Question

The latent heat of Vaporisation of a liquid at 500K and 1atm pressure is 30 Kcal/mole. What will be the change in internal energy of 3 mol of liquid at same temperature?

• A. 13 Kcal
• B. -13 Kcal
• C. 27Kcal
• D. -27Kcal

Answer 1

The correct option is C

27Kcal

$△$H=$△$u +$△$nRT

$△$H=30Kcal

3 $H_2$O $\to$ 3 $H_2$O

(l) (v)

$△$n=3-0=3

$\Rightarrow$ 30 =$△$U + 3$\times$2$\times$500$\times$ ${10}^{-3}$

$△$U=27Kcal