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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 ? |
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Answer» Solution :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 |
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