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 / \gamma}} d p=\left(c_{2}\right)^{1 / \gamma} \int p^{-1 / \gamma} d p \\ $

$$ \begin{aligned} &=\left(c_{2}\right)^{1 / \gamma}\left[\frac{p^{-\frac{1}{\gamma}+1}}{\left(-\frac{1}{\gamma}+1\right)}\right]=\frac{\left(c_{2}\right)^{1 / \gamma}(p)^{\left(\frac{\gamma-1}{\gamma}\right)}}{\left(\frac{\gamma-1}{\gamma}\right)}=\frac{\gamma}{\gamma-1}\left(c_{2}\right)^{1 / \gamma}(p)^{\left(\frac{\gamma-1}{\gamma}\right)} \\\\ &=\left(\frac{\gamma}{\gamma-1}\right)\left(\frac{p}{\rho^{\gamma}}\right)^{1 / \gamma}(p)^{\left(\frac{\gamma-1}{\gamma}\right)} \quad\left(\because c_{2}=\frac{p}{\rho^{\gamma}}\right) \\\\ &=\left(\frac{\gamma}{\gamma-1}\right)\left(\frac{p^{1 / \gamma}}{\gamma \times \frac{1}{\gamma}}\right)(p)^{\left(\frac{\gamma-1}{\gamma}\right)}\\\\ &=\left(\frac{\gamma}{\gamma-1}\right) \frac{(p)^{\left(\frac{1}{\gamma}+\frac{\gamma-1}{\gamma}\right)}}{\rho}=\left(\frac{\gamma}{\gamma-1}\right) \frac{p}{\rho} \end{aligned} \\ $$

Substituting the value of, $\int \frac{d p}{p}$ , we get

$$ \left(\frac{\gamma}{\gamma-1}\right) \frac{p}{\rho}+\frac{V^{2}}{2}+g z=\text { constant } \\ $$

Dividing both sides by $g$, we get, $$ \left(\frac{\gamma}{\gamma-1}\right) \frac{p}{\rho g}+\frac{V^{2}}{2 g}+z=\text { constant } \ $$

The Bernoulli's equation for compressible flow undergoing adiabatic process.