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

Answer

1.8 bar

Explanation

Let the partial pressure of `H_(2)` in the vessel be `p_(H_2).`
Now,
`p_(1)=0.8"bar" " "p_(2)=p_(H_2)` ?
`V_(1)=0.5 " "V_(2)=1L`
It is known that,
`p_(1)V_(1)=p_(2)V_(2)`
`rArr p_(2)=(p_(1)V_(1))/(V_(2))
`rArr p_(H_2)=(0.8xx0.5)/(1)
=0.4 bar
Now, let the partial pressure of `O_(2)` in the vessel be `p_(O_2).`
Now,
`p_(1)=0.7"bar" " "p_(2)=p_(o_2)=` ?
`V_(1)=2.0L " "V_(2)=1L`
`p_(1)V_(1)=p_(2)V_(2)`
`rArrp_(2)=(p_(1)V_(1))/(V_(2))`
`rArrp_(o_2)=(0.7xx20)/(1)`
= 0.4 bar
Total prassure of the gas mixture in the vessel can be obtained as:
`p_("total")=p_(H_2)+p_(O_2)`
= 0.4+1.4
=1.8 bar
Hence, the total preesure of the gaseous mixture in the vesses is 1.8 bar