1.

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 ?

Answer»

13.0 kcal
`-13.0 kcal`
27.0 kcal
`-27.0 kcal`

Solution :`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


Discussion

No Comment Found

Related InterviewSolutions