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The correct relation between the degrees of freedom $$f$$ and the ratio of specific heat $$\gamma$$ is:
For an ideal gas with $$f$$ degrees of freedom, the molar specific heat at constant volume is $$C_v = \frac{f}{2}R$$ and at constant pressure is $$C_p = C_v + R = \frac{f}{2}R + R = \frac{f+2}{2}R$$.
The ratio of specific heats is $$\gamma = \frac{C_p}{C_v} = \frac{(f+2)/2}{f/2} = \frac{f+2}{f} = 1 + \frac{2}{f}$$.
Solving for $$f$$: $$\gamma - 1 = \frac{2}{f}$$, so $$f = \frac{2}{\gamma - 1}$$.
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