NEET Sample Paper NEET Sample Test Paper-69

  • question_answer
    In an adiabatic change, the pressure and temperature of a monoatomic gas are related with relation as \[P\propto {{T}^{C}}\], where C is equal to    

    A) \[\frac{5}{4}\]                          

    B) \[\frac{5}{3}\]

    C) \[\frac{5}{2}\]                          

    D) \[\frac{3}{5}\] 

    Correct Answer: C

    Solution :

    Poisson?s equation for adiabatic process is given by \[{{\operatorname{PV}}^{\gamma }}=constant\]                                               For adiabatic process, Poisson?s equation is given by \[{{\operatorname{PV}}^{\gamma }}=constant\]                                               [a] Ideal gas relation is \[PV\,\,=\,\,\,\,RT\] \[\Rightarrow \,\,\,\,\,V=\frac{RT}{P}\]                                         [b] From Eqs. [a] and [2], we get \[P{{\left( \frac{RT}{P} \right)}^{\gamma }}=cons\tan t\] \[\Rightarrow \,\,\,\,\,\frac{T}{{{P}^{\gamma -1}}}=cons\tan t\]                           [c] where is the ratio of specific heats of the gas. Given, \[P\,\,\propto \,\,{{T}^{C}}\]                                           [d] On comparing with Eq. [c], we have \[C=\frac{\gamma }{\gamma -1}\] For a monoatomic gas \[\gamma =\frac{5}{3}\] \[\therefore \] We have \[C=\frac{\frac{5}{3}}{\frac{5}{3}-1}=\frac{5}{2}\,\,\,\,\,\,\,\,\,\]


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