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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2005

  • question_answer
    Molar heat of vaporisation of a liquid is\[6\,kJ\,mo{{l}^{-1}}\]. If die entropy change is \[16\,J\,mo{{l}^{-1}}{{K}^{-1}}\], the boiling point of the liquid is:

    A)  \[{{375}^{o}}C\]                              

    B)  375K

    C)  273 K                   

    D)  \[{{102}^{o}}C\]

    Correct Answer: B

    Solution :

    \[\Delta S=16\,J\,\,mo{{l}^{-1}}{{K}^{-1}},\Delta {{H}_{\upsilon }}=6\,kJ\,mo{{l}^{-1}}\] \[{{T}_{b.p.}}=\frac{\Delta {{H}_{vapour}}}{\Delta {{S}_{vapour}}}=\frac{6\times 1000}{16}=375\,K\]


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