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JEE Main & Advanced Chemistry Thermodynamics / रासायनिक उष्मागतिकी Question Bank 2nd and 3rd law of thermodynamics and Entropy

  • question_answer
    The entropy changed involved in the conversion of 1 mole of liquid water at 373 K to vapour at the same temperature will be \[[\Delta {{H}_{\text{vap}}}=2.257kJ/gm]\]                                         [MP PET 2002]

    A)                 0.119 kJ

    B)                 0.109 kJ

    C)                 0.129 kJ

    D)                 0.120 kJ

    Correct Answer: B

    Solution :

           \[{{H}_{2}}{{O}_{(l)}}\]⇌ \[{{H}_{2}}{{O}_{(g)}},\,\,\,\Delta S=\frac{\Delta {{H}_{vap}}}{T},\]            \[\Delta {{H}_{vap.}}=2.257\,KJ/g\]            or \[\Delta {{H}_{vap}}=2.257\times 18\,kJ/mol.\,=40.7\,KJ/mol\]                                 hence, \[\Delta S=\frac{40.7}{373}=0.109\,kJ/mol/K.\]


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