J.R. S. answered • 09/07/20
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At STP (standard temperature and pressure) one mole of an ideal gas occupies 22.4 Liters. (memorize this).
Thus 3.00 mol gas x 22.4 L/mol = 67.2 liters (3 sig. figs.)
J.R. S.
tutor
It does NOT have initial and final volumes except to say they are the same. Avogadro's Law says two dissimilar ideal gases OCCUPYING THE SAME VOLUME at a constant temperature and pressure must have the same number of moles of gas. The volume would only change if you added or removed argon. The law is V1/n1 = V2/n2
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09/07/20
J.R. S.
tutor
Is there more to the problem than you are presenting? As presented, initial and final volumes are the same and initial and final moles of gas is the same (3.00 moles).
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09/07/20
Kay B.
Oh okay, it makes much more sense now. Thank you so much, sir!
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09/07/20
J.R. S.
tutor
No problem. Glad I could help.
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09/07/20
Robert S.
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I agree with J.R. S. 67.2 liters of argon at STP. This is a really good number to remember [1 mole of any gas at STP is 22.4 liters). Also useful: 6.02x10E-23 and STP at 1 atm and 273 K, 273K = 0C.
The question is oddly worded, but one can only conclude that the initial and final values are the same. For fun, one could use the ideal gas law calculate the same answer, even though it is a waste of time.
PV=nRT
PV = 3RT at STP for 3 moles of any gas:
n=3 moles
P= 1 atm
R = 0.082057 L*atm*K^-1*mol^-1
T = 273 K
V = nRT/P
since P= 1: V=nRT
V = 3*0.082057*273 = 67.02 liters
[cancel all the units to be certain you use the correct R value]
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09/07/20
Kay B.
Does this problem not have the initial and final volume etc used in Avogadro's Law?09/07/20