The ratio of the adiabatic exponent $\gamma$ for a mixture of 1 mole of a monatomic gas and 2 moles of a diatomic gas is

Solution

1. For a monatomic gas, $C_{v1}=\frac{3}{2}R$ and $C_{p1}=\frac{5}{2}R$; for a diatomic gas, $C_{v2}=\frac{5}{2}R$ and $C_{p2}=\frac{7}{2}R$.
2. The total heat capacity at constant volume for the mixture is $$\begin{array}{rl}{C}_{v,mix}& =\frac{n_1{C}_{v1}+n_2{C}_{v2}}{n_1+n_2}\\ & =\frac{1(\frac{3}{2}R)+2(\frac{5}{2}R)}{1+2}\\ & =\frac{13R}{6}\end{array}$$.
3. The total heat capacity at constant pressure for the mixture is $$\begin{array}{rl}{C}_{p,mix}& =\frac{n_1{C}_{p1}+n_2{C}_{p2}}{n_1+n_2}\\ & =\frac{1(\frac{5}{2}R)+2(\frac{7}{2}R)}{1+2}\\ & =\frac{19R}{6}\end{array}$$.
4. The adiabatic exponent is $$\begin{array}{rl}{\gamma }_{mix}& =\frac{C_{p,mix}}{C_{v,mix}}\\ & =\frac{19R/6}{13R/6}\\ & =\frac{19}{13}\end{array}$$.
5. Hence the answer is (A).