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KVPY Sample Paper KVPY Stream-SX Model Paper-2

  • question_answer
    Work done for the conversion of 0.5 mole of water at \[100{}^\circ \]C to steam at 1 atm pressure is (heat of vaporision of water\[at\,100{}^\circ C\text{ }is\text{ }40670\text{ }J\text{ }mo{{l}^{-1}}~~\])

    A) \[-1.54\,kJ\]                   

    B) \[1.54\,kJ\]

    C) \[1.25\,kJ\]                    

    D) \[-1.35\,kJ\]

    Correct Answer: A

    Solution :

    [a]
    volume of 0.5 mole of steam at I atm pressure \[=\frac{nRT}{P}=\frac{0.5\times 0.0821\times 373}{1.0}=15.3L\]
    Change in volume = vol. of steam - vol. of water =\[15..3-negligible=15.3L\]  
    Work done by the system,
    \[w=P\times \]volume change
    \[=\overset{ext}{\mathop{1}}\,\times 15.3=15.3\]litre-atm
    =\[15.3\times 101.3J=1549.89J\]
    ?w? should be negative as the work has been done by the system on the surroundings.
    = -1549.89 J = -1.54 kJ


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