/*! 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 32 A scuba diver must decompress af... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A scuba diver must decompress after a deep dive to allow excess nitrogen to exit safely from his bloodstream. The length of time required for decompression depends on the total change in pressure that the diver experienced. Find this total change in pressure for a diver who starts at a depth of \(d=20.0 \mathrm{~m}\) in the ocean (density of seawater \(\left.=1024 \mathrm{~kg} / \mathrm{m}^{3}\right)\) and then travels aboard a small plane (with an unpressurized cabin) that rises to an altitude of \(h=5000 . \mathrm{m}\) above sea level.

Short Answer

Expert verified
Answer: The total change in pressure experienced by the diver is -248159 N/m².

Step by step solution

01

Calculate initial pressure at depth

We first determine pressure at the initial depth due to the water above the diver. To do this, we need to use the hydrostatic pressure formula: \(P = \rho g h\). Here, \(P\) is the pressure, \(\rho\) is the density of seawater, \(g\) is the gravitational acceleration (approximately \(9.81 \mathrm{m/s^2}\)), and \(h\) is the depth. \(P_{water} = (1024\,\mathrm{kg/m^3}) \times (9.81\,\mathrm{m/s^2}) \times (20.0\,\mathrm{m})\) \(P_{water} = 200960\,\mathrm{N/m^2}\) Next, we also need to add atmospheric pressure at sea level (approximately \(101325\, \mathrm{N/m^2}\)) to the initial pressure. \(P_{initial} = P_{water} + P_{atm}\) \(P_{initial} = 200960\,\mathrm{N/m^2} + 101325\,\mathrm{N/m^2}\) \(P_{initial} = 302285\,\mathrm{N/m^2}\)
02

Calculate final pressure at altitude

Now we need to find the atmospheric pressure at the end altitude (5000 meters above sea level). We know that the atmospheric pressure decreases with increasing altitude. We can use the barometric formula to find the pressure at the given altitude: \(P_{altitude} = P_{0} e^{(-\frac{mg}{kT} h)}\) Here, \(P_{0}\) is the sea level standard atmospheric pressure (\(101325\,\mathrm{N/m^2}\)), \(m\) is the molar mass of Earth's air (\(0.029\,\mathrm{kg/mol}\)), \(g\) is the gravitational acceleration (\(9.81\,\mathrm{m/s^2}\)), \(k\) is the Boltzmann constant (\(1.381\times10^{-23}\,\mathrm{J/K}\)), \(T\) is the temperature (assuming \(293\,\mathrm{K}\)), and \(h\) is the altitude. \(P_{altitude} = 101325 \times e^{(-\frac{0.029 \times 9.81}{1.381\times10^{-23} \times 293} \times 5000)}\) \(P_{altitude} \approx 54126\,\mathrm{N/m^2}\)
03

Calculate the total change in pressure

Now we can find the total change in pressure by subtracting the initial pressure from the final pressure: \(\Delta P = P_{altitude} - P_{initial}\) \(\Delta P = 54126\,\mathrm{N/m^2} - 302285\,\mathrm{N/m^2}\) \(\Delta P = -248159\,\mathrm{N/m^2}\) The total change in pressure is -\(248159\,\mathrm{N/m^2}\). Since the change in pressure is negative, this means that the total pressure experienced by the diver decreased from the starting depth to the end altitude.

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 91Ó°ÊÓ!

Key Concepts

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

Scuba Diving Decompression
Understanding scuba diving decompression is crucial for any diver venturing into the depths of the ocean. Decompression is the process divers use to safely adapt to the changes in pressure as they ascend from deep waters. When a diver submerges, water pressure increases, and gases like nitrogen dissolve into their bloodstream at higher rates. If the diver ascends too quickly, these gases can form bubbles in the body, a condition known as decompression sickness or 'the bends.' To prevent this, divers must follow a decompression schedule, which allows nitrogen to be released slowly and safely from their body.

In our example, the scuba diver experiences a significant pressure change from a depth of 20 meters to an altitude of 5000 meters above sea level. This extreme shift necessitates a careful decompression strategy, taking into account both the underwater pressure and the decreased atmospheric pressure at high altitudes. By calculating the total pressure change, the diver can determine the necessary decompression stops and duration required to avoid the bends.
Barometric Formula
The barometric formula is a key component in understanding how atmospheric pressure decreases with increasing altitude. This equation expresses the relationship between the pressure and the height above sea level. It is based on the principle that the atmosphere becomes less dense, and thus less pressurized, as altitude increases.

