The pressure is increased by $\Delta \mathrm{P}$, when volume is increased by $\Delta \mathrm{V}$ at each stroke.

$\therefore \mathrm{P}_{1} \mathrm{~V}_{1}^{\gamma}=\mathrm{P}_{2} \mathrm{~V}_{2}^{\gamma}$ (Initially)

$\therefore \mathrm{P}(\mathrm{V}+\Delta \mathrm{V})^{\gamma}=(\mathrm{P}+\Delta \mathrm{P}) \mathrm{V}^{\gamma}(\because$ volume is fixed $)$

$\mathrm{PV}^{\gamma}\left[1+\frac{\Delta \mathrm{V}}{\mathrm{V}}\right]^{\gamma}=\mathrm{P}\left[1+\frac{\Delta \mathrm{P}}{\mathrm{P}}\right] \mathrm{V}^{\gamma}$

As $\Delta \mathrm{V}<<\mathrm{V}$ so by using binomial theorem we get,

$\therefore \mathrm{PV}^{\gamma}\left(1+\gamma \frac{\Delta \mathrm{V}}{\mathrm{V}}\right)=\mathrm{PV}^{\gamma}\left(1+\frac{\Delta \mathrm{P}}{\mathrm{P}}\right)$ $\therefore \gamma \frac{\Delta \mathrm{V}}{\mathrm{V}}=\frac{\Delta \mathrm{P}}{\mathrm{P}}$ $\therefore \Delta \mathrm{V}=\frac{1}{\gamma} \cdot \frac{\mathrm{V}}{\mathrm{P}} \cdot \Delta \mathrm{P}$ but $\Delta \mathrm{V}$ and $\Delta \mathrm{P}$ are very small.

$d \mathrm{~V}=\frac{1}{\gamma} \cdot \frac{\mathrm{V}}{\mathrm{P}} \cdot d \mathrm{P}$

Hence, work done in increasing the pressure in tube from $\mathrm{P}_{1}$ to $\mathrm{P}_{2}$ is

$W=\int_{P_{1}}^{P_{2}} P d V=\int_{P_{1}} P \times \frac{1}{\gamma} \times \frac{V}{P} \cdot d P$

$=\frac{V}{\gamma} \int_{P_{1}}^{P_{2}} d P$

$=\frac{V}{\gamma}\left(P_{2}-P_{1}\right)$

$\therefore W =\frac{\left(P_{2}-P_{1}\right) V}{\gamma}$