Work Done in Adiabatic Radial Compression of a Spherical Bubble

To find the magnitude of the work done in compressing the spherical bubble, we calculate the net work done by an external agency to reduce the radius of the bubble from $R$ to $(R - a)$ under adiabatic conditions, against the pressure inside the bubble and the surrounding water pressure.

1. Volume and Pressure Relations

The initial radius of the bubble is $R$ and the final radius is $R - a$, where $a \ll R$.

The initial volume of the bubble is: $V_0 = \frac{4}{3} \pi R^3$

The final volume of the bubble is: $V_f = \frac{4}{3} \pi (R - a)^3 = \frac{4}{3} \pi R^3 \left(1 - \frac{a}{R}\right)^3$

Expanding $\left(1 - \frac{a}{R}\right)^3$ up to second order in $\frac{a}{R}$: $V_f \approx V_0 \left( 1 - \frac{3a}{R} + \frac{3a^2}{R^2} \right)$

Let $\delta = 1 - \frac{V_f}{V_0} = \frac{3a}{R} - \frac{3a^2}{R^2}$.

---

2. Change in Internal Energy of the Gas ($\Delta U$)

Since the process inside the bubble is adiabatic, no heat exchange takes place ($Q = 0$). By the First Law of Thermodynamics, the work done on the gas equals the change in its internal energy: $W_{\text{gas, on}} = \Delta U = \frac{P_f V_f - P_0 V_0}{\gamma - 1}$

For an adiabatic process, $P V^\gamma = \text{constant}$: $P_f = P_0 \left(\frac{V_0}{V_f}\right)^\gamma = P_0 (1 - \delta)^{-\gamma}$

Thus, the product $P_f V_f$ is given by: $P_f V_f = P_0 V_0 (1 - \delta)^{1 - \gamma}$

Using Taylor expansion $(1 - \delta)^{1 - \gamma} \approx 1 + (\gamma - 1)\delta + \frac{\gamma(\gamma - 1)}{2} \delta^2$: $\frac{P_f V_f - P_0 V_0}{\gamma - 1} = P_0 V_0 \left( \delta + \frac{\gamma}{2} \delta^2 \right)$

Substituting $\delta = \frac{3a}{R} - \frac{3a^2}{R^2}$ into the equation: $\delta + \frac{\gamma}{2} \delta^2 = \left( \frac{3a}{R} - \frac{3a^2}{R^2} \right) + \frac{\gamma}{2} \left( \frac{9a^2}{R^2} \right) = \frac{3a}{R} + \left( \frac{9\gamma}{2} - 3 \right) \frac{a^2}{R^2}$

Substituting $V_0 = \frac{4}{3} \pi R^3$: $\Delta U = p_0 \left(\frac{4}{3} \pi R^3\right) \left[ \frac{3a}{R} + \left( \frac{9\gamma}{2} - 3 \right) \frac{a^2}{R^2} \right] = 4\pi p_0 R^2 a + 4\pi p_0 R a^2 \left( \frac{3\gamma}{2} - 1 \right)$

---

3. Work Done by the Surrounding Water ($W_{\text{surr}}$)

The surrounding water exerts a constant pressure $p_0$. The work done by the surrounding water on the bubble is: $W_{\text{surr}} = p_0 (V_0 - V_f) = p_0 V_0 \delta$

Substituting $\delta = \frac{3a}{R} - \frac{3a^2}{R^2}$: $W_{\text{surr}} = p_0 \left(\frac{4}{3} \pi R^3\right) \left( \frac{3a}{R} - \frac{3a^2}{R^2} \right) = 4\pi p_0 R^2 a - 4\pi p_0 R a^2$

---

4. Net Work Done ($W_{\text{net}}$)

The net work done by an external agency to perform this compression is: $W_{\text{net}} = \Delta U - W_{\text{surr}}$

$W_{\text{net}} = \left[ 4\pi p_0 R^2 a + 4\pi p_0 R a^2 \left( \frac{3\gamma}{2} - 1 \right) \right] - \left[ 4\pi p_0 R^2 a - 4\pi p_0 R a^2 \right]$

$W_{\text{net}} = 4\pi p_0 R a^2 \left( \frac{3\gamma}{2} - 1 + 1 \right) = (4\pi p_0 R a^2) \left( \frac{3\gamma}{2} \right)$

---

5. Calculation of $X$

Comparing $W_{\text{net}}$ with the given expression $(4\pi p_0 R a^2) X$: $X = \frac{3\gamma}{2}$

Given $\gamma = \frac{41}{30}$: $X = \frac{3}{2} \times \frac{41}{30} = \frac{41}{20} = 2.05$