Given situation is

Let final temperature after opening the valve is $T_f$, then $\Delta W_{\text {ext }}=0$ and $\Delta Q_{\text {ext }}=0$ So, from first law of thermodynamics,

$\Delta U=0$

$\Rightarrow n_1 C_V T+n_2 C_V T=\left(n_1+n_2\right) C_V T_f$

$\Rightarrow T_f=T$

Now, by gas equation, we have

$\text { As, } n_1+n_2=n$

$\Rightarrow \frac{V}{R T}+\frac{4 V \times 0.5}{R T} =\frac{5 V \times p_1}{R T}$

$\Rightarrow p =0.6 \,atm$