Answer
$V_1=50.9L;p_2=6atm; V_3=153L;T_3=930K$
Work Step by Step
We know that
$V_1=\frac{m}{pM}RT$
We plug in the known values to obtain:
$V_1=\frac{8}{4(101300)}(8.31)(310)$
$\implies V_1=50.9L$
Now we can find $p_2$ as
$p_2=\frac{T_2}{T_1}p_1$
We plug in the known values to obtain:
$p_2=\frac{930}{310}(2)$
$\implies p_2=6atm$
The value of $V_3$ can be calculated as
$V_3=\frac{p_2}{p_3}V_2$
We plug in the known values to obtain:
$V_3=\frac{6}{2}(50.9)$
$\implies V_3=153L$
We know that the temperature at state (3) is equal to the temperature at state (2), thus $T_2=T_3=930K$
