`1g` of water at atmospheric pressure and at `100^(@)C, 1671 cm^(3)` of water vapour is converted into ` cm^(3)` water. Calculate (i) work doen against the surroundigs and (ii) increases in the internal energy.

Explanation

`DeltaV=V_(2)-V_(1)=(1671-1) cm^(3)=1670 xx100^(-6)J=1693171J`
Work done `=P Delta V=(1.013xx10^(5))(1670xx10^(-6))J=1693171J`
Work done against the surroundings `=169.2J=(169.2)/(4.18)=40.48` cal.
(ii) `dQ=mL=540` cals/g
From the law of thermodynamics,
`dQ=dU+dW`
`:. du=dQ-dW=540-40.48`
`dU=499.52` cal
Change in internal energy `=499.52` cal/g.