Aflexible plastic container contains 0.860g of helium has in a volume of 19.2l if 0.205g of helium is removed at contact pressure and temperature what will be the new volume

Given:

Initial volume of He, V1 = 19.2 L

Initial mass of He, m1 = 0.0860 g

Mass of He removed = 0.205 g

To determine:

The new volume of He i.e V2

Explanation:

Based on Avogadro's law:

Volume of a gas is directly proportional to the # moles of the gas

Volume (V) α moles (n) (1)

Atomic mass of He = 4 g/mol

Initial moles of He, n1 = 0.860 g/4 g.mol-1 = 0.215 moles

Final moles of He, n2 = (0.860-0.205)g/4 g.mol-1 = 0.164 moles

Based on eq(1) we have:

V1/V2 = n1/n2

V2 = V1 n2/n1 = 19.2 L * 0.164 moles/0,215 moles = 14.6 L

Ans: New volume is 14.6 L

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