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The compression factor (compressibility factor) for one mole of a van der Waals' gas at 0^(@)C and 100 atmosphere pressure is found to be 0.5. Assuming that the volume of a gas molecule is negligible, calculate the van der Waal's constant 'a'. |
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Answer» Solution :`Z=(PV)/(RT) =0.5` `therefore (100xxV)/(0.082xx273)=0.5` `therefore V=0.112` litre Now, USING van der Waals. equation `[P+(a)/(V^(2))][V]=RT` or `[P+(a)/(V^(2))]=(RT)/(V)" "(therefore` b is negligible) `therefore [100+(a)/((0.112)^(2))]=(0.082xx273)/(0.112)` `=199.88` `therefore (a)/((0.112)^(2))=99.88` `therefore a=1.253" litre"^(2)" mole"^(-2)` atm. |
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