JEE Main & Advanced Physics Thermodynamical Processes Question Bank Self Evaluation Test - Thermodynamics

  • question_answer
    One mole of an ideal gas is taken from state A to state B by three different processes, (i) ACB (ii) ADB (iii) AEB as shown in the P-V diagram. The heat absorbed by the gas is-  

    A) greater in process (ii) than in (i)

    B) the least in process (ii)

    C) the same in (i) and (iii)

    D) less in (iii) than in (ii)

    Correct Answer: D

    Solution :

    [d] Heat absorbed by gas in three processes is given by \[{{Q}_{ACB}}=\Delta U+{{W}_{ACB}}\] \[{{Q}_{ACB}}=\Delta U\] \[{{Q}_{ACB}}=\Delta U+{{W}_{AEB}}\] The change in internal energy in all the three cases is same and \[{{W}_{ACB}}\] is \[+ve,\] \[{{W}_{AEB}}\] is\[-ve\]. Hence \[{{Q}_{ACB}}>{{Q}_{ADB}}>{{Q}_{AEB}}\]


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