Assuming that water vapour is an ideal gas, the internal energy change (ΔE) when 1 mole of water is

Q: Why do we only consider gaseous moles for $\Delta n_g$?

The volume occupied by liquids and solids is negligible compared to gases. Therefore, the expansion work ($P\Delta V$) is almost entirely due to the change in the number of moles of gas.

Q: Why is $\Delta E$ less than $\Delta H$?

When a liquid vaporises, it expands against the atmospheric pressure. Some of the heat supplied ($\Delta H$) is used to do this expansion work, leaving slightly less energy to increase the internal energy ($\Delta E$).

Q: Can I use $R = 0.0821$ in this calculation?

No! The value $R = 0.0821$ has units of $\text{L atm K}^{-1}\text{mol}^{-1}$. Since we are calculating energy in Joules, we must use $R = 8.314$ (or $8.3$ as given in the question) which has units of $\text{J K}^{-1}\text{mol}^{-1}$.

Q: What is the difference between $\Delta E$ and $\Delta U$?

There is no difference! Both $\Delta E$ and $\Delta U$ are standard symbols used to represent the change in internal energy of a system. Different textbooks just prefer one over the other.

Q: Why did we use $T = 373\text{ K}$?

The vaporisation of water at standard atmospheric pressure (1 bar) occurs at its boiling point, which is $100^\circ\text{C}$. To use it in thermodynamic equations, we must convert it to the absolute Kelvin scale by adding 273, giving us $373\text{ K}$.

Question

Assuming that water vapour is an ideal gas, the internal energy change ($\Delta E$) when 1 mole of water is vaporised at 1 bar pressure and $100^\circ	ext{C}$, (Given : molar enthalpy of vaporisation of water at 1 bar and 373 K = $41	ext{ kJ mol}^{-1}$ and $R = 8.3	ext{ J mol}^{-1}	ext{K}^{-1}$) will be

Options:

(A) $4.100	ext{ kJ mol}^{-1}$

(B) $3.7904	ext{ kJ mol}^{-1}$

(C) $37.904	ext{ kJ mol}^{-1}$

(D) $41.00	ext{ kJ mol}^{-1}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Boiling water is an everyday phenomenon, but thermodynamically, it is a fascinating battle between internal energy and atmospheric pressure. When liquid water turns into water vapour, it undergoes a phase change.

Let's write down the chemical equation for this process:
$$ 	ext{H}_2	ext{O}(l) ightarrow 	ext{H}_2	ext{O}(g) $$

Notice that we are starting with zero moles of gas (since water is a liquid) and ending up with one mole of gas (water vapour). The change in the number of gaseous moles, denoted by $\Delta n_g$, is a crucial parameter. 
$$ \Delta n_g = n_{	ext{products}(g)} - n_{	ext{reactants}(g)} = 1 - 0 = 1 $$

### The Master Equation

To connect the heat supplied to the system with the change in its internal energy, we invoke the **First Law of Thermodynamics** at constant pressure. The heat supplied at constant pressure is known as the enthalpy change ($\Delta H$).

The law states that the enthalpy change is the sum of the internal energy change ($\Delta E$) and the expansion work done by the gas ($P\Delta V$, which can be written as $\Delta n_g RT$ for ideal gases).
$$ \Delta H = \Delta E + \Delta n_g RT $$

Since we are interested in finding the internal energy change, we can rearrange this equation:
$$ \Delta E = \Delta H - \Delta n_g RT $$

### Final Calculation

Now, we must carefully substitute the given values into our rearranged equation. This is where many students fall into a classic trap: **unit inconsistency**. The enthalpy of vaporisation is given in kilojoules, while the gas constant $R$ is in joules. We must convert $\Delta H$ to joules!

Given values:
$$ \Delta H = 41 	ext{ kJ mol}^{-1} = 41000 	ext{ J mol}^{-1} $$
$$ R = 8.3 	ext{ J mol}^{-1}	ext{K}^{-1} $$
$$ T = 100^\circ	ext{C} = 373 	ext{ K} $$

Substituting these into the equation:
$$ \Delta E = 41000 - (1 	imes 8.3 	imes 373) $$

First, let's calculate the expansion work term:
$$ 	ext{Work} = 1 	imes 8.3 	imes 373 = 3095.9 	ext{ J mol}^{-1} $$

Now, subtract this work from the total enthalpy:
$$ \Delta E = 41000 - 3095.9 = 37904.1 	ext{ J mol}^{-1} $$

Finally, converting this back to kilojoules to match our options:
$$ \Delta E \approx 37.904 	ext{ kJ mol}^{-1} $$

### The Physical Significance

Why is $\Delta E$ less than $\Delta H$? The heat we supply ($\Delta H$) serves two distinct purposes. A major portion of it goes into increasing the internal energy ($\Delta E$), which physically means breaking the strong intermolecular hydrogen bonds holding the liquid water molecules together.

However, as the water turns into a gas, it expands massively. To create space for itself, the newly formed gas must push against the surrounding atmospheric pressure. This requires energy, which is the expansion work ($\Delta n_g RT$). Therefore, the actual increase in internal energy is always slightly less than the total heat supplied during vaporisation.

Select Answer:

Visualized Solution

$Chemical Equation$

$First Law of Thermodynamics$

$Substitution of Values$

$Calculating Expansion Work$

$Final Internal Energy Change$

$Physical Significance$

The Sigma Insight: First Law of Thermodynamics

Analyzing the Setup

The Master Equation

Final Calculation

The Physical Significance

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