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The air in your room is composed mostly of oxygen \(\left(\mathrm{O}_{2}\right)\) and nitrogen \(\left(\mathrm{N}_{2}\right)\) molecules. The oxygen molecules are more massive than the nitrogen molecules. (a) Is the rms speed of the \(\mathrm{O}_{2}\) molecules greater than, less than, or equal to the rms speed of the \(\mathrm{N}_{2}\) molecules? (b) Choose the best explanation from among the following: I. The more massive oxygen molecules have greater momentum and therefore greater speed. II. Equal temperatures for the oxygen and nitrogen molecules imply they have equal rms speeds. III. The temperature is the same for both molecules, and hence their average kinetic energies are equal. As a result, the more massive oxygen molecules have lower speeds.

Short Answer

Expert verified
(a) Less than; (b) Explanation III.

Step by step solution

01

Understanding the rms speed formula

The root-mean-square speed (rms speed) of a gas molecule is given by the formula \( v_{rms} = \sqrt{\frac{3kT}{m}} \), where \( k \) is the Boltzmann constant, \( T \) is the temperature, and \( m \) is the mass of the molecule. This formula indicates that rms speed is inversely proportional to the square root of the molecular mass.
02

Comparing molecular masses of O2 and N2

Oxygen molecules \( (\mathrm{O}_2) \) have a higher molecular mass than nitrogen molecules \( (\mathrm{N}_2) \). Specifically, \( \mathrm{O}_2 \) has a molecular mass of 32 g/mol and \( \mathrm{N}_2 \) has a molecular mass of 28 g/mol.
03

Analyzing the effect on rms speed

Since the rms speed is inversely proportional to the square root of the mass, the \( \mathrm{O}_2 \) molecules will have a lower rms speed than \( \mathrm{N}_2 \) molecules because they are more massive. Thus, the rms speed of \( \mathrm{O}_2 \) is less than that of \( \mathrm{N}_2 \).
04

Evaluating explanation options

Option I is incorrect because greater mass does not result in greater speed; it results in greater momentum at the same speed, but not greater speed itself. Option II is incorrect because equal temperatures mean equal kinetic energies, not equal rms speeds. Option III accurately states that equal temperatures imply equal kinetic energies, leading to lower speeds for more massive molecules. Thus, the correct explanation is III.

