Q.If water vapour is assumed to be a perfect gas, molar enthalpy change for vapourisation of 1 mol of water at 1 bar and 100 °C is 41 kJ mol$^{-1}$. Calculate the internal energy change, when 1 mol of water is vapourised at 1 bar pressure and 100 °C.

The key idea is that for a phase change at constant pressure, the enthalpy change $\Delta H$ and internal energy change $\Delta U$ are related by $\Delta H = \Delta U + \Delta (PV)$. For a perfect gas, $\Delta (PV) = \Delta n_g RT$, where $\Delta n_g$ is the change in moles of gas.

Step 1: Write the relation for constant pressure and temperature:

Since the volume of liquid water is negligible compared to vapour, $\Delta V \approx V_{\text{vapour}}$.

Step 2: Treating water vapour as a perfect gas, $P \Delta V = \Delta n_g RT$. Here, 1 mol of liquid gives 1 mol of vapour, so $\Delta n_g = 1$. …

The key idea is that when a liquid vaporises at constant pressure, the enthalpy change $\Delta H$ includes both the energy needed to overcome intermolecular forces (the internal energy change $\Delta U$) and the work done by the system as it expands against the external pressure. For a perfect gas, that expansion work is $P\Delta V$, and since the volume of liquid is negligible compared to vapour, it simplifies to $RT$ per mole of gas produced.

Let’s walk through it carefully.

1. Recall the fundamental relation between $\Delta H$ and $\Delta U$ Enthalpy is defined as $H = U + PV$. For a change at constant pressure,

Here $P\Delta V$ is the work done by the system (pressure–volume work) during the expansion. So $\Delta H$ always exceeds $\Delta U$ when the system expands ($\Delta V > 0$).

2. What is $\Delta V$ for vaporisation? We start with 1 mol of liquid water at 100 °C and 1 bar. Its volume is tiny — about 18 mL (since density ≈ 1 g/mL). After vaporisation, we get 1 mol of steam. Assuming steam behaves as a perfect gas (the problem says so), its volume is

So $\Delta V = V_\text{vapour} - V_\text{liquid} \approx 31.0\ \text{L} - 0.018\ \text{L} \approx 31.0\ \text{L}$. The liquid volume is negligible — a common and safe approximation.

3. Compute $P\Delta V$ in consistent units $P = 1\ \text{bar}$, $\Delta V \approx 31.0\ \text{L}$.

Convert to kJ: $1\ \text{L bar} = 100\ \text{J}$, so