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Chapter 8: Problem 8

The simplest model of the atmosphere is isothermal: The atmosphere has one temperature throughout it. A better approximation, the adiabatic atmosphere, relaxes this assumption and incorporates the adiabatic gas law: $$p V^{\gamma} \propto 1$$ where \(p\) is atmospheric pressure, \(V\) is the volume of a parcel of air, and \(\gamma \equiv c_{p} / c_{v}\) is the ratio of the two specific heats in the gas. (For dry air, \(\gamma=1.4\).) Imagine an air parcel rising up a mountain. As the parcel rises into air with a lower. pressure, it expands, and its volume and temperature change according to a combination of the adiabatic and ideal gas laws. a. What easy case of \(\gamma\) reproduces the isothermal atmosphere? b. For \(\gamma=1.4\) (dry air), will air temperature decrease with, increase with, or be independent of height?

Short Answer

Expert verified
a. The isothermal atmosphere is reproduced when \(\gamma = 1\). b. For \(\gamma=1.4\) (dry air), the air temperature will decrease with height.

Step by step solution

01

Recall the adiabatic gas law and ideal gas law

Adiabatic gas law: \(pV^{\gamma} \propto 1\), where \(p\) is pressure, \(V\) is volume, and \(\gamma\) is the specific heats ratio (\(c_p / c_v\)). Ideal gas law: \(pV = nRT\), where \(n\) is the number of moles, \(R\) is the gas constant, and \(T\) is the temperature.
02

Combine the adiabatic gas law and ideal gas law

First, we express the volume from the ideal gas law as \(V = \frac{nRT}{p}\) and substitute it into the adiabatic gas law: \(p\left(\frac{nRT}{p}\right)^{\gamma} \propto 1\) Now, simplify the equation: (\(\frac{nR^{\gamma}T^{\gamma}}{p^{\gamma-1}} \propto 1\))
03

Analyze the case when 纬 reproduces the isothermal atmosphere

To find the case when adiabatic gas law exhibits isothermal behavior, we need to find the \(\gamma\) value that makes the temperature independent of pressure and volume. This means the temperature remains constant throughout the atmosphere. In the equation, \(\frac{nR^{\gamma}T^{\gamma}}{p^{\gamma-1}} \propto 1\), we need to find a value of \(\gamma\) that makes the equation independent of \(p\). For this to be the case, the exponent of \(p\) must be zero, which means that \(\gamma - 1 = 0\), so \(\gamma = 1\). So the easy case when 纬 reproduces the isothermal atmosphere is when \(\gamma = 1\).
04

Analyze the case when 纬=1.4 (dry air) and the temperature with height

We have 纬=1.4 for dry air. We will use this value in the equation \(\frac{nR^{\gamma}T^{\gamma}}{p^{\gamma-1}} \propto 1\) to analyze the temperature with height. Using the given value for 纬, we have: \(\frac{nR^{1.4}T^{1.4}}{p^{0.4}} \propto 1\) From this equation, we see that the temperature is not independent of pressure, as it was when 纬=1. As the air parcel rises in height, the pressure decreases, causing the exponent \(p^{0.4}\) to increase and therefore the temperature to decrease. So for dry air with 纬=1.4, the air temperature will decrease with height. Answers: a. The easy case of 纬 that reproduces the isothermal atmosphere is when 纬=1. b. For 纬=1.4 (dry air), the air temperature will decrease with height.

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Key Concepts

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

Adiabatic Gas Law
The Adiabatic Gas Law is a fascinating concept in thermodynamics, describing how the pressure and volume of a gas change when it expands or compresses without exchanging heat with its surroundings. It is mathematically expressed as: \[ p V^{\gamma} \propto 1 \] where:
  • \( p \) is the atmospheric pressure.
  • \( V \) is the volume of the gas.
  • \( \gamma \equiv c_{p} / c_{v} \) represents the ratio of specific heats, where \( c_{p} \) is the specific heat at constant pressure and \( c_{v} \) is the specific heat at constant volume.
For dry air, \( \gamma \) is typically 1.4. When a parcel of air rises, it moves into an area of lower pressure, causing it to expand. According to the adiabatic process, this expansion results in both a change in volume and a decrease in temperature. Understanding this law helps predict how temperature will vary with altitude, especially in adiabatic processes where there is no heat transfer between the air parcel and its environment.
Isothermal Atmosphere
An Isothermal Atmosphere is a simplified model where the entire atmosphere maintains a constant temperature regardless of the height. This scenario allows for easier calculations and predictions since the temperature does not change with altitude. Mathematically, an isothermal condition arises when the atmospheric behavior aligns such that:
  • The gas experiences negligible temperature variation with changes in pressure and volume.
In our exercise, we found that the atmosphere can be considered isothermal if \( \gamma = 1 \). When \( \gamma \) equals 1, the variations in pressure and volume do not affect the temperature, resulting in an unchanged isothermal state throughout atmospheric layers. However, in real-world conditions, this model considerably oversimplifies atmospheric dynamics, as temperature usually decreases with height due to the adiabatic effect.
Ideal Gas Law
The Ideal Gas Law is another crucial principle for understanding atmospheric conditions. It relates the pressure, volume, and temperature of an ideal gas using the equation: \[ pV = nRT \] where:
  • \( p \) is pressure.
  • \( V \) is volume.
  • \( n \) is the number of moles of the gas.
  • \( R \) is the universal gas constant.
  • \( T \) is the temperature.
This law combines with the Adiabatic Gas Law to describe how an air parcel behaves as it moves through different atmospheric pressures without exchanging heat. When a gas packet rises in altitude, it experiences lower pressure and expands, leading to a temperature decrease if no heat is added. This characteristic is crucial for understanding meteorological phenomena and predicting weather patterns involving rising or sinking air parcels.

