Question

Two spherical soap bubbles of radii (a) and (b) in vacuum coalesce under isothermal conditions. The resulting bubble has a radius given by:
\[A.\quad \dfrac{\left( a+b \right)}{2}\]
\[B.\quad \dfrac{ab}{a+b}\]
\[C.\quad \sqrt{{{a}^{2}}+{{b}^{2}}}\]
\[D.\quad a+b\]

Answer 1

Hint: This question needs the knowledge of surface tension and basic thermodynamics to solve this problem. The surface tension formula is: \[{{T}_{s}}=\dfrac{{{F}_{ST}}}{l}\] and the formula for finding the pressure inside a soap bubble of radius R is: ${{P}_{i}}=\dfrac{4{{T}_{s}}}{R}$. The ideal gas equation is: $PV=nRT$ and for isothermal condition: ${{P}_{1}}{{V}_{1}}={{P}_{2}}{{V}_{2}}$.

Step by step solution:
Let’s start by understanding what is surface tension after all. As the word suggests, it is linked to a surface of the liquid. The surface tension is a physical property of the liquids only, which happens when the molecules from all the sides are drawn closer. This inward attracting force by all the molecules is known as the surface tension $\left( {{F}_{ST}} \right)$, which is given by the dragging force per unit length. That is the surface tension $\left( {{T}_{s}} \right)$ is, \[{{T}_{s}}=\dfrac{{{F}_{ST}}}{l}\].
Let’s consider the pressure inside the liquid drop to be \[{{P}_{i}}\] and the pressure outside the drop to be ${{P}_{o}}$. The excess pressure inside the liquid drop is given by: $\Delta P={{P}_{i}}-{{P}_{o}}$. Due to this excess pressure, the drop tries to expand outwards. We know that the pressure inside a soap bubble of radius R is: ${{P}_{i}}=\dfrac{4{{T}_{s}}}{R}$.
From the question, we have that the system is in vacuum. Hence, the pressure outside is zero. Therefore; $\Delta P={{P}_{i}}-{{P}_{o}}\Rightarrow \Delta P={{P}_{i}}-0$. That is, the pressure inside the soap bubbles will be; ${{P}_{i}}=\dfrac{4{{T}_{s}}}{R}$. Further, the initial radii of the bubbles are (a) and (b) respectively. Let’s consider the radius of the coalesced bubble to be R. Below is a diagram of the system as per the problem.

Using the Isothermal condition given by: ${{P}_{1}}{{V}_{1}}={{P}_{2}}{{V}_{2}}$, we understand that the sum of the products before coalescing will be the same as after coalescing. That is; ${{P}_{1}}{{V}_{1}}(a)+{{P}_{2}}{{V}_{2}}(b)={{P}_{3}}{{V}_{3}}(R)\to (1)$.
Using the known value of pressure from above and the volume of a sphere$\left( {{V}_{sphere}}=\dfrac{4}{3}\pi {{R}^{3}} \right)$, equation (1) becomes: $\dfrac{4{{T}_{s}}}{a}\left( \dfrac{4}{3}\pi {{a}^{3}} \right)+\dfrac{4{{T}_{s}}}{b}\left( \dfrac{4}{3}\pi {{b}^{3}} \right)=\dfrac{4{{T}_{s}}}{R}\left( \dfrac{4}{3}\pi {{R}^{3}} \right)$. All the soap bubbles have the same surface tension value.
Therefore; $\dfrac{4{{T}_{s}}}{a}\left( \dfrac{4}{3}\pi {{a}^{3}} \right)+\dfrac{4{{T}_{s}}}{b}\left( \dfrac{4}{3}\pi {{b}^{3}} \right)=\dfrac{4{{T}_{s}}}{R}\left( \dfrac{4}{3}\pi {{R}^{3}} \right)\Rightarrow {{a}^{2}}+{{b}^{2}}={{R}^{2}}\Rightarrow R=\sqrt{{{a}^{2}}+{{b}^{2}}}$.
Hence, the radius of the coalesced bubble through Isothermal condition is: $R=\sqrt{{{a}^{2}}+{{b}^{2}}}$, which is given by Option C.

Note:
It’s important to remember that the pressure difference inside a liquid drop is: $\Delta P=\dfrac{2{{T}_{s}}}{R}$ as per the derivation above. However, the pressure difference inside a soap bubble is: $\Delta P=\dfrac{4{{T}_{s}}}{R}$. This value is twice that of the pressure difference of a liquid drop.
The reason due to this difference is, in a water drop, the inner surface consists of water and the surface also consists of water and the outer surface of the other medium such as air. However, for a soap bubble, the inner surface is of air then the surface of the liquid making the bubble and then the outer surface again consists of any other medium such as air and in the question it was vacuum.