The correct relation between the degree of fr | vedclass

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(B) The ratio of specific heats $\gamma$ is defined as the ratio of molar specific heat at constant pressure $C_p$ to molar specific heat at constant

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) The ratio of specific heats $\gamma$ is defined as the ratio of molar specific heat at constant pressure $C_p$ to molar specific heat at constant volume $C_v$,i.e.,$\gamma = \frac{C_p}{C_v}$.According to the equipartition theorem,$C_v = \frac{f}{2}R$ and $C_p = C_v + R = (\frac{f}{2} + 1)R$.Therefore,$\gamma = \frac{(\frac{f}{2} + 1)R}{\frac{f}{2}R} = \frac{\frac{f+2}{2}}{\frac{f}{2}} = 1 + \frac{2}{f}$.Rearranging the equation: $\gamma - 1 = \frac{2}{f}$.Solving for $f$,we get $f = \frac{2}{\gamma - 1}$.

The correct relation between the degree of freedom $f$ and the ratio of specific heat $\gamma$ is

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