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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

If you watch water falling from a faucet, you will notice that the flow decreases in radius as the water falls. This can be explained by the equation of continuity, since the cross-sectional area of the water decreases as the speed increases. If the water flows with an initial velocity of $0.62 \mathrm{m} / \mathrm{s}\( and a diameter of \)2.2 \mathrm{cm}$ at the faucet opening, what is the diameter of the water flow after the water has fallen $30 \mathrm{cm} ?$
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