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.]

Step 1: Identify the given parameters.

The chamber pressure, radii of the bubbles, and surface tension of the soap solution are given.

$$P_0 = 8\,\text{N/m}^{2}$$ $$R_A = 2\,\text{cm} = 0.02\,\text{m}$$ $$R_B = 4\,\text{cm} = 0.04\,\text{m}$$ $$S = 0.04\,\text{N/m}$$

Step 2: Calculate the internal pressure of bubble A.

Since a soap bubble has two surfaces, the excess pressure is $\Delta P = \frac{4S}{r}$.

$$P_A = P_0 + \frac{4S}{R_A}$$ $$P_A = 8 + \frac{4(0.04)}{0.02}$$ $$P_A = 8 + 8 = 16\,\text{N/m}^{2}$$

Step 3: Calculate the internal pressure of bubble B.

Apply the same excess pressure formula for bubble B.

$$P_B = P_0 + \frac{4S}{R_B}$$ $$P_B = 8 + \frac{4(0.04)}{0.04}$$ $$P_B = 8 + 4 = 12\,\text{N/m}^{2}$$

Step 4: Relate the number of moles to pressure and volume.

Using the ideal gas law, the number of moles $n$ is determined by pressure and volume at a constant temperature $T$.

$$n = \frac{PV}{RT}$$

Since the volume of a spherical bubble is $V = \frac{4}{3}\pi r^{3}$, the number of moles is proportional to $P r^{3}$.

$$n \propto P r^{3}$$

Step 5: Set up the ratio of the number of moles.

Express the ratio of the moles in bubble B to bubble A.

$$\frac{n_B}{n_A} = \frac{P_B}{P_A}\left(\frac{R_B}{R_A}\right)^{3}$$

Step 6: Calculate the final ratio.

Substitute the calculated pressures and given radii into the ratio formula.

$$\frac{P_B}{P_A} = \frac{12}{16} = \frac{3}{4}$$ $$\left(\frac{R_B}{R_A}\right)^{3} = \left(\frac{4}{2}\right)^{3} = 8$$ $$\frac{n_B}{n_A} = \frac{3}{4} \times 8 = 6$$

Answer:

The ratio $\dfrac{n_B}{n_A}$ is $6$.