Animated Solution for Chemistry - Chemical Thermodynamics: The internal energy change (in J) when $90\text{ g}$ of water undergoes complete evaporation at $100^{\circ }\text{C}$ is ........., (Given : $\Delta H_{\text{vap}}$ for water at $373\text{ K}=41\text{ kJ/mol}$, $R=8.314{\text{ J K}}^{-1}{\text{mol}}^{-1}$)

Imagine a beaker filled with exactly $90\text{ g}$ of liquid water, sitting on a hot plate. The temperature is a steady $100^{\circ }\text{C}$ ($373\text{ K}$). As you watch, the water begins to boil, transforming from a dense liquid into an expansive gas. This isn't just a physical transformation; it is a thermodynamic journey.

During this process, we are constantly pumping heat into the system. This heat is known as the enthalpy of vaporization, $\Delta H_{\text{vap}}$. But where does all this energy go? Does it all stay inside the water molecules, or is some of it spent doing something else? To answer this, we need to dive into the First Law of Thermodynamics and calculate the true change in the system's internal energy, $\Delta U$.

In the realm of chemistry, mass is just a stepping stone. The true currency of thermodynamics is the mole. The problem gives us the enthalpy of vaporization as $41\text{ kJ/mol}$. This means it takes $41\text{ kJ}$ of energy to vaporize exactly one mole of water.

But we don't have one mole; we have $90\text{ g}$. To find out how many moles we are dealing with, we divide the given mass by the molar mass of water ($18\text{ g/mol}$):

So, our system consists of exactly $5\text{ moles}$ of water undergoing a phase change.

To understand the expansion work, we must look at the chemical equation for evaporation:

Initially, we have $5\text{ moles}$ of liquid water. Finally, we have $5\text{ moles}$ of water vapor. The change in the number of gaseous moles, denoted as $\Delta n_g$, is crucial because gases occupy significantly more volume than liquids.

This tells us that the system has generated $5\text{ moles}$ of new gas, which will push against the atmosphere, doing work.

The First Law of Thermodynamics connects the heat added at constant pressure (Enthalpy, $\Delta H$) to the change in internal energy ($\Delta U$) and the expansion work done by the system ($p\Delta V$). For ideal gases, we can replace $p\Delta V$ with $\Delta n_{g}RT$. This gives us our master equation:

We want to find the internal energy change, so we rearrange the equation:

This equation tells a beautiful story: The change in internal energy is the total heat supplied minus the energy spent by the gas pushing the atmosphere away.

Before we plug numbers into our master equation, we need the total enthalpy change, $\Delta H_{\text{total}}$. The problem gives us the molar enthalpy, so we must scale it up for our $5\text{ moles}$:

Because the universal gas constant $R$ is given in Joules ($8.314{\text{ J K}}^{-1}{\text{mol}}^{-1}$), we must convert our enthalpy into Joules to avoid a catastrophic unit mismatch:

Now, we have all the pieces of the puzzle. Let's substitute them into our rearranged First Law equation:

The term $(5\times 8.314\times 373)$ represents the expansion work. Calculating this gives:

Finally, we subtract this work from the total heat supplied:

The question asks us to round off to the nearest integer.

And there we have it! Out of the $205,000\text{ J}$ of heat we pumped into the water, about $15,505\text{ J}$ was spent pushing the atmosphere away to make room for the steam, leaving exactly $189,494\text{ J}$ to increase the internal energy of the water molecules.