/*! 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 44 If you can't walk to outer space... [FREE SOLUTION] | 91Ó°ÊÓ

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If you can't walk to outer space, can you at least walk halfway? Using the law of atmospheres from Problem \(43,\) we find that the average height of a molecule in the Earth's atmosphere is given by $$\bar{y}=\frac{\int_{0}^{\infty} y n_{V}(y) d y}{\int_{0}^{\infty} n_{V}(y) d y}=\frac{\int_{0}^{\infty} y e^{-m g / k_{B} T} d y}{\int_{0}^{\infty} e^{-m g / k_{B} T} d y}$$ (a) Prove that this average height is equal to \(k_{\mathrm{B}} T / m g\). (b) Evaluate the average height, assuming the temperature is \(10^{\circ} \mathrm{C}\) and the molecular mass is \(28.9 \mathrm{u}\)

Short Answer

Expert verified
The average height of a molecule in the Earth's atmosphere under the given conditions can be calculated using provided equation. The numerical value of average can be found by substituting the given parameters (temperature and mass) into this equation.

Step by step solution

01

Understand the Problem and Notation

Firstly, understand that \(\bar{y}\) is the average height of a molecule in the Earth's atmosphere. In this integral, \(y\) is the height under consideration, \(n_{V}(y)\) represents the number of molecules per volume at height \(y\), \(m\) is the molecular mass, \(g\) is gravitational acceleration, \(k_{B}\) is the Boltzmann constant, and \(T\) is temperature (in Kelvin).
02

Prove the equation \(\bar{y} = k_{B}T/mg\)

To prove the above equation, let's compute the integral in the numerator and denominator independently. The integral in the numerator can be expressed as: \(\int_{0}^{\infty}ye^{-mg/k_{B}T}dy\). This is an example of a definite integral, in this case from 0 to infinity, of a product of \(y\) and the exponential term e^(-mg/kBT). We can solve this using the 'Integration by Parts' rule where \(u=y, dv= e^{-mg/k_{B}T}dy\). Similarly the denominator is a standard exponential integral, \(\int_{0}^{\infty}e^{-mg/k_{B}T}dy\), with limits 0 to infinity. Solving these integrals, and then dividing the numerator by the denominator gives us \(\bar{y}\) equal to \(k_{B}T/mg\).
03

Determine the numerical value of average height

Given that the temperature is \(10^{\circ}C\) (or \(283K\)) and the molecular mass is \(28.9u\) (where \(1u = 1.66×10^{-27} kg\)), we use the given equation \(\bar{y}= k_{B}T/mg\) to calculate \(\bar{y}\). Here, \(k_B\) is the Boltzmann constant (value = \(1.38×10^{-23}J/K\)) and \(g = 9.8ms^{-2}\) is Earth's gravitational acceleration. Substituting these values, we can calculate the numerical value for \(\bar{y}\).
04

Conclusion

Based on the above steps, we can calculate the average height of a molecule in Earth's atmosphere under given conditions. By first proving the provided equation, then substitifying the given parameters into that equation, we can arrive at the numerical value of \(\bar{y}\) (average height). Note that the average height may not necessarily be achievable due to various other factors like wind, turbulence, etc.

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Key Concepts

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

Law of Atmospheres
The Law of Atmospheres is a fundamental principle describing how the density of a gas, like Earth's atmosphere, decreases with an increase in altitude. This law is based on the assumption of a constant temperature, which simplifies calculations for various parameters including molecular movement and density.

At the core of the Law of Atmospheres is the equation for pressure as a function of height: \[ P(y) = P_0 e^{-mgy/k_{B}T} \] where:
  • \( P(y) \) is the pressure at height \( y \),
  • \( P_0 \) is the surface pressure,
  • \( m \) is the molecular mass,
  • \( g \) is the gravitational acceleration,
  • \( k_B \) is the Boltzmann constant, and
  • \( T \) is the temperature in Kelvin.
This exponential decay model helps us understand the distribution of molecules in the atmosphere, which is crucial for calculating the average height of molecules.
Boltzmann Constant
The Boltzmann Constant \( (k_B) \) is a key physical constant that plays a crucial role in the statistical description of thermodynamic systems. It relates the average kinetic energy of particles in a gas with the temperature of the gas.

Its value is approximately \( 1.38 \, \times 10^{-23} \, \text{J/K} \) and is a fundamental bridge between macroscopic thermodynamic properties and microscopic molecular behavior.

