Question

If two soap bubbles A and B of radii \(r_1\) and \(r_2\) respectively are kept in vaccum at constant temperature, then the ratio of masses of air inside the bubbles A and B is

• A. \(r_2^2 : r_1^2 \)
• B. \(r_1^2 : r_2^2 \)
• C. \(r_1 : r_2 \)
• D. \(r_2 : r_1 \)

Hint:

To solve problems involving soap bubbles and gas properties: 1. Pressure inside a bubble: Remember that the excess pressure inside a soap bubble (two surfaces) is \(\frac{4T}{r}\), while for a liquid drop (one surface) it's \(\frac{2T}{r}\). In vacuum, this excess pressure is the internal pressure. 2. Ideal Gas Law: Use \(PV = nRT\) or \(PV = \frac{m}{M}RT\). 3. Volume of a sphere: \(V = \frac{4}{3}\pi r^3\). Combine these equations and look for proportionalities to find the desired ratio.

Answer 1

The Correct Option is B

Step 1: Determine the pressure inside a soap bubble in vacuum.
For a soap bubble, there are two liquid-air interfaces. The excess pressure inside a soap bubble compared to the outside pressure is given by: \[ \Delta P = P_{in} - P_{out} = \frac{4T}{r} \] where \(T\) is the surface tension of the soap solution and \(r\) is the radius of the bubble.
Since the bubbles are kept in vacuum, the outside pressure \(P_{out} = 0\). Therefore, the pressure inside the bubble \(P_{in}\) is: \[ P_{in} = \frac{4T}{r} \] Step 2: Apply the ideal gas law to the air inside the bubble.
Assume the air inside the soap bubble behaves as an ideal gas. The ideal gas law is: \[ PV = nRT \] where \(P\) is the pressure, \(V\) is the volume, \(n\) is the number of moles, \(R\) is the ideal gas constant, and \(T\) is the absolute temperature.
The number of moles \(n\) can be expressed as \(n = \frac{m}{M}\), where \(m\) is the mass of the gas and \(M\) is the molar mass of the gas. So, the ideal gas law becomes: \[ P V = \frac{m}{M} RT \] We want to find the ratio of masses, so rearrange this equation to solve for \(m\): \[ m = \frac{P V M}{RT} \] Step 3: Substitute the pressure and volume of the bubble into the mass equation.
The volume of a spherical bubble is \(V = \frac{4}{3}\pi r^3\).
Substitute \(P_{in} = \frac{4T}{r}\) and \(V = \frac{4}{3}\pi r^3\) into the equation for \(m\): \[ m = \frac{\left(\frac{4T}{r}\right) \left(\frac{4}{3}\pi r^3\right) M}{RT_{temp}} \] (Note: Using \(T_{temp}\) for temperature to avoid confusion with surface tension \(T\)).
Simplify the expression: \[ m = \frac{16 \pi T M}{3RT_{temp}} \cdot \frac{r^3}{r} \] \[ m = \left(\frac{16 \pi T M}{3RT_{temp}}\right) r^2 \] Since the bubbles are at a constant temperature, and \(T\) (surface tension), \(M\) (molar mass of air), and \(R\) (gas constant) are constants, the term in the parenthesis is a constant. Let's call it \(K\). \[ m = K r^2 \] This shows that the mass of air inside the bubble is directly proportional to the square of its radius. 
Step 4: Determine the ratio of masses for bubbles A and B.
For bubble A with radius \(r_1\) and mass \(m_A\): \[ m_A = K r_1^2 \] For bubble B with radius \(r_2\) and mass \(m_B\): \[ m_B = K r_2^2 \] The ratio of their masses \(m_A : m_B\) is: \[ \frac{m_A}{m_B} = \frac{K r_1^2}{K r_2^2} = \frac{r_1^2}{r_2^2} \] Thus, the ratio of masses is \(r_1^2 : r_2^2\). The final answer is $\boxed{r_1^2 : r_2^2}$.