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

A spherical bubble inside water has radius R. Take the pressure inside the bubble and the water pressure to be p0. The bubble now gets compressed radially in an adiabatic manner so that its radius becomes (R-a). For a≪R the magnitude of the work done in the process is given by 4πP0Ra2X, where X is a constant and γ=Cp/CV=41/30. The value of X is______

Solution

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

Asked in: JEE Advanced 2020 (Paper 2)

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