If one mole of a monoatomic gas () is mixed with one mole of a diatomic gas (), the value of for the — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

$C_{v(mix)} = \frac{n_1 C_{v1} + n_2 C_{v2}}{n_1 + n_2}$. Monoatomic $C_{v1} = \frac{R}{5/3-1} = 1.5R$. Diatomic $C_{v2} = \frac{R}{7/5-1} = 2.5R$. $C_{v(mix)} = \frac{1(1.5R) + 1(2.5R)}{2} = 2R$. The mixture's $C_{p(mix)} = C_{v(mix)} + R = 3R$. The adiabatic index $\gamma_{mix} = \frac{C_{p(mix)}}{C_{v(mix)}} = \frac{3R}{2R} = 1.5$.