Question

A mixture with H2 and He exerts a total pressure of 0.48 atm. If there is 1.0 g of H2 and 1.0 g of He in the mixture, what is the partial pressure of the helium gas?

Answer 1

Answer is: the partial pressure of the helium gas is 0.158 atm.

p(mixture) = 0.48 atm; total pressure.

m(H₂) = 1.0 g; mass of hydrogen gas.

n(H₂) = m(H₂) ÷ M(H₂).

n(H₂) = 1.0 g ÷ 2 g/mol.

n(H₂) = 0.5 mol; amount of hydrogen.

m(He) = 1.0 g; mass of helium.

n(He) = 1 g ÷ 4 g/mol.

n(He) = 0.25 mol; amount of helium.

χ(H₂) = 0.5 mol ÷ 0.75 mol.

χ(H₂) = 0.67; mole fraction of hydrogen.

χ(He) = 0.25 mol ÷ 0.75 mol.

χ(He) = 0.33; mole fraction of helium.

p(He) = 0.33 · 0.48 atm.

p(He) = 0.158 atm; the partial pressure of the helium gas.