Adiabatic Index Of A Mixture

For a mixture, the effective molar specific heats are number-weighted averages of the components' values, because internal energy is additive while the total number of moles fixes the per-mole capacity. With the monatomic gas $C_{v1}=\frac{3}{2}R$ and the diatomic gas $C_{v2}=\frac{5}{2}R$, the mixture's $C_v$ is $C_v=\frac{n_1{C}_{v1}+n_2{C}_{v2}}{n_1+n_2}=\frac{2(\frac{3}{2}R)+3(\frac{5}{2}R)}{5}=\frac{(3+7.5)R}{5}=2.1R$. Then $C_p=C_v+R=3.1R$, because Mayer's relation still holds for the mixture as a whole, so $\gamma =\frac{C_p}{C_v}=\frac{3.1}{2.1}\approx 1.48$. The value 1.40 ignores the monatomic component and treats the mixture as purely diatomic. The value 1.67 treats it as purely monatomic. The value 1.50 mishandles the mole weighting. This uses the NCERT result that specific heats of an ideal-gas mixture combine by mole fraction, an idea frequently tested in JEE Advanced thermodynamics problems. An equivalent and often quicker route is to compute the mixture's total degrees of freedom per mole, $f=(2\times 3+3\times 5)/5=4.2$, and then apply $\gamma =1+2/f=1+2/4.2\approx 1.48$, which reproduces the same answer. As a plausibility check, the answer 1.48 lies between the monatomic limit 1.67 and the diatomic limit 1.40, and sits closer to the diatomic value because diatomic moles outnumber monatomic ones, confirming the calculation is self-consistent and physically reasonable for such a blended gas.