WB JEE Medical WB JEE Medical Solved Paper-2014

  • question_answer
    The enthalpy of vaporisation of a certain liquid at its boiling point of \[35{{\,}^{o}}C\]is \[24.64\,kJ\,mo{{l}^{-1}}.\] The value of change in entropy for the process is

    A)  \[704\,J{{K}^{-1}}\,mo{{l}^{-1}}\]

    B)  \[80\,J{{K}^{-1}}\,mo{{l}^{-1}}\]

    C)  \[24.64\,J{{K}^{-1}}mo{{l}^{-1}}\]

    D)  \[7.04\,J{{K}^{-1}}\,mo{{l}^{-1}}\]

    Correct Answer: B

    Solution :

     Entropy of vaporisation \[(\Delta S)\] \[\text{=}\frac{\text{enthalopyof}\,\text{vaporisation}\,\text{( }\!\!\Delta\!\!\text{  }\!\!\Eta\!\!\text{ )}}{\text{boiling}\,\text{point}\,\text{(in}\,\text{K)}}\] Given, AH = 24.64 kJ mol1 and\[\Delta H=24.64\,\text{KJ}\,\text{mo}{{\text{l}}^{-1}}\]and  boiling point = 35 + 273 = 308 K \[\therefore \] \[\Delta S=\frac{24.64\times {{10}^{3}}\,J\,mo{{l}^{-1}}}{308\,K}\] \[=80\,\text{J}{{\text{K}}^{-1}}\,\text{mo}{{\text{l}}^{-1}}\]


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