Assuming ideal behavior, how many moles of argon would you need to fill a 12.5×12.0×10.0 ft room?
Assume atmospheric pressure of 1.00 atm, a room temperature of 20.0∘ C, and 28.2 L/ft
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Assume atmospheric pressure of 1.00 atm, a room temperature of 20.0∘ C, and 28.2 L/ft
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12.5×12.0×10.0 ft = 1500 ft^3 = 42475.3 L
n = PV / RT = (1.00 atm) × (42475.3 L) /
((0.08205746 L atm/K mol) × (20.0 + 273.15 K)) =
1766 mol Ar