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CP Blo The Effect of Altitude on the Lungs. (a) Calculate the change in air pressure you will experience if you climb a 1000 -m mountain, assuming that the temperature and air density do not change over this distance and that they were \(22^{\circ} \mathrm{C}\) and \(1.2 \mathrm{kg} / \mathrm{m}^{3},\) respectively, at the bottom of the mountain. (Note that the result of Example 18.4 doesn't apply, since the expression derived in that example accounts for the variation of air density with altitude and we are told to ignore that in this problem.) If you took a \(0.50-\) breath at the foot of the mountain and managed to hold it until you reached the top, what would be the volume of this breath when you exhaled it there?

Short Answer

Expert verified
The pressure change is -11760 Pa, and the breath volume increases to approximately 0.566 liters.

Step by step solution

01

Understanding Pressure Change with Altitude

The change in air pressure with height can be determined using the hydrostatic pressure formula: \[ \Delta P = -\rho \cdot g \cdot h \] where \( \Delta P \) is the change in pressure, \( \rho = 1.2 \, \text{kg/m}^3 \) is density of air, \( g = 9.8 \, \text{m/s}^2 \) is acceleration due to gravity, and \( h = 1000 \, \text{m} \) is the change in height.
02

Calculate the Pressure Change

Substitute the given values into the hydrostatic pressure formula to find the change in pressure:\[ \Delta P = -(1.2 \, \text{kg/m}^3)(9.8 \, \text{m/s}^2)(1000 \, \text{m}) \]Simplifying gives:\[ \Delta P = -11760 \, \text{Pa (Pascal)} \] This result represents the decrease in air pressure as you ascend 1000 meters.
03

Understanding the Breath Volume Change

Assuming ideal gas behavior, the change in the volume of air due to pressure change can be described by Boyle's Law: \[ P_1 V_1 = P_2 V_2 \] where \( P_1 \) and \( V_1 \) are the initial pressure and volume, and \( P_2 \) and \( V_2 \) are the final pressure and volume, respectively. We assume the initial volume \( V_1 = 0.50 \, \text{l (liters)} \) and need to find \( V_2 \).
04

Calculate Final Volume with Boyle’s Law

First, express \( P_2 \) in terms of \( P_1 \):\[ P_2 = P_1 + \Delta P \]Assuming the initial pressure \( P_1 \approx 101325 \, \text{Pa} \) (sea level standard atmospheric pressure), then:\[ P_2 = 101325 \, \text{Pa} - 11760 \, \text{Pa} = 89565 \, \text{Pa} \]Using Boyle’s Law, solve for \( V_2 \):\[ V_2 = \frac{P_1 V_1}{P_2} = \frac{101325 \, \text{Pa} \times 0.50 \, \text{l}}{89565 \, \text{Pa}} \approx 0.566 \, \text{l} \]
05

Verify the Calculation

Ensure all unit conversions and assumptions (like constant temperature and ideal gas behavior) remain valid throughout the problem to confirm the result is \( 0.566 \, \text{l} \).

