One moles of strem is compressed reversibly of water at boiling point `100^(@)C`. The heat of vapourisation of water at `100^(@)C` and `1atm` is `540cal g^(-1)`. Calculate `DeltaU` and `DeltaH`.

Explanation

`W = PDeltaV =- nRT = 1 xx 2xx 373 =- 746 cal`
[`:' n = 1,R = 2 cal, T = (273 +100)]`
Now, `q =- 18 xx540 cal =- 9720 cal ( :' M_(H_(2O)) = 18)`
`:. DeltaU = q -W =- 9720-(-746) =- 8974 cal`
Again, `DeltaH = DeltaU + PDeltaV =- 8974 - 746 cal =- 9720cal`
`DeltaU =- 8974 ca`l and `DeltaH =- 9720 cal`