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JEE Main & Advanced Physics Thermodynamical Processes Question Bank Critical Thinking

  • question_answer
    \[1c{{m}^{3}}\] of water at its boiling point absorbs 540 calories of heat to become steam with a volume of \[1671c{{m}^{3}}\].If the atmospheric pressure = \[1.013x{{10}^{5}}N/{{m}^{2}}\] and the mechanical equivalent of heat = \[4.19J/calorie\], the energy spent in this process in overcoming intermolecular forces is [MP PET 1999, 2001; Orissa JEE 2002]

    A)            540 cal                                    

    B)            40 cal

    C)            500 cal                                    

    D)            Zero

    Correct Answer: C

    Solution :

                       \[\Delta Q=\Delta U+\Delta W\]                    \[\therefore \]\[\Delta U=\Delta Q-\Delta W=540-\frac{P({{V}_{2}}/{{V}_{1}})}{J}\]                    \[=540-\frac{1.013\times {{10}^{5}}\times [(1671-1)\times {{10}^{-6}}]}{4.2}\]            \[=540-39.7=500\ calories\]


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