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Chhattisgarh PMT Chhattisgarh PMT Solved Paper-2006

  • question_answer
    \[1c{{m}^{3}}\] of water at its boiling point absorbs 540 cal of heat to become steam with a volume of \[1671\text{ }c{{m}^{3}}.\] If the atmospheric pressure \[=1.013\times {{10}^{5}}N/{{m}^{2}}\]  and  the  mechanical equivalent of heat = 4.19 J/cal, the energy spent in this process in overcoming intermolecular forces is

    A)  540 cal                                 

    B)  40cal

    C)  500 cal                                 

    D)  zero

    Correct Answer: C

    Solution :

    Change in volume \[=1671-1=1670c{{m}^{3}}\] \[\therefore \]  Work done \[(W)=p.dV\]                 \[=(1.013\times {{10}^{5}}\times 1670)J\]                 \[=\frac{1670\times 1.013\times {{10}^{5}}}{4.2}cal\]                 \[=39.7cal\] Heat given \[(Q)=540cal.\] From first law of thermodynamics,                 \[\Delta U=Q-W\]                 \[=540-39.7\]                 \[=500.3cal\]                 \[\approx 500cal\]


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