The volume of one mole of an ideal gas, at 202kPa and 273 K, is?
I've been trying to use the ideal gas law, but I just can't get what I need. My teacher said that the answer is 11.2 L, but I can't come to that. Any help please?
2007-12-20 07:43:22 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
PV = nRT
V = nRT/P
V = [(1)(8.314472)(273)]/202
V = 11.2 L
You need to make sure you are using the right value for the gas constant, actually they are all the same but you need to make sure you have the right units.
2007-12-20 07:50:13 · answer #1 · answered by Brian K² 6 · 1⤊ 0⤋
PV = nRT. Solve for R. R = PV/nT = 202*11.2/1*273 = 8.3
(roughly), the value to take R at (energy per kelvin per mole).
2007-12-20 08:44:38 · answer #2 · answered by BB 7 · 0⤊ 0⤋
the main solution to this problem is that you MUST have the correct UNITS for all VARIABLES!!! PV = nRT
R = gas constant (there are alot of them... depends on which units you want to cancel out) ... 8.314472 L·kPa·K-1·mol-1
V = ???V (Liters)
P = kPa
T = 273 Kelvin
n = 1 moles
V = nRT/P
plug in... then you should get the answer..
2007-12-20 07:52:36 · answer #3 · answered by J 3 · 0⤊ 0⤋
I got 11.2 right away using the ideal gas law. Are you using R=8.314 (I can't remember the units right now)
PV=nRT
V= (1mol)(8.314)(273K) / 202kPa
2007-12-20 07:53:05 · answer #4 · answered by b 3 · 0⤊ 0⤋
i think of the vol. would be a million liter simply by fact its a liter manometer than u assume to locate mol. Wt. different smart it would be inconsistent answer as there are 2 variables ie. P and Mt yet basically one eq that's of suitable gas eq. Do tell me how u arise with 30.39........
2016-11-23 17:56:08 · answer #5 · answered by Anonymous · 0⤊ 0⤋
PV = nRT
V = nRT/P = (1mol)(8.31kPa-L/mol-K)(273K) / (202kPa) = 11.2L
2007-12-20 07:49:47 · answer #6 · answered by steve_geo1 7 · 0⤊ 0⤋