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Punjab Medical Punjab - MET Solved Paper-2004

  • question_answer
    One mole of water at \[{{100}^{o}}C\] is converted into steam at \[{{100}^{o}}C\] at a constant pressure of \[1\,\,atm\]. The change in entropy is [heat of vaporisation of water at\[{{100}^{o}}C=540cal/gm\]]:

    A) \[8.74\]               

    B) \[18.76\]

    C) \[24.06\]                             

    D) \[26.06\]

    Correct Answer: D

    Solution :

    The entropy change                 \[=\frac{heat\,\,of\,\,vaporisation}{temperature}\] Here, heat of vaporisation\[=540cal/gm\] \[=540\times 18\,\,cal\,\,mo{{l}^{-1}}\] Temperature of water\[=100+273=373K\] \[\therefore \]entropy change\[=\frac{540\times 18}{373}\] \[=26.06\,\,cal\,\,mo{{l}^{-1}}{{\text{K}}^{-1}}\]


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