A vessel of volume `2 xx 10^(-2) m^(3)` contains a mixture of hydrogen and helium at `47^(@)C` temperature and `4.15 xx 10^(5) N//m^(2)` Pressure. The mass of the mixture is `10^(-2) kg`. Calculate the masses of hydrogen and helium in the given mixture.

Explanation

Let mass of `H_(2) is m_(1)` and `He` is `m_(2)`
`:. M_(1) +m_(2) = 10^(-2) kg =10 xx 10^(-3) kg ……….(1)`
Let `P_(1),P_(2)` are partial pressure of `H_(2)` and `He`
`P_(1) +P_(2) = 4.15 xx 10^(5) N//m^(2)`
for the mixture
`(P_(1)+P_(2))V = ((m_(1))/(M_(1))+(m_(2))/(M_(2))) RT`
`rArr 4.15 xx 10^(5) xx2xx 10^(-2) = ((m_(1))/(2xx10^(-3))+(m_(2))/(4xx10^(-3))) 8.31 xx 320`
`rArr (m_(1))/(2) +(m_(2))/(4) = (4.15 xx2)/(8.31 xx 320) = 0.00312 = 3.12 xx 10^(-3)`
`rArr 2m_(1) +m_(2) = 12.48 xx 10^(-3) kg ..........(2)`
Solving (1) and (2)
`m_(1) = 2.48 xx 10^(-3) kg ~= 2.5 xx 10^(-3) kg`
and `m_(2) = 7.5 xx 10^(-3) kg`.