Latent heat of vaporisation of a liquid at 500 K and 1 atm pressure is `10 kcal//mol`. What will be the change in internal energy `(DeltaU)` of 3 mol of liquid at the same temperature ?

Options

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

Answer

C

Explanation

`Delta H` = 10.0 kcal for one mole
`Delta H` = 30.0 kcal for three moles
`"Liquid" hArr "Vapours"`.
`Delta U = Delta H - Delta n_(g) RT`
`= 30.0 - 3 xx 1.98 xx 10^(-3) xx 500`
= 30 - 3
= 27.0 kcal