$n _2=1 \text { mole } C _{ v _2}=\frac{5 R }{2}$

$\left(C_v\right)_m=\frac{n_1 C_{v_1}+n_2 C_{v_1}}{n_1+n_2}=\frac{5 \times \frac{3 R}{2}+1 \times \frac{5 R}{2}}{6}=\frac{5 R}{3}$

$\gamma _{ m }=\frac{\left( c _{ P }\right)_{ m }}{\left( c _{ v }\right)_{ m }}=\frac{8}{5}$

$\therefore$ Option $4$ is correct

$\left( C _{ P }\right)_{ m }=\frac{5 R }{3}+ R =\frac{8 R }{3}$

$(1)$ $P _0 V _0^\gamma= P \left(\frac{ V _0}{4}\right)^\gamma \Rightarrow P = P _0(4)^{8 / 5}=9.2 P _0$ which is between $9 P _0$ and $10 P _0$

$(2)$

Average $K.E.=5 \times \frac{3}{2} R T+1 \times \frac{5 R T}{2}$

$=10 R T$

To calculate $T$

$\frac{ P _0 V _0}{ T _0}=9.2 P _0 \times \frac{ V _0}{4 \times T }$

$50$

$T=\frac{9.2}{4} T_0$

Now average $KE =10 R \times 9.2 \frac{ T _0}{4}=23 RT _0$

$(3)$ $W =\frac{ P _1 V _1- P _2 V _2}{\gamma-1}$

$=\frac{ P _0 V _0-9.2 P _0 \times \frac{ V _0}{4}}{3 / 5}=-13 RT _0$