Question

Latent heat of vaporization of a liquid at 500K and 1atm pressure is 10.0Kcal/mol. What will be the change in internal energy of 3 mol of liquid at same temperature and pressure?

• A. 13.0Kcal
• B. -13.0Kcal
• C. 27.0Kcal
• D. -27.0Kcal

Answer 1

The correct option is C 27.0Kcal

Vaporization of 3 moles of $H_{2}O$ vapors is:
$3H_{2}O\left(l\right)\to 3H_{2}O$ (g)
$\Delta$ n = 3-0 = 3
therefore,
$\Delta U=\Delta H-\Delta nRT$
$=(3\times 10)-3\left(0.002\right)\left(500\right)$
change in internal energy is:
$\Delta U=27Kcal.$