1 mole of ice at 0^(@)C and 4.6 mmHg pressure is converted to water vapour at a constant temperature and pressure. Find DeltaH and DeltaU 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) is neglected.

1 mole of ice at 0^(@)C and 4.6 mmHg pressure is converted to water va

1.	1 mole of ice at 0^(@)C and 4.6 mmHg pressure is converted to water vapour at a constant temperature and pressure. Find DeltaH and DeltaU 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) is neglected.
Answer» Solution :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)`