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RAJASTHAN PMT Rajasthan - PMT Solved Paper-2009

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

    A)  540 cal                                 

    B)  40 cal

    C)  500 cal                 

    D)         zero

    Correct Answer: C

    Solution :

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


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