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JIPMER Jipmer Medical Solved Paper-2002

  • question_answer
    During an experiment, an ideal gas is found to obey an additional law \[V{{P}^{2}}=\] constant. The gas is initially at a temperature T, and volume V. When it expands to a volume 2V, the temperature becomes:

    A)  \[\sqrt{2}T\]                                     

    B)  T                            

    C)  4T                         

    D)         2T                         

    Correct Answer: A

    Solution :

    For an ideal gas, the relation is \[PV=RT\]                           ?(i)                 Given    \[V{{P}^{2}}=K\]                               ?(ii) Squaring equation (i) we get \[{{P}^{2}}{{V}^{2}}={{R}^{2}}{{T}^{2}}\]                               ?(iii) Now dividing equation (ii) by (iii) \[\frac{1}{V}=\frac{K}{{{R}^{2}}{{T}^{2}}}\] If volume V expands to volume 2V so           \[{{T}^{2}}\propto V\] Hence   \[\frac{T_{1}^{2}}{T_{2}^{2}}=\frac{V}{2V}=\frac{1}{2}\] so,          \[{{T}_{2}}=\sqrt{2}\,{{T}_{1}}\]


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