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 $(4\pi P_0 R a^2)\ X$ where $X$ is a constant and $\gamma = C_p/C_v = 41/30$. The value of $X$ is ________.

Step 1: Define the initial and final states of the bubble.

The initial volume $V_0$ and final volume $V_1$ of the spherical bubble are determined by its initial radius $R$ and final radius $R - a$.

$$V_0 = \frac{4}{3}\pi R^3$$ $$V_1 = \frac{4}{3}\pi(R-a)^3 = V_0\left(1-\frac{a}{R}\right)^3$$

Step 2: Set up the expression for the work done during adiabatic compression.

For an adiabatic process, the relation between pressure and volume is $P V^\gamma = \text{constant}$. The work done on the gas during compression is given by the integral of $P \, dV$.

$$W = \int_{V_1}^{V_0} P\,dV$$ $$W = \frac{P_0V_0 - P_1V_1}{\gamma-1}$$ $$W = \frac{P_0V_0}{\gamma-1}\left[\left(\frac{V_0}{V_1}\right)^{\gamma-1}-1\right]$$

Step 3: Apply binomial expansion for $a \ll R$.

Express the volume ratio in terms of $a/R$.

$$\frac{V_0}{V_1} = \left(1-\frac{a}{R}\right)^{-3}$$

Substitute this ratio into the work equation.

$$W = \frac{P_0V_0}{\gamma-1}\left[\left(1-\frac{a}{R}\right)^{-3(\gamma-1)}-1\right]$$

Using the binomial expansion $(1-x)^{-n} \approx 1 + nx + \frac{n(n+1)}{2}x^2$ for $x \ll 1$:

$$\left(1-\frac{a}{R}\right)^{-3(\gamma-1)} \approx 1 + 3(\gamma-1)\frac{a}{R} + \frac{3(\gamma-1)(3\gamma-2)}{2}\frac{a^2}{R^2}$$

Step 4: Simplify the expression for work done.

Substitute the expansion back into the equation for $W$.

$$W \approx \frac{P_0V_0}{\gamma-1}\left[3(\gamma-1)\frac{a}{R} + \frac{3(\gamma-1)(3\gamma-2)}{2}\frac{a^2}{R^2}\right]$$ $$W \approx P_0V_0\left[\frac{3a}{R} + \frac{3(3\gamma-2)}{2}\frac{a^2}{R^2}\right]$$

Substitute $V_0 = \frac{4}{3}\pi R^3$ into the equation.

$$W \approx P_0\left(\frac{4}{3}\pi R^3\right)\left[\frac{3a}{R} + \frac{3(3\gamma-2)}{2}\frac{a^2}{R^2}\right]$$ $$W \approx 4\pi P_0 R^2 a + \left(\frac{3\gamma-2}{2}\right)(4\pi P_0 R a^2)$$

Step 5: Solve for the constant $X$.

Comparing the second-order term of the work done to the given expression $(4\pi P_0 R a^2) X$:

$$X = \frac{3\gamma-2}{2}$$

Substitute the given value of $\gamma = \frac{41}{30}$.

$$X = \frac{3\left(\frac{41}{30}\right)-2}{2}$$ $$X = \frac{\frac{41}{10}-2}{2}$$ $$X = \frac{21}{20}$$

Answer: The value of $X$ is $\dfrac{21}{20}$ (or $1.05$).