Calculate the total pressure in a mixture of 8 g os dioxygen and 4 g of dihydrogen confined in a vessel of `1 dm^(3)` at `27^(@)C`.
(R = 0.083 bar `dm^(3)K^(-1)mol^(-1)`)

Explanation

Calculation of dioxygen mole `(n_(1))` : Molecular mass of dioxygen `= 2(16)=32 g mol^(-1)`
Weight of mass = 8 g
`therefore` Mole of dioxygen `= ("Mass")/("Molecular mass")`
`=(8g)/(32 g mol^(-1))=0.25 mol = n_(1)`
Calculation of dihydrogen mole `(n_(2))` : Molecular mass of dihydrogen `= 2.0 g mol^(-1)`
Weight of mass = 4.0 g
`therefore` Mole of dihydrogen `=("Mass")/("Molecular mass")`
`= (4.0 g)/(2.0 g mol^(-1))=2.0 mol = n_(2)`
Total mole `(n)=(n_(1)+n_(2))`
`= 0.25 mol O_(2)+2.0 mol H_(2)`
= 2.25 mol
where, Volume of vessel `(V) = 1dm^(3)`
Temperature `(T)=(27+273)K = 300K`
R = 0.083 bar `dm^(3)K^(-1)mol^(-1)`
pV = nRT and `p=(nRT)/(V)`
`therefore p =((2.25 mol)(0.083" bar dm"^(3)K^(-1)mol^(-1))(300 K))/(1 dm^(3))`
= 56.025 bar pressure.