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Chapter 1: Problem 88

A liquid with a surface tension of \(0.072 \mathrm{~N} / \mathrm{m}\) is used to form 60 -mm-diameter bubbles in air. What is the difference between the air pressure inside the bubble and the air pressure outside the bubble?

Short Answer

Expert verified
The pressure difference is 9.6 N/m虏.

Step by step solution

01

Understand the Concept

To find the pressure difference across the bubble, we use the formula for excess pressure in a bubble: \( \Delta P = \frac{4\sigma}{r} \), where \( \sigma \) is the surface tension, and \( r \) is the radius of the bubble.
02

Convert Diameter to Radius

The given diameter of the bubble is 60 mm. First, convert the diameter from millimeters to meters (since SI units are required) and then find the radius: \( r = \frac{60 \text{ mm}}{2} = 30 \text{ mm} = 0.03 \text{ m} \).
03

Substitute the Values

Now that you have the values, substitute \( \sigma = 0.072 \mathrm{~N} / \mathrm{m} \) and \( r = 0.03 \mathrm{~m} \) into the formula \( \Delta P = \frac{4\sigma}{r} \).
04

Calculate the Pressure Difference

Calculate \( \Delta P = \frac{4 \times 0.072}{0.03} = \frac{0.288}{0.03} = 9.6 \text{ N/m}^2 \).
05

Conclusion

The pressure difference between the inside and the outside of the bubble is \( 9.6 \text{ N/m}^2 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Surface Tension in Bubble Formation
Surface tension is a physical phenomenon that occurs at the surface of a liquid due to molecular forces. It is the force that makes it easier for the bubble to maintain its shape, as the molecules on the surface create a kind of skin.
This tension acts along the surface, making the liquid droplet behave as if it were covered with an elastic membrane.
  • The value of surface tension is crucial in determining the pressure required to keep a bubble stable and intact.
  • The higher the surface tension, the more force it takes to expand the bubble further.
  • In the exercise, the surface tension of the liquid used is given as \(0.072 \text{ N/m}\), which will influence the pressure inside the bubble.
Understanding surface tension helps us grasp why a certain amount of pressure is needed within a bubble and how bubbles behave when immersed in air or another gaseous environment.
Calculating the Pressure Difference Inside a Bubble
When explaining pressure differences in bubbles, it's vital to understand that the pressure inside the bubble must be greater than the pressure outside to maintain shape. This phenomenon is described by the formula \( \Delta P = \frac{4\sigma}{r} \).
Here, \( \Delta P \) represents the pressure difference, \( \sigma \) is the surface tension, and \( r \) is the radius of the bubble.
  • This relationship was applied in the exercise, where with a surface tension of \(0.072 \text{ N/m} \) and a radius of \(0.03 \text{ m}\), the pressure difference was calculated as \(9.6 \text{ N/m}^2\).
  • Bubbles in air need this internal overpressure; otherwise, they would collapse due to atmospheric pressures surrounding them.
  • This principle helps in understanding practical applications like the behavior of soap bubbles and the design of medical bubbles used in treatments.
Understanding this pressure difference is essential for anyone studying fluid mechanics or working with bubble dynamics.
Converting Radius from Diameter in Calculations
Converting measurements accurately is crucial in scientific work, particularly when transitioning from one set of units to another. In the exercise, students were required to convert the diameter of the bubble into a radius to use in the pressure difference formula. The diameter was given as \(60 \text{ mm}\), which was converted into meters, resulting in a radius of \(0.03 \text{ m}\).
This conversion involved a few simple steps:
  • Divide the diameter by two to get the radius, since a radius is half the diameter.
  • Convert the unit from millimeters to meters, as consistent unit usage is essential in formulas for accuracy. This included changing \(30 \text{ mm} \) into \(0.03 \text{ m}\), ensuring SI units were used throughout.
Correct conversion is vital in calculations because any discrepancy can lead to significant errors in derived values like pressure difference or force.

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Most popular questions from this chapter

Use the standard atmosphere to estimate the temperature and pressure at the top of Mount Everest, which is approximately \(8840 \mathrm{~m}\) above sea level. Compare your estimated values with values reported in the open literature.

A 1.5-mm-diameter capillary tube is inserted in a liquid, and it is observed that the liquid rises \(15 \mathrm{~mm}\) in the tube and has a contact angle of \(15^{\circ}\) with the surface of the glass tube. If a hydrometer indicates that the liquid has a specific gravity of \(0.8,\) what is the surface tension of the liquid? Would you expect that this same surface tension would be found if the experiment was done using a tube material other than glass? Explain.

A fluid with a specific gravity of 0.92 has a kinematic viscosity of \(5 \times 10^{-4} \mathrm{~m}^{2} / \mathrm{s}\) What is the dynamic viscosity of the fluid?

If \(4 \mathrm{~L}\) of a liquid with density \(1020 \mathrm{~kg} / \mathrm{m}^{3}\) is mixed with \(6 \mathrm{~L}\) of a liquid with density \(940 \mathrm{~kg} / \mathrm{m}^{3},\) what is the density of the mixture?

A car tire has a volume of \(20 \mathrm{~L}\) and has a recommended (gauge) inflation pressure of \(210 \mathrm{kPa}\) at a temperature of \(25^{\circ} \mathrm{C}\). (a) If driving on the highway on a hot day causes the temperature of the air in the tire to increase to \(65^{\circ} \mathrm{C},\) what will be the (gauge) air pressure in the tire? (b) If under the condition in part (a) air is let out of the tire to restore the tire pressure to \(210 \mathrm{kPa}\), what will be the (gauge) air pressure in the tire when the air cools down to \(25^{\circ} \mathrm{C}\) ? Assume that the tire volume remains constant and that atmospheric pressure is \(101.3 \mathrm{kPa}\).
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