NEET AIPMT SOLVED PAPER MAINS 2010

  • question_answer
    For vaporisation of water at 1 atm pressure, the values of \[\Delta H\]and \[\Delta S\] are40.63\[\text{kJ}\,\text{mo}{{\text{l}}^{\text{-1}}}\]and 108.8\[\text{J}{{\text{K}}^{\text{-1}}}\,\text{mo}{{\text{l}}^{\text{-1}}}\], respectively. The temperature when Gibbs energy change \[(\Delta G)\] for this transformation will be zero, is                                                                           

    A)  273.4 K                

    B)  393.4 K                

    C)  373.4 K                

    D)  293.4 K              

    Correct Answer: C

    Solution :

    \[\Delta G=\Delta H-T\Delta S\] \[\Delta G=0\](given) \[\Delta H=T\Delta S,\] \[T=\frac{40.63\times {{10}^{3}}}{108.8}=373.4K\]


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