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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: jennielynn_1980 on May 14, 2006, 03:16:55 PM

Title: ideal gas law equation - need help to solve
Post by: jennielynn_1980 on May 14, 2006, 03:16:55 PM
How many moles of oxygen gas are present if the volume of the gas is 274mL at a pressure of 120 kPa and a temperature of 75 oC.

V= 274 mL = 0.274L

n = ?

R = 8.31 kPa x L/mol x K

T = 75 = 348 K

PV = nRT

therefore n = PV/RT

n = 120 kPa x 0.274 L
      --------------------------
       8.31 kPa x L x mol -1 x K -1 x 348 K


(insert missing steps which is what I cannot figure out)

n = 0.0114 mol

I have never taken chemistry before so I really have no idea about conversions and I have very limited knowledge about the ideal gas law.

Thanks!

Jennie

Title: Re: ideal gas law equation - need help to solve
Post by: Albert on May 14, 2006, 03:28:05 PM
 ???

Everything is ok: what's the matter?
Title: Re: ideal gas law equation - need help to solve
Post by: jennielynn_1980 on May 14, 2006, 03:33:26 PM
n = 120 kPa x 0.274 L
      --------------------------
       8.31 kPa x L x mol -1 x K -1 x 348 K

When I go to work out this part of the equation to get to the final answer, I am not sure if I am subing the correct numbers because I dont' come out with the correct answer

So kPa = 101.3
       L= 22.4
       mol -1 = 1 ?
       K -1 1/273 ?

Are these the correct substitutions?
Title: Re: ideal gas law equation - need help to solve
Post by: Borek on May 14, 2006, 03:39:49 PM
These are units, you don't substitute any values for these as they define units of your answer. You have to do two things:

1. Multiply and divide all numbers present in the formula - it will give 0.0114 (watch significant digits number).

2. Cancel out units to check what is unit of your answer.
Title: Re: ideal gas law equation - need help to solve
Post by: jennielynn_1980 on May 14, 2006, 03:46:37 PM
Thank you!  I understand now :)
Title: Re: ideal gas law equation - need help to solve
Post by: Donaldson Tan on May 15, 2006, 05:45:48 AM
Quote from: jennielynn_1980 on May 14, 2006, 03:33:26 PM
8.31 kPa x L x mol -1 x K -1 x 348 K

Why is the unit of R kPa.L/mol.K? R ought to be 8.314 J/mol.K or 8.314 kJ/kmol.K
Title: Re: ideal gas law equation - need help to solve
Post by: billnotgatez on May 15, 2006, 06:04:36 AM
PV=nRT
R=PV/nT

Approximately
R = ( 101.325 kPa * 22.414 L ) / ( 1 mol * 273.15 K )


Title: Re: ideal gas law equation - need help to solve
Post by: Borek on May 15, 2006, 06:05:43 AM
Quote from: geodome on May 15, 2006, 05:45:48 AM
Why is the unit of R kPa.L/mol.K?

Why not?

Quote
R ought to be 8.314 J/mol.K or 8.314 kJ/kmol.K

0.08205783 L*atm/(K*mol)
8.314510 kPa*dm3/(K*mol)
8,314472 L*kPa/(K*mol)
8.314472 J/(mol*K)
62,3637 L*mmHg/(K*mol)
83,14472 L*mbar/(K*mol)
1.987216 cal/(K*mol)

Use whichever fits units given in the question ;)
Title: Re: ideal gas law equation - need help to solve
Post by: Donaldson Tan on May 15, 2006, 03:16:41 PM
Oh well.. I thought 8.314 is only applicable to J/mol.K or kJ/kmol.K
Title: Re: ideal gas law equation - need help to solve
Post by: rctrackstar2007 on May 15, 2006, 07:23:11 PM
Quote from: Borek on May 15, 2006, 06:05:43 AM

83,14472 L*mbar/(K*mol)


what is this one? like what is mbar?
Title: Re: ideal gas law equation - need help to solve
Post by: Borek on May 15, 2006, 07:25:48 PM
1 bar = 100 000 pascals (Pa)
Title: Re: ideal gas law equation - need help to solve
Post by: rctrackstar2007 on May 15, 2006, 07:50:23 PM
Quote from: Borek on May 15, 2006, 07:25:48 PM
1 bar = 100 000 pascals (Pa)

oh ok thanks  :)

i had no idea about so many unit derevations for R, i was only aware of the ones for 8.314 and .0821  :P
Title: Re: ideal gas law equation - need help to solve
Post by: jennielynn_1980 on May 18, 2006, 01:14:45 PM
So if I have learned anything  ;) this would be right:

P1V1 = P2V2
 T1         T2

P = 0.75 atm
V = ?
T = -30?C = 243K

P = 1.25 atm
V = 255 L
T = 40.0 ?C = 313K

0.75 atm ? = 1.25atm x 255 L
  243K                 313K

? = 243K x 1.25 atm x 255 L             
                     0.75 atm x 313K

? = 77456.25
        234.75

? = 330 L

The volume of gas at -30? C and 0.75 atm would be 330L if the same gas had a volume of 255 L at 40.0?C and 1.25 atm.

Title: Re: ideal gas law equation - need help to solve
Post by: Albert on May 18, 2006, 02:03:50 PM
Looks correct.
Title: Re: ideal gas law equation - need help to solve
Post by: Borek on August 23, 2006, 03:32:33 PM
Quote from: Borek on May 15, 2006, 06:05:43 AM
0.08205783 L*atm/(K*mol)
8.314510 kPa*dm3/(K*mol)
8,314472 L*kPa/(K*mol)
8.314472 J/(mol*K)
62,3637 L*mmHg/(K*mol)
83,14472 L*mbar/(K*mol)
1.987216 cal/(K*mol)

Oh man, that's nothing...

http://www.katmarsoftware.com/gconvals.htm