Considering ideal gas behavior, the expansion work done (in kJ) when 144 g of water is electrolyzed

Q: Why do we ignore the volume of liquid water?

The volume of a liquid is negligible compared to the volume of the same amount of substance in the gaseous state. The expansion work is overwhelmingly dominated by the creation of gases.

Q: Why is the formula $W = -\Delta n_g RT$ instead of just $P\Delta V$?

At constant pressure and temperature, the ideal gas law $PV = nRT$ gives us $P\Delta V = \Delta n_g RT$. Substituting this into the work definition $W = -P\Delta V$ gives the extremely useful form $W = -\Delta n_g RT$.

Q: Why is the final answer positive when the formula gives a negative value?

The formula $W = -\Delta n_g RT$ calculates the work done *on* the system. A negative result means the system did work *on* the surroundings (expansion). The question asks for the "expansion work done", which implies the magnitude of this work, hence the positive 29.88 kJ.

Q: What if the reaction was carried out in a closed, rigid container?

In a rigid container, the volume cannot change ($\Delta V = 0$). Therefore, no expansion work can be done ($W = 0$), regardless of how much gas is produced. The pressure would simply increase.

Q: How did we get 12 moles of gas from 8 moles of water?

Look at the stoichiometry: $H_2O \rightarrow H_2 + \frac{1}{2}O_2$. One mole of water yields 1.5 moles of gas (1 mole of $H_2$ + 0.5 moles of $O_2$). Therefore, 8 moles of water yield $8 \times 1.5 = 12$ moles of gas.

Question

Considering ideal gas behavior, the expansion work done (in kJ) when 144 g of water is electrolyzed completely under constant pressure at 300 K is _______.
Use: Universal gas constant (R) = 8.3 J K$^{-1}$ mol$^{-1}$; Atomic mass (in amu): H = 1, O = 16

Sigma Percentile · Accepted Answer

The concept of work in thermodynamics often feels abstract, but it is deeply rooted in physical reality. When a system expands, it must push against the surrounding atmosphere. This pushing requires energy, and we call this energy transfer **expansion work**. In this problem, we are looking at the electrolysis of water—a process that violently transforms a dense liquid into expansive gases.

### Analyzing the Setup

Imagine a beaker filled with $144 	ext{ g}$ of liquid water. When we pass an electric current through it, the water molecules are torn apart.

The balanced chemical equation for this transformation is:
$$H_2O(l) ightarrow H_2(g) + \frac{1}{2}O_2(g)$$

Notice the phase states. We are starting with a liquid and ending with gases. Because gases occupy roughly 1000 times more volume than liquids, this reaction will cause a massive expansion. The system will have to do work to push the atmosphere out of the way to make room for these new gases.

### The Gaseous Expansion

To calculate the work done, we first need to know exactly how much gas is produced. We start by finding the number of moles of water we are electrolyzing. The molar mass of water is $18 	ext{ g/mol}$.

$$n_{H_2O} = \frac{144 	ext{ g}}{18 	ext{ g/mol}} = 8 	ext{ mol}$$

According to our balanced equation, every $1 	ext{ mol}$ of liquid water produces $1 	ext{ mol}$ of hydrogen gas and $0.5 	ext{ mol}$ of oxygen gas. Therefore, $8 	ext{ mol}$ of water will produce:
- $8 	ext{ mol}$ of $H_2(g)$
- $4 	ext{ mol}$ of $O_2(g)$

The total change in the number of gaseous moles, denoted as $\Delta n_g$, is the moles of gaseous products minus the moles of gaseous reactants. Since water is a liquid, its gaseous contribution is zero.

$$\Delta n_g = (8 + 4) - 0 = 12 	ext{ mol}$$

### The Master Equation

The fundamental definition of work done at a constant pressure is $W = -P\Delta V$. However, we don't know the pressure or the change in volume directly. This is where the ideal gas law, $PV = nRT$, comes to the rescue. At constant pressure and temperature, we can write $P\Delta V = \Delta n_g RT$.

Substituting this into our work equation gives us a incredibly powerful tool:
$$W = -\Delta n_g RT$$

### Final Calculation

Now, we simply substitute our known values into the master equation. We found $\Delta n_g = 12 	ext{ mol}$, the universal gas constant is given as $R = 8.3 	ext{ J K}^{-1} 	ext{ mol}^{-1}$, and the temperature is $T = 300 	ext{ K}$.

$$W = -12 	imes 8.3 	imes 300$$
$$W = -29880 	ext{ J}$$

The negative sign is crucial. In thermodynamics, a negative work value means the system is doing work *on* the surroundings. It is losing energy to push the atmosphere away.

The question asks for the "expansion work done" in $	ext{kJ}$. When asked for "expansion work", we typically provide the magnitude of the work done by the system.

$$	ext{Expansion Work} = |-29880 	ext{ J}| = 29.88 	ext{ kJ}$$

And there we have it! The system expended $29.88 	ext{ kJ}$ of energy just to make room for the newly created hydrogen and oxygen gases.

Enter Numerical Value:

Visualized Solution

\text{Electrolysis of Water}

\text{Balanced Chemical Equation}

\text{Moles of Reactant}

\text{Moles of Reactant}

\text{Change in Gaseous Moles } (\Delta n_g)

\text{Calculating } \Delta n_g

\text{Calculating } \Delta n_g

\text{Work Done Formula}

\text{Substituting Values}

\text{Calculating Work}

\text{Expansion Work in kJ}

\text{What if?}

The Sigma Insight: First Law of Thermodynamics

Analyzing the Setup

The Gaseous Expansion

The Master Equation

Final Calculation

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Enter Numerical Value:

Visualized Solution

\text{Electrolysis of Water}

\text{Balanced Chemical Equation}

\text{Moles of Reactant}

\text{Moles of Reactant}

\text{Change in Gaseous Moles } (\Delta n_g)

\text{Calculating } \Delta n_g

\text{Calculating } \Delta n_g

\text{Work Done Formula}

\text{Substituting Values}

\text{Calculating Work}

\text{Expansion Work in kJ}

\text{What if?}

The Sigma Insight: First Law of Thermodynamics

Analyzing the Setup

The Gaseous Expansion

The Master Equation

Final Calculation

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