The pressure of air in a soap bubble of 0.7 cm diameter is 8 mm of water above the atmospheric pressure calculate the surface tension of soap solution. (take `g=980cm//sec^(2))`

Answer

To solve the problem, we need to calculate the surface tension of the soap solution using the given parameters. Let's break it down step by step.

### Step 1: Understand the Given Data
- Diameter of the soap bubble (d) = 0.7 cm
- Radius of the soap bubble (r) = d/2 = 0.7 cm / 2 = 0.35 cm
- Excess pressure (P) = 8 mm of water

### Step 2: Convert Excess Pressure to CGS Units
Since we need to work in CGS units, we convert the excess pressure from mm of water to cm of water:
- 8 mm of water = 0.8 cm of water

### Step 3: Calculate the Pressure in Dynes
The pressure in dynes can be calculated using the formula:
\[ P = h \cdot \rho \cdot g \]
Where:
- \( h \) = height of the water column (0.8 cm)
- \( \rho \) = density of water = 1 g/cm³ (in CGS units)
- \( g \) = acceleration due to gravity = 980 cm/s²

Thus, the excess pressure in dynes is:
\[ P = 0.8 \, \text{cm} \times 1 \, \text{g/cm}^3 \times 980 \, \text{cm/s}^2 = 784 \, \text{dyne/cm}^2 \]

### Step 4: Apply the Formula for Excess Pressure in a Soap Bubble
The formula for the excess pressure inside a soap bubble is given by:
\[ P = \frac{4T}{r} \]
Where \( T \) is the surface tension and \( r \) is the radius of the bubble.

### Step 5: Rearranging the Formula to Solve for Surface Tension
Rearranging the formula to find \( T \):
\[ T = \frac{P \cdot r}{4} \]

### Step 6: Substitute the Values
Substituting the values we have:
- \( P = 784 \, \text{dyne/cm}^2 \)
- \( r = 0.35 \, \text{cm} \)

So,
\[ T = \frac{784 \cdot 0.35}{4} \]

### Step 7: Calculate the Surface Tension
Calculating the above expression:
\[ T = \frac{274.4}{4} = 68.6 \, \text{dyne/cm} \]

### Final Answer
The surface tension of the soap solution is approximately **68.6 dyne/cm**.

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