Given: Moles of monoatomic gas \(=\alpha\), moles of diatomic gas is \(\beta\).

Solution:

The mixture is behaving as diatomic gas. If we neglect the vibrational mode of freedom then degree of freedom ( \(\left.f_{\text {mix }}\right)\) of mixture is \(3\) (for translational) \(+2\) (for rotational \(=5\).

We know that,

\(f _{\text {mix }}=\frac{ f _1 n _1+ f _2 n _2}{ n _1+ n _2}\)

\(\Rightarrow 5=\frac{3 \alpha+6 \beta}{\alpha+\beta}\)

\(\Rightarrow 5 \alpha+5 \beta=3 \alpha+6 \beta\)

\(\therefore 2 \alpha=\beta\)