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JEE Main & Advanced Physics Thermodynamical Processes Question Bank Isothermal Process

  • question_answer
    When 1 gm of water at \[{{0}^{o}}C\] and \[1\times {{10}^{5}}\ N/{{m}^{2}}\] pressure is converted into ice of volume \[1.091\ c{{m}^{2}}\], the external work done will be

    A)            0.0091 joule                          

    B)            0.0182 joule

    C)            ? 0.0091 joule                      

    D)            ? 0.0182 joule

    Correct Answer: A

    Solution :

                       It is an isothermal process. Hence work done \[=P({{V}_{2}}-{{V}_{1}})\]            \[=1\times {{10}^{5}}\times (1.091-1)\times {{10}^{-6}}=0.0091\ J\]


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