Question

A spherical bubble inside water has radius$\mathrm{R}$. Take the pressure inside the bubble and the water pressure to be${\mathrm{p}}_0$. The bubble now gets compressed radially in an adiabatic manner so that its radius becomes $(\mathrm{R}-\mathrm{a})$. For $𝑎<<\mathrm{R}$ the magnitude of the work done in the process is given by $\left(4{\mathrm{\pi P}}_0{\mathrm{R}}^2\mathrm{a}\right)\mathrm{X}$, where $\mathrm{X}$ is a constant and$\mathrm{\gamma }=𝐶𝑝⁄𝐶𝑉=41⁄30$. The value of$\mathrm{X}$ is________.

Answer 1

The value of$\mathrm{X}$ is $\mathbf{1}$.

Explanation:

$$\begin{gathered} \mathbf{Step}\mathbf{}\mathbf{1}\mathbf{:}\mathbf{}\mathbf{Calculation}\mathbf{}\mathbf{of}\mathbf{}\mathbf{pressure} \\ {\mathrm{P}}_1{{\mathrm{V}}_1}^{\mathrm{\gamma }}={\mathrm{P}}_2{{\mathrm{V}}_2}^{\mathrm{\gamma }} \\ {\mathrm{P}}_0{\left(\frac{4}{3}{\mathrm{\pi R}}^3\right)}^{\mathrm{\gamma }}={\mathrm{P}}_2{\left[\frac{4}{3}\mathrm{\pi }{(\mathrm{R}-\mathrm{a})}^3\right]}^{\mathrm{\gamma }} \\ {\mathrm{P}}_0(\frac{\mathrm{R}}{\mathrm{R}-\mathrm{a}}{)}^{3\mathrm{\gamma }}={\mathrm{P}}_2 \\ \mathbf{Step}\mathbf{}\mathbf{2}\mathbf{:}\mathbf{}\mathbf{Calculation}\mathbf{}\mathbf{of}\mathbf{}\mathbf{magnitude}\mathbf{}\mathbf{of}\mathbf{}\mathbf{work}\mathbf{}\mathbf{done}\mathbf{} \\ \mathrm{W}=\frac{{\mathrm{P}}_1{\mathrm{V}}_1-{\mathrm{P}}_2{\mathrm{V}}_2}{\mathrm{\gamma }-1} \\ =\frac{{\mathrm{P}}_0\frac{4}{3}{\mathrm{\pi R}}^3-{\mathrm{P}}_0{\left(\frac{\mathrm{R}}{\mathrm{R}-\mathrm{a}}\right)}^{3\mathrm{\gamma }}\frac{4}{3}\mathrm{\pi }{(\mathrm{R}-\mathrm{a})}^3}{({\displaystyle \frac{41}{30}}-1)}[3\mathrm{\gamma }=\frac{41}{10}] \\ =\frac{{\mathrm{P}}_0\frac{4}{3}\mathrm{\pi }}{{\displaystyle \frac{11}{30}}}[{\mathrm{R}}^3-{\mathrm{R}}^{41/10}{(\mathrm{R}-\mathrm{a})}^{3-\frac{41}{10}}] \\ =\frac{30{\mathrm{P}}_0}{11}\frac{4}{3}\mathrm{\pi }[{\mathrm{R}}^3-{\mathrm{R}}^{41/10}{(\mathrm{R}-\mathrm{a})}^{-\frac{11}{10}}] \\ =\frac{10{\mathrm{P}}_{0}4\mathrm{\pi }}{11}[{\mathrm{R}}^3-{\mathrm{R}}^{41/10}{\mathrm{R}}^{-11/10}{(1-\frac{\mathrm{a}}{\mathrm{R}})}^{\frac{-11}{10}}] \\ =\frac{10{\mathrm{P}}_{0}4\mathrm{\pi }}{11}[{\mathrm{R}}^3-{\mathrm{R}}^3{(1-\frac{\mathrm{a}}{\mathrm{R}})}^{\frac{-11}{10}}] \\ =\frac{10{\mathrm{P}}_{0}4\mathrm{\pi }}{11}{\mathrm{R}}^3[1-1+\frac{11\mathrm{a}}{10\mathrm{R}}] \\ =(4{\mathrm{\pi P}}_0{\mathrm{R}}^2\mathrm{a}) \\ \mathrm{Given}\mathrm{work}\mathrm{done}=(4{\mathrm{\pi P}}_0{\mathrm{R}}^2\mathrm{a})\mathrm{X} \\ \mathrm{Therefore},\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{X}\mathbf{=}\mathbf{1} \end{gathered}$$