Molar mass of `O_(2)`=32 g `mol^(-1)" ":. 8 g O_(2)=(8)/(32)mol=0.25 mol`
Molar mass of `H_(2)=2 g mol^(-1)" " :. 4 g O_(2)=(4)/(2)=2 mol`
`:.` Total number of moles (n)=2+0.25=2.25
V=1 `dm^(3)`, T=`27^(@)C=300 K,R=0.083 bar dm^(3)K^(-1)mol^(-1)`
PV=nRT or P`=(nRT)/(V)=((2.25 mol)(0.083 bar dm^(3)K^(-1)mol^(-1))(300 K))/(1 dm^(3))=56.025 bar`