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Ch. 17 - Temperature, Thermal Expansion, and the Ideal Gas Law
Giancoli Douglas - Physics for Scientists and Engineers 5th edition
Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Not the one you use?Change textbook
All textbooksGiancoli Douglas 5th editionCh. 17 - Temperature, Thermal Expansion, and the Ideal Gas LawProblem 97
Chapter 17, Problem 97

Why snorkels are not 4 feet long. Snorkelers breathe through short tubular “snorkels” while swimming under water very near the surface (Fig. 17–24). One end of the snorkel is in the snorkeler’s mouth and the other end protrudes just above the water’s surface. Unfortunately, snorkels cannot support breathing to any great depth: it is said that a typical snorkeler below a water depth of only about 30 cm cannot draw a breath through a snorkel. Based on this observation, what is the approximate change in a typical person’s lung pressure (in atm) when drawing a breath? (Note that your diaphragm muscles, which expand your lungs, must work also against the extra water pressure.)
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Understand the problem: The question asks us to determine the approximate change in lung pressure (in atm) when a snorkeler tries to breathe at a depth of 30 cm underwater. The key concept here is that the diaphragm must work against the water pressure exerted at that depth.
Recall the formula for pressure due to a fluid column: The pressure exerted by a fluid column is given by \( P = \rho g h \), where \( \rho \) is the density of the fluid (water in this case), \( g \) is the acceleration due to gravity, and \( h \) is the depth of the fluid column.
Substitute the known values into the formula: Use \( \rho = 1000 \; \text{kg/m}^3 \) (density of water), \( g = 9.8 \; \text{m/s}^2 \), and \( h = 0.30 \; \text{m} \) (30 cm converted to meters). This will give the pressure in Pascals (Pa).
Convert the pressure from Pascals to atmospheres: 1 atmosphere (atm) is approximately equal to \( 101325 \; \text{Pa} \). Divide the pressure calculated in the previous step by \( 101325 \; \text{Pa} \) to express the result in atm.
Interpret the result: The calculated pressure difference represents the additional pressure the diaphragm must overcome to draw a breath at a depth of 30 cm. This explains why snorkels cannot be much longer, as the diaphragm muscles would struggle to overcome the increasing water pressure at greater depths.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Hydrostatic Pressure

Hydrostatic pressure is the pressure exerted by a fluid at equilibrium due to the force of gravity. It increases with depth, calculated as the product of the fluid's density, gravitational acceleration, and depth. For snorkelers, this means that as they dive deeper, the water pressure increases, making it harder to breathe through a snorkel, which is limited by the pressure difference between the lungs and the surrounding water.
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Lung Pressure and Breathing Mechanics

Lung pressure refers to the pressure inside the lungs during the breathing process. When a snorkeler inhales, the diaphragm contracts, creating a negative pressure that allows air to flow in. However, at depths greater than 30 cm, the external water pressure can exceed the lung pressure, making it difficult for the snorkeler to draw air through the snorkel, as the diaphragm must work against this additional pressure.
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Atmospheric Pressure

Atmospheric pressure is the pressure exerted by the weight of the atmosphere above a given point, typically measured at sea level as 1 atm. When snorkelers are at the surface, their lungs are at atmospheric pressure. As they descend, the increase in water pressure must be countered by the lung pressure to facilitate breathing, which is why snorkels are not effective at greater depths where the pressure difference becomes significant.
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Related Practice
Textbook Question

A brass lid screws tightly onto a glass jar at 15°C. To help open the jar, it can be placed into a bath of hot water. After this treatment, the temperatures of the lid and the jar are both 55°C. The inside diameter of the lid is 7.0 cm. Find the size of the gap (difference in radius) that develops by this procedure.

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Textbook Question

A helium balloon has volume V₀ and temperature T₀ at sea level where the pressure is P₀ and the air density is ρ₀. The balloon is allowed to float up in the air to altitude y where the temperature is T₁. Show that the buoyant force does not depend on altitude y. Assume that the skin of the balloon maintains the helium pressure at a constant factor of 1.05 times greater than the outside pressure. [Hint: Assume that the pressure change with altitude is P = P₀ e⁻ᶜʸ , Eq. 13–6c in Chapter 13.]

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Textbook Question

A copper wire sags 54.0 cm between two utility poles 30.0 m apart when the temperature is -15° C. Estimate the amount of sag when the temperature is + 35° C. [Hint: An estimate can be made by assuming the shape of the wire is approximately an arc of a circle; hard equations can sometimes be solved by guessing values.]

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