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The maximum pressure most organisms can survive is about 1000 times atmospheric pressure. Only small, simple organisms such as tadpoles and bacteria can survive such high pressures. What then is the maximum depth at which these organisms can live under the sea (assuming that the density of seawater is \(1025 \mathrm{kg} / \mathrm{m}^{3}\) )?

Short Answer

Expert verified
The maximum depth is approximately 10,091 meters.

Step by step solution

01

Understand the Problem

Maximum tolerable pressure is 1000 times the atmospheric pressure. Atmospheric pressure is approximately 101,325 Pa (Pascals). Thus, the pressure the organisms can withstand is \(1000 \times 101325 = 101,325,000 \text{ Pa}\).
02

Define the Pressure Equation

Pressure due to a column of liquid is given by the formula: \( P = \rho gh \), where \( \rho \) is the density of the liquid (1025 kg/m³ for seawater), \( g \) is the acceleration due to gravity (9.81 m/s²), and \( h \) is the depth of the liquid column.
03

Solve for Depth

Rearrange the formula \( P = \rho gh \) to solve for depth \( h \): \( h = \frac{P}{\rho g} \). Substitute the maximum pressure \( P = 101,325,000 \text{ Pa} \), \( \rho = 1025 \text{ kg/m}^3 \) and \( g = 9.81 \text{ m/s}^2 \) into the equation to find: \[ h = \frac{101325000}{1025 \times 9.81} \approx 10,091 \text{ meters} \].

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Key Concepts

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

Atmospheric Pressure
Atmospheric pressure is the force exerted by the weight of the atmosphere on a surface. It is typically measured at sea level and is approximately 101,325 Pascals (Pa). This measure is crucial for understanding how pressure varies with altitude or depth. When discussing pressure in fluids like air or seawater, atmospheric pressure serves as a baseline or reference point.

- For small, simple organisms like tadpoles and bacteria, atmospheric pressure is just part of their environment on land. - Certain organisms can survive environments with pressures much higher than normal atmospheric pressure. For them, tolerating up to 1000 times the atmospheric pressure is possible.

Atmospheric pressure affects everything on Earth’s surface and is necessary for calculating how much pressure increases when you dive underwater. It illustrates how organisms need special adaptations to endure extreme environments.
Density of Seawater
The density of seawater is a critical factor when discussing underwater pressure, as it determines how pressure increases with depth. Seawater has a typical density of about 1025 kilograms per cubic meter (kg/m³), slightly denser than freshwater due to the presence of dissolved salts.

- This density is significant because it influences buoyancy and affects the way objects and organisms rise or sink in ocean water. - Higher density means that pressure increases more rapidly with depth.

Understanding the density of seawater helps us predict how pressure builds up in the ocean. The higher the density, the greater the weight of the water column above, increasing pressure as you dive deeper.
Pressure-Depth Relationship
The pressure-depth relationship explains how pressure increases as you go deeper under the sea. The pressure at a particular depth is caused by the weight of the water above that point.

This relationship is represented by the equation: \[ P = \rho gh \]where:
  • \( P \) is the pressure at depth,
  • \( \rho \) is the density of the liquid (for seawater, it is 1025 kg/m³),
  • \( g \) is the acceleration due to gravity (~9.81 m/s²), and
  • \( h \) is the depth.

With this formula, we can calculate how much pressure increases with every meter you descend.

- This explains why pressure increases dramatically within a few kilometers of depth.- In the case of organisms that can withstand 1000 times atmospheric pressure, the maximum depth they can survive is around 10,091 meters.
Understanding the pressure-depth relationship is essential for studying life under the sea and the adaptations needed for survival in high-pressure environments.

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Most popular questions from this chapter

If the cardiac output of a small dog is $4.1 \times 10^{-3} \mathrm{m}^{3} / \mathrm{s}\( the radius of its aorta is \)0.50 \mathrm{cm},$ and the aorta length is \(40.0 \mathrm{cm},\) determine the pressure drop across the aorta of the dog. Assume the viscosity of blood is $4.0 \times 10^{-3} \mathrm{Pa} \cdot \mathrm{s}$
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