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Manipal Medical Manipal Medical Solved Paper-2000

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

    A)  8.74               

    B)  18.76

    C)  24.06              

    D)  26.06

    Correct Answer: D

    Solution :

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


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