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Question

If the surface tension at the soap air interface is 0.09 N/m, then estimate the internal pressure (in N/m2 above atmospheric pressure) in a soap bubble of 0.03 m diameter.

The correct answer is

24

Understanding Soap Bubble Internal Pressure

This question asks us to estimate the internal pressure inside a soap bubble, specifically the pressure above the surrounding atmospheric pressure. We are given the surface tension of the soap-air interface and the diameter of the soap bubble.

Surface Tension and Soap Bubbles

Surface tension is a property of liquid surfaces that causes them to behave like an elastic membrane under tension. It is defined as the force per unit length acting perpendicular to a line drawn in the surface of the liquid. In the context of bubbles and drops, surface tension creates an inward force that results in higher pressure inside the bubble or drop compared to the outside.

A key distinction is made between a liquid drop and a soap bubble:

  • A liquid drop (like a raindrop) has only one free surface (the outer surface exposed to air).
  • A soap bubble has two free surfaces (an inner surface and an outer surface, both in contact with air). This is because a soap bubble is a thin film of liquid with air inside and air outside.

Because of these two surfaces, the excess pressure formula for a soap bubble is different from that of a liquid drop.

Given Parameters for Soap Bubble Calculation

Let's list the values provided in the question:

  • Surface tension of the soap-air interface (\(\sigma\)) = \(0.09 \, \text{N/m}\)
  • Diameter of the soap bubble (D) = \(0.03 \, \text{m}\)

Calculating Internal Pressure in a Soap Bubble

The excess pressure (P) inside a soap bubble above the external atmospheric pressure is given by the formula:

\[P = \frac{4\sigma}{R}\]

Where:

  • P is the excess internal pressure.
  • \(\sigma\) (sigma) is the surface tension of the liquid film.
  • R is the radius of the soap bubble.

Since the diameter (D) is given, we know that the radius \(R = \frac{D}{2}\).

Substituting \(R = \frac{D}{2}\) into the formula, we get:

\[P = \frac{4\sigma}{D/2} = \frac{8\sigma}{D}\]

Step-by-Step Internal Pressure Calculation

Now, let's substitute the given values into the formula to find the internal pressure:

  • Surface tension (\(\sigma\)) = \(0.09 \, \text{N/m}\)
  • Diameter (D) = \(0.03 \, \text{m}\)

Applying the formula for excess pressure in a soap bubble:

\[P = \frac{8\sigma}{D}\]

\[P = \frac{8 \times 0.09 \, \text{N/m}}{0.03 \, \text{m}}\]

First, calculate the numerator:

\(8 \times 0.09 = 0.72\)

So, the equation becomes:

\[P = \frac{0.72 \, \text{N}}{0.03 \, \text{m}^2}\]

Now, perform the division:

\(P = \frac{0.72}{0.03} = \frac{72}{3} = 24\)

The unit for pressure is Newtons per square meter (\(\text{N/m}^2\)), which is also known as Pascal (Pa).

Therefore, the internal pressure above atmospheric pressure is \(24 \, \text{N/m}^2\).

Final Internal Pressure Result

Based on the calculations, the internal pressure above atmospheric pressure in the soap bubble is \(24 \, \text{N/m}^2\).

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Important Questions from Surface Tension

  1. Which of the following statements is NOT correct about surface tension?

  2. If a liquid droplet and a soap bubble are formed from the same liquid and have the same radius $R$, how does the excess pressure inside the soap bubble ($\Delta P_{bubble}$) compare to the excess pressure inside the liquid droplet ($\Delta P_{droplet}$)?

  3. Mercury does NOT wet the glass. This is due to the property of the liquid known as

  4. ________ is a surface phenomenon.

  5. A liquid drop is spherical in shape due to

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