`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

No. of mole `=1 "mole"`
`T=273 K`
`Pv=nRT`
`(4.6)/(760)v=1xx0.0821xx273`
`v=3699 "lit"~~[3700 "lit"]`
latent heat of fusion `= 80 cal//gram` Latent heat of vaporisation`=596 cal //"gram"`
`Delta H=80xx18+596xx18[80+596]xx18`
`Delta H=12168 cal`
`Delta H=Delta E+P[V_(2)-V_(1)]`
`12168=Delta E+(4.6)/(760)[3699]xx24.24`
`rArr Delta E=12168-(4.6)/(760)xx3699xx24.24`
`=12168-542.72=11625.28 cal`