Is the Ideal Gas Law Supposed to Be in kPa?

AI Thread Summary
The discussion centers on the appropriate gas constant (R) to use in the ideal gas law and Van der Waals equation, with options being 8.3145 J/(K·mol) or 0.082058 (L·atm)/(K·mol). Users express confusion over pressure calculations, noting discrepancies when using different values of R. It is emphasized that consistency in units is crucial, as using the correct gas constant aligned with the pressure unit (kPa or atm) is necessary for accurate results. Ultimately, both constants can be used, but the choice depends on the specific application and the desired units for pressure. The importance of unit conversion between atm and kPa is also highlighted.
leaf345
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For the ideal gas law and van der waals eq'n, R can be a) 8.3145J/(k*mol) or b) 0.082058(L*atm)/(k*mol) . But if I use the first value of R, and try to solve for P my value always comes out as about 1000 smaller than it should be in Pa(ie/ it should be 2000 but it gives me 2). If I use the 2nd value, my value comes out in atm which is okay. But I can't use this value in another eq'n I need, so I don't want to use it. Is the ideal gas law supposed to be in kPa?
 
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leaf345 said:
For the ideal gas law and van der waals eq'n, R can be a) 8.3145J/(k*mol) or b) 0.082058(L*atm)/(k*mol) . But if I use the first value of R, and try to solve for P my value always comes out as about 1000 smaller than it should be in Pa(ie/ it should be 2000 but it gives me 2). If I use the 2nd value, my value comes out in atm which is okay. But I can't use this value in another eq'n I need, so I don't want to use it. Is the ideal gas law supposed to be in kPa?

hey,

i think since the gas constant is being used to calculate pressure...it should be the .082058(L*atm)/(k*mol)

the other one is for thermodynamics...it's units include the energy (J)

correct me if I am wrong
 
gas constants

the gas constant is either 0.08206L*atm/(K*mol) or 8.314L*kPa/(K*mol)

you should be able to use either or. i personally like to use kPa, but if you want to use atm, you could always convert from atm to kPa and the other way around using the equality 1atm = 101.3kPa.
 
gradeaswimr said:
the gas constant is either 0.08206L*atm/(K*mol) or 8.314L*kPa/(K*mol)

you should be able to use either or. i personally like to use kPa, but if you want to use atm, you could always convert from atm to kPa and the other way around using the equality 1atm = 101.3kPa.

i disagree because one gas constant is used for thermo problems (heat and energy) and the other is for pressure probs...There is a definite difference between the two numbers
 
cheechnchong said:
i disagree because one gas constant is used for thermo problems (heat and energy) and the other is for pressure probs...There is a definite difference between the two numbers

WRONG! Units of the gas constant are energy per amount of substance per unit temperature; pressure x volume = energy. So long as one is consistent with units throughout a problem, one gets the "same" answer regardless of "which" gas constant one uses.
 
Bystander said:
WRONG! Units of the gas constant are energy per amount of substance per unit temperature; pressure x volume = energy. So long as one is consistent with units throughout a problem, one gets the "same" answer regardless of "which" gas constant one uses.

well, yeah that's what i meant when i posted previous response...I was definitely implying units of measurements!--if they give a pressure (atm, kpa, etc) or energy (J, kj...) keep consistent by using the appropriate gas constant

thinking back at my prev response...i can see it's sort of half wrong because i didnt set a stipulation for using a gas constant--my bad :redface: :wink:...
they're not different because they're both gas constants
PS: i haven't had much experience with vander's law, but I've definitely plugged things into it...
 
Last edited:
0.08205783 L*atm/(K*mol)
8.314472 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)

or something :wink:
 
Borek said:
0.08205783 L*atm/(K*mol)
8.314472 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)

or something :wink:

Borek
--
General Chemistry Software
www.pH-meter.info

pwned


or something...
 
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