A spherical soap bubble inside an air chamber at pressure P0 = 10^5 Pa has a certain radius so that

Q: Why do we ignore $\Delta P$ when applying Boyle's Law?

Because $144 \text{ Pa}$ is negligible compared to $10^5 \text{ Pa}$. Including it would make the calculation unnecessarily complex without significantly changing the final answer. The problem explicitly tells us to assume $\Delta P$ is much smaller than the chamber pressure to encourage this approximation.

Q: Why is the excess pressure $\frac{4T}{r}$ and not $\frac{2T}{r}$?

A soap bubble is essentially a thin film of liquid with air on both the inside and the outside. Therefore, it has two surface areas (inner and outer) contributing to the surface tension forces. An air bubble inside a liquid only has one surface, so its excess pressure is $\frac{2T}{r}$.

Q: Does the surface tension $T$ change during this process?

No. Surface tension is primarily a function of temperature for a given liquid. Since the problem states that all temperatures remain unchanged, the surface tension $T$ remains constant.

Q: Why did the bubble expand?

When the external chamber pressure was reduced, the pressure inside the bubble was temporarily higher than the outside pressure. This imbalance caused the gas inside to expand, increasing the volume and radius until the internal pressure dropped enough to restore equilibrium.

Q: What if the temperature was not constant?

If the temperature changed, we could no longer use Boyle's Law ($P_1V_1 = P_2V_2$). We would have to use the combined ideal gas law $\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$, and we would need to know the initial and final temperatures to solve the problem.

Question

A spherical soap bubble inside an air chamber at pressure $P_0 = 10^5 	ext{ Pa}$ has a certain radius so that the excess pressure inside the bubble is $\Delta P = 144 	ext{ Pa}$. Now, the chamber pressure is reduced to $\frac{8P_0}{27}$ so that the bubble radius and its excess pressure change. In this process, all the temperatures remain unchanged. Assume air to be an ideal gas and the excess pressure $\Delta P$ in both the cases to be much smaller than the chamber pressure. The new excess pressure $\Delta P$ in $	ext{Pa}$ is

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing inside a massive air chamber, observing a delicate soap bubble floating in equilibrium. The pressure inside the chamber is a hefty $P_0 = 10^5 	ext{ Pa}$. The bubble maintains its shape because the pressure inside it is slightly higher than the outside pressure. This difference is the **excess pressure**, given as $\Delta P = 144 	ext{ Pa}$.

Now, the problem introduces a critical assumption: $\Delta P \ll P_0$. This means the total pressure inside the bubble, which is $P_{in} = P_0 + \Delta P$, is practically just $P_0$. This tiny approximation is the secret key that unlocks the entire problem without getting bogged down in messy algebra.

### The Master Equation

Suddenly, the chamber pressure is reduced to $\frac{8P_0}{27}$. Because the temperature remains constant, the air trapped inside the bubble must obey **Boyle's Law**, which states that for an isothermal process, $P 	imes V = 	ext{constant}$.

Let's set up our master equation by equating the initial and final states of the air inside the bubble:

$$P_1 V_1 = P_2 V_2$$

Substituting our known values, we get:

$$P_0 \left(\frac{4}{3}\pi r_1^3ight) = \left(\frac{8P_0}{27}ight) \left(\frac{4}{3}\pi r_2^3ight)$$

Notice the beautiful cancellation! The $P_0$ and the $\frac{4}{3}\pi$ terms vanish from both sides, leaving us with a pure geometric relationship:

$$r_1^3 = \frac{8}{27} r_2^3$$

Taking the cube root of both sides reveals how much the bubble has expanded:

$$r_2 = \frac{3}{2} r_1$$

### Final Calculation

Now that we know the new radius, we need to find the new excess pressure. For a soap bubble, which has two surfaces (inner and outer) in contact with air, the excess pressure is governed by the surface tension $T$:

$$\Delta P = \frac{4T}{r}$$

This tells us that excess pressure is inversely proportional to the radius. Let's find the new excess pressure $\Delta P'$:

$$\Delta P' = \frac{4T}{r_2} = \frac{4T}{\frac{3}{2}r_1}$$

By pulling the fraction out, we can express the new excess pressure in terms of the old one:

$$\Delta P' = \frac{2}{3} \left(\frac{4T}{r_1}ight) = \frac{2}{3} \Delta P$$

Finally, we substitute the original excess pressure value of $144 	ext{ Pa}$:

$$\Delta P' = \frac{2}{3} 	imes 144 = 96 	ext{ Pa}$$

The bubble expanded, and as a result, its excess pressure dropped to **$96 	ext{ Pa}$**. A perfect harmony of thermodynamics and fluid mechanics!

Enter Numerical Value:

Visualized Solution

$P_{0}, Δ P, r_{1}$

$P_{in} \approx P_{c hamb er}$

$P_{1} V_{1} = P_{2} V_{2}$

$r_{2} = \frac{3}{2} r_{1}$

$Δ P^{'} = \frac{4 T}{r _{2}}$

$Δ P^{'} = 96$

$Δ P \propto \frac{1}{r}$

The Sigma Insight: Surface Tension and Capillary Action

Analyzing the Setup

The Master Equation

Final Calculation

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P0​,ΔP,r1​

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P1​V1​=P2​V2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r2​=23​r1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ΔP′=r2​4T​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ΔP′=96

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ΔP∝r1​