Question

The ratio C_p / C_v for a gas is 1.29. What is the degree of freedom of the molecules of this gas?

Answer 1

For the molecules of a gas, $\gamma =\frac{C_{\mathrm{p}}}{C_{\mathrm{v}}}=1+\frac{2}{f}$,
where f is the degree of freedom.
$$\begin{gathered} \mathrm{Given}:\gamma =1.29 \\ \Rightarrow 1+\frac{2}{f}=1.29=\frac{9}{7} \\ \Rightarrow \frac{2}{f}=\frac{9}{7}-1 \\ \Rightarrow \frac{2}{f}=\frac{2}{7} \\ \Rightarrow f=7 \end{gathered}$$
Therefore, the molecules of this gas have 7 degrees of freedom.
But in reality, no gas can have more than 6 degrees of freedom.