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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` ? |
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Answer» Correct Answer - 1.8 bar 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 |
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