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

Explanation

No. of moles of hydrogen `(n_(1)) = ("Mass of hydrongen")/("Molar mass") = ((8.0 g))/((32.0 g mol^(-1))) = 0.25` mol
No. of moles of hydrogen `(n_(2)) = ("Mass of hydrogen")/("Molar mass") = ((4.0 g))/((2.0 g mol^(-1))) = 2` mol
Partial pressure of oxygen i.er., `Po_(2) = (n_(1)RT)/(V)`
`= ((0.25 mol) xx (0.083 "bar" dm^(3) K^(-1) mol^(-1)) xx 300 K)/((1 dm^(3))) = 6.225 "bar"`
Partial pressure of hydrogen i.e., `P_(H_(2)) = (n_(2)RT)/(V)`
`= ((2 mol) xx (0.083 "bar" dm^(3)K^(-1) mol^(-1)) xx 300 K)/((1 dm^(3))) = 49.8 "bar"`
Total pressure of gaseous mixture `= (6.25 + 49.8) = 56.025 "bar"`