Barometric Formula Explanation:

At sea level, we experience standard atmospheric pressure, but as we ascend, say in a small plane with an unpressurized cabin, the pressure decreases exponentially. The reduction in pressure at a specific altitude is calculated using the formula:\[P_{altitude} = P_{0} e^{(-\frac{mg}{kT} h)}\]where the variables represent sea-level atmospheric pressure \(P_{0}\), molar mass of air \(m\), gravitational acceleration \(g\), Boltzmann constant \(k\), temperature \(T\), and altitude \(h\). This formula helps us understand how atmospheric conditions change, which is particularly critical for activities such as aviation, mountaineering, and indeed, for planning the decompression phases in scuba diving.
Pressure Change Calculation
Calculating the change in pressure that a scuba diver experiences is a pivotal part of planning a safe dive and successful ascent. To figure out this change, we first need to determine the hydrostatic pressure caused by the weight of water above the diver at a certain depth. This is done using the formula:\[P = \rho g h\]where \(P\) is the pressure, \(\rho\) is the density of the fluid (seawater in this scenario), \(g\) is the acceleration due to gravity, and \(h\) is the depth underwater.

Once we've established the pressure underwater, we need to calculate the pressure at the altitude the diver will ascend to. For this, we can employ the barometric formula discussed above. These two values give us the initial and final pressures, which we can then compare to deduce the total change in pressure as:\[\text{Pressure change, } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text: \text{ } \text{ } \text{ equal to the final pressure minus the initial pressure.\]

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

Brass weights are used to weigh an aluminum object on an analytical balance. The weighing is done one time in dry air and another time in humid air, with a water vapor pressure of \(P_{\mathrm{h}}=2.00 \cdot 10^{3} \mathrm{~Pa}\). The total atmospheric pressure \(\left(P=1.00 \cdot 10^{5} \mathrm{~Pa}\right)\) and the temperature \(\left(T=20.0^{\circ} \mathrm{C}\right)\) are the same in both cases. What should the mass of the object be to be able to notice a difference in the balance readings, provided the balance's sensitivity is \(m_{0}=0.100 \mathrm{mg}\) ? (The density of aluminum is \(\rho_{\mathrm{A}}=2.70 \cdot 10^{3} \mathrm{~kg} / \mathrm{m}^{3} ;\) the density of brass is \(\left.\rho_{\mathrm{B}}=8.50 \cdot 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\right)\)

A water-powered backup sump pump uses tap water at a pressure of \(3.00 \mathrm{~atm}\left(p_{1}=3 p_{\mathrm{atm}}=\right.\) \(3.03 \cdot 10^{5} \mathrm{~Pa}\) ) to pump water out of a well, as shown in the figure \(\left(p_{\text {well }}=p_{\text {ttm }}\right)\). This system allows water to be pumped out of a basement sump well when the electric pump stops working during an electrical power outage. Using water to pump water may sound strange at first, but these pumps are quite efficient, typically pumping out \(2.00 \mathrm{~L}\) of well water for every \(1.00 \mathrm{~L}\) of pressurized tap water. The supply water moves to the right in a large pipe with cross-sectional area \(A_{1}\) at a speed \(v_{1}=2.05 \mathrm{~m} / \mathrm{s}\). The water then flows into a pipe of smaller diameter with a cross-sectional area that is ten times smaller \(\left(A_{2}=A_{1} / 10\right)\). a) What is the speed \(v_{2}\) of the water in the smaller pipe, with area \(A_{2} ?\) b) What is the pressure \(p_{2}\) of the water in the smaller pipe, with area \(A_{2} ?\) c) The pump is designed so that the vertical pipe, with cross-sectional area \(A_{3}\), that leads to the well water also has a pressure of \(p_{2}\) at its top. What is the maximum height, \(h,\) of the column of water that the pump can support (and therefore act on ) in the vertical pipe?

A tourist of mass \(60.0 \mathrm{~kg}\) notices a chest with a short chain attached to it at the bottom of the ocean. Imagining the riches it could contain, he decides to dive for the chest. He inhales fully, thus setting his average body density to \(945 \mathrm{~kg} / \mathrm{m}^{3}\), jumps into the ocean (with saltwater density = \(1020 \mathrm{~kg} / \mathrm{m}^{3}\) ), grabs the chain, and tries to pull the chest to the surface. Unfortunately, the chest is too heavy and will not move. Assume that the man does not touch the bottom. a) Draw the man's free-body diagram, and determine the tension on the chain. b) What mass (in kg) has a weight that is equivalent to the tension force in part (a)? c) After realizing he cannot free the chest, the tourist releases the chain. What is his upward acceleration (assuming that he simply allows the buoyant force to lift him up to the surface)?

A water pipe narrows from a radius of \(r_{1}=5.00 \mathrm{~cm}\) to a radius of \(r_{2}=2.00 \mathrm{~cm} .\) If the speed of the water in the wider part of the pipe is \(2.00 \mathrm{~m} / \mathrm{s}\), what is the speed of the water in the narrower part?

Which of the following assumptions is not made in the derivation of Bernoulli's Equation? a) Streamlines do not cross. c) There is negligible friction. b) There is negligible d) There is no turbulence. viscosity. e) There is negligible gravity.

See all solutions

Recommended explanations on Physics Textbooks

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.