1. Given data

• Mass of water $= 90\,\text{g}$
• Molar mass of water $= 18\,\text{g mol}^{-1}$
• Therefore, moles of water: $n = \frac{90}{18} = 5\,\text{mol}$
• Enthalpy of vaporization at $373\,\text{K}$: $\Delta H_{\text{vap}} = 41\,\text{kJ mol}^{-1} = 41000\,\text{J mol}^{-1}$
• Gas constant: $R = 8.314\,\text{J K}^{-1}\text{mol}^{-1}$
• Temperature: $T = 373\,\text{K}$

2. Relation between enthalpy change and internal energy change

For vaporization, $\Delta H = \Delta U + \Delta n_g RT$

Hence, $\Delta U = \Delta H - \Delta n_g RT$

For 1 mole of water evaporating:

• Liquid water produces 1 mole of gaseous water.
• So, $\Delta n_g = 1$

Thus, per mole: $\Delta U_{\text{vap}} = 41000 - (8.314)(373)$

3. Calculate $RT$

$RT = 8.314 \times 373 = 3101.122\,\text{J mol}^{-1}$

So, $\Delta U_{\text{vap}} = 41000 - 3101.122 = 37898.878\,\text{J mol}^{-1}$

4. For 5 moles of water

$\Delta U = 5 \times 37898.878$ $\Delta U = 189494.39\,\text{J}$

5. Final integer answer

$\boxed{189494\,\text{J}}$

6. Comparison with stored correct answer

Stored correct answer: $189494$ to $189495$

Our calculated answer $189494.39\,\text{J}$ lies in this range, so it agrees with the stored answer.