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Black smokers are hot volcanic vents that emit smoke deep in the ocean floor. Many of them teem with exotic creatures, and some biologists think that life on earth may have begun around such vents. The vents range in depth from about 1500 m to 3200 m below the surface. What is the gauge pressure at a 3200-m deep vent, assuming that the density of water does not vary? Express your answer in pascals and atmospheres.

Short Answer

Expert verified
The gauge pressure at a 3200-m deep vent is 32,064,000 Pa or approximately 316.4 atm.

Step by step solution

01

Understand the Problem

We need to find the gauge pressure at a depth of 3200 meters below the ocean surface, assuming constant water density. Gauge pressure is defined as the pressure relative to atmospheric pressure.
02

Formula for Pressure Calculation

The pressure at a depth in a fluid is given by the formula: \[ P = \rho g h \]where \( P \) is the pressure,\( \rho \) is the density of the fluid (water),\( g \) is the acceleration due to gravity, and \( h \) is the depth of the fluid.
03

Input Known Values

For sea water, the density \( \rho \) is approximately 1025 kg/m³, gravitational acceleration \( g \) is approximately 9.81 m/s², and the depth \( h \) is 3200 meters.
04

Perform the Calculation

Substitute the known values into the pressure equation:\[ P = 1025 \times 9.81 \times 3200 \]Calculate:\[ P = 32,064,000 \text{ Pa} \]This is the gauge pressure in pascals.
05

Convert Pressure to Atmospheres

To convert pascals to atmospheres, use the conversion factor: 1 atm = 101,325 Pa. Thus,\[ \text{Pressure in atm} = \frac{32,064,000}{101,325} \approx 316.4 \text{ atm} \].

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Key Concepts

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

Fluid Mechanics
Fluid mechanics is a branch of physics that focuses on the behavior and properties of fluids, which include liquids and gases. It covers how fluids move and the forces that act upon them. One important aspect of fluid mechanics is understanding pressure, which is vital when analyzing phenomena like oceanic vents. In the context of deep-sea environments, pressure increases with depth. This is key to solving problems involving fluid dynamics beneath the ocean surface.
To investigate fluid pressure, equations account for the fluid's density, gravitational force, and depth. Understanding these principles helps explain and predict various physical phenomena occurring within fluid systems, such as the operation of hydraulic systems or the movement of liquids through pipes.
  • Pressure in fluids is determined mainly by their weight and density.
  • It is fundamental in assessing natural features like underwater vents.
By understanding fluid mechanics, we can better comprehend the forces affecting both living and inanimate objects submerged in water.
Density of Water
Density describes how much mass an object or substance has within a certain volume. In fluid mechanics, the density of water is a critical parameter in calculations related to pressure and buoyancy. For most calculations involving sea water, a constant density is assumed. For example, in the provided exercise, we use a density of 1025 kg/m³.
While the density of water can actually vary due to temperature, salinity, and pressure, assuming a fixed density simplifies many problems. This approximation is often sufficiently accurate for problems like computing hydrostatic pressure.
  • Density is critical to determining how pressure increases with depth.
  • In most oceanic computations, a standard density is used for simplicity.
Understanding the density of water helps explain why pressures are significantly greater in deep-sea environments, influencing marine life and underwater structures.
Pressure Conversion
The conversion of pressure units is an essential aspect of fluid mechanics calculations, especially when dealing with high pressures like those found in deep oceanic depths. Pressure is commonly measured in pascals (Pa) or atmospheres (atm). To reflect different contexts and simplify communication, converting between these units becomes necessary.
In the exercise, you observe the conversion of pressure from pascals to atmospheres, using the relation that 1 atm equals 101,325 Pa. The conversion allows an easier comparison to the atmospheric pressure experienced at the Earth's surface.
  • Pressure in pascals is often converted into atmospheres for better understanding.
  • This conversion makes it easier to relate underwater pressures to atmospheric conditions.
Understanding how to switch between these units aids in solving and interpreting pressure-related problems, both academic and real-world applications.
Oceanic Depth
The oceanic depth plays a fundamental role in determining the pressure experienced by underwater structures or features like black smokers. As one descends deeper into the ocean, the weight of the water above increases, leading to greater pressure. This is a crucial factor for both engineering applications and biological studies.
At 3200 meters, which is the depth mentioned in the original problem, pressure reaches remarkably high values. These pressures affect everything from the behavior of chemical reactions at the vent sites to the structures of deep-sea organisms.
  • Pressure grows approximately linearly with depth.
  • Deep ocean environments are subjected to intense pressures due to the immense depth.
By understanding how depth impacts pressure, we can develop better technologies for exploring and harnessing resources from these hard-to-reach parts of our planet.

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Most popular questions from this chapter

The piston of a hydraulic automobile lift is 0.30 m in diameter. What gauge pressure, in pascals, is required to lift a car with a mass of 1200 kg? Also express this pressure in atmospheres.

A cube 5.0 cm on each side is made of a metal alloy. After you drill a cylindrical hole 2.0 cm in diameter all the way through and perpendicular to one face, you find that the cube weighs 6.30 N. (a) What is the density of this metal? (b) What did the cube weigh before you drilled the hole in it?

A firehose must be able to shoot water to the top of a building 28.0 m tall when aimed straight up. Water enters this hose at a steady rate of 0.500 m\(^3\)/s and shoots out of a round nozzle. (a) What is the maximum diameter this nozzle can have? (b) If the only nozzle available has a diameter twice as great, what is the highest point the water can reach?

(a) What is the \(difference\) between the pressure of the blood in your brain when you stand on your head and the pressure when you stand on your feet? Assume that you are 1.85 m tall. The density of blood is 1060 kg/m\(^3\). (b) What effect does the increased pressure have on the blood vessels in your brain?

The Environmental Protection Agency is investigating an abandoned chemical plant. A large, closed cylindrical tank contains an unknown liquid. You must determine the liquid's density and the height of the liquid in the tank (the vertical distance from the surface of the liquid to the bottom of the tank). To maintain various values of the gauge pressure in the air that is above the liquid in the tank, you can use compressed air. You make a small hole at the bottom of the side of the tank, which is on a concrete platform\(-\)so the hole is 50.0 cm above the ground. The table gives your measurements of the horizontal distance \(R\) that the initially horizontal stream of liquid pouring out of the tank travels before it strikes the ground and the gauge pressure \({p_g}\) of the air in the tank. (a) Graph \({R^2}\) as a function of \({p_g}\). Explain why the data points fall close to a straight line. Find the slope and intercept of that line. (b) Use the slope and intercept found in part (a) to calculate the height \(h\) (in meters) of the liquid in the tank and the density of the liquid (in kg/m\(^3\)). Use \(g\) \(=\) 9.80 m/s\(^2\). Assume that the liquid is nonviscous and that the hole is small enough compared to the tank's diameter so that the change in h during the measurements is very small.

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