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

Q: Why do we use the area of a triangle for work?

For very small changes in volume, the P-V curve can be approximated as a straight line. The extra work done by the external agent is the integral of $(P - p_0) dV$, which geometrically forms a triangle above the constant pressure line $p_0$.

Q: Why isn't the work just $p_0 \Delta V$?

$p_0 \Delta V$ is the work done by the surrounding water. However, as the bubble is compressed, its internal pressure increases. An external agent must do *additional* work to overcome this increasing pressure, which is what the question is asking for.

Q: How did we get $dV = -4\pi R^2 a$?

We can use differentials. Since $V = \frac{4}{3}\pi R^3$, differentiating gives $dV = 4\pi R^2 dR$. Here, the change in radius $dR$ is $-a$, so the change in volume is $-4\pi R^2 a$.

Q: Why is $dP = -\gamma \frac{P}{V} dV$?

This comes from differentiating the adiabatic condition $PV^\gamma = \text{constant}$. Using the product rule, we get $dP \cdot V^\gamma + P \cdot \gamma V^{\gamma-1} dV = 0$. Rearranging this gives the relation.

Q: What does the constant $X$ represent physically?

The constant $X$ is directly related to the adiabatic bulk modulus of the gas. It tells us how much energy is stored in the gas due to its compressibility and the specific thermodynamic process (adiabatic) it undergoes.

Question

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________.

Sigma Percentile · Accepted Answer

### The Physical Setup: A Bubble Under Pressure

Imagine a spherical bubble resting deep underwater. The water exerts a constant pressure, $p_0$, on it. Now, we compress this bubble slightly, reducing its radius from $R$ to $R - a$. This compression happens adiabatically, meaning no heat is exchanged with the surrounding water. Our goal is to find the magnitude of the work done in this process.

### The Subtlety of "Work Done"

When the bubble is compressed, the surrounding water naturally does some work, which is simply $p_0 \Delta V$. But the question asks for the magnitude of the work done *in the process* to compress it adiabatically. This refers to the **extra work** done by an external agent.

Geometrically, if we plot this on a P-V graph, the total work done on the gas is the area under the adiabatic curve. The work done by the water is the rectangular area below the constant pressure line $p_0$. The external work is the area of the small triangle *above* the constant pressure line.

$$W_{ext} \approx \frac{1}{2} dP |dV|$$

### The Mathematics of Small Changes

First, let's find the change in volume. The initial volume is $V = \frac{4}{3}\pi R^3$. Since the compression $a$ is very small compared to $R$ ($a \ll R$), we can use differentials to find the change in volume.

$$dV = 4\pi R^2 dR$$

Since the radius decreases by $a$, we have $dR = -a$. Therefore, the magnitude of the decrease in volume is simply the surface area of the bubble multiplied by the small thickness $a$:

$$|dV| = 4\pi R^2 a$$

### The Adiabatic Constraint

Now, how does the pressure change? The problem states the compression is adiabatic. So, the equation $PV^\gamma = 	ext{constant}$ governs the process. If we differentiate this equation, we get a direct relationship between small changes in pressure and volume:

$$V^\gamma dP + \gamma P V^{\gamma-1} dV = 0$$

$$dP = -\gamma \frac{P}{V} dV$$

Let's substitute our known values into this differential equation. The initial pressure is $p_0$, the volume is $\frac{4}{3}\pi R^3$, and $dV$ is $-4\pi R^2 a$.

$$dP = -\gamma \frac{p_0}{\frac{4}{3}\pi R^3} (-4\pi R^2 a)$$

Notice how beautifully the $\pi$ and $R$ terms cancel out! The increase in pressure simplifies to:

$$dP = \frac{3\gamma p_0 a}{R}$$

### Bringing It All Together

We have our $dP$ and our $dV$. Let's plug them back into our work equation. The external work is half times $dP$ times $|dV|$.

$$W_{ext} = \frac{1}{2} \left( \frac{3\gamma p_0 a}{R} ight) (4\pi R^2 a)$$

Let's rearrange this to match the format given in the question. We group $(4\pi p_0 R a^2)$ together.

$$W_{ext} = (4\pi p_0 R a^2) \left( \frac{3\gamma}{2} ight)$$

By comparing this with the given expression $(4\pi p_0 R a^2)X$, it's crystal clear that our unknown constant $X$ is exactly equal to $\frac{3\gamma}{2}$.

### The Final Calculation

We are at the finish line! The problem gives us the ratio of specific heats, $\gamma$, as $\frac{41}{30}$. Substituting this into our expression for $X$:

$$X = \frac{3}{2} 	imes \frac{41}{30}$$

$$X = \frac{41}{20} = 2.05$$

And that is our final answer! The elegance of this problem lies in recognizing that the work asked for is the second-order term in the Taylor expansion of the energy, represented geometrically by the tiny triangular area on the P-V diagram.

Enter Numerical Value:

Visualized Solution

\text{The Physical Setup}

\text{Defining the Work Done}

\text{Change in Volume } (dV)

\text{Adiabatic Process Condition}

\text{Calculating } dP

\text{Calculating External Work}

\text{Simplifying the Expression}

\text{Final Value of } X

\text{The Way Forward}

The Sigma Insight: Thermodynamic Processes

The Physical Setup

The Subtlety of "Work Done"

The Mathematics of Small Changes

The Adiabatic Constraint

Bringing It All Together

The Final Calculation

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