Answer
$T_2=50°C, P_2=400\ kPa$
Work Step by Step
Since the tank is insulated and the partition is removed (no boundary work), the energy balance reduces to:
$\Delta U = 0 \rightarrow \Delta T=0$
The final temperature is: $T_2=T_1=50°C$
From ideal gas law:
$PV=mRT$
And since the final volume is twice the original:
$V_2=2V_1 \rightarrow \frac{2mRT_1}{P_1} = \frac{mRT_2}{P_2}$
$\frac{2}{P_1}=\frac{1}{P_2}\rightarrow P_2=400\ kPa$
