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Question

(B) The ratio of specific heats $\gamma$ is defined as $\gamma = \frac{C_p}{C_v}$.For an ideal gas,the molar specific heat at constant volume is $C_v

Vedclass · Accepted Answer

$\frac{2}{\gamma-1}$

Vedclass · Answer

(B) The ratio of specific heats $\gamma$ is defined as $\gamma = \frac{C_p}{C_v}$.For an ideal gas,the molar specific heat at constant volume is $C_v = \frac{f}{2}R$,where $f$ is the number of degrees of freedom.The molar specific heat at constant pressure is $C_p = C_v + R = (\frac{f}{2} + 1)R$.Therefore,$\gamma = \frac{C_p}{C_v} = \frac{(\frac{f}{2} + 1)R}{\frac{f}{2}R} = \frac{\frac{f+2}{2}}{\frac{f}{2}} = \frac{f+2}{f} = 1 + \frac{2}{f}$.Rearranging this equation to solve for $f$:$\gamma - 1 = \frac{2}{f}$.$f = \frac{2}{\gamma - 1}$.Thus,the correct option is $B$.

If the ratio of specific heats of a gas at constant pressure and at constant volume is $\gamma$,then the number of degrees of freedom of the rigid molecules of the gas is

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