Question

The latent heat of vaporization of liquid at 500 K and 1 atmospheric pressure is 10.0 kcal/mol. What will be the change in internal energy of 3 moles of the liquid at the same temperature aud pressure?

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

Answer 1

The correct option is A 27.0 kcal

The latent heat of vaporization of liquid at 500 K and 1 atmospheric pressure is 10.0 kcal/mol.
The change in internal energy of 3 moles of the liquid at the same temperature and pressure will be 27.0 kcal.
The calculations are as shown below.

The reaction showing the transformation is:

$3H_{2}O\left(l\right)⟶3H_{2}O\left(g\right):\Delta n_g=3$

The relationship between the enthalpy change and the change in the internal energy is :

$\Delta H=\Delta U+\Delta n_{g}RT$

$\Delta U=\Delta H-\Delta n_{g}RT$

Substitute values in the above expression.

$30$ = $\Delta U$ - $(3\times 2\times {10}^{-3}\times 500)$

$\Delta U=27kcal$