91Ó°ÊÓ

The average temperature of the atmosphere near the surface of the earth is about \(20^{\circ} \mathrm{C}\). (a) What is the root-mean-square speed of hydrogen molecules, \(\mathrm{H}_{2}\), at this temperature? (b) The escape speed from the earth is about \(11 \mathrm{~km} / \mathrm{s}\). Is the average \(\mathrm{H}_{2}\) molecule moving fast enough to escape? (c) Compare the rms speeds of oxygen \(\left(\mathrm{O}_{2}\right)\) and nitrogen \(\left(\mathrm{N}_{2}\right)\) with that of \(\mathrm{H}_{2}\). (d) So why has the hydrogen been able to escape the earth's gravity, but the heavier gases (such as \(\mathrm{O}_{2}\) and \(\mathrm{N}_{2}\) ) have not, even though none of these gases has an rms speed equal to the escape speed of the earth? (Hint: Do all the molecules have the same rms speed, or are some moving faster? since the rms speed for \(\mathrm{H}_{2}\) is greater than that of \(\mathrm{O}_{2}\) and \(\mathrm{N}_{2}\), which gas would have a higher percentage of its molecules moving fast enough to escape?)

Step by step solution

01

Understanding Root-Mean-Square Speed Formula

The root-mean-square (rms) speed for a gas is given by the formula \( v_{rms} = \sqrt{\frac{3kT}{m}} \), where \( k \) is Boltzmann's constant \( (1.38 \times 10^{-23} \text{ J/K}) \), \( T \) is the temperature in Kelvin, and \( m \) is the mass of one molecule of the gas.

02

Convert Temperature to Kelvin

The given temperature is \( 20^{\circ} \text{C} \). To convert Celsius to Kelvin, use the formula \( T_{K} = T_{C} + 273.15 \). Substituting, \( T = 20 + 273.15 = 293.15 \text{ K} \).

03

Calculate Molecular Masses

For hydrogen \( (\text{H}_2) \), the molecular mass is about \(2 \text{ u} = 2 \times 1.66 \times 10^{-27} \text{ kg} \). For oxygen \( (\text{O}_2) \), it's about \(32 \text{ u} = 32 \times 1.66 \times 10^{-27} \text{ kg} \), and for nitrogen \( (\text{N}_2) \), it's about \(28 \text{ u} = 28 \times 1.66 \times 10^{-27} \text{ kg} \).

04

Calculate RMS Speed for Hydrogen

Substitute the values for hydrogen into the rms speed formula: \[ v_{rms(\text{H}_2)} = \sqrt{\frac{3 \times 1.38 \times 10^{-23} \times 293.15}{2 \times 1.66 \times 10^{-27}}} \text{ m/s}\]. Perform the calculation and find \( v_{rms(\text{H}_2)} \approx 1938 \text{ m/s}.\)

05

Compare with Escape Speed

The escape speed from Earth is \( 11,000 \text{ m/s} \). Compare this with the calculated \( v_{rms(\text{H}_2)} \). Since \( 1938 \text{ m/s} \) is much less than \( 11,000 \text{ m/s} \), an average hydrogen molecule is not moving fast enough to escape.

06

Calculate RMS Speeds for Oxygen and Nitrogen

Repeat the calculation for \( \text{O}_2 \) and \( \text{N}_2 \) using their molecular masses. For \( \text{O}_2 \): \[ v_{rms(\text{O}_2)} = \sqrt{\frac{3 \times 1.38 \times 10^{-23} \times 293.15}{32 \times 1.66 \times 10^{-27}}} \text{ m/s} \approx 489 \text{ m/s} \]. For \( \text{N}_2 \): \[ v_{rms(\text{N}_2)} = \sqrt{\frac{3 \times 1.38 \times 10^{-23} \times 293.15}{28 \times 1.66 \times 10^{-27}}} \text{ m/s} \approx 515 \text{ m/s} \].

