Answer
$ \frac{p_4}{p_1}= 0.165$
Work Step by Step
Process $4 \rightarrow 1$ is also adiabatic, so $p_4V_4^\gamma = p_1V_1^\gamma$
Where $V_4 = 4.00 V_1$
$\gamma = 1.30$
Substitute all the values and solve for $\frac{p_4}{p_1}$
$p_4 = [\frac{V_1}{V_4}]^{\gamma}p_1$
$p_4 = [\frac{V_1}{4.00 V_1}]^{1.30}p_1$
$ \frac{p_4}{p_1}= [\frac{1}{4.00}]^{1.30}$
$ \frac{p_4}{p_1}= 0.165$
