Question

Calculate the internal energy change when $9{\text{ gms}}$ of water is vaporized at $1{\text{ bar}}$ pressure and $373{\text{ K}}$ . Molar enthalpy of vaporization of water at $1{\text{ bar}}$ pressure and $373{\text{ K}}$ is $41{\text{ kJ/mol}}$
(A) $18.95{\text{ kJ/mol}}$
(B) ${\text{25}}{\text{.52 kJ/mol}}$
(C) ${\text{34}}{\text{.65 kJ/mol}}$
(D) $37.904{\text{ kJ/mol}}$

Answer 1

To answer this question, you must recall the first law of thermodynamics and its formula. The first law of thermodynamics is a modified version of the law of conservation of energy for thermodynamic processes. Its basic principle in the conservation of energy, i.e. energy can neither be created nor be destroyed in a thermodynamic process.
Formula used:
$\Delta {\text{H}} = \Delta {\text{U}} + \Delta {\text{W}}$
And W = pdV
Where, $\Delta {\text{U}}$ represents the change in the internal energy of the system during the given process
$\Delta {\text{H}}$ represents the change in enthalpy of the system during the given process
And, $\Delta {\text{W}}$ represents the work done by the system during the given process.

Complete step by step solution:
We need to find the change in the internal energy of the system and we can calculate it using the first law of thermodynamics which gives us a relation between the enthalpy change, internal energy and work done by the system during a thermodynamic process.
We know the formula for first law of thermodynamics as $\Delta {\text{H}} = \Delta {\text{U}} + \Delta {\text{W}}$
Substituting the known values, we get,
$41 \times {10^3} = \Delta {\text{U}} + {\text{nRT}}$
$\Rightarrow 41 \times {10^3} = \Delta {\text{U}} + \left( {1 \times 8.314 \times 373} \right)$
Simplifying:
$\Rightarrow \Delta {\text{U}} = 41 \times {10^3} - 3.1 \times {10^3}$
$\therefore \Delta {\text{U}} = 37.9{\text{ kJ/mol}}$
The correct answer is D.

Note:
The work done by a gas is given by the formula $\Delta \left( {PV} \right)$ . In the above question, we are not provided with any information regarding the change in volume. The pressure and the temperature of the system are given as constant. The only change occurring in the system is that in the number of moles of the gaseous substances present in the mixture.
We know that the work done for an isobaric process is given as $W = P\Delta V$
Using the ideal gas equation, at the given constant pressure and temperature conditions, we can write, $W = P\Delta V = \Delta nRT$