Q 11-12-057JEE MainJEE Main 2024 (30 Jan, Shift 2)Medium
If three moles of a monoatomic gas $\left(\gamma = \frac53\right)$ are mixed with two moles of a diatomic gas $\left(\gamma = \frac75\right)$, the value of the adiabatic exponent $\gamma$ for the mixture is:
Answer: (C) $1.52$
$$C_V = \frac{3\times\frac32R + 2\times\frac52R}{5} = \frac{9.5R}{5} = 1.9R,\qquad C_P = 2.9R$$
$$\gamma = \frac{2.9}{1.9} \approx 1.52$$
Solution by Sreeraj P, M.Sc Physics