A vessel of volume `20 L` contains a mixture o hydrogen and helium at temperature of `27^(@)C` and pressure `2.0 atm` The mass of the mixture is `5 g`. Assuming the gases to be ideal, the ratio of the mass of hydrogen to heat of helium in the given mixture will be

Options

A. `1 :2`
B. `2 : 3`
C. `2 : 1`
D. `2 : 5`

Answer

D

Explanation

Suppose there are `n_(1)` moles of hydrogen and `n_(2)` moles of helium in the given mixture. Then the pressure of the mixture will be
`P = (n_(1) RT)/(V) + (n_(2) RT)/(V) = (n_(1) + n_(2)) (RT)/(V)` (i)
`2 xx 101.3 xx 10^(3) = (n_(1) + n_(2)) ((8.3)(300))/(20 xx 10^(-3))`
`(n_(1) + n_(2)) = (2 xx 101.3 xx 3 xx 10^(3) xx 20 xx 10^(-3))/((8.3) (300)) = 1.62` (ii)
The mass of the mixture is (in grams):
`n_(1) xx 2 + n_(2) xx 4 = 5`
`(n_(1) + 2 n_(2)) = 2.5`
Solving Eqs. (i) and (ii), `n_(1) = 0.74, n_(2) = 0.88`
Hence `(m_(H))/(m_(He)) = (0.74 xx 2)/(0.88 xx 4) = (1.48)/(3.52) rarr 2 : 5`