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Most popular questions from this chapter

The terminal speed \(v\) of a raindrop with radius \(r\) can be written in the following dimensionless form: $$\frac{v^{2}}{r g}=f\left(\frac{\rho_{\text {water }}}{\rho_{\text {air }}}\right)$$ In this problem, you use easy cases of \(x=\rho_{\text {waler }} / \rho_{\text {air }}\) to guess how buoyancy affects this result. (Imagine that you may vary the density of air or water as needed.) In dimensional analysis, including the buoyant force requires including \(\rho_{\text {air }}, g\), and \(r\) in order to compute the weight of the displaced fluid (which is the buoyant force) - but those variables are already included in the dimensional analysis. Therefore, including buoyancy doesn't require a new dimensionless group. So it must change the form of the dimensionless function \(f\). a. Before you account for buoyancy: What is the dimensionless function \(f(x)\) ? Assume spherical raindrops and that \(c_{\mathrm{d}} \approx\) \(0.5\) b. What would be the effect of buoyancy if \(\rho_{\text {water }}\) were equal to \(r h o_{\text {air }}\) ? This thought experiment is the easy case \(x=1\). Therefore, find \(f(1)\). c. Guess the general form of \(f\) with buoyancy, and thereby find \(v\) including the effect of buoyancy. d. Explain physically the difference between \(v\) without and with buoyancy. Hint: How does buoyancy affect \(g\) ?

Once in a while, there are four interesting easy-cases regimes. An example is orbits. The dimensionless parameter \(\beta\) that characterizes the type of an orbit is \(\beta=\frac{\text { kinetic energy }}{\mid \text { gravitational potential energy } \mid}\) where the absolute value handles a possibly negative potential energy. A related dimensionless parameter is the orbit eccentricity \(\epsilon .\) In terms of the eccentricity, a planet's orbit in polar coordinates is $$r(\theta)=\frac{l}{1+\epsilon \cos \theta}$$ where the Sun is at the origin, and \(l\) is the length scale of the orbit ( \(l\) is diagrammed in Problem 8.24). Sketch and classify the four orbit shapes according to their values of \(\beta\) and \(\epsilon\) (giving a point value or a range, as appropriate): (a) circle, (b) ellipse, (c) parabola, and (d) hyperbola.

A liquid's surface tension, usually denoted \(\gamma\), is the energy required to create new surface divided by the area of the new surface. For a falling raindrop, surface tension and drag compete: the drag force flattens the raindrop, and surface tension keeps it spherical. If the drop gets too flat, it can lower its surface energy by breaking into smaller and more spherical droplets. The fluid dynamics is complicated, but we don't need to know it. Instead, use competition reasoning (easy cases) to estimate the maximum size of raindrops.

A fixed-term, fixed-interest-rate loan has four important parameters: the principal \(P\) (the amount borrowed), the interest rate \(r\), the repayment interval \(\tau\), and the number of payments \(n\). The loan is repaid in \(n\) equal payments over the loan term \(n \tau\). Each payment consists of a principal and an interest portion. The interest portion is the interest accumulated on the principal outstanding during the term; the principal portion reduces the outstanding principal. The dimensionless quantity determining the type of loan is \(\beta \equiv n \tau r . .\) a. Estimate the payment (the amount per term) in the easy case \(\beta=0\), in terms of \(P, n\), and \(\tau\). (The term \(n \tau\) and the repayment interval \(\tau\) don't vary that much \(-\tau\) is usually 1 month and \(n \tau\) is somewhere between 3 to 30 years - so \(\beta \ll 1\) is usually reached by lowering the interest rate \(r\).) b. Estimate the payment in the slightly harder case where \(\beta \ll 1\) (which includes the \(\beta=0\) case). In this regime, the loan is called an installment loan. c. Estimate the payment in the easy case \(\beta \gg 1 .\) In this regime, the loan is called an annuity. (This regime is usually reached by increasing \(n\).)

Find the dimensionless function \(f\) based on the simpler, but less symmetric, dimensionless group \(m_{1} / m_{2}\) : $$\frac{a}{g}=f\left(\frac{m_{1}}{m_{2}}\right)$$ Compare it to the dimensionless function that results from choosing the symmetric form \(\left(m_{1}-m_{2}\right) /\left(m_{1}+m_{2}\right)\) as the independent variable.
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