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An oxygen molecule is moving near the earth's surface. Another oxygen molecule is moving in the ionosphere (the uppermost part of the earth's atmosphere) where the Kelvin temperature is three times greater. Determine the ratio of the translational rms speed in the ionosphere to the translational rms speed near the earth's surface.

Short Answer

Expert verified
The ratio of rms speeds is \( \sqrt{3} \).

Step by step solution

01

Understand the Problem

To solve this problem, we need to determine the ratio of the translational root mean square (rms) speed of an oxygen molecule in the ionosphere to that near the earth's surface. We know that the temperature in the ionosphere is three times greater than that near the earth's surface.
02

Formula for rms Speed

The formula for the translational rms speed of a gas is given by\[ v_{rms} = \sqrt{\frac{3kT}{m}} \]where \( v_{rms} \) is the root mean square speed, \( k \) is the Boltzmann constant, \( T \) is the temperature, and \( m \) is the mass of a gas molecule.
03

Relation of Speeds to Temperature

According to the formula, the rms speed is proportional to the square root of the absolute temperature:\[ v_{rms} \propto \sqrt{T} \]Therefore, if the temperature increases, the rms speed increases as the square root of the temperature ratio.
04

Calculate the Ratio of Speeds

Let \( T_1 \) be the temperature near the earth's surface and \( T_2 = 3T_1 \) be the temperature in the ionosphere. The ratio of the rms speeds is given by:\[ \frac{v_{rms, 2}}{v_{rms, 1}} = \frac{\sqrt{3T_1}}{\sqrt{T_1}} = \sqrt{3} \]
05

Conclusion

The ratio of the translational rms speeds of an oxygen molecule in the ionosphere to that near the earth's surface is \( \sqrt{3} \).

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Key Concepts

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

rms speed
Root mean square (rms) speed is a key concept in thermal physics, giving us insight into how fast molecules are moving in a gas. It's derived from the kinetic theory of gases, which assumes that gas molecules are in constant, random motion. The rms speed is essentially a measure of the average speed of these molecules. It is not identical to the average speed itself because it accounts for the square of the velocities, ensuring that we do not end up with negative values.

The rms speed is calculated using the formula: \[ v_{rms} = \sqrt{\frac{3kT}{m}} \] Here, \( v_{rms} \) is the root mean square speed, \( k \) is the Boltzmann constant, \( T \) is the absolute temperature, and \( m \) is the mass of a molecule.

This equation indicates that the rms speed increases with temperature and decreases with an increase in molecular mass. Hence, lighter molecules at higher temperatures have higher rms speeds.
  • Makes it easier to see how fast molecules are moving.
  • At higher temperatures, molecules move faster.
  • Lighter molecules also move faster compared to heavier ones at the same temperature.
Understanding rms speed allows scientists to predict the behavior of gases under different conditions, which is crucial in fields such as meteorology and engineering.
Boltzmann constant
The Boltzmann constant is a fundamental constant in physics, symbolized by \( k \), with a value of approximately \( 1.38 \times 10^{-23} \text{ J/K} \). It acts as a bridge between the macroscopic and microscopic worlds.

In the formula for rms speed, it helps quantify the effect of temperature on the motion of molecules. The Boltzmann constant essentially relates the average kinetic energy of particles in a gas with the temperature of the gas. This relationship is given by:\[ \text{Average kinetic energy per molecule} = \frac{3}{2}kT \]

This expression shows that the kinetic energy of molecules is directly proportional to the temperature in Kelvin, and it also reflects the commitment of thermal physics towards explaining the microscopic behavior of particle systems using macroscopic measurements like temperature.
  • Connects temperature with kinetic energy
  • Aids in understanding microscopic behaviors through macroscopic concepts
  • Fundamental in thermodynamics and statistical mechanics
Its utility extends beyond gases, influencing theories and applications in fields such as solid-state physics and cosmology.
root mean square speed
The root mean square speed is a specialized way of considering the speed of particles in a gas that takes into account the speeds of all individual particles. It provides more insight than simply looking at average speed because it squares the individual velocities before averaging them, giving more weight to higher speeds. It is an important quantity because it reflects the energy contained in the movement of gas molecules.

The expression for root mean square speed follows the formula:\[ v_{rms} = \sqrt{\frac{3kT}{m}} \]This equation incorporates both the Boltzmann constant \( k \) and the temperature \( T \), making it clear that the rms speed is temperature-dependent. As temperature rises, molecules move faster on average.

Using the concept of root mean square speed, scientists can extrapolate the dynamic behaviors of gases and understand how they interact with their environment. This has practical applications in understanding atmospheric phenomena, studying chemical reactions at the molecular level, and improving technologies that rely on gas dynamics like aerospace engineering.
  • Provides deeper insight into molecular speeds
  • Essential for interpreting gas behavior
  • Important for applications in multiple scientific fields
Exploring the root mean square speed offers a window into the intrinsic properties of gases, enabling more details on the nature of molecular movement.

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Most popular questions from this chapter

What is the density (in \(\mathrm{kg} / \mathrm{m}^{3}\) ) of nitrogen gas (molecular mass \(=28 \mathrm{u})\) at a pressure of 2.0 atmospheres and a temperature of \(310 \mathrm{K} ?\)

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A runner weighs \(580 \mathrm{N}\) (about \(130 \mathrm{lb}\) ), and \(71 \%\) of this weight is water. (a) How many moles of water are in the runner's body? (b) How many water molecules ( \(\mathrm{H}_{2} \mathrm{O}\) ) are there?

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