Physics - Heat & Thermo Question with Solution | TestHub

Let $n_1,n_2$ be the number of moles of gases $1$ and $2$.
We have $\frac{C_{p_1}}{C_{v_1}}=\frac{5}{3}$ and We know $C_{p_1}=C_{v_1}+R$
$\therefore \frac{5C_{v_1}}{3}=C_{v_1}+R,\frac{2C_{v_1}}{3}{R}_1,C_{v_1}=\frac{3R}{2}$
$A$ and $C_{p_1}=C_{v_1}+R=\frac{5R}{2}$
Also, $\frac{C_{p_2}}{C_{v_2}}=\frac{4}{3}$ and
$C_{p_2}=C_{v_2}+R$
$\therefore \frac{4C_{v_2}}{3}=C_{v_2}+R,\frac{C_{v_2}}{3}=R,C_{v_2}=3R$
$C_{p_2}=C_{v_2}+R=4R$
Now $C_{v_{mix}}=\frac{n_1{C}_{v_1}+n_2{C}_{v_2}}{n_1+n_2}=\frac{2\times \frac{3R}{2}+3\times 3R}{2+3}$
$=\frac{3R+9R}{5}=\frac{12R}{5}$
$C_{p_{mix}}=\frac{n_1{C}_{p_1}+n_2{C}_{p_2}}{n_1+n_2}=\frac{2\times \frac{5R}{2}+3\times 4R}{2+3}$
$=\frac{17R}{5}$
$l_{mix}=\frac{C_{p_{mix}}}{C_{v_{mix}}}=\frac{17}{12}=1.4166=1.42$.