The latent heat of vapourization of a liquid at $500 \mathrm{~K}$ and 1 atm pressure is $10.0 \mathrm{kcal}…

The latent heat of vapourization of a liquid at $500 \mathrm{~K}$ and 1 atm pressure is $10.0 \mathrm{kcal} / \mathrm{mol}$. What will be the change in internal energy $(\Delta \mathrm{U})$ of 3 moles of liquid at the same temperature
  1. $13.0 \mathrm{kcal} / \mathrm{mol}$
  2. $-13.0 \mathrm{kcal} / \mathrm{mol}$
  3. $27.0 \mathrm{kcal}$
  4. $-7.0 \mathrm{kcal} / \mathrm{mol}$

Solution

$3 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) ightarrow 3 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})$
$\Delta \mathrm{n}=3, \Delta \mathrm{E}=\Delta \mathrm{H}-\Delta \mathrm{nRT}$
$=30-3 \times \frac{2}{1000} \times 500=27 \mathrm{kcal}$

Asked in: JEE-TOPICTESTS-CHEMISTRY

Practice more THERMODYNAMICS questions on Aicharya