Animated Solution for Chemistry - Chemical Thermodynamics: For water ${\Delta }_{\text{vap}}H=41{\text{ kJ mol}}^{-1}$ at $373\text{ K}$ and $1\text{ bar}$ pressure. Assuming that water vapour is an ideal gas that occupies a much larger volume than liquid water, the internal energy change during evaporation of water is ......... ${\text{kJ mol}}^{-1}$. [Use $R=8.3{\text{ J mol}}^{-1}{\text{K}}^{-1}$]

Imagine you are boiling water in a beaker. As heat is supplied, the liquid water transforms into water vapour. This phase change is called evaporation. The heat supplied at constant pressure is the enthalpy of vaporization, denoted by ${\Delta }_{\text{vap}}H$. The problem asks us to find the change in internal energy, $\Delta U$, during this process.

To connect enthalpy and internal energy, we use the First Law of Thermodynamics tailored for chemical reactions and phase changes at constant pressure:

Here, $\Delta n_g$ represents the change in the number of moles of gas. Let's look at the chemical equation for evaporation:

In this reaction, we have $1$ mole of gas on the product side and $0$ moles of gas on the reactant side. Therefore, $\Delta n_g=1-0=1$.

Before we plug in the numbers, we must ensure all units are consistent. This is where many students make a silly mistake! The enthalpy $\Delta H$ is given in ${\text{kJ mol}}^{-1}$, but the universal gas constant $R$ is given in ${\text{J mol}}^{-1}{\text{K}}^{-1}$. We must convert $R$ to ${\text{kJ mol}}^{-1}{\text{K}}^{-1}$ by multiplying it by $10^{-3}$.

Now, let's substitute the values into our master equation:

First, calculate the work done term ($\Delta n_{g}RT$):

Finally, solve for $\Delta U$:

Rounding off to the nearest integer, we get our final answer:

This result tells us that out of the $41\text{ kJ}$ of heat supplied, approximately $38\text{ kJ}$ goes into increasing the internal energy of the water molecules, while the remaining $\sim 3\text{ kJ}$ is used to do work against the atmospheric pressure as the water expands into vapour.