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.]

Options

A. 2
B. 4
C. 6
D. 8

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: Determine the Pressure Inside Each Bubble

The pressure inside a soap bubble can be calculated using the formula:
\[ P_{\text{inside}} = P_{\text{outside}} + \frac{4S}{R} \]
where:
- \( P_{\text{outside}} \) is the external pressure,
- \( S \) is the surface tension,
- \( R \) is the radius of the bubble.

Given:
- \( P_{\text{outside}} = 8 \, \text{N/m}^2 \)
- \( S = 0.04 \, \text{N/m} \)
- For bubble \( A \) (radius \( R_A = 2 \, \text{cm} = 0.02 \, \text{m} \)):
\[ P_{A} = 8 + \frac{4 \times 0.04}{0.02} = 8 + 8 = 16 \, \text{N/m}^2 \]
- For bubble \( B \) (radius \( R_B = 4 \, \text{cm} = 0.04 \, \text{m} \)):
\[ P_{B} = 8 + \frac{4 \times 0.04}{0.04} = 8 + 4 = 12 \, \text{N/m}^2 \]

### Step 2: Calculate the Volume of Each Bubble

The volume \( V \) of a sphere is given by:
\[ V = \frac{4}{3} \pi R^3 \]
- For bubble \( A \):
\[ V_A = \frac{4}{3} \pi (0.02)^3 = \frac{4}{3} \pi (8 \times 10^{-6}) = \frac{32\pi}{3} \times 10^{-6} \, \text{m}^3 \]
- For bubble \( B \):
\[ V_B = \frac{4}{3} \pi (0.04)^3 = \frac{4}{3} \pi (64 \times 10^{-6}) = \frac{256\pi}{3} \times 10^{-6} \, \text{m}^3 \]

### Step 3: Apply the Ideal Gas Law

Using the ideal gas law \( PV = nRT \), we can express the number of moles \( n \) for each bubble:
\[ n = \frac{PV}{RT} \]
Since \( R \) and \( T \) are constants, we can write the ratio of the number of moles:
\[ \frac{n_B}{n_A} = \frac{P_B V_B}{P_A V_A} \]

### Step 4: Substitute Values into the Ratio

Substituting the values we calculated:
\[ \frac{n_B}{n_A} = \frac{12 \times \frac{256\pi}{3} \times 10^{-6}}{16 \times \frac{32\pi}{3} \times 10^{-6}} \]
The \( \frac{\pi}{3} \) and \( 10^{-6} \) cancel out:
\[ \frac{n_B}{n_A} = \frac{12 \times 256}{16 \times 32} \]

### Step 5: Simplify the Ratio

Calculating the ratio:
\[ \frac{n_B}{n_A} = \frac{12 \times 256}{16 \times 32} = \frac{3072}{512} = 6 \]

Thus, the final result is:
\[ \frac{n_B}{n_A} = 6 \]

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

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