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

  • question_answer
    Following data is known about melting of a compound AB. \[\Delta H=9.2\,kJ\,mo{{l}^{-1}}\], \[\Delta S=0.008\,kJ\,{{K}^{-1}}mo{{l}^{-1}}\]. Its melting point is               [Pb. PMT 2000; AIIMS 2000]

    A)                 736 K    

    B)                 1050 K

    C)                 \[1150\,K\]        

    D)                 \[{{1150}^{o}}C\]

    Correct Answer: C

    Solution :

              \[{{T}_{m}}=\frac{\Delta {{H}_{\text{fusion}}}}{\Delta {{S}_{\text{fusion}}}}=\frac{9.2}{0.008}=1150\,K\].


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