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 of calculating the surface tension of a soap solution in a soap bubble, we will follow these steps:

### Step 1: Convert the given diameter to radius
The diameter of the soap bubble is given as 0.7 cm. The radius \( R \) is half of the diameter.

\[ R = \frac{D}{2} = \frac{0.7 \, \text{cm}}{2} = 0.35 \, \text{cm} \]

### Step 2: Convert the excess pressure from mm of water to Pascals
The excess pressure \( \Delta P \) is given as 8 mm of water. We need to convert this to Pascals.

Using the conversion factor:
\[ 1 \, \text{mm of water} = \frac{1.01 \times 10^5 \, \text{Pa}}{760 \, \text{mm}} \approx 133.322 \, \text{Pa} \]

Thus, the excess pressure in Pascals is:
\[ \Delta P = 8 \, \text{mm} \times \frac{133.322 \, \text{Pa}}{1 \, \text{mm}} = 1066.576 \, \text{Pa} \]

### Step 3: Use the formula for excess pressure in a soap bubble
The formula for the excess pressure in a soap bubble is given by:
\[ \Delta P = \frac{4T}{R} \]
Where \( T \) is the surface tension and \( R \) is the radius of the bubble.

### Step 4: Rearrange the formula to solve for surface tension \( T \)
Rearranging the formula gives us:
\[ T = \frac{\Delta P \cdot R}{4} \]

### Step 5: Substitute the values into the equation
Now we can substitute the values we have calculated:
\[ T = \frac{1066.576 \, \text{Pa} \cdot 0.35 \, \text{cm}}{4} \]
Note: Convert \( R \) from cm to meters for consistency in units:
\[ R = 0.35 \, \text{cm} = 0.0035 \, \text{m} \]

Now substituting:
\[ T = \frac{1066.576 \, \text{Pa} \cdot 0.0035 \, \text{m}}{4} = \frac{3.733016}{4} = 0.933254 \, \text{N/m} \]

### Step 6: Final answer
The surface tension \( T \) of the soap solution is approximately:
\[ T \approx 0.933 \, \text{N/m} \]

### Summary
The surface tension of the soap solution is approximately \( 0.933 \, \text{N/m} \).

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