07

Explain Hydrogen Escape

Despite none of the gases having an rms speed equal to the escape speed, hydrogen escapes because the velocity distribution means some molecules have speeds significantly above the average. This means a more significant fraction of lighter molecules, such as \( \text{H}_2 \), will exceed the escape velocity compared to heavier gases.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Root-mean-square speed

The root-mean-square (rms) speed is a statistical measure of the speed of particles in a gas. It represents an average speed value, useful for understanding the kinetic energy of gas molecules. This speed is crucial in physics because it helps in comparing the kinetic energy of different molecules at the same temperature. The formula to calculate the rms speed is given by:

• \[ v_{rms} = \sqrt{\frac{3kT}{m}} \]
• Here, \( k \) is Boltzmann's constant \(1.38 \times 10^{-23} \mathrm{J/K}\), \( T \) is the absolute temperature in Kelvin, and \( m \) is the mass of a single molecule in kilograms.

The rms speed increases with temperature and decreases with an increase in molecular mass. This implies that lighter molecules at the same temperature move faster than heavier ones.

Escape velocity

Escape velocity is the minimum speed an object must have to break free from the gravitational attraction of a planet without further propulsion. For Earth, this velocity is about 11 km/s. Comparing escape velocity with the molecular speeds of different gases helps explain why some gases can escape a planet's atmosphere over time. Although the rms speed of a gas like hydrogen might be less than Earth's escape velocity, some molecules in a distribution will naturally exceed this speed. As a result, lighter gases such as hydrogen are more likely to reach escape velocity than heavier gases, enabling them to leave the atmosphere over long periods.

Boltzmann's constant

Boltzmann's constant \( k \) is a fundamental physical constant that connects the average kinetic energy of particles in a gas with the temperature of the gas. Its value is \(1.38 \times 10^{-23} \mathrm{J/K}\). In the context of kinetic theory, it's used to quantify the energy per degree of freedom of a molecule.

• It plays a crucial role in calculating the rms speed: \( v_{rms} = \sqrt{\frac{3kT}{m}} \).
• Boltzmann's constant acts as a bridge between macroscopic and microscopic physics, linking molecular-level behavior with thermodynamic temperature.

Understanding Boltzmann's constant helps explain how temperature influences the energy and motion of molecules in gases, crucial for the kinetic theory of gases.

Molecular mass and velocity

Molecular mass significantly influences the speed of gas particles. According to the formula for rms speed, as molecular mass \( m \) increases, the rms speed \( v_{rms} \) decreases. Let's see how the masses of different gases affect their velocities:

• For hydrogen \( (\text{H}_2) \), with a molecular mass of about \(2 \text{ u}\), the rms speed is approximately \(1938 \text{ m/s} \).
• For oxygen \( (\text{O}_2) \) and nitrogen \( (\text{N}_2) \), with larger molecular masses of \(32 \text{ u}\) and \(28 \text{ u}\) respectively, their rms speeds are significantly lower, around \(489 \text{ m/s} \) and \(515 \text{ m/s} \) respectively.

So, lighter molecules such as hydrogen move faster than heavier molecules like oxygen and nitrogen at the same temperature. This concept is essential for understanding the behavior and distribution of gases in an atmosphere, affecting how they escape planetary gravitational fields over time.

Gas laws and behavior

The behavior of gases is described by a set of laws that relate parameters like pressure, volume, temperature, and molecular speed. These include Boyle's Law, Charles's Law, and Avogadro's Law. The ideal gas law, given by \( PV = nRT \), is a fundamental equation combining these relationships, where \( P \) is pressure, \( V \) is volume, \( n \) is number of moles, \( R \) is the ideal gas constant, and \( T \) is temperature.In connection with the kinetic theory of gases, these laws help explain:

• How changes in temperature can affect the speed and energy of gas molecules.
• How pressure and volume changes are interconnected with molecular motion and speeds.

Understanding gas laws provides insight into predicting and explaining how gases will behave under different environmental conditions, essential for many applications in science and engineering.