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Question

1 mole of ice at 0 and 4.6 mm Hg pressure is converted to water vapor at a constant temperature and pressure. Find H and E if the latent heat of fusion of ice is 80Cal/gm and latent heat of vaporisation of liquid water at 0 is 536 Cal per gram and the volume of ice in comparison to that of water(vapour) is neglected.

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Solution

Hence ice get converted to vapour
ΔH=ΔHf+ΔHv=80×18+596×18
ΔH=12168cal/mol
ΔH=ΔU+PΔV According to first law of thermodynamics
Now PV=nRT
PΔV=nRT=1×8314×2780.18=543cal
ΔU=ΔHPΔV
ΔU=12168543cal
ΔU=11625cal
Hence, ΔH is 12168cal and ΔU is 11625cal.

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