Answer
$\frac{p_2}{p_1} = 4$.
Work Step by Step
Along both processes, the change in internal energy is the same, as the start and end points are the same. Using the first law of thermo, we find that , $5{p_1}V_1-2p_1V_1 = 5.5p_1V_1-\frac{p_1+p_2}{2} *2V_1$. Dividing out $V_1$ and simplifying, we get $5p_1 =p_1+p_2, p_1 = 0.25p_2, \frac{p_2}{p_1} = 4$.
