What will be the pressure of the gaseous mixture when 0.5 L of `H_(2)` at 0.8 bar and 2.01 of dioxygen at 0.7 bar are introduced in a 1L vessel at `27^(@)C` ?

Explanation

Moles of `H_(2)i.e., n_(H_(2)) = (pV)/(RT) = (0.8 xx 0.5)/(RT) = (0.40)/(RT)`
Moles of `O_(2), no_(2)=(0.7xx 2)/(RT) = (1.4)/(RT)`
Total number of moles `=(1.40)/(RT) +(1.4)/(RT) =(1.8)/(RT)`
Total pressure, `p_("Total")=(nRT)/(V) =(1.8 xx RT)/(RT xx 1)=1.8` atm.