Answer
$ \frac{p_3}{p_1}= 0.495$
Work Step by Step
Process $2 \rightarrow 3$ is also adiabatic, so $p_2V_2^\gamma = p_3V_3^\gamma$
Where $V_3 = 4.00 V_1$
$V_2 = V_1$
$p_2 = 3.00p_1$
$\gamma = 1.30$
Substitute all the values and solve for $\frac{p_3}{p_1}$
$p_3 = [\frac{V_2}{V_3}]^{\gamma}p_2$
$p_3 = [\frac{V_1}{4.00 V_1}]^{1.30}3.00p_1$
$ \frac{p_3}{p_1}= [\frac{1}{4.00}]^{1.30}3.00$
$ \frac{p_3}{p_1}= 0.495$
