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WB JEE Medical WB JEE Medical Solved Paper-2006

  • question_answer
    One mole of an ideal gas at an initial temperature of T K does 6R joules of work adiabatically. If the ratio of specific heats of this gas at constant pressure and at constant volume is 5/3, then final temperature of the gas will be :

    A)  (T - 4) K                               

    B)  (T + 4) K

    C)  (T - 2.4) K           

    D)  (T + 2.4) K

    Correct Answer: A

    Solution :

                     Work done by gas in adiabatic process \[W=\frac{nR({{T}_{i}}-{{T}_{f}})}{\gamma -1}\]                 \[\therefore \]  \[6R=\frac{1\times R(T-{{T}_{2}})}{{{(5/3)}^{-1}}}\] or                            \[6=\frac{T-{{T}_{2}}}{2/3}\] or            \[T-{{T}_{2}}=4\]                 \[{{T}_{2}}=(T-4)k\]


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