Two soap bubbles `A` and `B` are kept in a closed chamber where the air is maintained at pressure `8 N//m^(2)`. The radii of bubbles `A` and `B` are `2 cm` and `4 cm`, respectively. Surface tension of the soap. Water used to make bubbles is `0.04 N//m`. Find the ratio `n_(B)//n_(A)`, where `n_(A)` and `n_(B)` are the number of moles of air in bubbles `A` and `B` respectively. [Neglect the effect of gravity.]

Explanation

To solve the problem of finding the ratio \( \frac{n_B}{n_A} \) for the two soap bubbles \( A \) and \( B \), we will follow these steps:

### Step 1: Calculate the pressure inside bubble A
The pressure inside a soap bubble is given by the formula:
\[ P_A = P_0 + \frac{4T}{R_A} \]
Where:
- \( P_0 = 8 \, \text{N/m}^2 \) (external pressure)
- \( T = 0.04 \, \text{N/m} \) (surface tension)
- \( R_A = 0.02 \, \text{m} \) (radius of bubble A, converted from cm to m)

Substituting the values:
\[ P_A = 8 + \frac{4 \times 0.04}{0.02} \]
\[ P_A = 8 + \frac{0.16}{0.02} = 8 + 8 = 16 \, \text{N/m}^2 \]

### Step 2: Calculate the pressure inside bubble B
Using the same formula for bubble B:
\[ P_B = P_0 + \frac{4T}{R_B} \]
Where \( R_B = 0.04 \, \text{m} \) (radius of bubble B).

Substituting the values:
\[ P_B = 8 + \frac{4 \times 0.04}{0.04} \]
\[ P_B = 8 + \frac{0.16}{0.04} = 8 + 4 = 12 \, \text{N/m}^2 \]

### Step 3: Calculate the volume of bubbles A and B
The volume \( V \) of a bubble is given by the formula:
\[ V = \frac{4}{3} \pi R^3 \]

For bubble A:
\[ V_A = \frac{4}{3} \pi (0.02)^3 = \frac{4}{3} \pi (0.000008) = \frac{32\pi}{3} \times 10^{-6} \, \text{m}^3 \]

For bubble B:
\[ V_B = \frac{4}{3} \pi (0.04)^3 = \frac{4}{3} \pi (0.000064) = \frac{256\pi}{3} \times 10^{-6} \, \text{m}^3 \]

### Step 4: Apply the ideal gas law for both bubbles
Using the ideal gas law \( PV = nRT \):
For bubble A:
\[ P_A V_A = n_A RT \]
\[ 16 \cdot \frac{32\pi}{3} \times 10^{-6} = n_A RT \]

For bubble B:
\[ P_B V_B = n_B RT \]
\[ 12 \cdot \frac{256\pi}{3} \times 10^{-6} = n_B RT \]

### Step 5: Find the ratio \( \frac{n_B}{n_A} \)
Dividing the equations for \( n_A \) and \( n_B \):
\[ \frac{n_A}{n_B} = \frac{16 \cdot \frac{32\pi}{3} \times 10^{-6}}{12 \cdot \frac{256\pi}{3} \times 10^{-6}} \]
The \( \frac{3\pi}{3\pi} \) and \( 10^{-6} \) cancel out:
\[ \frac{n_A}{n_B} = \frac{16 \cdot 32}{12 \cdot 256} \]
Simplifying:
\[ \frac{n_A}{n_B} = \frac{512}{3072} = \frac{1}{6} \]
Thus,
\[ \frac{n_B}{n_A} = 6 \]

### Final Answer
The ratio \( \frac{n_B}{n_A} = 6 \).

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