200 ml of He at 0.66 atm pressure and 400 ml of $O_2$ at 0.52 atm pressure are mixed in a 400 ml vessel at ${25}^{\circ }$C. The partial pressure of He and $O_2$ will be

$$\begin{gathered} Volume of He (V_{1 )}=200 ml \\ Pressure of He (P_1)=0.66 atm \\ Volume of O_{2 }(V_{2 })=400 ml \\ Pressure of O_{2_{ }} (P_2)=0.52 atm \\ We know, P\propto \frac{1}{V} \\ Final volume of gas = volume of container \\ \Rightarrow volume of He increases from 200 ml to 400 ml. \\ \Rightarrow Pressure will decrease in the same ratio. \\ \Rightarrow P_{Final\left(He\right)}= \frac{P_{initial}}{2} = \frac{0.66}{2 } = 0.33 atm \\ and for O_2 remains cons\tan t. \end{gathered}$$

Hence, Partial pressure of He = 0.33 atm

and   Partial pressure of O₂ = 0.52 atm