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) $1.5$
$C_v$ of the mixture $= \dfrac{1 \times \frac{3}{2}R + 1 \times \frac{5}{2}R}{2} = 2R$.
$C_p = C_v + R = 3R$, so $\gamma = \dfrac{3R}{2R} = 1.5$.
Solution by Sreeraj P, M.Sc Physics