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

Explanation

To calculate the number of moles of unmixed `H_(2)` and `O_(2)` gases, we apply ideal gas equation PV=nRT
In case of `H_(2)`, P=0.8 bar, V=0.5 L and T=(273+27)K=300K and in case of `O_(2),P=0.7` bar,
V=2.0L and T=(273+27)K=300K
therefore, `n_(H_(2))=(PV)/(RT)=(0.8xx0.5)/(Rxx300)=(0.4)/(300R)L*`bar
and `n_(O_(2))=(PV)/(RT)=(2xx0.7)/(Rxx300)=(1.4)/(300R)L*`bar
In the mixture of `H_(2)` and `O_(2)`, toal number of mol
`=n_(H_(2))+n_(O_(2))=(1)/(300R)(0.4+1.4)L` bar`=(1.8)/(300R)L*`bar
Ffor this mixture `V=1L and T=(273+27)K=300K`
If the pressure of the mixture be P, then
`P=(nRT)/(V)=(1.8)/(300R)xx(Rxx300)/(1)`bar=1.8bar