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Chemistry · THERMODYNAMICS
हिन्दीOne mole of liquid water is vaporised at 1 bar pressure and \( 100^\circ\text{C} \). Given that the enthalpy of vaporisation of water is \( 41.00\text{ kJ mol}^{-1} \) and \( R = 8.3\text{ J mol}^{-1}\text{ K}^{-1} \), the change in internal energy (\( \Delta U \)) for the vaporisation of 2.0 moles of water under the same conditions is:
Correct answer · Option 2
75.81 kJ
Explanation
For 2 moles, \( \Delta n_g = 2 \). \( \Delta H = 2 \times 41.00 = 82.00\text{ kJ} \), and \( \Delta n_g RT = 2 \times 8.3 \times 373 \times 10^{-3} = 6.192\text{ kJ} \). Thus, \( \Delta U = \Delta H - \Delta n_g RT = 82.00 - 6.192 = 75.808\text{ kJ} \approx 75.81\text{ kJ} \).
