Help with Ideal Gas Law.... (1 Viewer)

[JohnE]

OK, I know there are some science minded people around here and I'm hoping one of you can help me out with the ideal gas law.

I know how to set up the problem, for example, PV=nRT, where "R" is the constant, and if I have to solve for "T", it should read T=PV/nR correct? And R is .0821 L-atm/mol-K, correct?

I think what's giving me the trouble is the math. I"m not sure if I'm reading "R" correct? Is it supposed to read, R x (L x atm/mol x K)? Or am I looking at this wrong?

An example problem giving me trouble...

I'm trying to find out how many moles of Nitrogen there are here. I'm given...

P=1.20 atm
V=20.0 L
T=298K
I need to solve for "n". And I keep getting the wrong answer.

I set it up like this, n=PV/RT, or n=1.20 x 20.0/.0821 x (20.0 x 1.20/1.0 x 298) where I'm using the STP to get the 1.0 for n in the "R", but it's just not working out right!!! Grrrr!!! Like I said, I think I'm doing the math for "R" wrong.

Any help or examples would really be appreciated. Hope this makes some sense. =) Thanks.

 

[Jim_F]

I should probably butt out since I'm tired and it's been so long since I've looked at this stuff, but if I understand correctly, you're plugging in values where you shouldn't for the units of "R".

so,

n = 1.2 x 20/0.0821 x 298
n = 24/24.5
n = 0.98 Moles

correct?

 

[Holadem]

Seems pretty straight forward to me... why do you feel the need to plug any values into the unit of R? Because the unit is an expression (L-atm/mol-K) doesn't make any different from say, L or K. They simply haven't created a name for a L-atm/mol-K, that is all.

--
Holadem

 

[David-S]

yeah, they're right

R is a constant, so that (l*atm/mol*K) is just defining what "units" R has (it's basically so the units all even out in the equation...)

 

[JohnE]

So when I'm using R in the equation, the only number I should use for it is, 0.0821?

Jim, yes that is correct, but do did you have to amke it look so easy you creep?

And thank's.

 

[Leo Hinze]

Ahhh, our old friend 'dimensional analysis' rears it's ugly head.

I rarely do problems like this (chemistry, engineering, etc.) without also writing the units. Before, I would often get burned by the wrong temperature scale, or the wrong constant, or the wrong amount of material. Once I started doing dimensional analysis, my results improved dramatically. Ask me later to tell the story of how I passed second-semester engineering physics by only using dimensional analysis and the square root of 2!

 

[Holadem]

Ask me later to tell the story of how I passed second-semester engineering physics by only using dimensional analysis and the square root of 2!

Please do tell!

--
Holadem

 

[David-S]

"dimensional analysis" is basically just analyzing the "dimensions" (units) of a problem... at least that's what i always thought of it as...

and yes john, just use .0821

 

[Darren Davis]

Dimensional analysis (I guess that's what it's called. I've never had a name for it) is really a great thing. If you write the dimensions out then the right ones should cancel or make up the units you desire. For example: distance = rate x time (d=rt), rate in meters per second (m/s) and time in seconds (s). The (s) in the denominator of the rate will cancel with the (s) in the numerator with the time, thus yeilding the unit (m), which is what you want. My AP Physics class is better with little things like this. Good luck!

 

[Kris McLaughlin]

Dimensional analysis, eh? My high school teachers just called it "unit analysis", which doesn't sound nearly as impressive.

 

[Leo Hinze]

Please do tell!

Okay, so maybe I built the story up to make it sound just a wee bit more interesting than it is.... For me, the 2nd semester of engineering physics was the electrical stuff, i.e. Maxwell's equations. I didn't understand the material all that well, and I definitely didn't understand some of the formulas. Luckily, our tests were all multiple choice. I found that proper use of dimensional analysis really helped - I would just find a formula that required the given units. If the actual number didn't quite work out, an occasional division or multiplication by the square root of two would usually do the trick!

 

[Morgan Jolley]

We just finished using this formula in school, so I have a few tidbits that might help.

First of all, if your teacher is as mean as ours, then they might use R to be different units. You might have to convert it to something like (mL*kPa)/(mol*K). Also, at our school, we're taught it as:

PV=nRT

which is a bit easier to use.

 

[JohnE]

Well thanks for all the help guys. I think I've got a handle on it now. Damn math is always my weak point!