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Manipal Medical Manipal Medical Solved Paper-2007

  • question_answer
    The work done during the expansion of a gas from a volume of\[\text{4 }d{{m}^{3}}\]to \[6\text{ }d{{m}^{3}}\]against a constant external pressure of 3 atm, is

    A)  \[-6\text{ }J\]          

    B)  \[-608\text{ }J\]

    C)  \[+304\text{ }J\]        

    D)  \[-304\,J\]

    Correct Answer: B

    Solution :

     Work done\[(W)=-{{P}_{ext}}({{V}_{2}}-{{V}_{1}})\] \[=-3\times (6-4)=-6L-atm\] \[=-6\times 101.32\text{ }J\]          (\[\therefore \] 1 L-atm\[=101.32\text{ }J\]) \[=-607.92\approx 608\text{ }J\]


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