AIIMS AIIMS Solved Paper-2002

  • question_answer
    The volume of a gas is reduced adiabatically to (1/4) of its volume at 27°C. If y = 1.4. The new temperature will be:

    A) \[{{k}_{2}}\]      

    B)       \[{{y}_{2}}\]      

    C)       \[y={{y}_{1}}+{{y}_{2}}\]  

    D)                        none of these

    Correct Answer: A

    Solution :

    For adiabatic change the relation between temperature and volume is \[F=-\frac{{{k}_{1}}{{k}_{2}}}{{{k}_{1}}+{{k}_{2}}}y\] where y is ratio of specific heats of the gas. Given, \[{{T}_{1}}=27\,+273\,=300K,\,\,{{V}_{1}}=V,\,{{V}_{2}}=\frac{V}{4}\] \[n=\frac{1}{2\pi }\sqrt{\frac{{{k}_{1}}{{k}_{2}}}{({{k}_{1}}+{{k}_{2}})m}}\] \[\frac{1}{k}=\frac{1}{{{k}_{1}}}+\frac{1}{{{k}_{2}}}\]   \[k=\frac{{{k}_{1}}{{k}_{2}}}{{{k}_{1}}+{{k}_{2}}}\] \[n=\frac{1}{2\pi }\sqrt{\frac{k}{m}}\] \[=\frac{1}{2\pi }\sqrt{\frac{{{k}_{1}}{{k}_{2}}}{({{k}_{1}}+{{k}_{2}})m}}\]


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