A mixture of ideal gas containing 5 moles of monatomic gas and 1 mole of rigid diatomic gas is initially at pressure $P_0$, volume $V_0$, and temperature $T_0$. If the gas mixture is adiabatically compressed to a volume $V_0$/4, then the correct statement(s) is/are, (Given $2^{1.2}$ = 2.3 ; $2^{3.2}$ = 9.2 ; R is gas constant)

$P_1=\frac{5R{T}_o}{V_o}$

$P_2=\frac{R{T}_0}{V}$

$\begin{array}{l}{P}_1+P_2=P_o=\frac{6R{T}_0}{V_0}\\ Y_{\text{mixture }}=\frac{5C_P^1+\left(1\right)C_P^2}{5C_V^1+\left(1\right)C_V^2}=\frac{5\times \frac{5R}{2}+\left(1\right)\left(\frac{7R}{2}\right)}{5\times \frac{3R}{2}+\left(1\right)\left(\frac{5R}{2}\right)}=\frac{32}{20}=1.6\\ P{V}^{\gamma }=\text{ constant }\\ \Rightarrow P_0{V}_0^{1.6}=P{\left[V_0/4\right]}^{1.6}\\ \Rightarrow P=P_0(4{)}^{1.6}=P_o\left(2^{3.2}\right)\\ \Rightarrow P=9.2P_0\end{array}$

$\begin{array}{l}\text{ Work }=W=\frac{P_2{V}_2-P_1{V}_1}{1-\gamma }=\frac{\left(9.2P_0\right)\frac{V_0}{4}-P_0{V}_0}{1-1.6}=\frac{(2.3-1)P_o{V}_0}{-0.6}=\frac{1.3\times 6R{T}_0}{-0.6}\\ \Rightarrow W=-13R{T}_0\\ \Rightarrow \left|W\right|=13R{T}_0\\ T{V}^{\gamma -1}=\text{ constant }\\ \Rightarrow T_i{V}_i^{\gamma -1}=T_f{V}_f^{\gamma -1}\\ \Rightarrow \left(T_0\right){\left(V_0\right)}^{0.6}=T_f{\left(\frac{V_0}{4}\right)}^{0.6}\\ \Rightarrow T_f=(2{)}^{1.2}{T}_0=2.3T_0\\ \text{ Kinetic Energy }=n{C}_{Vmix}{T}_f=\left(6\right)\left(\frac{R}{1.6-1}\right)\left(2.3T_0\right)=23R{T}_0\end{array}$