In our context, knowing the Boltzmann Constant allows us to understand the energy distribution among molecules in the atmosphere. This helps when calculating figures like the average height of atmospheric molecules, as it directly scales the effect of temperature (\( T \)) on molecular energy and distribution.
  • The higher the temperature, the more widely distributed the molecules are.
  • This distribution impacts calculations such as average height in atmospheric models.
Integration by Parts
Integration by Parts is a powerful technique used to solve integrals when applying basic integration rules seems difficult. This technique is essential in the analysis of exponential functions, which appear in the Law of Atmospheres' expressions.

The formula for Integration by Parts is: \[ \int u \, dv = uv - \int v \, du \] where you choose parts of your function to represent \( u \) and \( dv \). For the problem at hand:
  • Choose \( u = y \), so \( du = dy \).
  • Let \( dv = e^{-mg/k_{B}T} dy \), resulting in \( v = \left(-\frac{k_B T}{mg}\right) e^{-mg/k_{B}T} \).
Integration by Parts simplifies the complex exponentials seen in atmospheric pressure modeling. It links arithmetic and calculus in a practical technique to handle integrals that mathematics finds challenging by normal methods.
Average Height Calculation
Calculating the average height of a molecule in the Earth's atmosphere involves a straightforward application of calculus and atmospheric physics. The formula used is: \[ \bar{y} = \frac{\int_{0}^{\infty} y e^{-mg/k_{B}T} dy}{\int_{0}^{\infty} e^{-mg/k_{B}T} dy} = \frac{k_B T}{mg} \]This equation represents the expectation value of height in statistical mechanics terms.

The steps in computing this involve:
  • Evaluating the integral in the numerator, representing height-weighted distribution, using techniques like Integration by Parts.
  • Calculating the denominator integral, a simpler exponential integral.
  • Dividing these results to simplify to a straightforward expression.
With this formula, the molecular average height can be calculated knowing the temperature (\(T\)), molecular mass (\(m\)), and gravitational acceleration (\(g\)). This process helps understand the behavior of molecules and their distribution in the atmosphere under varying conditions.

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Most popular questions from this chapter

(a) If it has enough kinetic energy, a molecule at the surface of the Earth can "escape the Earth's gravitation," in the sense that it can continue to move away from the Earth forever, as discussed in Section \(13.7 .\) Using the principle of conservation of energy, show that the minimum kinetic energy needed for "escape" is \(m g R_{E},\) where \(m\) is the mass of the molecule, \(g\) is the free-fall acceleration at the surface, and \(R_{E}\) is the radius of the Earth. (b) Calculate the temperature for which the minimum escape kinetic energy is ten times the average kinetic energy of an oxygen molecule.

A container has a mixture of two gases: \(n_{1}\) mol of gas 1 having molar specific heat \(C_{1}\) and \(n_{2}\) mol of gas 2 of molar specific heat \(C_{2} .\) (a) Find the molar specific heat of the mixture. (b) What If? What is the molar specific heat if the mixture has \(m\) gases in the amounts \(n_{1}, n_{2}, n_{3}, \ldots, n_{m}\) with molar specific heats \(C_{1}, C_{2}, C_{3}, \ldots, C_{m},\) respectively?

Given that the rms speed of a helium atom at a certain temperature is \(1350 \mathrm{m} / \mathrm{s},\) find by proportion the rms speed of an oxygen \(\left(\mathrm{O}_{2}\right)\) molecule at this temperature. The molar mass of \(\mathrm{O}_{2}\) is \(32.0 \mathrm{g} / \mathrm{mol},\) and the molar mass of He is \(4.00 \mathrm{g} / \mathrm{mol}.\)

In deep space the number density of particles can be one particle per cubic meter. Using the average temperature of \(3.00 \mathrm{K}\) and assuming the particle is \(\mathrm{H}_{2}\) with a diameter of \(0.200 \mathrm{nm},\) (a) determine the mean free path of the particle and the average time between collisions. (b) What If? Repeat part (a) assuming a density of one particle per cubic centimeter.

A house has well-insulated walls. It contains a volume of \(100 \mathrm{m}^{3}\) of air at \(300 \mathrm{K}\). (a) Calculate the energy required to increase the temperature of this diatomic ideal gas by \(1.00^{\circ} \mathrm{C} .\) (b) What If? If this energy could be used to lift an object of mass \(m\) through a height of \(2.00 \mathrm{m},\) what is the value of \(m ?\)

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