/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 8 Black smokers are hot volcanic v... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 12: Problem 8

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 is approximately 31,392,000 Pa or 309.71 atm.

Step by step solution

01

Understanding the Problem

We need to calculate the gauge pressure at a depth of 3200 meters in the ocean. Gauge pressure is the pressure relative to atmospheric pressure and is often calculated for submerged objects using the formula \( P = \rho g h \), where \( \rho \) is the density of water, \( g \) is the acceleration due to gravity, and \( h \) is the depth.
02

Parameters and Constants

The density of water \( \rho \) is typically about \( 1000 \, \text{kg/m}^3 \). The acceleration due to gravity \( g \) is approximately \( 9.81 \, \text{m/s}^2 \). The depth \( h \) is given as 3200 meters.
03

Calculate Gauge Pressure in Pascals

Using the formula \( P = \rho g h \), substitute the values: \( \rho = 1000 \, \text{kg/m}^3 \), \( g = 9.81 \, \text{m/s}^2 \), and \( h = 3200 \, \text{m} \).\[ P = 1000 \times 9.81 \times 3200 = 31,392,000 \, \text{Pa} \] This is the gauge pressure in pascals.
04

Convert Gauge Pressure to Atmospheres

1 atmosphere is equivalent to approximately \( 101,325 \, \text{Pa} \). Convert the calculated pressure from pascals to atmospheres:\[ P_{atm} = \frac{31,392,000}{101,325} \approx 309.71 \, \text{atm} \] This is the gauge pressure in atmospheres.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Density of Water
The density of water is a fundamental property that plays a crucial role in various calculations involving fluids, especially in physics and engineering. Here, it's essential for determining the pressure exerted by water at different depths in the ocean.
  • The standard value for the density of water is typically \(1000 \, \text{kg/m}^3\), which allows us to perform calculations with quite a bit of precision.
  • This value assumes pure water at a temperature around 4掳C, where water is densest. Real-world conditions like salinity and temperature can affect density, but for basic problem-solving, \(1000 \, \text{kg/m}^3\) is a convenient approximation.
  • Understanding water density is crucial because it impacts buoyancy, pressure under water, and even how sound travels through water.

For calculations like those involving gauge pressure at ocean depths, this density helps calculate the weight of a column of water above a point. This weight directly impacts the pressure exerted at that depth.
Acceleration Due to Gravity
The acceleration due to gravity, denoted as \(g\), is the rate at which objects accelerate towards Earth. It's a critical component in calculating various forces and pressures, including those underwater.
  • The commonly accepted value of \(g\) is \(9.81 \, \text{m/s}^2\), representing the acceleration of an object in freefall near the Earth's surface.
  • This value can slightly vary depending on your location on Earth because the planet is not a perfect sphere and differences in altitude can affect gravity.
  • In underwater pressure calculations, such as for gauge pressure, \(g\) helps determine the force exerted by the water's weight.

Incorporating \(g\) into the formula for pressure \(P = \rho g h\) means we account for how quickly objects, including water, are pulled towards the Earth, influencing how pressure increases with depth.
Pascal to Atmosphere Conversion
In many scientific and engineering problems, it's essential to convert between different units of pressure to understand or communicate results effectively. One common conversion is from pascals to atmospheres.
  • A pascal (Pa) is the SI unit of pressure, defined as one newton per square meter. It's suitable for large-pressure calculations but is often more informative to express pressures in atmospheres (atm) for everyday understanding.
  • One atmosphere is defined as \(101,325 \, \text{Pa}\), which approximates the average atmospheric pressure at sea level on Earth.
  • To convert from pascals to atmospheres, you divide the pressure value in pascals by \(101,325\).

This conversion is evident in the solution of the gauge pressure at a deep-sea vent, where an understanding of both units helps clarify the extreme conditions deep underwater compared to surface pressures.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Water is flowing in a pipe with a varying cross-sectional area, and at all points the water completely fills the pipe. At point 1 the cross-sectional area of the pipe is \(0.070 \mathrm{m}^{2},\) and the magnitude of the fluid velocity is 3.50 \(\mathrm{m} / \mathrm{s}\) (a) What is the fluid speed at points in the pipe where the cross-sectional area is (a) 0.105 \(\mathrm{m}^{2}\) and \((\mathrm{b}) 0.047 \mathrm{m}^{2} ?\) (c) Calculate the volume of water discharged from the open end of the pipe in 1.00 hour.

A hollow plastic sphere is held below the surface of a freshwater lake by a cord anchored to the bottom of the lake. The sphere has a volume of 0.650 \(\mathrm{m}^{3}\) and the tension in the cord is 900 \(\mathrm{N}\) . (a) Calculate the buoyant force exerted by the cord on the sphere. (b) What is the mass of the sphere? (c) The cord breaks and the sphere rises to the surface. When the sphere comes to rest, what fraction of its volume will be submerged?

In a lecture demonstration, a professor pulls apart two hemispherical steel shells (diameter \(D )\) with ease using their attached handles. She then places them together, pumps out the air to an absolute pressure of \(p,\) and hands them to a bodybuilder in the back row to pull apart. (a) If atmospheric pressure is \(p_{0},\) how much force must the bodybuilder exert on each shell? (b) Evaluate your answer for the case \(p=0.025\) atm, \(D=10.0 \mathrm{cm} .\)

A cube 5.0 \(\mathrm{cm}\) on each side is made of a metal alloy. After you drill a cylindrical hole 2.0 \(\mathrm{cm}\) in diameter all the way through and perpendicular to one face, you find that the cube weighs 7.50 \(\mathrm{N}\) . (a) What is the density of this metal? (b) What did the cube weigh before you drilled the hole in it?

A uniform lead sphere and a uniform aluminum sphere have the same mass. What is the ratio of the radius of the aluminum sphere to the radius of the lead sphere?
See all solutions

Recommended explanations on Physics Textbooks

Further Mechanics and Thermal Physics

Read Explanation

Scientific Method Physics

Read Explanation

Radiation

Read Explanation

Quantum Physics

Read Explanation

Kinematics Physics

Read Explanation

Space Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.