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:                                                            [AIPMT 2000]

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,  \[\gamma =\frac{{{C}_{p}}}{{{C}_{v}}}\]

\[=\frac{\left( \frac{1}{2}f+1 \right)R}{\frac{1}{2}f\,R}\]

\[=\frac{\left( \frac{1}{2}f+1 \right)}{\frac{1}{2}f}=1+\frac{2}{f}\]