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

  • 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 \[{{100}^{o}}C=540\,cal/gm\]]       [Pb. PMT 2004]

    A)                 8.74       

    B)                 18.76

    C)                 24.06    

    D)                 26.06

    Correct Answer: D

    Solution :

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


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