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NEET AIPMT SOLVED PAPER MAINS 2012

  • question_answer
    An ideal gas goes from state A to state B via three different processes  as  indicated in the p-V diagram If \[{{Q}_{1}},{{Q}_{2}},{{Q}_{3}}\] indicate the   heat absorbed by the gas along the three processes and \[\Delta {{U}_{1}},\Delta {{U}_{2}},\Delta {{U}_{3}}\]indicate the change in internal energy along the three processes respectively, then                                                          

    A) \[{{Q}_{1}}>{{Q}_{2}}>{{Q}_{3}}and\Delta {{U}_{1}}=\Delta {{U}_{2}}=\Delta {{U}_{3}}\]

    B) \[{{Q}_{3}}>{{Q}_{2}}>{{Q}_{1}}and\Delta {{U}_{1}}=\Delta {{U}_{2}}=\Delta {{U}_{3}}\]

    C) \[{{Q}_{1}}={{Q}_{2}}={{Q}_{3}}and\Delta {{U}_{1}}>\Delta {{U}_{2}}>\Delta {{U}_{3}}\]

    D) \[{{Q}_{3}}>{{Q}_{2}}>{{Q}_{1}}and\Delta {{U}_{1}}>\Delta {{U}_{2}}>\Delta {{U}_{3}}\]

    Correct Answer: A

    Solution :

    For all process 1, 2 and 3 \[\Delta U={{U}_{B}}-{{U}_{A}}\]is same                 \[\therefore \]  \[\Delta {{U}_{1}}=\Delta {{U}_{2}}=\Delta {{U}_{3}}\]                 Now,     \[\Delta Q=\Delta U+\Delta W\] Now, \[\Delta W=\] work done by the gas \[\therefore \]  \[\Delta {{Q}_{1}}=\Delta {{Q}_{2}}>\Delta {{Q}_{3}}\]


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