Asked by Avonlea
what is the gas constant (R) for mm Hg?
8.31 is for atm
.08206 is for kPa
right?
8.31 is for atm
.08206 is for kPa
right?
Answers
Answered by
DrBob222
You have them reversed.
0.08205 for atm
8.31 for kPa.
0.08205 for atm
8.31 for kPa.
Answered by
Avonlea
Then my teacher is messed up. So, for mm Hg it would be...?
By the way, Happy Halloween!!!!!
By the way, Happy Halloween!!!!!
Answered by
DrBob222
If you have pressure listed in mm mercury, change to atm by dividing by 760 mm/1 atm, then use R as 0.08206 L*atm/mol*K. If pressure is in kPa, use R as 8.3145 J/mol*K (kPa*L/mol*K).
Pressure in mm Hg may be changed to kPa by multiplying mm Hg x 0.13332 and using R of 8.1345 BUT I've not seen that done. You can try this yourself.
PV = nRT
Let's have P = 760 mm, V = 22.415 L, n = 1, R =?? and T = 273.15 K.
Solve for R.
R = PV/nT = (760/760)*22.415/1*273.15 = 0.08206
Let P = 101.325 kPa, V = 22.415 L, n = 1, R = ??, and T = 273.15 K.
R = 101.325*22.415/1*273.15 = 8.315
Let P = 760 mm, V = 22.415 L, n = 1, R = ?? and T = 273.15 K.
R = 760*22.415*0.13332/1*273.15 = 8.315
Here is a site that talks a little more about it but the conversion from mm Hg to kPa is not listed. http://en.wikipedia.org/wiki/Gas_constant
Pressure in mm Hg may be changed to kPa by multiplying mm Hg x 0.13332 and using R of 8.1345 BUT I've not seen that done. You can try this yourself.
PV = nRT
Let's have P = 760 mm, V = 22.415 L, n = 1, R =?? and T = 273.15 K.
Solve for R.
R = PV/nT = (760/760)*22.415/1*273.15 = 0.08206
Let P = 101.325 kPa, V = 22.415 L, n = 1, R = ??, and T = 273.15 K.
R = 101.325*22.415/1*273.15 = 8.315
Let P = 760 mm, V = 22.415 L, n = 1, R = ?? and T = 273.15 K.
R = 760*22.415*0.13332/1*273.15 = 8.315
Here is a site that talks a little more about it but the conversion from mm Hg to kPa is not listed. http://en.wikipedia.org/wiki/Gas_constant
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.