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Key Concepts

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

Root-Mean-Square Speed
In kinetic theory, the concept of root-mean-square speed (\(v_{rms}\)) is vital for understanding how molecules in a gas move. It is essentially the square root of the average of the squares of the speeds of the gas molecules. This gives us a kind of "average" speed that is useful for describing the motion of gas particles. The formula for \(v_{rms}\) is given by: \[ v_{rms} = \sqrt{\frac{3kT}{m}} \]where
  • \( k \) is the Boltzmann constant, a key value that relates the average kinetic energy of particles in a gas with the temperature.
  • \( T \) is the absolute temperature in Kelvin.
  • \( m \) is the molecular mass of the gas particles.
Notably, this formula shows that the root-mean-square speed is inversely proportional to the square root of the molecular mass. Consequently, if two different gases are at the same temperature, the gas with the lighter molecules will generally move faster. This is why nitrogen (\(\mathrm{N}_2\)) molecules move faster than oxygen (\(\mathrm{O}_2\)) molecules at room temperature. This concept is significant in explaining various gas behaviors and dynamics.
Molecular Mass
Molecular mass is fundamentally a measure of the mass of a molecule. It is the sum of the masses of the individual atoms that make up the molecule. For example, in our context, oxygen (\(\mathrm{O}_2\)) has a molecular mass of 32 g/mol, which is higher than that of nitrogen (\(\mathrm{N}_2\)) with a molecular mass of 28 g/mol.
The molecular mass plays a crucial role in determining the root-mean-square speed of gas molecules. As seen in the formula for \(v_{rms}\), molecular mass is in the denominator under the square root, indicating that as the molecular mass increases, \(v_{rms}\) decreases. This means heavier molecules move more slowly than lighter ones at the same temperature.
Understanding molecular mass helps in predicting how different gases will behave under similar conditions, such as in a room filled with both oxygen and nitrogen.
In many practical applications, knowing the molecular mass aids in the calculation of other properties and behaviors of gases, like diffusion rates and pressure inside a container.
Kinetic Energy
Kinetic energy is an essential concept in the kinetic theory of gases. It refers to the energy possessed by an object due to its motion, and in gases, it is a reflection of the movement of particles. The average kinetic energy of gas molecules can be expressed as: \[ KE_{avg} = \frac{3}{2}kT \]where:
  • \( k \) is the Boltzmann constant,
  • \( T \) is the absolute temperature.
This equation tells us that kinetic energy depends only on the temperature of the gas, not on the type or mass of the molecules. Therefore, at the same temperature, different gases have the same average kinetic energy.
This is why, even though the oxygen (\(\mathrm{O}_2\)) is more massive than nitrogen (\(\mathrm{N}_2\)), they have the same kinetic energy at a given temperature. However, because of their different masses, they have different speeds—lighter nitrogen molecules move faster than heavier oxygen molecules at equal kinetic energy levels.
Understanding kinetic energy is crucial for grasping how temperature influences the motion and speed of gas particles.
Boltzmann Constant
The Boltzmann constant (\( k \)) is a fundamental physics constant that plays a crucial role in the kinetic theory of gases. It serves as a bridge between the macroscopic and microscopic worlds, relating the temperature of a gas to the average kinetic energy of its particles.
The value of the Boltzmann constant is approximately \( 1.38 \times 10^{-23} \text{ J/K} \).
In essence, this constant is part of the connection between energy at the molecular level marked by temperature and energy in a thermal system overall. By using the Boltzmann constant in the formulas for root-mean-square speed and kinetic energy, it becomes possible to accurately predict and understand the behavior of gases in various conditions.
  • The Boltzmann constant allows kinetic energy to be expressed in terms of temperature, simplifying calculations and providing clear insights into the energy dynamics of gas molecules.
  • In the context of statistical mechanics, it provides a scale for temperature relative to energy per particle.
Its role is indispensable in connecting molecular motion to observable thermal properties, making it a key element in thermodynamics and statistical mechanics across scientific disciplines.

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Most popular questions from this chapter

A balloon is filled with helium at a pressure of \(2.4 \times 10^{5} \mathrm{Pa}\). The balloon is at a temperature of \(18^{\circ} \mathrm{C}\) and has a radius of \(0.25 \mathrm{m} .\) (a) How many helium atoms are contained in the balloon? (b) Suppose we double the number of helium atoms in the balloon, keeping the pressure and the temperature fixed. By what factor does the radius of the balloon increase? Explain.

If you put a helium-filled balloon in the refrigerator, (a) will its volume increase, decrease, or stay the same? (b) Choose the best explanation from among the following: I. Lowering the temperature of an ideal gas at constant pressure results in a reduced volume. II. The same amount of gas is in the balloon; therefore, its volume remains the same. III. The balloon can expand more in the cool air of the refrigerator, giving an increased volume.

A piano wire \(0.82 \mathrm{m}\) long and \(0.93 \mathrm{mm}\) in diameter is fixed on one end. The other end is wrapped around a tuning peg \(3.5 \mathrm{mm}\) in diameter. Initially the wire, whose Young's modulus is \(2.4 \times 10^{10} \mathrm{N} / \mathrm{m}^{2},\) has a tension of \(14 \mathrm{N} .\) Find the tension in the wire after the tuning peg has been turned through one complete revolution.

The molecules in a tank of hydrogen have the same rms speed as the molecules in a tank of oxygen. State whether each of the following statements is true, false, or unknowable with the given information: (a) the pressures are the same; (b) the hydrogen is at the higher temperature; \((c)\) the hydrogen has the higher pressure; (d) the temperatures are the same; (e) the oxygen is at the higher temperature.

An automobile tire has a volume of \(0.0185 \mathrm{m}^{3} .\) At a temperature of \(294 \mathrm{K}\) the absolute pressure in the tire is \(212 \mathrm{kPa}\). How many moles of air must be pumped into the tire to increase its pressure to \(252 \mathrm{kPa}\), given that the temperature and volume of the tire remain constant?

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