A mixture of helium (monatomic) and hydrogen (rigid diatomic), both ideal, has $C_{P} / C_{V} = 1.52$ . Find the mole percent of helium in the mixture (nearest integer).

Numerical answer type, answer in %, tolerance 1.

Answer: 58 %

Let the mole fraction of He be

x

. Then

C_{V} = 1.5 x + 2.5 (1 - x) = (2.5 - x) R

and

C_{P} = C_{V} + R = (3.5 - x) R

. Setting

(3.5 - x) / (2.5 - x) = 1.52

gives

3.5 - x = 3.8 - 1.52 x

, so

0.52 x = 0.30

and

x = 0.5769

. The mixture is 57.7 mol % helium, about 58.