/*! 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 47 (a) Calculate the \(change\) in ... [FREE SOLUTION] | 91Ó°ÊÓ

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(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\(^\circ\)C and 1.2 kg/m\(^3\), respectively, at the bottom of the mountain. (Note: 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 here.) (b) If you took a 0.50-L 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 air pressure decreases by 11760 Pa and the breath expands to 0.565 L.

Step by step solution

01

Determine the air pressure change

To find the change in air pressure over the climb, we use the formula for hydrostatic pressure change:\[\Delta P = \rho \cdot g \cdot \Delta h\]where \( \rho = 1.2 \, \text{kg/m}^3 \) is the air density (constant), \( g = 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity, and \( \Delta h = 1000 \, \text{m} \) is the change in height.Substituting these values, we get:\[\Delta P = 1.2 \, \text{kg/m}^3 \times 9.8 \, \text{m/s}^2 \times 1000 \, \text{m} = 11760 \, \text{Pa}\]
02

Calculate final air pressure at the top

Assuming standard air pressure at sea level is \( P_0 = 101325 \, \text{Pa} \), the air pressure at the top of the mountain is:\[P_{top} = P_0 - \Delta P = 101325 \, \text{Pa} - 11760 \, \text{Pa} = 89565 \, \text{Pa}\]
03

Use Boyle's Law to determine volume change

Boyle's Law states that for a given mass of gas at constant temperature, the pressure and volume are inversely proportional:\[P_1 V_1 = P_2 V_2\]At the foot of the mountain \( P_1 = 101325 \, \text{Pa} \) and \( V_1 = 0.50 \, \text{L} \). At the mountain top, \( P_2 = 89565 \, \text{Pa} \). Substituting into the equation to find \( V_2 \):\[101325 \, \text{Pa} \times 0.50 \, \text{L} = 89565 \, \text{Pa} \times V_2\]Solving for \( V_2 \):\[V_2 = \frac{101325 \, \text{Pa} \times 0.50 \, \text{L}}{89565 \, \text{Pa}} \approx 0.565 \, \text{L}\]
04

Conclusion of calculations

The change in air pressure experienced during the ascent is 11760 Pa, and the volume of the air breathed expands to approximately 0.565 L at the top of the mountain.

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Key Concepts

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

Hydrostatic Pressure Change
When climbing up a mountain, the air pressure changes as you ascend. This change in pressure is known as hydrostatic pressure change. It's essentially the difference in atmospheric pressure at the base of the mountain compared to its peak. To calculate this change, we use the formula:
  • \( \Delta P = \rho \cdot g \cdot \Delta h \)
Here:
  • \( \rho \) is the air density, which is the mass of air per unit volume.
  • \( g \) stands for the acceleration due to gravity (approximately 9.8 m/s² on Earth).
  • \( \Delta h \) is the height change, i.e., how far up the mountain you've climbed.
In our problem, assuming constant air density and temperature, when you climb 1000 meters, the drop in air pressure is 11760 Pascals. This is vital for understanding how air behaves under different altitudes, like the thinner air at the top of a mountain.
Boyle's Law
Boyle's Law is a fundamental principle in physics that defines how the pressure and volume of gases relate when the temperature remains unchanged. Simply put, it tells us that the product of pressure and volume is constant for a fixed amount of gas at a constant temperature. The formula is:
  • \( P_1 V_1 = P_2 V_2 \)
Where:
  • \( P_1 \) and \( V_1 \) are the initial pressure and volume.
  • \( P_2 \) and \( V_2 \) are the final pressure and volume.
In our scenario, as you climb higher up the mountain and pressure decreases, the volume of a breath (if kept at a constant amount and temperature) actually increases. Holding a 0.50 L breath at the bottom translates to approximately 0.565 L at the top, due to decreased pressure. Boyle's Law is a handy tool to understand how air behaves under pressure changes, which is important, for example, in breathing or any process involving gases under changing conditions.
Air Density
Air density is an important concept when calculating pressure changes with altitude. It refers to how much air mass exists in a given volume. At sea level, air density is relatively high, which is indicated by it being at 1.2 kg/m³ in our exercise. This density tells us how many molecules are packed into a space and directly affects air pressure. The greater the air density, the higher the pressure.
  • Air density decreases with altitude, but in our exercise, we've assumed it remains constant for simplicity.
Understanding air density helps explain why high-altitude environments, like mountain tops, feel different—you might notice thinner air, meaning fewer molecules available in each breath. While we assumed constant density here, in reality, it typically decreases with elevation, affecting various factors such as how we breathe or weather conditions.

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Most popular questions from this chapter

The rate of \(effusion\)-that is, leakage of a gas through tiny cracks-is proportional to \(v_{rms}\) . If tiny cracks exist in the material that's used to seal the space between two glass panes, how many times greater is the rate of \(He\) leakage out of the space between the panes than the rate of \(Xe\) leakage at the same temperature? (a) 370 times; (b) 19 times; (c) 6 times; (d) no greater-the \(He\) leakage rate is the same as for \(Xe\).

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