Answer
$V_{2}=1.33m^3$
$P_{f}=263.96kPa$
Work Step by Step
For the first tank:
$V_{1}=1m^3$
$m_{1}=\frac{P_{1}V_{1}}{RT_{1}}=\frac{350kPa*1m^3}{0.287\frac{kPam^3}{kgK}*(10+273.15)K}=4.31kg$
Fro the second tank:
$m_{2}=3kg$
$V_{2}=\frac{m_{2}RT_{2}}{P_{2}}=\frac{3kg*0.287\frac{kPam^3}{kgK}*(35+273.15)K}{200kPa}=1.33m^3$
At the equilibrium conditions:
$m_{t}=4.31kg+3kg=7.31kg$
$V_{t}=1m^3+1.33m^3=2.33m^3$
$P_{f}=\frac{m_{t}RT_{f}}{V_{t}}=\frac{7.31kg*0.287\frac{kPam^3}{kgK}*(20+273.15)K}{2.33m^3}=263.96kPa$
