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There is a maximum depth at which a diver can breathe through a snorkel tube (Figure 13.41 ) because as the depth increases, so does the pressure difference, which tends to collapse the diver's lungs. Since the snorkel connects the air in the lungs to the atmosphere at the surface, the pressure inside the lungs is atmospheric pressure. What is the external-internal pressure difference when the diver's lungs are at a depth of \(6.1 \mathrm{~m}\) (about \(20 \mathrm{ft}\) )? Assume that the diver is in freshwater. (A scuba diver breathing from compressed air tanks can operate at greater depths than can a snorkeler, since the pressure of the air inside the scuba diver's lungs increases to match the external pressure of the water.)

Step by step solution

01

Understand the Problem

The problem involves calculating the pressure difference between the outside and inside of a diver's lungs at a certain depth. The pressure inside the lungs is atmospheric, while the external pressure is determined by the weight of the water above the diver at that depth.

02

Identify Given Values

We are given that the diver is at a depth of 6.1 meters in freshwater. The density of freshwater (\(\rho\)) is typically \(1000 \, \text{kg/m}^3\), and gravitational acceleration (\(g\)) is \(9.8 \, \text{m/s}^2\). The atmospheric pressure is approximately \(101325 \, \text{Pa}\).

03

Calculate External Water Pressure

The pressure exerted by the water at a certain depth can be calculated using the formula: \[P_{ ext{water}} = \rho gh\]where \(h\) is the depth. Substituting given values: \[P_{ ext{water}} = 1000 \, \text{kg/m}^3 \times 9.8 \, \text{m/s}^2 \times 6.1 \, \text{m}\].

04

Solve for External Water Pressure

Carry out the calculation:\[P_{ ext{water}} = 1000 \times 9.8 \times 6.1 = 59780 \, \text{Pa}\].

05

Calculate Pressure Difference

The external-internal pressure difference is the difference between the water pressure at the depth and the atmospheric pressure. So:\[\Delta P = P_{ ext{water}} - P_{ ext{atm}} = 59780 \, \text{Pa} - 101325 \, \text{Pa}\].

06

Solve for Pressure Difference

Calculating the difference:\[\Delta P = 59780 - 101325 = -41545 \, \text{Pa}\].However, a negative result indicates that the external pressure exerted by the water is less than the atmospheric pressure. Thus, in terms of magnitude of difference, \(41545 \, \text{Pa}\) is the pressure difference imposed by the water depth.

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Key Concepts

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

Pressure Difference Calculation

Calculating the pressure difference is essential in fluid dynamics, especially for divers who need to understand the forces acting on their bodies. In this context, the pressure inside the diver's lungs remains at the atmospheric level since they are connected to the surface air via a snorkel. However, the surrounding external pressure is impacted by the water's weight above the diver. The formula to determine the external water pressure is given by \[ P_{\text{ext}} = \rho gh \]where:

• \( \rho \) is the freshwater density, \( 1000 \, \text{kg/m}^3 \)
• \( g \) is the gravitational acceleration, \( 9.8 \, \text{m/s}^2 \)
• \( h \) is the depth, \( 6.1 \, \text{m} \)

This formula helps us calculate how much pressure the water exerts at a given depth.

Atmospheric Pressure

Atmospheric pressure is the force exerted on an area by the weight of air above it. At sea level, this is approximately \( 101325 \, \text{Pa} \). When a diver breathes through a snorkel, the lungs' internal pressure matches the atmospheric pressure since the snorkel connects directly to the air at the surface. Maintaining internal atmospheric pressure is crucial for divers. As the water depth increases, the external pressure rises, potentially leading to a collapse of the lungs if the pressure becomes imbalanced. Hence, understanding atmospheric pressure ensures safety and stability for divers exploring various depths.

Water Pressure

Water pressure is an important concept in fluid dynamics, particularly in understanding underwater environments. This pressure increases linearly with depth due to the cumulative weight of the water molecules above. For divers, knowing the water pressure is crucial for safety, as it helps determine the forces acting on the body.For a diver at 6.1 meters in depth, the pressure can be calculated using the formula:\[ P_{\text{ext}} = \rho gh \]Substituting the values, we calculate the external water pressure as \( 59780 \, \text{Pa} \). This figure is critical as it helps in assessing whether the external pressure exceeds what a diver’s lung can handle.

Hydrostatic Pressure

Hydrostatic pressure is the pressure exerted by a fluid at equilibrium due to the force of gravity. In the context of diving, it becomes an important factor as it addresses the pressure experienced by the diver due to water depth.The deeper one goes, the greater the hydrostatic pressure. This pressure is calculated through the relationship:\[ P = \rho gh \]Understanding hydrostatic pressure is essential for divers to comprehend the potential risks involved at increased depths. It informs decisions about safe diving practices and equipment needed to counterbalance these pressures, such as advanced breathing apparatuses for scuba divers.

Freshwater Density

Freshwater density is a key factor when calculating water pressure in fluid dynamics. It is typically considered to be \( 1000 \, \text{kg/m}^3 \). This value is crucial for determining how a liquid behaves under various conditions, notably involving pressure calculations.The freshwater density, alongside gravitational acceleration and water depth, contributes to the understanding of how much pressure will act upon a diver at any given depth. Hence, knowing this value allows us to predict and prepare for the physical stress experienced when submerged, ensuring safety and comfort while diving.