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Step-by-Step Solution
Step 1: Write the Vaporization Process
The vaporization of liquid water at 100 °C can be represented as:
$$
\text{H}_2\text{O}(l) \xrightarrow{100^\circ\text{C}} \text{H}_2\text{O}(g)
$$
Step 2: Note the Given Data
Enthalpy of vaporization, $ \Delta H^\circ_\text{vap} = 40.66 \text{ kJ mol}^{-1} $
Gas constant, $ R = 8.314 \times 10^{-3} \text{ kJ mol}^{-1} \text{K}^{-1} $
Temperature, $ T = 100^\circ\text{C} = 373 \text{ K} $
Mole change in gas, $ \Delta n_g = 1 $ (since one mole of liquid water forms one mole of water vapor)
Step 3: Use the Relation Between Enthalpy and Internal Energy
The relationship between enthalpy change ($ \Delta H^\circ $) and internal energy change ($ \Delta U^\circ $) is given by:
$$
\Delta H^\circ = \Delta U^\circ + \Delta n_g \, R \, T
$$
Step 4: Substitute the Values
Substitute the known values into the formula:
$$
40.66 = \Delta U^\circ + (1) \times \bigl(8.314 \times 10^{-3}\bigr) \times 373
$$
Step 5: Calculate the Internal Energy Change
First, compute the product $ \Delta n_g \, R \, T $:
$$
(1) \times (8.314 \times 10^{-3}) \times 373 \approx 3.10 \text{ kJ mol}^{-1}
$$
Then, solve for $ \Delta U^\circ $:
$$
\Delta U^\circ = 40.66 - 3.10 = 37.56 \text{ kJ mol}^{-1}
$$
Final Answer
The internal energy of vaporization of water at 100 °C is
$$
+37.56 \text{ kJ mol}^{-1}.
$$
