A flexible plastic container contains 0.833 g of helium gas in a volume of 18.3 L. If 0.203 g of helium is removed at constant pressure and temperature, what will be the new volume?

Answer 1

Let n = moles and V = Volume
n ∝ V => n(1)/n(2) = V(1)/V(2)
n(1) = (0.833 gms / 4 gms/mol) = 0.2083 mol
n(2) = [(0.833 – 0.203) gms / 4 gms/mol) = 0.1575 mol
V(1) = 18.3 Liters
V(2) = X
(0.2083 mol / 0.1575 mol) = (18.3 Liters / X)
X = [(0.1575 mol)(18.3 Liters)/ (0.5083 mol)] = 13.8 Liters