A bubble has surface tension S. The ideal gas inside the bubble has ratio of specific heats gamma =

Q: Why is the excess pressure $4S/r$ and not $2S/r$?

A bubble floating in the air is essentially a thin film of liquid enclosing a gas. This film has two surfaces that are in contact with air: the inner surface and the outer surface. Both surfaces contribute to the surface tension force, doubling the excess pressure compared to a solid liquid drop, which only has one surface.

Q: Why does 'perfect heat conductor' mean the process is isothermal?

If the bubble's surface conducts heat perfectly, any infinitesimal change in the gas's temperature will immediately cause heat to flow between the gas and the vast atmosphere until their temperatures equalize. Since the atmosphere is huge, its temperature doesn't change, forcing the gas to remain at a constant temperature throughout the process.

Q: In Option B, why isn't the total energy conserved?

The First Law of Thermodynamics states that energy is conserved for the *entire universe*, but for the bubble system, work is being done. As the bubble expands or contracts, the gas does work against the external atmospheric pressure ($P_a dV$). This means energy is transferred between the bubble and the atmosphere, so the sum of the gas's internal energy and the surface energy cannot remain constant.

Q: How did we get the relation $P^{1-\gamma}T^\gamma = \text{constant}$?

We start with the standard adiabatic equation $PV^\gamma = \text{constant}$. From the ideal gas law, we know $V \propto T/P$. Substituting this into the adiabatic equation gives $P(T/P)^\gamma = \text{constant}$, which simplifies to $P^{1-\gamma}T^\gamma = \text{constant}$.

Q: Why did Option A look so tempting but turn out wrong?

Option A perfectly executes the adiabatic volume-pressure relation mathematically, arriving at the correct exponents. However, it sets a physical trap by using $2S/r$ for the excess pressure. This is the formula for a liquid drop, not a bubble. It's a classic JEE trap testing if you are paying attention to the physical system while doing the math!

Question

A bubble has surface tension $S$. The ideal gas inside the bubble has ratio of specific heats $\gamma = \frac{5}{3}$. The bubble is exposed to the atmosphere and it always retains its spherical shape. When the atmospheric pressure is $P_{a1}$, the radius of the bubble is found to be $r_1$ and the temperature of the enclosed gas is $T_1$. When the atmospheric pressure is $P_{a2}$, the radius of the bubble and the temperature of the enclosed gas are $r_2$ and $T_2$, respectively.
Which of the following statement(s) is(are) correct?

Options:

(A) If the surface of the bubble is a perfect heat insulator, then $\left(\frac{r_1}{r_2}\right)^5 = \frac{P_{a2} + \frac{2S}{r_2}}{P_{a1} + \frac{2S}{r_1}}$.

(B) If the surface of the bubble is a perfect heat insulator, then the total internal energy of the bubble including its surface energy does not change with the external atmospheric pressure.

(C) If the surface of the bubble is a perfect heat conductor and the change in atmospheric temperature is negligible, then $\left(\frac{r_1}{r_2}\right)^3 = \frac{P_{a2} + \frac{4S}{r_2}}{P_{a1} + \frac{4S}{r_1}}$.

(D) If the surface of the bubble is a perfect heat insulator, then $\left(\frac{T_2}{T_1}\right)^{\frac{5}{2}} = \frac{P_{a2} + \frac{4S}{r_2}}{P_{a1} + \frac{4S}{r_1}}$.

Sigma Percentile · Accepted Answer

### The Beautiful Intersection of Fluids and Thermodynamics

Imagine a delicate soap bubble floating gently in the air. It seems so simple, yet it is a battleground of immense physical forces. On the outside, the vast atmosphere is pressing inward. On the inside, the trapped ideal gas is pushing outward. And right at the boundary, the liquid film itself is trying to collapse under its own surface tension.

This problem is a masterpiece because it forces us to bridge two entirely different worlds of physics: the mechanical properties of fluids and the laws of thermodynamics. Let's break down this beautiful system step by step.

### Analyzing the Setup: The Pressure Inside

Before we even look at the options, we must establish the fundamental mechanical equilibrium of the bubble. A bubble floating in the air is not a solid drop of liquid; it is a thin film enclosing a gas. This means it has **two surfaces**—an inner surface in contact with the trapped gas, and an outer surface in contact with the atmosphere.

Because surface tension acts on both of these surfaces, the excess pressure inside the bubble is double that of a liquid drop. The formula for this excess pressure is:
$$ \Delta P = \frac{4S}{r} $$

Therefore, the absolute pressure of the ideal gas inside the bubble, $P_{gas}$, must balance both the atmospheric pressure $P_a$ and this excess pressure:
$$ P_{gas} = P_a + \frac{4S}{r} $$

This is our master equation for the pressure state of the gas at any given radius $r$.

### The Isothermal Journey: Perfect Heat Conductor

Let's evaluate Option C. It proposes a scenario where the bubble's surface is a **perfect heat conductor**. What does this mean physically?

If the surface conducts heat perfectly, any slight change in the temperature of the gas inside will immediately cause heat to flow to or from the surrounding atmosphere. Since the atmosphere is infinitely large, its temperature remains constant. Consequently, the gas inside is forced to remain at that exact same constant temperature.

