Excess pressure inside a soap bubble of radius r and surface tension T is

Options

A. `(2T)/r`
B. `(4T)/r`
C. `T/(2r)`
D. `T/(4r)`

Answer

To find the excess pressure inside a soap bubble of radius \( r \) and surface tension \( T \), we can follow these steps:

### Step 1: Understand the Concept of Excess Pressure
A soap bubble has two surfaces (the inner and outer surfaces), and the pressure inside the bubble is greater than the pressure outside due to surface tension. The excess pressure \( P \) inside the bubble can be defined as the difference between the internal pressure and the external pressure.

### Step 2: Define the Pressures
Let:
- \( P_0 \) = external pressure
- \( P \) = internal pressure

The excess pressure can be expressed as:
\[ P = P_{\text{inside}} - P_{\text{outside}} = P - P_0 \]

### Step 3: Relate Excess Pressure to Surface Tension
For a soap bubble, the excess pressure is related to the surface tension \( T \) and the radius \( r \) of the bubble. The formula for excess pressure inside a soap bubble is given by:
\[ P = \frac{4T}{r} \]
This is because there are two surfaces contributing to the pressure difference.

### Step 4: Derive the Formula
1. **Work Done by Surface Tension**: When the bubble expands by a small amount \( dr \), the increase in surface area is \( 2 \times 4\pi r^2 \) (since there are two surfaces).
2. **Force due to Excess Pressure**: The force due to the excess pressure acting on the area is given by:
\[ F = P \times \text{Area} = P \times 4\pi r^2 \]
3. **Work Done**: The work done in expanding the bubble is equal to the increase in surface energy due to the surface tension:
\[ \text{Work Done} = 2T \times \Delta \text{Area} = 2T \times (4\pi (r + dr)^2 - 4\pi r^2) \]
Simplifying this, we find:
\[ \Delta \text{Area} \approx 8\pi r dr \]
Therefore, the work done becomes:
\[ \text{Work Done} = 2T \times 8\pi r dr = 16T\pi r dr \]

### Step 5: Equate Work Done to Force
Setting the work done equal to the force times the distance gives:
\[ P \times 4\pi r^2 = 16T\pi r dr \]
Dividing both sides by \( 4\pi r \):
\[ P = \frac{16T}{4r} = \frac{4T}{r} \]

### Conclusion
Thus, the excess pressure inside a soap bubble of radius \( r \) and surface tension \( T \) is given by:
\[ P = \frac{4T}{r} \]

### Final Answer
The excess pressure inside a soap bubble is \( \frac{4T}{r} \).

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