A neon `(Ne)` -dioxygen `(O_(2))` mixture contains `160 g` neon and `96 g` dioxygen . If the pressure of the mixture of gases in the container is `30` bar, calculate the partial pressure of neon and dioxygen in the mixture.
Strategy : Calculate the number of moles of each component to get their mole fractions. Then use Eq.(5.24) to determine the partial pressure of any component and substract it from the total pressure to find the partial pressure of the component .

Explanation

`n_("Ne") = ("Mass"_("Ne"))/("Molar mass"_("Ne")) = (160 g)/(20 g "mol"^(-1)) = 8 "mol"`
`n_(O_(2)) = ("Mass"_(O_(2)))/("Molar mass"_(O_(2))) = (96 g)/(32 g "mol"^(-1)) = 3 "mol"`
`x_("Ne") = (n_("Ne"))/(n_(N"e" + n_(O_(2)))) = (8)/(8 + 3) = (8)/(11) = 0.73`
According to Eq.(5.24),
`"Partial pressure"._(Ne) = ("Mole fraction"._(Ne)) ("Total pressure")`
= (0.73) (30 bar)
= 21.9 bar
Thus , partial pressure of `O_(2) = (p_("total") - p_(H_(2)O)`
= (30 bar) - (21.9 bar)
= 8.1 bar