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Haryana PMT Haryana PMT Solved Paper-2011

  • question_answer
    Temperature of one mole of an ideal gas contained in a closed container is increased from \[300K\]to\[301K\]. Pressure-volume work obtained is

    A)  \[8.314J\]

    B)  \[2.494\text{ }kJ\]

    C)  \[0.8314\text{ }J\]

    D)  \[83.14\text{ }J\]

    Correct Answer: A

    Solution :

                    \[PV=RT\]      at temperature \[TK\] \[p(V.+\Delta V)=R(T+1)\] at temperature \[(T+1)K\]              \[\therefore \]    \[p.\Delta V=R\]                            \[(pV=RT)\] \[\therefore \] Mechanical work (pressure-volume work)                 \[=p.\Delta V\]                 \[p.\Delta V=R=8.314J\]


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