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Determine γ from degree of freedom.

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Suppose a polyatomic gas molecule has n degree of freedom.

Total energy associated with one gram molecule of the gas, i.e.,

E = n × \(\frac{1}{2}\)RT × 1

=\(\frac{n}{2}\)RT

As,

Cv \(\frac{d}{dT}\)

= \(\frac{d}{dT}\)(\(\frac{n}{2}\)RT)

= \(\frac{n}{2}\)R

Cp = Cv + R

Cp = \(\frac{n}{2}\)R + R

= (\(\frac{n}{2}\)+1)R

γ= \(\frac{C_p}{C_V}\)

γ= \(\frac{(\frac{n}{2}+1)R}{\frac{n}{2}R}\)

∴ = \(\frac{2}{n}\)(\(\frac{n}{2}\)+1)

γ= 1+\(\frac{2}{n}\)

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