Question

Three moles of an ideal gas having $\gamma =1.67$ are mixed with $2moles$ of another ideal gas having $\gamma =1.4$. The equivalent value of $\gamma$ for the mixture is approximately $\frac{x}{2}$. Find $x$.

Answer 1

Using Dulong Petit's law for a mixture of gases, ${\gamma }_{mix}=\frac{C_{p_{mix}}}{C_{v_{mix}}}$
where $C_{v_{mix}}=\frac{n_1{C}_{v_1}+n_2{C}_{v_2}}{n_1+n_2}$ and $C_{p_{mix}}=R+C_{v_{mix}}$
here, ${\gamma }_1=\frac{C_{p_1}}{C_{v_1}}$
or $1.67=\frac{R+C_{v_1}}{C_{v_1}}\Rightarrow C_{v_1}=1.49R$

similarly, $1.4=\frac{R+C_{v_2}}{C_{v_2}}\Rightarrow C_{v_2}=2.5R$
now, $C_{v_{mix}}=\frac{3\left(1.49R\right)+2\left(2.5\right)R}{3+2}=1.89R$ and $C_{p_{mix}}=R+1.89R=2.89R$
$\therefore {\gamma }_{mix}=\frac{2.89R}{1.89R}=1.5=3/2$
thus, $x=3$