A spherical bubble inside water has radius R . Take the pressure...

$$\begin{aligned} &\mathrm{W}=(\Delta \mathrm{P})_{\text{avg }} \times 4 \pi \mathrm{R}^2 \mathrm{a} \\ &=\left|\frac{\mathrm{dP}}{2} \cdot 4 \pi \mathrm{R}^2 \mathrm{a}\right| \end{aligned}$$) (for small change ((\Delta \mathrm{P})_{\text{avg }}<\mathrm{P}>) arithmetic mean) (=\mathrm{PV} \text{ gamma }=\mathrm{c} \Rightarrow \mathrm{dP}=-\gamma \frac{\mathrm{P}}{\mathrm{V}} \mathrm{dV}=-\frac{\gamma \mathrm{P}_0}{\mathrm{ V}} 4 \pi \mathrm{R}^2 \mathrm{a}) (=\frac{\gamma \mathrm{P}_0}{2 \mathrm{ V}} \times 4 \pi \mathrm{R}^2 \mathrm{a} \times 4 \pi \mathrm{R}^2 \mathrm{a}) (=\frac{\gamma \mathrm{P}_0}{2 \times 4 \pi \mathrm{R}^3} 4 \pi \mathrm{R}^2 \mathrm{a} \times 4 \pi \mathrm{R}^2 \mathrm{a}) (=\left(4 \mathrm{pRP} \times \mathrm{a}^2\right) \frac{3 \gamma}{2}) (\therefore \mathrm{x}=2.05)$