Two spherical soap bubbles of radii r1 and r2 in vacuum combine under isothermal conditions. The res

Q: Why can't we just say the volumes add up, so $V_1 + V_2 = V$?

Volumes only add up if the pressure remains exactly the same throughout the process. Here, the pressure inside each bubble depends on its radius ($p = 4T/r$). Since the radii are different, the pressures are different, so you cannot simply add the volumes.

Q: Why is the pressure inside the bubble just $4T/r$? Shouldn't we add atmospheric pressure?

The problem explicitly states the bubbles are in a 'vacuum'. In a vacuum, the external atmospheric pressure $p_0$ is zero. Therefore, the total internal pressure is just the excess pressure $4T/r$.

Q: Why do we use $4T/r$ instead of $2T/r$?

A soap bubble has two liquid surfaces (an inner surface and an outer surface) with air trapped in between. Each surface contributes $2T/r$ to the excess pressure, making the total excess pressure $4T/r$. A liquid drop only has one surface, so its excess pressure is $2T/r$.

Q: The solution in some books says 'by conservation of energy $p_1V_1 + p_2V_2 = pV$'. Is that correct?

That is a common misconception. The equation $p_1V_1 + p_2V_2 = pV$ comes from the conservation of moles ($n_1 + n_2 = n$) combined with the ideal gas law at constant temperature. It is NOT conservation of energy. In fact, surface energy is lost when bubbles merge because the total surface area decreases.

Q: What would happen if this experiment was done in the open atmosphere instead of a vacuum?

If it were in the atmosphere, the pressure inside would be $p_0 + 4T/r$. The mole conservation equation would then become $(p_0 + 4T/r_1)r_1^3 + (p_0 + 4T/r_2)r_2^3 = (p_0 + 4T/r)r^3$. This would result in a much more complex cubic equation to solve for $r$.

Question

Two spherical soap bubbles of radii $r_1$ and $r_2$ in vacuum combine under isothermal conditions. The resulting bubble has a radius equal to

Options:

(A) $\frac{r_1 r_2}{r_1 + r_2}$

(B) $\sqrt{r_1 r_2}$

(C) $\sqrt{r_1^2 + r_2^2}$

(D) $\frac{r_1 + r_2}{2}$

Sigma Percentile · Accepted Answer

Imagine this... two soap bubbles floating in a vacuum. They have radii $r_1$ and $r_2$. Now, they gently collide and merge into a single, larger bubble of radius $r$. How do we find this new radius?

### The Trap: Is Surface Energy Conserved?
Many students fall into the trap of assuming that the total surface energy of the bubbles is conserved. However, when bubbles merge, the total surface area decreases, which means surface energy is actually released (often as heat). Therefore, we cannot equate the initial and final surface energies.

### The Core Principle: Conservation of Moles
The problem states this happens under **isothermal conditions**. This means the temperature $T$ remains constant throughout the process. Furthermore, since the bubbles are in a vacuum, the air trapped inside these initial bubbles cannot escape when they form the final bubble.

This leads us to our master equation: the total number of moles of gas is conserved.

$$n_1 + n_2 = n$$

From the ideal gas law, we know that the number of moles is equal to pressure times volume divided by $RT$ ($n = \frac{pV}{RT}$). Since temperature $T$ is constant, we can rewrite our conservation equation as:

$$\frac{p_1 V_1}{RT} + \frac{p_2 V_2}{RT} = \frac{pV}{RT}$$

Multiplying throughout by $RT$, we get a beautiful relationship:

$$p_1 V_1 + p_2 V_2 = pV$$

### Analyzing the Setup
Now, what is the pressure inside a soap bubble in a vacuum? The total pressure inside a bubble is the sum of the outside pressure and the excess pressure due to surface tension. Since it's in a vacuum, the outside pressure $p_0 = 0$.

Therefore, the pressure inside is simply the excess pressure for a soap bubble (which has two surfaces):

$$p = \frac{4T}{r}$$

And the volume of a spherical bubble is:

$$V = \frac{4}{3}\pi r^3$$

### Final Calculation
Let's substitute these values into our conservation equation. For the first bubble, the second bubble, and the final bubble, we get:

$$\left(\frac{4T}{r_1}\right)\left(\frac{4}{3}\pi r_1^3\right) + \left(\frac{4T}{r_2}\right)\left(\frac{4}{3}\pi r_2^3\right) = \left(\frac{4T}{r}\right)\left(\frac{4}{3}\pi r^3\right)$$

Look closely at the equation... the terms $4T$ and $\frac{4}{3}\pi$ are common on both sides and will cancel out perfectly. Additionally, dividing $r^3$ by $r$ leaves us with $r^2$ for each term.

This simplifies our massive equation down to:

$$r_1^2 + r_2^2 = r^2$$

Taking the square root, we get the radius of the new bubble:

**$$r = \sqrt{r_1^2 + r_2^2}$$**

And that is our final answer! The physics elegantly boils down to a Pythagorean-like relationship between the radii.

Animated Solution for Physics - Properties of Solids and Liquids: Two spherical soap bubbles of radii $r_{1}$ and $r_{2}$ in vacuum combine under isothermal conditions. The resulting bubble has a radius equal to

Select Answer:

Visualized Solution

$r_{1}, r_{2} \to r$

$n_{1} + n_{2} = n$

$p_{1} V_{1} + p_{2} V_{2} = p V$

$p = \frac{4 T}{r}$

$Substitution$

$r_{1}^{2} + r_{2}^{2} = r^{2}$

$r = r_{1}^{2} + r_{2}^{2}$

The Sigma Insight: Surface Tension and Capillary Action

The Trap

The Core Principle

Analyzing the Setup Now, what is the pressure inside a soap bubble in a vacuum? The total pressure inside a bubble is the sum of the outside pressure and the excess pressure due to surface tension

Final Calculation Let's substitute these values into our conservation equation

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Select Answer:

Visualized Solution

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n1​+n2​=n

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p1​V1​+p2​V2​=pV

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substitution

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