When1 mole of ice melt at `0^(@)` C and at constant pressure of 1 atm . 1440 calories of heat are absorbed by the system. The molar volumes of ice and water are 0.0196 and 0.0180 litres respectively. Calculate `DeltaH` and `DeltaE` for the reaction.

Explanation

Since `DeltaH=q_(p)`
= heat absorbed by the system at constant pressure
`=1440 "calories"`.
In the equation `DeltaH=DeltaU+pDeltaV`
`pDeltaV=76xx13.6xx981(18-19.6)"ergs"`
`=-(76xx13.6xx981xx1.6)/(4.18xx10^(7))"calories"`
`=-0.039 "calories"`.
`(p=1atm=76xx13.6xx981"dynes"//cm^(2),V_(2)=18c c,V_(1)=19.6 c c)`
Since `pDeltaV` is very small compared to `DeltaU, pDeltaV` can be neglected.
Thus, `DeltaH=DeltaU=1440 "calories"`.