Ideal Gas Law Practice Problems with Molar Mass

Now we are going to look at some

ideal

gas law

problems

where we also have to use

molar

mass

to convert between gam and moles of a gas. Take this problem, for example, calculate the volume that 12.5 G of CO2 will occupy at a temperature of 40° C and a pressure of 1.0 ATM. I went ahead and filled in several of the variables that we're going to use, as you can see here, what we're going to solve for is the volume when we do that. This the first thing I always check is to make sure the units of the variables match the units in R.

Well, we have an ATM that has a great temperature, as you know, it has to be in Kelvin to work with gases, so I went ahead . and I already did that because I'm sure you guys are probably familiar with how we add 273 to degrees Celsius to get Calin, so I'm going to change 40° CSI here to 33 Kelvin. The next thing we have to do is Look here at the amount of gas we have n right now, it is expressed in grams, 12.5 G, but to use it in the

ideal

gas law it must be in moles, which means we will have than using the

molar

mass

of CO2.

To convert gr of CO2 to moles of CO2, the first thing we have to do is determine the molar mass of carbon dioxide. It has one carbon and two oxygens, so we take the mass of carbon plus 2 times the mass of oxygen and add up. these together and it gives us 44.0 G for each Mo, so now what we are going to do is take 12.5 G of CO2 times. I want to cancel gr, so I'm going to reverse this fraction. 1 mole / 440 G and the method I'm going to do is 12.5 * 1 divided 44. I'm going to round that to three significant figures one mole, this is an exact counting number so we don't worry about the significant figures for this. three here three here so I'm going to round this to three significant figures and I'm going to get 0.284 moles of CO2 so we can cancel grams and show that we've changed this to 0.284 moles now all I have The terms of my variables match a value that I have in R, so I'll take PV equal to the ideal gas law and calculate and rearrange this to get V on its own, so all I have to do is divide both sides by P. then the PS cancels out and I get that V is equal to nrt/P, so I went ahead and plugged most of these values ​​into the equation.

I have N, I have t and p as before, I left R until the end and here it is now. let's cancel out the units ATM here ATM up here moles here moles here Kelvin there Kelvin there which leaves me with liters which makes sense since I'm solving for the volume now the calculations I'm going to do I'm going to do this multiplied this multiplied All this then divided by 1.20 ATM and I will get 6 6.81 169. How many significant figures do I round it to? I round it up to three, so I'm left with six. I'll take the zero and I'll take the eight. see if I round it up or down, I keep it the same, so my final answer will be 6.08, what are my units?

The units are what I was left with after canceling 6.08L and that's my final answer for this, okay, here's another one. problem a certain gas has a molar mass of 28.0 G per mole how many grams of this gas would fit in a 3.00 l container at 182 kPa and 47.2 C I completed my variables here n how many grams of this gas how much gas is that We are solving as before. I'm going to check these variables and make sure they match the units in R Celsius. It is always a problem when we deal with gases. I think, as you know, I want to take any Celsius temperature with gases and add 273 to it to get Kelvin here.

I also had to round using my addition rules for significant figures, so I get 320 Kelvin, which will be the new temperature I'm going to use. so now Calin agrees. I have liters here, but look at this. I have kilop pascals and I have ATM, so my pressure units don't match. There are two things you could do. One of these things you could do is use a new R I. I said before that you want to be able to use a different R so that your pressure units match, so you could use 8.31 which has kPa instead of ATM, that would be totally fine if you want to do that, what I'm going to do is I'm going to stick with the r that I'm using now and instead I'm going to convert KPA to ATM so that the units match. 1 atm is equal to 101.3 kPa so we can do these calculations kPa up here kPa down there cancel out and I'm going to round them to three significant figures 1.80 ATM, so this expressed in atmospheres is 1.80, so now everything I have matches the units in R and I'm ready to solve this, so PV is equal to nrt I.

I'm solving for n, I divide both sides by RT, now those guys cancel. I get PV/RT equal to n or if you want to reverse it, I can have n equal PV/RT, so this is what I get when I enter these values. I have pressure and I have volume, I have temperature and I left R until the end because R is at the bottom. I take the value I have here and flip it to have moles Kelvin and then the number of liters ATM, okay, cancel out the units. and I have moles left and when I do these calculations this multiplied by this divided by this multiplied by this final answer that I get rounded to three significant figures is 0206 and there are my mole units but we are not done with the problem yet because If you see here, we are solving GRS, not moles, so how do we go from moles to grams?

We use molar mass. We know that this gas has a mass m of 28 G per mole, so what I can do is take my moles. and multiply this number by 28.0 G I divided a Mo moles up here moles down here so that they cancel and the final answer that I am going to round again to three significant figures a mole does not count because it is a counting number goes to give me 5.77 G and this is how we solve ideal gas law

problems

using molar mass to convert grams and moles