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Calculate w and DeltaU for the conversion of 0.5 mole of water at 100^(@)C to steam at 10 atm pressure. Heat of vaporisation of water at 100^(@)C is 40670Jmol^(-1) |
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Answer» Solution :Volume of 0.5 MOLE of steam at 1 atm pressure `=(nRT)/(P)=(0.5xx0.0821xx373)/(1.0)=15.3L` Change in volume `=Vol. ` of steam-vol. of water `=15.3-"negligible"=15.3L` WORK done by the system `W=P_(ext)xx`volume change `=1xx15.3=15.3` litre-atm `=15.3xx101.3J=1549.89J` 'w' shoud be negative as the work has been done by the system on the surroundings. `w=-1549.89J` heat requird to convert 0.5 MOLEOF water at `100^(@)C` to steam `=0.5xx40670J=20335J` Accoridng to first LAWOF thermodynamics. `DeltaU=q+w=20335-1549.89=18785.11J` |
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