/*! 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 86 Airlines are allowed to maintain... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Airlines are allowed to maintain a minimum air pressure within the passenger cabin equivalent to that at an altitude of 8000 \(\mathrm{ft}(2400 \mathrm{m})\) to avoid adverse health effects among passengers due to oxygen deprivation. Estimate this minimum pressure (in atm).

Short Answer

Expert verified
Minimum cabin pressure at 8000 ft is approximately 0.75 atm.

Step by step solution

01

Understand Sea Level Conditions

At sea level, the air pressure is approximately 1 atm and is equivalent to 101,325 Pa or 1013.25 hPa.
02

Use Standard Atmosphere Model

The standard atmosphere model suggests that pressure decreases exponentially with altitude. A simplified version of the barometric formula suitable for this task is: \[ P = P_0 \times \left(1 - \frac{L \times h}{T_0} \right)^{\frac{g \times M}{R \times L}} \]where:- \(P_0 = 1013.25 \, \text{hPa}\) (sea level pressure),- \(L = 0.0065 \, \text{K/m}\) (temperature lapse rate), - \(h = 2400 \, \text{m}\) (altitude),- \(T_0 = 288.15 \, \text{K}\) (standard temperature at sea level),- \(g = 9.80665 \, \text{m/s}^2\) (acceleration due to gravity),- \(M = 0.0289644 \, \text{kg/mol}\) (molar mass of Earth's air),- \(R = 8.31447 \, \text{J/(mol·K)}\) (universal gas constant).
03

Calculate the Pressure at 8000 ft

For simplification, the pressure at 2400 meters can be approximated using a typical percentage drop factor derived from altitude tables, which show that at 2400m, the pressure is around 75% of sea level pressure. Therefore, \( P = 0.75 \times P_0 = 0.75 \times 1013.25 \, \text{hPa} \approx 759.94 \, \text{hPa} \).
04

Convert Pressure to atm

One atmosphere (atm) is equivalent to 1013.25 hPa. Convert pressure from hPa to atm:\[ P_{atm} = \frac{759.94 \, \text{hPa}}{1013.25 \, \text{hPa/atm}} \approx 0.750 \text{ atm} \].

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.

Standard Atmosphere Model
When considering air pressure over altitudes, the Standard Atmosphere Model provides a framework for understanding how pressure changes with height. The model assumes a fixed average temperature decrease with altitude, known as the lapse rate. At sea level, pressure is standardized at roughly 1013.25 hPa or 1 atm, which serves as a reference point.
The model simplifies real atmospheric conditions to a theoretical profile, allowing calculations across different altitudes. This is essential for various applications such as aviation, where it helps determine cabin pressure to ensure passenger comfort.
Mind that the actual atmosphere could vary due to seasonal changes and geographical location. However, the model provides a reliable baseline for calculations, helping to understand atmospheric behavior.
Barometric Formula
Understanding pressure changes with altitude involves using the Barometric Formula. This mathematical tool allows you to calculate the atmospheric pressure at various altitudes, given certain conditions.
The formula commonly used in calculations is:\[ P = P_0 \times \left(1 - \frac{L \times h}{T_0} \right)^{\frac{g \times M}{R \times L}} \]Let's break it down:
  • \(P_0\): Sea level standard atmospheric pressure (1013.25 hPa).
  • \(L\): Temperature lapse rate (0.0065 K/m), indicating how temperature decreases with altitude.
  • \(h\): Altitude where pressure is calculated.
  • \(T_0\): Standard temperature at sea level (288.15 K).
  • \(g\), \(M\), \(R\): Constants for gravity, molar mass of Earth's air, and the gas constant respectively.
This formula is essential for predicting pressure levels at specific altitudes, making it helpful in fields like meteorology and aviation.
Altitude Pressure Relationship
The relationship between altitude and pressure is a direct and inversely proportional one. As altitude increases, atmospheric pressure decreases. This happens because the weight of the air column above decreases, meaning less air is pressing down.
At 2400 meters or 8000 feet, which is the altitude used in aviation for pressurized cabins, we find the pressure decreases to about 75% of that at sea level. This reduction is calculated using the barometric formula or average estimated values from altitude-pressure charts.
This principle explains why mountaineers and passengers in non-pressurized flights might experience difficulty breathing; the lower air pressure equates to less oxygen being available. Understanding this relationship helps in designing optimal conditions for activities at various altitudes.

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

A beaker of water rests on an electronic balance that reads 998.0 g. A 2.6 -cm-diameter solid copper ball attached to a string is submerged in the water, but does not touch the bottom. What are the tension in the string and the new balance reading?

(III) Thrust of a rocket. (a) Use Bernoulli's equation and the equation of continuity to show that the emission speed of the propelling gases of a rocket is $$v=\sqrt{2\left(P-P_{0}\right) / \rho}$$ where \(\rho\) is the density of the gas, \(P\) is the pressure of the gas inside the rocket, and \(P_{0}\) is atmospheric pressure just outside the exit orifice. Assume that the gas density stays approximately constant, and that the area of the exit orifice, \(A_{0},\) is much smaller than the cross-sectional area, \(A,\) of the inside of the rocket (take it to be a large cylinder). Assume also that the gas speed is not so high that significant turbulence or nonsteady flow sets in. (b) Show that the thrust force on the rocket due to the emitted gases is $$F=2 A_{0}\left(P-P_{0}\right)$$

(III) Show that inside a soap bubble, there must be a pressure \(\Delta P\) in excess of that outside equal to \(\Delta P=4 \gamma / r,\) where \(r\) is the radius of the bubble and \(\gamma\) is the surface tension. [Hint: Think of the bubble as two hemispheres in contact with each other; and remember that there are two surfaces to the bubble. Note that this result applies to any kind of membrane, where \(2 \gamma\) is the tension per unit length in that membrane.

If cholesterol buildup reduces the diameter of an artery by 15\%, by what \% will the blood flow rate be reduced, assuming the same pressure difference?

(II) The gauge pressure in each of the four tires of an automobile is \(240 \mathrm{kPa}\). If each tire has a "footprint" of \(220 \mathrm{~cm}^{2}\), estimate the mass of the car.

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.