A spherical bubble inside water has radius 𝑅. Take the pressure inside the bubble and the water pressure to be $$p_0$$. The bubble now gets compressed radially in an adiabatic manner so that its radius becomes (𝑅 − 𝑎). For $$a \ll R$$ the magnitude of the work done in the process is given by $$(4 \pi p_0 R a^2)X^2$$, where 𝑋 is a constant and $$\gamma = \frac{C_p}{C_V} = \frac{41}{30}$$. The value of X is ____________.

Initially the gas inside the bubble and the surrounding water are both at the same pressure $$p_0$$, so there is no net force on the interface.

When the bubble is compressed adiabatically from radius $$R$$ to $$R-a\;(a\ll R)$$, the extra (differential) work that must be supplied is obtained by integrating the excess pressure $$\left(p-p_0\right)$$ over the volume change:

$$W=\int_{V_i}^{V_f}\left(p-p_0\right)dV$$

Step 1 : Equation of state during the adiabatic process
For the enclosed ideal gas, $$pV^{\gamma}=K$$. Hence $$p=p_0\left(\dfrac{V_i}{V}\right)^{\gamma}$$, where $$V_i=\dfrac{4}{3}\pi R^3$$.

Step 2 : Express everything in terms of the small parameter $$\varepsilon$$
Write $$V=V_i(1+\varepsilon)\quad\Longrightarrow\quad dV=V_i\,d\varepsilon.$$ For the final state $$V_f=V_i\!\left(1+\varepsilon_f\right),\qquad \varepsilon_f=\dfrac{V_f-V_i}{V_i}.$$

For small $$a$$ (keep terms up to $$a^2$$), $$V_f=\dfrac{4}{3}\pi\left(R-a\right)^3 \approx \dfrac{4}{3}\pi\left(R^3-3R^2a\right) =V_i-4\pi R^2a.$$ Thus $$\varepsilon_f=\dfrac{-4\pi R^2a}{V_i} =\dfrac{-4\pi R^2a}{\tfrac{4}{3}\pi R^3} =-\,\dfrac{3a}{R}.$$ (The $$a^2$$ term in $$\varepsilon_f$$ can be ignored, because it only contributes to $$a^3$$ or higher in $$W$$.)

Step 3 : Expand the integrand for small $$\varepsilon$$
$$p-p_0=p_0\Big[(1+\varepsilon)^{-\gamma}-1\Big] =p_0\Big[-\gamma\varepsilon+\dfrac{\gamma(\gamma+1)}{2}\varepsilon^2+O(\varepsilon^3)\Big].$$

Step 4 : Integrate
$$W=p_0V_i\int_{0}^{\varepsilon_f} \Big[-\gamma\varepsilon+\dfrac{\gamma(\gamma+1)}{2}\varepsilon^2\Big]d\varepsilon =p_0V_i\left[-\dfrac{\gamma}{2}\varepsilon_f^{\,2} +O(\varepsilon_f^{\,3})\right].$$

Keeping only the leading $$\varepsilon_f^{\,2}$$ term (order $$a^2$$), $$|W|=\dfrac{\gamma}{2}\,p_0V_i\,\varepsilon_f^{\,2}.$$

Step 5 : Insert $$V_i$$ and $$\varepsilon_f$$
$$|W|=\dfrac{\gamma}{2}\,p_0\!\left(\dfrac{4}{3}\pi R^3\right) \left(\dfrac{3a}{R}\right)^2 =\dfrac{\gamma}{2}\,p_0\!\left(\dfrac{4}{3}\pi R^3\right) \dfrac{9a^2}{R^2} =6\gamma\,\pi p_0 R a^2.$$

Step 6 : Match with the required form
The problem states $$|W|=(4\pi p_0 R a^2)\,X^2.$$ Therefore $$X^2=\dfrac{|W|}{4\pi p_0 R a^2} =\dfrac{6\gamma\pi}{4\pi} =\dfrac{3\gamma}{2}.$$

Step 7 : Insert the given value of $$\gamma$$
$$\gamma=\dfrac{C_p}{C_V}=\dfrac{41}{30},\qquad X^2=\dfrac{3}{2}\cdot\dfrac{41}{30}=\dfrac{41}{20}.$$

Hence $$X=\sqrt{\dfrac{41}{20}} =\dfrac{\sqrt{41}}{2\sqrt{5}} \approx 1.43.$$

Answer (numerical): $$X=\sqrt{\dfrac{41}{20}}\;(\approx 1.43).$$