This means the gas undergoes an **Isothermal Process** ($T = 	ext{constant}$). For an isothermal process, Boyle's Law governs the relationship between pressure and volume:
$$ P_1 V_1 = P_2 V_2 $$

Let's substitute our master pressure equation and the volume of a sphere ($V = \frac{4}{3}\pi r^3$) into Boyle's Law:
$$ \left(P_{a1} + \frac{4S}{r_1}ight) \left(\frac{4}{3}\pi r_1^3ight) = \left(P_{a2} + \frac{4S}{r_2}ight) \left(\frac{4}{3}\pi r_2^3ight) $$

The geometric constant $\frac{4}{3}\pi$ beautifully cancels out from both sides. Rearranging the terms to isolate the ratio of the radii, we get:
$$ \left(\frac{r_1}{r_2}ight)^3 = \frac{P_{a2} + \frac{4S}{r_2}}{P_{a1} + \frac{4S}{r_1}} $$

This perfectly matches the expression in Option C! So, we have our first correct statement.

### The Adiabatic Journey: Perfect Heat Insulator

Now, let's shift our perspective to Options A and D. Here, the bubble's surface is a **perfect heat insulator**. This means no heat can enter or escape the bubble ($dQ = 0$). The gas is completely thermally isolated, which is the exact definition of an **Adiabatic Process**.

For an adiabatic process, the relationship between pressure and volume is governed by:
$$ P V^\gamma = 	ext{constant} $$

We are given the ratio of specific heats, $\gamma = \frac{5}{3}$. Let's test Option A by substituting our pressure and volume expressions:
$$ \left(P_{a1} + \frac{4S}{r_1}ight) \left(r_1^3ight)^{5/3} = \left(P_{a2} + \frac{4S}{r_2}ight) \left(r_2^3ight)^{5/3} $$

Notice how the exponent $3$ and $\frac{5}{3}$ multiply to give exactly $5$:
$$ \left(P_{a1} + \frac{4S}{r_1}ight) r_1^5 = \left(P_{a2} + \frac{4S}{r_2}ight) r_2^5 $$

Rearranging this gives:
$$ \left(\frac{r_1}{r_2}ight)^5 = \frac{P_{a2} + \frac{4S}{r_2}}{P_{a1} + \frac{4S}{r_1}} $$

Wait a minute! Look closely at Option A. It presents almost this exact equation, but it uses $\frac{2S}{r}$ instead of $\frac{4S}{r}$. This is a classic, devious trap set by the examiners. They are testing if you blindly apply the math or if you remember the physical reality that a bubble has two surfaces. Because of this subtle error, Option A is incorrect.

Let's move to Option D, which explores the temperature-pressure relationship for an adiabatic process:
$$ P^{1-\gamma} T^\gamma = 	ext{constant} $$

Substituting $\gamma = \frac{5}{3}$, the exponent for pressure becomes $1 - \frac{5}{3} = -\frac{2}{3}$.
$$ P_1^{-2/3} T_1^{5/3} = P_2^{-2/3} T_2^{5/3} $$

Let's group the temperatures and pressures:
$$ \left(\frac{T_2}{T_1}ight)^{5/3} = \left(\frac{P_2}{P_1}ight)^{2/3} $$

To isolate the temperature ratio, we raise both sides to the power of $\frac{3}{2}$:
$$ \left(\frac{T_2}{T_1}ight)^{5/2} = \frac{P_2}{P_1} $$

Finally, substituting our master pressure equation back into this ratio:
$$ \left(\frac{T_2}{T_1}ight)^{5/2} = \frac{P_{a2} + \frac{4S}{r_2}}{P_{a1} + \frac{4S}{r_1}} $$

This is a flawless match for Option D!

### The Energy Trap: Why Option B Fails

Finally, let's address Option B, which claims the total energy (internal + surface) remains constant during the adiabatic process.

According to the First Law of Thermodynamics, $dQ = dU + dW$. Since the process is adiabatic, $dQ = 0$, which means any change in internal energy must be perfectly balanced by the work done:
$$ dU_{gas} = -dW_{gas} $$

But what is the gas doing work against? As the bubble expands or contracts, the gas must push against the external atmospheric pressure, doing work equal to $P_a dV$. Simultaneously, it must stretch or compress the liquid film, changing its surface energy ($dU_{surface}$).

Therefore, the total work done by the gas is $dW_{gas} = P_a dV + dU_{surface}$. Substituting this back into our First Law equation:
$$ dU_{gas} = -(P_a dV + dU_{surface}) $$
$$ dU_{gas} + dU_{surface} = -P_a dV $$

The left side of this equation is the change in the total energy of the bubble system. Because the volume is changing ($dV 
eq 0$), the right side is not zero. The total energy of the bubble system is *not* conserved; energy is being exchanged with the atmosphere via mechanical work. Thus, Option B is incorrect.

### Final Thoughts

This problem is a brilliant exercise in maintaining situational awareness. It demands that you seamlessly switch between the mechanical reality of a two-surface film and the abstract mathematical laws of thermodynamics. By staying grounded in the physics, we successfully navigated the traps and arrived at the correct conclusions: Options C and D.

Select Answer:

Visualized Solution

$P_{g a s} = P_{a} + \frac{4 S}{r}$

$Perfect Heat Conductor$

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

$Simplifying Isothermal Equation$

$Perfect Heat Insulator$

$Checking Option A$

$Checking Option D$

$Checking Option B (Energy)$

$Conclusion$

The Sigma Insight: Surface Tension and Capillary Action

The Beautiful Intersection of Fluids and Thermodynamics

Analyzing the Setup

The Isothermal Journey

The Adiabatic Journey

The Energy Trap

Final Thoughts

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Pgas​=Pa​+r4S​

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

.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; } Simplifying Isothermal Equation

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Checking Option B (Energy)

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