Q 11-12-018NEETJEE MainMedium
One mole of a monatomic gas is mixed with one mole of a rigid diatomic gas. The ratio $\gamma = C_p/C_v$ for the mixture is
Answer: (D) $1.5$
Concept. For a mixture, $C_v$ is the mole-weighted average; $C_p = C_v + R$.
Step 1. $C_v$. $\dfrac{\frac32R + \frac52R}{2} = 2R$.
Step 2. $\gamma$. $C_p = 3R$, so $\gamma = \dfrac{3R}{2R} = 1.5$.
Answer. $1.5$ (Option D).
Solution by Sreeraj P, M.Sc Physics