When air is transferred into tyre adiabatically then let initial volume of air in tyre $V$, after pumping one stroke, if volume is increased by $\Delta V$ and pressure is increased by $\Delta p$. Then for just before and after an stroke, we can write
$$
\begin{aligned}
&p_1 V_1^\gamma=p_2 V_2^\gamma \\
&p(V+\Delta V)^\gamma=(p+\Delta p) V^\gamma(\because \text { volume is fixed) } \\
&p V^\gamma\left(1+\frac{\Delta V}{V}\right)^\gamma=p\left(1+\frac{\Delta p}{p}\right) V^\gamma \\
&\text { (As volume of tyre } V \text { remains constant) } \\
&p V^\gamma\left(1+\gamma \frac{\Delta V}{V}\right)=p V^\gamma\left(1+\frac{\Delta p}{p}\right)
\end{aligned}
$$
On expanding by binomial theorem and neglecting the higher terms of $\Delta V$ as $(\Delta V \ll V)$
$$
\begin{aligned}
&1+\gamma \frac{d V}{d V}=1+\frac{\Delta p}{p} \\
&\gamma \frac{\Delta V}{V}=\frac{\Delta p}{p} \Rightarrow \Delta V=\frac{1}{\gamma} \frac{V}{p} \Delta p \\
&d V=\frac{1}{\gamma} \frac{V}{p} d p
\end{aligned}
$$
Integrating both sides with work done limits $W_1$ to $W_2$ and pressure limit from $p_1$ to $p_2$.
$$
\begin{aligned}
&\int_{W_1}^{W_2}(d W)=\int_{p_1}^{p_2} p d V=\int_{p_1}^{p_2} p \times \frac{1}{\gamma} \frac{V}{p} d p\left[\because d V=\frac{1}{\gamma} \frac{V}{p} d p\right] \\
&=\frac{V}{\gamma} \int_{p_1}^{p_2} d p=\frac{V}{\gamma}\left(p_2-p_1\right) \\
&W=\frac{\left(p_2-p_1\right)}{\gamma} V
\end{aligned}
$$