# ideal gas law problem

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#### joonkimdds

##### Senior Member
10+ Year Member
7+ Year Member

this is number from kaplan blue book page 670 number 44

Q. A rigid container holds 3.00 moles of an ideal gas at 298K. How many moles of gas would need to be added to the container at constant temperature to increase the pressure from 1.00 atm to 1.80 atm?

the first solution says
pressure has slightly less than doubled.
temperature and volume are not changing so number of moles of gas needs to increase proportionally.

I thought that volume and pressure have inversely proportional relationship.
If pressure doubled, shouldn't volume be halved? and I believe the question didn't say that the volume isn't changing.

and also, the second solution says
p1/n1 = p2/n2 = RT/V = constant
n2= (n1/p1) x p2 = (3/1) x 1.8 = 5.40 moles
5.4 - 3 = 2.4

I don't understand how they come up with the formula above.
I only know p1v1=p2v2 or MW=DRT/P formulas for ideal gas problems
so I want to know what kind of formla they used.

#### tissy

##### Full Member
15+ Year Member
I think b/c everything else cancels out so they don't bother to show it..if you write out pv/nrt=pv/nrt and plug them in..you should be able to see it

#### tom_servo_dds

##### Senior Member
10+ Year Member
joonkimdds said:
this is number from kaplan blue book page 670 number 44

Q. A rigid container holds 3.00 moles of an ideal gas at 298K. How many moles of gas would need to be added to the container at constant temperature to increase the pressure from 1.00 atm to 1.80 atm?

the first solution says
pressure has slightly less than doubled.
temperature and volume are not changing so number of moles of gas needs to increase proportionally.

I thought that volume and pressure have inversely proportional relationship.
If pressure doubled, shouldn't volume be halved? and I believe the question didn't say that the volume isn't changing.

and also, the second solution says
p1/n1 = p2/n2 = RT/V = constant
n2= (n1/p1) x p2 = (3/1) x 1.8 = 5.40 moles
5.4 - 3 = 2.4

I don't understand how they come up with the formula above.
I only know p1v1=p2v2 or MW=DRT/P formulas for ideal gas problems
so I want to know what kind of formla they used.

Not a problem at all. What you're dealing with like all others in this class is still a PV=nRT problem. A great way to approach these types of problems when you have two containers and something's changing is to setup a PV=nRT equation for both containers and set them equal to each other. The first thing is to decide on both sides what is changing and what is staying the same.

So for instance with this problem they tell you that there is a container with 3.0 moles in it and that the pressure is 1.0 atm. Those are the only two variables in the ideal gas law that are changing for this container. All others you can cross out. So essentially what you should have is P1/n1 for this container. The volume didn't change (container is rigid). You don't need to include the Gas Constant R in this case (same in both containers), and the temperature remains the same (from the problem). Now do the same thing for the other container and you should end up with P1/n1 = P2/n2. So now you have 4 variables, three of which you know the values of so you just set it up and solve for the unknown one. Now that you have an answer you need to go back and make sure that you're answering what the question is asking. It asked for how much you need to add. Well, you started with 3.0 and ended up with 5.4 moles. That means what you'll need to add is 2.4 moles. Hope this explanation helped. Just remember, that the ideal gas law in itself is a combination of other laws just put together. It can be combined in any number of ways to solve for varying situations. Just look for what the question is asking for and what they give you. If they don't give you the information in it, then you won't need it in this type of equation. Best of luck!

#### joonkimdds

##### Senior Member
10+ Year Member
7+ Year Member
tom_servo_dds said:
Not a problem at all. What you're dealing with like all others in this class is still a PV=nRT problem. A great way to approach these types of problems when you have two containers and something's changing is to setup a PV=nRT equation for both containers and set them equal to each other. The first thing is to decide on both sides what is changing and what is staying the same.

So for instance with this problem they tell you that there is a container with 3.0 moles in it and that the pressure is 1.0 atm. Those are the only two variables in the ideal gas law that are changing for this container. All others you can cross out. So essentially what you should have is P1/n1 for this container. The volume didn't change (container is rigid). You don't need to include the Gas Constant R in this case (same in both containers), and the temperature remains the same (from the problem). Now do the same thing for the other container and you should end up with P1/n1 = P2/n2. So now you have 4 variables, three of which you know the values of so you just set it up and solve for the unknown one. Now that you have an answer you need to go back and make sure that you're answering what the question is asking. It asked for how much you need to add. Well, you started with 3.0 and ended up with 5.4 moles. That means what you'll need to add is 2.4 moles. Hope this explanation helped. Just remember, that the ideal gas law in itself is a combination of other laws just put together. It can be combined in any number of ways to solve for varying situations. Just look for what the question is asking for and what they give you. If they don't give you the information in it, then you won't need it in this type of equation. Best of luck!

