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Chemistry · THERMODYNAMICS

हिन्दी

For the vaporization of \( 1.0\text{ mol} \) of water at \( 100^\circ\text{C} \) and \( 1\text{ bar} \), the enthalpy of vaporization \( \Delta_{\text{vap}}H \) is \( 40.66\text{ kJ}\cdot\text{mol}^{-1} \). Assuming water vapor behaves as an ideal gas and the molar volume of liquid water is negligible, the change in internal energy \( \Delta_{\text{vap}}U \) is: (Take \( R = 8.314\text{ J}\cdot\text{K}^{-1}\cdot\text{mol}^{-1} \))

  1. \( 43.76\text{ kJ}\cdot\text{mol}^{-1} \)
  2. \( 39.83\text{ kJ}\cdot\text{mol}^{-1} \)
  3. \( 37.56\text{ kJ}\cdot\text{mol}^{-1} \)
  4. \( 40.66\text{ kJ}\cdot\text{mol}^{-1} \)

Correct answer · Option 3

\( 37.56\text{ kJ}\cdot\text{mol}^{-1} \)

Explanation

\( \text{H}_2\text{O}(l) \rightarrow \text{H}_2\text{O}(g) \), so \( \Delta n_g = 1 \). \( \Delta U = \Delta H - \Delta n_g RT = 40660 - (1 \times 8.314 \times 373.15) = 40660 - 3102.4 = 37557.6\text{ J} \approx 37.56\text{ kJ}\cdot\text{mol}^{-1} \).
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