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Latent heat of vaporisation of a liquid at `500K` and `1` atm pressure is `10.0 kcal//mol`. What will be the change in internal energy `(DeltaE)` of `3` mol of liquid at same temperature?A. `13.0 kcal`B. `-13.0 kcal`C. `27.0 kcal`D. `-27.0 kcal` |
Answer» Correct Answer - C Vaporisation of `3` moles of water is `3H_(2)O(l)to3H_(2)O(g)` Here `Deltan=3-0=3` Heat change for `3` moles of water to vapours `=3xx10=30 kJ` `:. DeltaE=DeltaH-DeltanRT` `=30-(3)(0.002)(500)=27 kcal` |
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