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)`
`therefore m _(1) +m _(2) = 10^(-2) kg =10xx10^(-3)….(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))/(n _(1))+ (m_(2))/(n _(2)))RT`
`implies 4.15 xx 10^(5) xx 2xx 10^(-2)`
`=((m_(1))/(2xx 10 ^(-3))+ (m_(2))/(4xx10^(-3)))8.31 xx320`
`implies(m_(1))/(2) + (m_(2))/(4) = (4.15 xx2)/(8.31xx320)`
`=0.00312 =3.12 xx10^(-3)`
`implies 2m_(1) + m_(2) =12.48xx10^(-3) kg " "....(2)`
Solving (1) and (2)
`m _(1) =2.48 xx10^(-3) kg = 2.5 xx 10^(-3) kg`
and ` m = 7.5 xx10^(-3) kg. `