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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 15

  • question_answer
    The equation of state for a gas is given by \[PV=nRT+\alpha V,\] where n is the number of moles and \[\alpha \] is a positive constant. The initial temperature and pressure of one mole of the gas contained in a cylinder are \[{{T}_{0}}\] and , respectively. \[{{P}_{0}}\]The work done by the gas when its temperature doubles isobarically will be

    A) \[{{P}_{0}}{{T}_{0}}R\,\,In\,\,2\]         

    B) \[\frac{{{P}_{0}}{{T}_{0}}R}{{{P}_{0}}-\alpha }\]

    C) \[{{P}_{0}}{{T}_{0}}R\]         

    D) \[\frac{{{P}_{0}}{{T}_{0}}R}{{{P}_{0}}+\alpha }\]

    Correct Answer: B

    Solution :

    [b] \[PV=nRT+\alpha V\] \[\Rightarrow \,\,\,\,\left( P-\alpha  \right)V=nRT\] Initially \[({{P}_{0}}-\alpha )V=nR{{T}_{0}}\]                    ?.(1) Finally \[({{P}_{0}}-\alpha )V=nR(2{{T}_{0}})\]                 ?.(2) Comparing eqs (1) and (2), \[V=2{{V}_{0}}\] Now, \[W=\int\limits_{{{V}_{0}}}^{2{{V}_{\theta }}}{{{P}_{0}}dV}={{P}_{0}}\int\limits_{{{V}_{0}}}^{2{{V}_{0}}}{dV={{P}_{0}}\left[ 2{{V}_{0}}-V \right]}\] \[W={{P}_{0}}{{V}_{0}}={{P}_{0}}\left( \frac{nR{{T}_{0}}}{{{P}_{0}}-\alpha } \right)\] Putting \[n=1,\] \[W=\frac{{{P}_{0}}{{T}_{0}}R}{{{P}_{0}}-\alpha }\]


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