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What would be the height of the atmosphere if the air density (a) were uniform and (b) decreased linearly to zero with height? Assume that at sea level the air pressure is \(1.0 \mathrm{~atm}\) and the air density is \(1.3 \mathrm{~kg} / \mathrm{m}^{3}\).

Short Answer

Expert verified
(a) ~7924 m for uniform density; (b) ~15848 m for density decreasing linearly to zero.

Step by step solution

01

Understand the Problem

We need to calculate the height of the atmosphere under two different conditions: (a) uniform air density, and (b) air density decreasing linearly to zero with height. Given data include the air pressure at sea level as \(1.0 \, \text{atm}\) and the air density as \(1.3 \, \text{kg/m}^3\).
02

Convert Units

Convert the air pressure from atm to Pascals (Pa) for consistency with the density units. Since \(1 \, \text{atm} = 101325 \, \text{Pa}\), the air pressure at sea level is \(101325 \, \text{Pa}\).
03

Calculate Height for Uniform Density (a)

For uniform density, the atmosphere's height \(h\) can be calculated using the formula \( P = \rho g h \), where \(P\) is pressure, \(\rho\) is density, and \(g\) is gravitational acceleration (\(9.81 \, \text{m/s}^2\)). Solve for \(h\): \[ h = \frac{P}{\rho g} = \frac{101325}{1.3 \times 9.81} \approx 7924.03 \, \text{m} \].
04

Set Up Equation for Linearly Decreasing Density (b)

Assume the density \(\rho\) decreases linearly with height and is given by \( \rho(h) = \rho_0 (1 - \frac{h}{H}) \), where \(H\) is the maximum height (when density is zero). Integrate the density over the height to find \(H\): - Set the integral of density over height equal to the sea level pressure: \[ \int_0^H \rho(h)gh \, dh = P \].
05

Integrate to Find Maximum Height

Use the linear density formula \(\rho(h) = \rho_0 (1 - \frac{h}{H})\) in the equation: \[ \int_0^H \rho_0 (1 - \frac{h}{H}) g \, dh = P \] This simplifies to: \[ \rho_0 g \left[ h - \frac{h^2}{2H} \right]_0^H = 101325 \].
06

Solve for Maximum Height (b)

Evaluate the integral and simplify:\[ \rho_0 g \left( H - \frac{H^2}{2H} \right) = 101325 \] leads to \[ \rho_0 g \left( \frac{H}{2} \right) = 101325 \]thus \[ H = \frac{2 \times 101325}{\rho_0 g} = \frac{2 \times 101325}{1.3 \times 9.81} \approx 15848.05 \, \text{m} \].
07

Final Review and Comparison

Review steps and compare results: (a) For uniform density: \(h \approx 7924.03 \, \text{m}\).(b) For linearly decreasing density to zero: \(H \approx 15848.05 \, \text{m}\).

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Key Concepts

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

Understanding Uniform Air Density
When we talk about uniform air density, it means that the density of air remains constant throughout a certain region. Imagine the atmosphere as a stack of bricks. Each brick is the same size and weight, just like how each layer of air would have the same density in this scenario.
This means that regardless of whether you are at sea level or way up high in the atmosphere, the air density stays at the same value. In our given problem, this constant density is provided as \(1.3\, \text{kg/m}^3.\)

Uniform density is a helpful simplification because it allows us to easily calculate the height of the atmosphere. By using the formula \(P = \rho g h\), where \(P\) is the atmospheric pressure, \(\rho\) is the density, \(g\) is the acceleration due to gravity (\(9.81 \text{m/s}^2\)), and \(h\) is the height, we can easily solve for \(h\) when the other values are known.
But remember, in reality, the atmosphere's density doesn't stay constant with height. Since the air gets thinner the higher we go, uniform air density is a simplified way to understand the basic concept of atmospheric pressure and density relationships. However, it is a great way for beginners to start understanding atmospheric physics!
  • Air behaves as if it had constant density in a small region.
  • Easy to calculate changes in pressure with height.
Exploring the Air Density Gradient
The air density gradient is a concept that looks at how air density decreases (or sometimes increases) as you move through different altitudes in the atmosphere. Think of it as the difference in how thick or thin the air feels as you climb a mountain or fly in an airplane.
In the actual atmosphere, density tends to decrease with increasing height. In our exercise, we are interested in a linear decrease, which means the density reduces evenly from a maximum value at the surface (\(\rho_0\)) to zero at a certain height \(H\). This is captured in the equation: \(\rho(h) = \rho_0 \left(1 - \frac{h}{H}\right).\)