Ok, Thank you very much for your awesome explanation, it really helped!

I want to ask you one more question though.

Whenever I see a problem related to the mole, pressure, volume, temperature all combined like the problem i posted, do i always think of PV=nRT and the problem is solved?

so....what I wanna know for any of the gas problems are

PV=nRT
molar weight = DRT/P
Volume and pressure have inversely proportional relationship
Volume and temperature have proportional relationship
d=mass/volume

these five things written above are all i need to know for gas related problems?
and will i be ready to rock n roll !?!?

sorry for this dumb question. I took general chemistry from community college about 2 years ago. Not only I forgot everything, i don't know what i study is good enough or i need to go more in detail.

#### tom_servo_dds

##### Senior Member
10+ Year Member
joonkimdds said:
Ok, Thank you very much for your awesome explanation, it really helped!

I want to ask you one more question though.

Whenever I see a problem related to the mole, pressure, volume, temperature all combined like the problem i posted, do i always think of PV=nRT and the problem is solved?

so....what I wanna know for any of the gas problems are

PV=nRT
molar weight = DRT/P
Volume and pressure have inversely proportional relationship
Volume and temperature have proportional relationship
d=mass/volume

these five things written above are all i need to know for gas related problems?
and will i be ready to rock n roll !?!?

sorry for this dumb question. I took general chemistry from community college about 2 years ago. Not only I forgot everything, i don't know what i study is good enough or i need to go more in detail.

Then the best thing to do my be to just crack open your old GChem text book and just read through a few chapters. Try to do the problems at the end of the chapter and check your answers with the key. This may be the most help to you and increase your confidence in solving these types of problems. Best of luck.

#### Immuno-guy

##### Member
10+ Year Member
OK, that works beautifully; but I hate math. I have realized that there is only a bare minimum of math in any gen chem problem. Here's how I'd do it:

3 moles = 1atm. You are looking to nearly double the pressure in a rigid container. So what's nearly double the 3 moles?

a) 2.4 moles
b) 16 moles
c) 4.2 moles
d) 56 moles
e) something else clearly too high or low.

Adding 2.4 would nearly double the number of moles, and therefore the pressure in the container. Quick huh?

If logic does not work I pick up my pencil and begin with pV=nRT or preferably p1v1=p2v2 which works much of the time. There really are very few calculatoins in this section. It is nearly always 1 mole or .1mole never anything that requires much math.

#### Envision

##### Envisioning...
10+ Year Member
7+ Year Member
tom_servo_dds said:
Then the best thing to do my be to just crack open your old GChem text book and just read through a few chapters. Try to do the problems at the end of the chapter and check your answers with the key. This may be the most help to you and increase your confidence in solving these types of problems. Best of luck.

I agree. I took gchem 4 years ago when i first entered college. i did alright back then, but when i took the DAT diagnostic i got a 13! so, i opened up my old gchem book and after a week of hard studying, i think I've regained most of my old gchem knowledge. i strongly suggest reading up on the sections you have questions on, like ideal gas law. Lots of people will tell you that Kaplan blue book is more than enough. this is true, however, the problem is Kaplan blue book assumes you have the basic foundation, and just reviews the major concepts. so a quick brush up will help a lot more than trying to extrapolate concepts from doing problems alone.

#### tom_servo_dds

##### Senior Member
10+ Year Member
Envision said:
I agree. I took gchem 4 years ago when i first entered college. i did alright back then, but when i took the DAT diagnostic i got a 13! so, i opened up my old gchem book and after a week of hard studying, i think I've regained most of my old gchem knowledge. i strongly suggest reading up on the sections you have questions on, like ideal gas law. Lots of people will tell you that Kaplan blue book is more than enough. this is true, however, the problem is Kaplan blue book assumes you have the basic foundation, and just reviews the major concepts. so a quick brush up will help a lot more than trying to extrapolate concepts from doing problems alone.

Well put!