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Ideal Gas Law Practice Problems with Density

Ideal Gas Law Practice Problems with Density
in this video we're going to do two

practice

problems

that have to do with using

density

and molar mass with the

ideal

gas law now so far we've been using this equation here PV equals NRT which lets us solve for pressure volume moles and temperature of a gas sample if we know three of these variables we can always figure out the fourth when we run into a problem when we try to do a question like this which asks what is the

density

of a particular gas sample well that's a problem
ideal gas law practice problems with density
because

density

isn't 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 which I've written here this lets us solve for slightly different variables particularly it lets us solve for

density

uppercase D and molar mass of a gas capital M as well as as well as pressure and temperature so when we when we're doing an

ideal

gas law question that asks us
about the

density

of a gas we have to use this alternate form of the

ideal

gas law equation instead of PV equals NRT now if you're interested in how I went from here to here I'll show you how to do that at the very end of the video you can watch it if you want but let's get started right now with how to actually solve this sort of a problem so I've laid out some of the variables right here obviously

density

is what we're going to be solving for and let's start with
temperature it's 65 degrees Celsius we know that won't work with gases Celsius temperatures are no good with gas we're going to have to convert it to Kelvin temperatures by adding 273 and I end up getting 338 Kelvin now that'll work with a gas equation let's take a look at pressure the units of pressure don't match my pressure units on our a problem have millimeters of mercury here and I have ATMs there so that means that I'm going to have to take my pressure multiply
it by a conversion factor in order to convert from millimeters of mercury to ATM so when I do that I get one point two three atm now the pressures match and I'm set with that now the last thing that I'll need to do is solve for the molar mass of the gas that I'm using which here is sulfur dioxide so2 in order to get the molar mass of so2 I'm going to have to take the molar mass of sulfur and add it to two times the molar mass of oxygen because I have two oxygens when I do that
I'm going to get sixty four point one grams per mole now I've got 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

ideal

gas law the the fractions that you get are very ugly you're going to see in just a minute that I'm gonna have to make a 4-story fraction there's no good way to avoid this when we're doing PV equals NRT I showed you a trick to avoid getting more than a top and bottom in a
ideal gas law practice problems with density
fraction but unfortunately there's just not a good way to get around it here so I'll slowly walk you through it and then most importantly I'll show you how to cancel units with one of these four-story fractions it's a pain but it's really not that hard so take this equation

density

equals pressure 1.23 ATM's 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's four different levels of this
fraction now what I want to do is I want to cancel the units and here's how I do it when I'm when I'm dealing with something looks like this the first thing is I just look at the top of the fraction are there any units that I can cancel here there aren't now I look at just the bottom of the fraction are there any units that I can cancel here there's one I have calvin up here and i have Calvin down there so those can cancel out all right now I look at the top of the top and
the top of the bottom I have ATM here and I have ATM there so they can cancel out and then I look at the bottom of the top and the bottom of the bottom moles here moles there I've canceled out everything I'm left with grams and liters so I plug this math into my calculator and the answer that I get out rounded to three significant digits is going to be two point eight four what are the units on this well I have grams on top of the fraction and I have liters on the bottom of fraction so
my final units are going to be grams per liter which makes perfect sense because that's how we report

density

as grams per liter so that's how we solve this sort of an equation let's go on let's go on to the next example okay so here's our second problem a gas has a

density

of three point zero one grams per liter at STP what is its 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 in it now instead of giving us a bunch of numbers this guy here just asks us for STP don't get freaked out by this some people do remember what STP is STP stands for standard temperature and pressure and it's a particular set of conditions zero degrees Celsius and 1 atm so we can go ahead and plug these right into our variables pressure 1 atm temperature 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 is
ideal gas law practice problems with density
going to be 273 Kelvin here are all of our variables laid out and where we're ready to go to work now I don't want to have to go through all the steps of rearranging the equation so I'll just show you what I'm going to do here I have this then I take our T I move it up onto this side of the fraction I divide both sides by P and then I just flip it so I'm left with M equals RT D divided by P and that's the form the equation that I'm going to be using 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've got to cancel units in this ugly triple-decker fraction okay let's look just to the top what can we cancel here liters up here liters down there Kelvin up here Kelvin down there all right that's it now we look at the top of the top and the bottom here ATM is there ATM is there the only thing that we're left is with is grams per mole
isn't that curious because it just turns out that molar mass is always reported in grams per mole so anyway I go through plug this into my calculator the math that I'm going to do is 0.0821 times 273 times 3.01 all of that divided by one and rounded to two three significant figures I'm going to get sixty seven point five grams per mole and that's how I solve for molar mass using

density

and the

ideal

gas law now finally if you're interested I'll show you how we can go
from PV equals NRT to the transformed version of the

ideal

gas law that uses

density

and molar mass so I'm going to 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 that I want to do is I want to get molar mass into the equation remember that moles in order to get moles we take the mass of a sample like 40 grams of a lump of carbon or something and divide it by the molar mass in order to find out how many moles
are so what I've written is moles here I can rewrite as mass divided by molar mass so I'm going to substitute little M divided by big m mass divided by molar mass in for n so mass divided by molar mass equals PV equals RT so there's molar mass that's how we got that into the equation how are we going to get

density

into the equation remember

density

is mass divided by volume so let's rearrange this to get mass over volume what I'm going to do is I'm going to move this
M up here and I move this video here so I'm going to get M divided by V equals P m RT now mass over volume I can rewrite as

density

because they're the same thing so now I have

density

equals pressure times molar mass divided by the universal gas constant times T and that is our transformed version of the

ideal

gas law equation