`1` mole of ice at `0^(@)C` and `4.6 mm Hg` pressure is converted to water vapour at a constant temperature and pressure. Find `Delta H` and `Delta E` if the latent heat of fusion of ice is `80 cal//g` and latent heat of vaporisation of liquid water at `0^(@)C` is `596 cal//g` and the volume of ice in comparison to that of water (vapour) in neglected.

Explanation

Latent heat of fusion of ice per "mole" `=80xx18=1440 cal`.
Latent heat of vaporisation of liquid water per mole `=596xx18`
`=10728 cal` .
`:.` total heat absorbed by 1 "mole" (18 grams) of ice in being converted to 1 mole of water vapour
`=1440+10728=12168"calories"`.
Since the conversion took place at a constant pressure,
`q_(p)=DeltaH=12168"calories"`
Now we have,
`DeltaU=DeltaH-pDeltaV=q_(p)-pDeltaV` ....(Eqn. 7)
As the volume of ice is to be neglacted, `V_(1)=0`
and `V_(2)`= volume of 1 mole of water vapour at `0^(@)C` and `4.6` mmHg pressure
`=22400xx(760)/(4.6)=3717000"mL"`. ( Charles.s law)
`:.DeltaV=(V_(2)-V_(1))=3717000"mL"`
`p=4.6 mm=0.46xx13.6xx981`
`=6137"dynes/cm"^(2)`
`:.pDeltaV=6137xx3717000"ergs"`
`=(6137xx3717000)/(4.18xx10^(7))"cal"`
`=545 cal`.
`:.DeltaU=12168-545=11623"calories"("from Eqn". 7)`