The latent heat of vapourisation of a liquid at `500K` and `1atm` pressure is `10kcal mol^(-1)`. What will be change in internal energy of `3mol` of liquid at same temperature?

Options

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

Answer

C

Explanation

`DeltaH=DeltaU+DeltanRT`, given `DeltaH=30 kcal`
for `3` mole
`Deltan=3` because, liquid `hArr`vapour,
`3.0=DeltaU+3xx2xx500xx10^(-3)`
`:. DeltaU=27 kcal`