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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. |
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Answer» 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`. |
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