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?

Options

A. `13.0 kcal`
B. `-13.0 kcal`
C. `27.0 kcal`
D. `-27.0 kcal`

Answer

C

Explanation

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`