What will be the pressure of the gaseous mixture when 0.5 L of `H_(2)` at 0.8 bar and 2.0 L of dioxygen at 0.7 bar are introduced in a 1 L vessel at `27^(@)C` ?

Explanation

Dihydrogen Gas `(H_(2))`
Volume `V_(H_(2))=0.5 L`
Pressure `p_(H_(2))=0.8` bar
Dihydrogen Gas `(O_(2))`
Volume `V_(O_(2))=2` Lit
Pressure `p_(O_(2))=0.7` bar
Temperature `= 27^(@)C=300 K` Which is constant
Volume of mixture `= V_("total")=1` Lit.
Calculation of Partial Presure of dil hydrogen gas `pH_(2)` of 1 Lit. Volume :
`p_(1)V_(1)=p_(2)V_(2)`
`therefore ("0.8 bar")(0.5 L) = p_(2)xx1L`
`therefore p_(2)=(("0.8 bar")(0.5 L))/(1L)`
`therefore p_(2)=0.4` bar `= p_(H_(2))`
where, At gas `H_(2)`
`p_(1)=0.8` bar
`V_(1)=0.5 L`
`V_(2)=1L`
`p_(2)=(?)`
Caslculation of Partial Pressure of di Oxygen gas `(PO_(2))` of 1 lit volume :
`p_(1)V_(1)=p_(2)V_(2)`
`therefore ("0.7 bar")(2.0 L)=(p_(2))(1L)`
`therefore p_(2)=(0.7xx2.0)/(1)` bar
`therefore p_(2)=1.4` bar `= p_(O_(2))`
where, At gas `O_(2)`
`p_(1)=0.7` bar
`V_(1)=2.0L`
`V_(2)=1.0L`
`p_(2)=(?)`
`p_("total")=p_(H_(2))+p_(O_(2))`
= 0.4 bar + 1.4 bar
= 1.8 bar