A spherical soap bubble inside an air chamber at pressure P_0=10^5 ~Pa has a certain radius so that the excess pressure inside the bubble is \Delta P=144 ~Pa . Now, the chamber pressure is reduced to 8 P_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 Pa is \underline\hspace2cm .

Heat and Thermodynamics PYQ — JEE Advanced 2024 · Shift 2 · Q47 with Solution

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Correct answer: 96

Initial condition for a soap bubble

For a soap bubble, excess pressure is

rac{4T}{R}= ext{excess pressure}= ext{inside pressure} - ext{outside pressure}.

Initially,

ext{outside pressure}=P_0=10^5\,\text{Pa},

ext{excess pressure}=\Delta P_1=144\,\text{Pa}.

So the inside pressure initially is

P_{\text{in},1}=P_0+\Delta P_1.

Final chamber pressure

The chamber pressure is reduced to

P_2=\frac{8P_0}{27}.

Let the new excess pressure be $\Delta P_2$ and new radius be $R_2$ . Then final inside pressure is

P_{\text{in},2}=P_2+\Delta P_2.

Use ideal gas law under isothermal change

Temperature remains unchanged, and the amount of gas inside the bubble remains constant. Hence

P_{\text{in}}V=\text{constant}.

Since the bubble is spherical,

V\propto R^3.

For a soap bubble,

\Delta P=\frac{4T}{R}\implies R\propto \frac{1}{\Delta P}.

Therefore,

V\propto R^3\propto \frac{1}{(\Delta P)^3}.

So,

P_{\text{in}}\cdot \frac{1}{(\Delta P)^3}=\text{constant}.

Thus,

\frac{P_{\text{in},1}}{(\Delta P_1)^3}= \frac{P_{\text{in},2}}{(\Delta P_2)^3}.

Use the approximation $\Delta P \ll$ chamber pressure

Given $\Delta P$ is much smaller than chamber pressure in both cases, we can take

P_{\text{in},1}\approx P_0, \qquad P_{\text{in},2}\approx P_2=\frac{8P_0}{27}.

Hence,

\frac{P_0}{(144)^3}= \frac{\frac{8P_0}{27}}{(\Delta P_2)^3}.

Cancel $P_0$ :

\frac{1}{144^3}= \frac{8/27}{(\Delta P_2)^3}.

So,

(\Delta P_2)^3=144^3\cdot \frac{8}{27}.

Now,

\frac{8}{27}=\left(\frac{2}{3}\right)^3.

Therefore,

\Delta P_2=144\cdot \frac{2}{3}=96\,\text{Pa}.

Final answer

\boxed{96}

The derived answer matches the stored correct answer.

More from Heat and Thermodynamics

View written solutionFree

Correct answer: 96

Initial condition for a soap bubble

For a soap bubble, excess pressure is

rac{4T}{R}= ext{excess pressure}= ext{inside pressure} - ext{outside pressure}.

Initially,

ext{outside pressure}=P_0=10^5\,\text{Pa},

ext{excess pressure}=\Delta P_1=144\,\text{Pa}.

So the inside pressure initially is

P_{\text{in},1}=P_0+\Delta P_1.

Final chamber pressure

The chamber pressure is reduced to

P_2=\frac{8P_0}{27}.

Let the new excess pressure be $\Delta P_2$ and new radius be $R_2$ . Then final inside pressure is

P_{\text{in},2}=P_2+\Delta P_2.

Use ideal gas law under isothermal change

Temperature remains unchanged, and the amount of gas inside the bubble remains constant. Hence

P_{\text{in}}V=\text{constant}.

Since the bubble is spherical,

V\propto R^3.

For a soap bubble,

\Delta P=\frac{4T}{R}\implies R\propto \frac{1}{\Delta P}.

Therefore,

V\propto R^3\propto \frac{1}{(\Delta P)^3}.

So,

P_{\text{in}}\cdot \frac{1}{(\Delta P)^3}=\text{constant}.

Thus,

\frac{P_{\text{in},1}}{(\Delta P_1)^3}= \frac{P_{\text{in},2}}{(\Delta P_2)^3}.

Use the approximation $\Delta P \ll$ chamber pressure

Given $\Delta P$ is much smaller than chamber pressure in both cases, we can take

P_{\text{in},1}\approx P_0, \qquad P_{\text{in},2}\approx P_2=\frac{8P_0}{27}.

Hence,

\frac{P_0}{(144)^3}= \frac{\frac{8P_0}{27}}{(\Delta P_2)^3}.

Cancel $P_0$ :

\frac{1}{144^3}= \frac{8/27}{(\Delta P_2)^3}.

So,

(\Delta P_2)^3=144^3\cdot \frac{8}{27}.

Now,

\frac{8}{27}=\left(\frac{2}{3}\right)^3.

Therefore,

\Delta P_2=144\cdot \frac{2}{3}=96\,\text{Pa}.

Final answer

\boxed{96}

The derived answer matches the stored correct answer.

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