Solution:

In case of an adiabatic process,

$$ \begin{aligned} &p v^{\gamma}=\text { constant or } \frac{p}{\rho^{\gamma}}=\text { constant }=c_{2} \text { (say) } \\\\ &\rho^{\gamma}=\frac{p}{c_{2}} \text { or } \rho=\left(\frac{p}{c_{2}}\right)^{1 / \gamma} \end{aligned}\\ $$

Hence, $ \int \frac{d p}{\rho}=\int \frac{d p}{\left(p / c_{2}\right)^{1 / \gamma}}==\left(c_{2}\right)^{1 / \gamma} \int \frac{1}{p^{1 / …