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)`

Explanation

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`