Consider a cycle tyre being filled with air by a pump. Let $V$ be the volume of the tyre (fixed) and at each stroke of the pump $\Delta V$ ( < < V) of air is transferred to the tube adiabatically. What is the work done when the pressure in the tube is increased from $P_1$ to $P_2$?

(N/A) The pressure increases by $\Delta P$ when a volume $\Delta V$ of air is added to the tyre at each stroke.
For an adiabatic process, $P V^{\gamma} = \text{constant}$.
Considering the state before and after one stroke:
$P (V + \Delta V)^{\gamma} = (P + \Delta P) V^{\gamma}$
$P V^{\gamma} (1 + \frac{\Delta V}{V})^{\gamma} = (P + \Delta P) V^{\gamma}$
Using the binomial approximation $(1 + x)^n \approx 1 + nx$ for small $x$:
$P (1 + \gamma \frac{\Delta V}{V}) = P + \Delta P$
$P + \gamma P \frac{\Delta V}{V} = P + \Delta P$
$\gamma P \frac{\Delta V}{V} = \Delta P \implies \Delta V = \frac{V}{\gamma P} \Delta P$
In the limit of infinitesimal changes, $dV = \frac{V}{\gamma P} dP$.
The work done $W$ in increasing the pressure from $P_1$ to $P_2$ is given by:
$W = \int P dV = \int_{P_1}^{P_2} P \left( \frac{V}{\gamma P} dP \right)$
$W = \frac{V}{\gamma} \int_{P_1}^{P_2} dP$
$W = \frac{V}{\gamma} (P_2 - P_1)$