Moles of He Question

Question

A container at 1.00 atm of pressure is 22.4 L. The gas is at room temperature and there is a mixture of gases. 50% is He, 30% is Ne, and 20% is Ar. Approximately how many moles of He are in the container?

Answer 1

Room temperature can be assumed to be 20°C, which is equal to 293K.

22.4 L x 0.50 = 11.2 L of He

PV = nRT and solving for moles of He we have...

n = PV/RT = (1.00 atm)(11.2 L)/(0.0821 Latm/Kmole)(293K)

n = 0.466 moles of He