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Key Concepts

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

Boyle's Law
Boyle's Law is an important principle to understand when studying how gases react to pressure changes. It is a fundamental aspect of gas physics. Boyle's Law states that for a given mass of gas at constant temperature, the pressure is inversely proportional to the volume. In simpler terms, if you increase the pressure on a gas, its volume decreases, and vice versa, as long as the temperature does not change. This relationship is expressed mathematically as:\[ P_1 V_1 = P_2 V_2 \]where, \( P_1 \) and \( P_2 \) are the initial and final pressures, and \( V_1 \) and \( V_2 \) are the initial and final volumes respectively.This formula helps us to predict how the volume of air in our lungs will change as we ascend or descend a mountain. For example, if you start with an air volume of 0.50 liters at sea level and ascend to high altitudes where the pressure is lower, you apply Boyle's Law to find the new volume. The result will be an increase in volume as you climb, because the pressure is less at higher altitudes, allowing the air to expand.
ideal gas behavior
The concept of ideal gas behavior underpins much of the logic behind Boyle's Law. An ideal gas is a theoretical gas composed of many randomly moving point particles that interact only through elastic collisions. These gases are idealized models behaving according to the Ideal Gas Law, which combines relationships between pressure, volume, temperature, and amount of gas.The Ideal Gas Law is given as:\[ PV = nRT \]where:
  • \( P \) stands for pressure,
  • \( V \) is volume,
  • \( n \) is the amount of gas in moles,
  • \( R \) is the ideal gas constant, and
  • \( T \) is temperature in Kelvin.
In real-world applications, gases often show ideal behavior under normal conditions, such as standard temperature and pressure (STP). These conditions are generally where the gas molecules don't interact strongly with each other. The assumption of ideal gas behavior simplifies calculations significantly when dealing with gases, like calculating how the volume of a breath changes when climbing a mountain.
hydrostatic pressure
Hydrostatic pressure is the pressure exerted by a fluid at equilibrium due to gravity. It is a critical concept for understanding the changes in air pressure with altitude. Unlike gases, which are compressible, liquids are typically considered incompressible; however, gases follow similar principles when subjected to gravitational forces.The formula for calculating hydrostatic pressure is:\[ P = \rho g h \]where:
  • \( P \) is the pressure difference,
  • \( \rho \) is the fluid density,
  • \( g \) is the acceleration due to gravity, and
  • \( h \) is the height of the fluid column.
For example, when you climb a 1000-meter mountain, the density of air above you decreases. Using the hydrostatic pressure formula, we can calculate that the air pressure at the top of the mountain is lower than at the base. Understanding this drop in pressure is crucial, as it affects everything from how hard your heart needs to pump to the volume changes in a held breath. These calculations help predict physical effects experienced while mountain climbing.

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Most popular questions from this chapter

If a certain amount of ideal gas occupies a volume \(V\) at STP on earth, what would be its volume (in terms of \(V )\) on Venus, where the temperature is \(1003^{\circ} \mathrm{C}\) and the pressure is 92 atm?

At what temperature is the root-mean-square speed of nitrogen molecules equal to the root-mean-square speed of hydrogen molecules at \(20.0^{\circ} \mathrm{C} ?\) (Hint: The periodic table in Appendix D shows the molar mass (in \(\mathrm{g} / \mathrm{mol}\) ) of each element under the chemical symbol for that element. The molar mass of \(\mathrm{H}_{2}\) is twice the molar mass of hydrogen atoms, and similarly for \(\mathrm{N}_{2} .\) )

A flask with a volume of \(1.50 \mathrm{L},\) provided with a stop cock, contains ethane gas \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\) at 300 \(\mathrm{K}\) and atmospheric pressure \(\left(1.013 \times 10^{5} \mathrm{Pa}\right) .\) The molar mass of ethane is 30.1 \(\mathrm{g} / \mathrm{mol}\) . The system is warmed to a temperature of \(490 \mathrm{K},\) with the stop cock open to the atmosphere. The stopcock is then closed, and the flask is cooled to its original temperature. (a) What is the final pressure of the ethane in the flask? (b) How many grams of ethane remain in the flask?

Meteorology. The vapor pressure is the pressure of the vapor phase of a substance when it is in equilibrium with the solid or liquid phase of the substance. The relative humidity is the partial pressure of water vapor in the air divided by the vapor pressure of water at that same temperature, expressed as a percentage. The air is saturated when the humidity is 100\(\%\) (a) The vapor pressure of water at \(20.0^{\circ} \mathrm{C}\) is \(2.34 \times 10^{3}\) Pa. If the air temperature is \(20.0^{\circ} \mathrm{C}\) and the relative humidity is \(60 \%,\) what is the partial pressure of water vapor in the atmosphere (that is, the pressure due to water vapor alone)? (b) Under the conditions of part (a), what is the mass of water in 1.00 \(\mathrm{m}^{3}\) of air? (The molar mass of water is 18.0 \(\mathrm{g} / \mathrm{mol}\) . Assume that water vapor can be treated as an ideal gas.)

Hydrogen on the Sun. The surface of the sun has a temperature of about 5800 \(\mathrm{K}\) and consists largely of hydrogen atoms. (a) Find the rms speed of a hydrogen atom at this temperature. (The mass of a single hydrogen atom is \(1.67 \times 10^{-27}\) kg. (b) The escape speed for a particle to leave the gravitational influence of the sun is given by \((2 G M / R)^{1 / 2},\) where \(M\) is the sun's mass, \(R\) its radius, and \(G\) the gravitational constant \((\) see Example 13.5 of Section 13.3\() .\) Use the data in Appendix \(F\) to calculate this escape speed. (c) Can appreciable quantities of hydrogen escape from the sun? Can any hydrogen escape? Explain.

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