Mole Ratio of Air Inside Two Connected Soap Bubbles

To find the ratio of the number of moles of air inside the two soap bubbles, $\frac{n_B}{n_A}$, we use the ideal gas equation:
$P V = n R T$

Assuming both bubbles are in thermal equilibrium with the chamber at the same temperature $T$:
$n \propto P V$

Therefore, the ratio of the number of moles is:
$\frac{n_B}{n_A} = \frac{P_B V_B}{P_A V_A} = \frac{P_B}{P_A} \left(\frac{r_B}{r_A}\right)^3$

A soap bubble has two liquid-gas interfaces (an inner surface and an outer surface). Hence, the excess pressure inside a soap bubble of radius $r$ is given by:
$\Delta P = \frac{4T}{r}$

where $T$ is the surface tension.

The absolute pressure inside each bubble is the sum of the chamber's atmospheric pressure $P_0$ and the excess pressure:
$P = P_0 + \frac{4T}{r}$

Given:

• Chamber pressure, $P_0 = 8\text{ N/m}^2$
• Surface tension, $T = 0.04\text{ N/m}$
• Radius of bubble A, $r_A = 2\text{ cm} = 0.02\text{ m}$
• Radius of bubble B, $r_B = 4\text{ cm} = 0.04\text{ m}$

Now, we calculate the pressure inside bubble A ($P_A$):
$P_A = 8 + \frac{4 \times 0.04}{0.02} = 8 + 8 = 16\text{ N/m}^2$

Next, we calculate the pressure inside bubble B ($P_B$):
$P_B = 8 + \frac{4 \times 0.04}{0.04} = 8 + 4 = 12\text{ N/m}^2$

Now, substituting these pressures and the radii into the mole ratio formula:
$\frac{n_B}{n_A} = \left(\frac{P_B}{P_A}\right) \left(\frac{r_B}{r_A}\right)^3 = \left(\frac{12}{16}\right) \left(\frac{4}{2}\right)^3 = \frac{3}{4} \times 8 = 6$