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NEET AIPMT SOLVED PAPER 2000

  • question_answer
                    A gas is formed of molecules each molecule possessing \[f\] degrees of freedom, then the value of \[\gamma =\frac{{{C}_{P}}}{{{C}_{V}}}\] is equal to:                                                                                                                                                       

    A)                 \[\frac{2}{f}\]                   

    B)                 \[1+\frac{2}{f}\]

    C)                 \[1+\frac{f}{2}\]              

    D)                 \[f+\frac{1}{2}\]

    Correct Answer: B

    Solution :

                                    According to law of equipartition of energy, the internal energy associated per degree of freedom is \[\frac{1}{2}\]= kT, where k is the Boltzmann's constant.                 Thus, internal energy associated per molecule                 \[=f\frac{1}{2}kT\]                 If \[{{N}_{A}}\] is Avagadro's number, then internal energy of one mole of an ideal gas is                 \[U={{N}_{A}}\,f\,\frac{1}{2}\,kT\]                 \[=\frac{1}{2}f\,({{N}_{A}}k)T\]                 \[=\frac{1}{2}\,f\,RT\]                 where \[R={{N}_{A}}k=\]gas constant                 \[{{C}_{V}}=\frac{dU}{dT}\]                 \[=\frac{d}{dT}\left( \frac{1}{2}fRT \right)\]                 \[=\frac{1}{2}fR\]                 Molar heat capacity at constant pressure                 \[{{C}_{p}}={{C}_{v}}+R(Mayor's\,relation)\]                 \[=\frac{1}{2}f\,R+R\]                 \[=\left( \frac{1}{2}f+1 \right)R\]                 \[Hence,\gamma =\frac{{{C}_{p}}}{{{C}_{v}}}\]                 \[=\frac{\left( \frac{1}{2}f+1 \right)R}{\frac{1}{2}f\,R}\]                 \[=\frac{\left( \frac{1}{2}f+1 \right)}{\frac{1}{2}f}=1+\frac{2}{f}\]


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