Question

Latent heat of vaporization of a liquid at 500 K and at 1 atm pressure is 10 kcal/mol. What will be the change in internal energy ($\Delta U$) of 3 moles of liquid at the same temperature?
(A) 13 kcal
(B) -13 kcal
(C) 27 kcal
(D) -27kcal

Answer 1

In order to find the latent heat of vaporization, we must first know what a latent heat of vaporization is. It is the amount of heat absorbed or released, when the matter disintegrates, the matter will change its phase from liquid to gas at a constant temperature. We can apply the values in the following equation.
$\Delta H = \Delta U + \Delta {n_g}RT$

Complete Solution :
First let us understand what a latent heat is. Latent heat is the heat which is absorbed or released during the change of phase change can occur from the gas to liquid or liquid to solid and vice versa. Latent heat is related to the enthalpy. Now let us move onto the latent heat of vaporization. It is the amount of heat absorbed or released, when the matter disintegrates, the matter will change its phase from liquid to gas at a constant temperature. Latent heat of vaporization is also known as the enthalpy of vaporization and it is denoted as $\Delta {H_V}$.
We know that
$\Delta H = \Delta U + P\Delta V$……. (1)
We know that
$P\Delta V = \Delta {n_g}RT$
Therefore, equation (1) becomes,
$\Delta H = \Delta U + \Delta {n_g}RT$……... (2)
3 moles of the liquid are converted into 3 moles of the gas
$3A(l) \to 3A(g)$
Therefore, $\Delta H$value for 3 moles of the liquid will become
$\Delta H = 3 \times 10 = 30kcals$
$\Delta {n_g}$ is given as the difference between the number of moles of product and the number of moles of reactant.
$\Delta {n_g} = 3 - 0 = 3$

By using the equation (3), we can write that
$\Delta H = \Delta U + \Delta {n_g}RT$
Where $\Delta H = 30kcals$
$\Delta {n_g} = 3$
$R = 2$
$T = 500 \times {10^{ - 3}}$

From equation (2)
$\Delta H = \Delta U + \Delta {n_g}RT$
$\Delta U = \Delta H - \Delta {n_g}RT$
$\Delta U = 30 - 3 \times 2 \times 500 \times {10^{ - 3}}$
$\Delta U = 30 - 3 = 27$
$\Delta U = 27kcals$.
So, the correct answer is “Option C”.

Note: Let us first see what a specific latent heat is. Specific latent heat is the measure of heat which is consumed or discharged by a body, when it experiences steady temperature process.
$L = \dfrac{Q}{M}$
L = specific latent heat
Q = Heat retained or discharged
M = mass of substance