the pressure times the molar mass divided by the universal gas constant times T and that is our transformed version of the ideal gas law equation.