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Chapter 13: Problem 51

A diver rises quickly to the surface from a 5.0 -m depth. If she did not exhale the gas from her lungs before rising, by what factor would her lungs expand? Assume the temperature to be constant and the pressure in the lungs to match the pressure outside the diver's body. The density of seawater is $1.03 \times 10^{3} \mathrm{kg} / \mathrm{m}^{3}$

Short Answer

Expert verified
Question: Calculate the expansion factor of a diver's lungs if she does not exhale the gas from her lungs while rising from a 5.0 m depth in seawater with a density of \(1.03 \times 10^3 \mathrm{kg/m^3}\). Answer: ___ (Round your answer to two decimal places.)

Step by step solution

01

Determine the initial pressure at 5.0 m depth

To find the pressure at 5.0 m depth, we'll need to use the formula for pressure in a fluid column: \(P = P_0 + \rho gh\), where \(P_0\) is atmospheric pressure, \(\rho\) is the fluid density, \(g\) is the acceleration due to gravity, and \(h\) is the depth. Use the given density of seawater, \(1.03 \times 10^3 \mathrm{kg} / \mathrm{m}^{3}\), and the standard value for atmospheric pressure, \(1.01 \times 10^5 \mathrm{Pa}\). \(P = (1.01 \times 10^5 \mathrm{Pa}) + (1.03 \times 10^3 \mathrm{kg/m^3})(9.81 \mathrm{m/s^2})(5.0 \mathrm{m})\) Calculate the value of \(P\).
02

Determine the final pressure at the surface

The final pressure at the surface will be equal to atmospheric pressure, \(P_0\). Therefore: \(P_2 = P_0 = 1.01 \times 10^5 \mathrm{Pa}\)
03

Apply Boyle's Law to find the expansion factor

Using Boyle's Law, \(P_1V_1 = P_2V_2\), we can find the expansion factor, \(V_2/V_1\), by dividing both sides of the equation by \(P_1V_1\): \(\frac{V_2}{V_1} = \frac{P_1}{P_2}\) Substitute the values of \(P_1\) (from Step 1) and \(P_2\) (from Step 2) into the equation, and calculate the value of the expansion factor: \(\frac{V_2}{V_1} = \frac{P_1}{P_2}\)

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Most popular questions from this chapter

In plants, water diffuses out through small openings known as stomatal pores. If \(D=2.4 \times 10^{-5} \mathrm{m}^{2} / \mathrm{s}\) for water vapor in air, and the length of the pores is \(2.5 \times 10^{-5} \mathrm{m},\) how long does it take for a water molecule to diffuse out through the pore?
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