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Chapter 9: Problem 69

An underwater air bubble has an excess inside pressure of \(10 \mathrm{Pa}\). What is the excess pressure inside an air bubble with twice the radius?

Short Answer

Expert verified
Answer: The excess pressure inside the second air bubble, with twice the radius of the initial bubble, is 5 Pa.

Step by step solution

01

Understand the Young-Laplace Equation

The Young-Laplace equation states that the excess pressure inside an air bubble is given by: $$P = 2 \frac{T}{R}$$ where \(P\) is the excess pressure inside the bubble, \(T\) is the surface tension of the liquid, and \(R\) is the radius of the bubble.
02

Set up the equation for the initial bubble

We are given that the initial excess inside pressure of the underwater air bubble is \(10 \mathrm{Pa}\). To find the proportionality constant between the excess pressure and the inverted radius, we can rewrite the equation as: $$T = \frac{1}{2} P \cdot R$$
03

Set up the equation for the second bubble

Let's denote the excess pressure inside the second air bubble (with twice the radius) as \(P_2\). The equation for this bubble is: $$P_2 = 2 \frac{T}{2R}$$ where the radius is doubled.
04

Solve for \(P_2\)

To find the excess pressure inside the second air bubble, we can use the proportionality constant from Step 2 and substitute it into the equation for the second bubble: $$P_2 = 2 \frac{\frac{1}{2} P \cdot R}{2R}$$ Now, we can simplify the expression as: $$P_2 = \frac{P \cdot R}{2R}$$ Since we know that \(P = 10 \mathrm{Pa}\), we can substitute it into the equation: $$P_2 = \frac{10 \mathrm{Pa} \cdot R}{2R}$$
05

Calculate the final excess pressure

Now that we have the equation set up, the radius of the bubble, \(R\), will cancel out: $$P_2 = \frac{10 \mathrm{Pa} \cdot R}{2R} = \frac{10 \mathrm{Pa}}{2}$$ So, the excess pressure inside the second air bubble (with twice the radius) is: $$P_2 = 5 \mathrm{Pa}$$

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Most popular questions from this chapter

A lid is put on a box that is \(15 \mathrm{cm}\) long, \(13 \mathrm{cm}\) wide, and \(8.0 \mathrm{cm}\) tall and the box is then evacuated until its inner pressure is \(0.80 \times 10^{5} \mathrm{Pa} .\) How much force is required to lift the lid (a) at sea level; (b) in Denver, on a day when the atmospheric pressure is \(67.5 \mathrm{kPa}\) ( \(\frac{2}{3}\) the value at sea level)?
A garden hose of inner radius \(1.0 \mathrm{cm}\) carries water at $2.0 \mathrm{m} / \mathrm{s} .\( The nozzle at the end has radius \)0.20 \mathrm{cm} .$ How fast does the water move through the nozzle?
A water tower supplies water through the plumbing in a house. A 2.54 -cm- diameter faucet in the house can fill a cylindrical container with a diameter of \(44 \mathrm{cm}\) and a height of \(52 \mathrm{cm}\) in 12 s. How high above the faucet is the top of the water in the tower? (Assume that the diameter of the tower is so large compared to that of the faucet that the water at the top of the tower does not move. \()\)
A hydraulic lift is lifting a car that weighs \(12 \mathrm{kN}\). The area of the piston supporting the car is \(A\), the area of the other piston is \(a,\) and the ratio \(A / a\) is \(100.0 .\) How far must the small piston be pushed down to raise the car a distance of \(1.0 \mathrm{cm} ?[\text {Hint}:\) Consider the work to be done.]

(a) What is the density of an object that is \(14 \%\) submerged when floating in water at \(0^{\circ} \mathrm{C} ?\) (b) What percentage of the object will be submerged if it is placed in ethanol at \(0^{\circ} \mathrm{C} ?\)
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