AFMC AFMC Solved Paper-2001

  • question_answer
    The change in entropy for the fusion of 1 mole of ice is [melting point of ice = 273 K, molar enthalpy of fusion for ice\[=6.0\,kJ\,mo{{l}^{-1}}\]]:

    A) \[11.73\,J{{K}^{-1}}\,mo{{l}^{-1}}\]         

    B) \[18.84\,J{{K}^{-1}}mo{{l}^{-1}}\]

    C) \[21.97J{{K}^{-1}}mo{{l}^{-1}}\]

    D) \[24.47\,J{{K}^{-1}}\,mo{{l}^{-1}}\]

    Correct Answer: C

    Solution :

    Entropy change of fusion \[\Delta {{S}_{f}}\] \[\Delta {{H}_{f}}=6.0\times {{10}^{3}}J,T=273\,K\] \[\Delta {{S}_{f}}=\frac{6000}{273}=21.97J/K\,mol\]


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