This setup allows us to determine the height at which the density will vanish. In this case, we use calculus to integrate the density formula across the height to calculate the maximum height where pressure equals the initial given pressure at sea level.
This understanding helps us tie together various concepts such as:
  • How pressure, density, and altitude interrelate.
  • Why airplanes use pressurization - to combat the natural thinning of air.
The gradient concept is more realistic than uniform density and provides a useful approximation for atmospheric conditions.
Air Pressure and Its Atmospheric Role
Air pressure is a fundamental concept in understanding atmospheric physics. It's the force that the air exerts on surfaces it comes into contact with. At sea level, this pressure is given as \(1.0\, \text{atm}\), or \(101325\, \text{Pa}\).
As altitude increases, air pressure decreases because there is less atmosphere above weighing down. This change is because air is more compressed at lower altitudes due to gravity. With less air above as you climb higher, the pressure lessens.

This variation of air pressure with altitude is not just a physical curiosity; it affects life significantly. It's why high-altitude locations might feel different when you're breathing - there's less atmospheric pressure and hence less oxygen per breath.
Understanding air pressure also relates directly back to the earlier concepts of air density:
  • Pressure is proportional to density and helps explain the calculation of atmospheric height.
  • It provides real-world applications like meteorology and engineering.
In summary, while pressure is a simple concept, its implications span across weather prediction, climate science, and even our daily experiences.

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Most popular questions from this chapter

An open tube of length \(L=2.3 \mathrm{~m}\) and cross-sectional area \(A=9.2 \mathrm{~cm}^{2}\) is fixed to the top of a cylindrical barrel of diameter \(D=1.2 \mathrm{~m}\) and height \(H=2.3 \mathrm{~m}\). The barrel and tube are filled with water (to the top of the tube). Calculate the ratio of the hydrostatic force on the bottom of the barrel to the gravitational force on the water contained in the barrel. Why is that ratio not equal to \(1.0\) ? (You need not consider the atmospheric pressure.)

Shows two sections of an old pipe system that runs through a hill, with distances \(d_{A}=d_{B}=40 \mathrm{~m}\) and \(D=110 \mathrm{~m}\). On each side of the hill, the pipe radius is \(2.00 \mathrm{~cm}\). However, the radius of the pipe inside the hill is no longer known. To determine it, hydraulic engineers first establish that water flows through the left and right sections at \(2.50 \mathrm{~m} / \mathrm{s}\). Then they release a dye in the water at point \(A\) and find that it takes \(88.8 \mathrm{~s}\) to reach point \(B\). What is the average radius of the pipe within the hill?

A venturi meter is used to measure the flow speed of a fluid in a pipe. The meter is connected between two sections of the pipe (Fig. 14-43); the cross- sectional area \(A\) of the entrance and exit of the meter matches the pipe's cross-sectional area. Between the entrance and exit, the fluid flows from the pipe with speed \(V\) and then through a narrow "throat" of crosssectional area \(a\) with speed \(v\). A manometer connects the wider portion of the meter to the narrower portion. The change in the fluid's speed is accompanied by a change \(\Delta p\) in the fluid's pressure, which causes a height difference \(h\) of the liquid in the two arms of the manometer. (Here \(\Delta p\) means pressure in the throat minus pressure in the pipe.) (a) By applying Bernoulli's equation and the equation of continuity to points 1 and 2 in Fig. 14-43, show that $$ V=\sqrt{\frac{2 a^{2} \Delta p}{\rho\left(a^{2}-A^{2}\right)}} $$ where \(\rho\) is the density of the fluid. (b) Suppose that the fluid is fresh water, that the cross-sectional areas are \(60 \mathrm{~cm}^{2}\) in the pipe and \(32 \mathrm{~cm}^{2}\) in the throat, and that the pressure is \(55 \mathrm{kPa}\) in the pipe and \(41 \mathrm{kPa}\) in the throat. What is the rate of water flow in cubic meters per second?

A large aquarium of height \(5.00 \mathrm{~m}\) is filled with fresh water to a depth of \(2.00 \mathrm{~m}\). One wall of the aquarium consists of thick plastic \(9.00 \mathrm{~m}\) wide. By how much does the total force on that wall increase if the aquarium is next filled to a depth of \(4.00 \mathrm{~m}\) ?

A wood block (mass \(3.67 \mathrm{~kg}\), density \(600 \mathrm{~kg} / \mathrm{m}^{3}\) ) is fitted with lead (density \(1.14 \times 10^{4} \mathrm{~kg} / \mathrm{m}^{3}\) ) so that it floats in water with \(0.700\) of its volume submerged. Find the lead mass if the lead is fitted to the block's (a) top and (b) bottom.

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