JEE Main & Advanced JEE Main Paper (Held on 09-4-2019 Morning)

  • question_answer Following figure shows two processes A and B for a gas. If \[\Delta {{Q}_{A}}\]and \[\Delta {{Q}_{B}}\]are the amount of heat absorbed by the system in two cases, and \[\Delta {{U}_{A}}\]and \[\Delta {{U}_{B}}\]are changes in internal energies, respectively, then : [JEE Main 9-4-2019 Morning]

    A) \[\Delta {{Q}_{A}}=\Delta {{Q}_{B}};\Delta {{U}_{A}}=\Delta {{U}_{B}}\]

    B) \[\Delta {{Q}_{A}}>\Delta {{Q}_{B}};\Delta {{U}_{A}}=\Delta {{U}_{B}}\]

    C) \[\Delta {{Q}_{A}}>\Delta {{Q}_{B}};\Delta {{U}_{A}}>\Delta {{U}_{B}}\]

    D) \[\Delta {{Q}_{A}}<\Delta {{Q}_{B}};\Delta {{U}_{A}}<\Delta {{U}_{B}}\]

    Correct Answer: B

    Solution :

    Initial and final states for both the processes are same. \[\therefore \]\[\Delta {{U}_{A}}=\Delta {{U}_{B}}\] Work done during process A is greater than in process B. By First Law of thermodynamics \[\Delta Q=\Delta U+W\]\[\Rightarrow \]\[\Delta {{Q}_{A}}>\Delta {{Q}_{B}}\]


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