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Ch 20: The Micro/Macro Connection
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
Chapter 20, Problem 61b

A nitrogen molecule consists of two nitrogen atoms separated by 0.11 nm, the bond length. Treat the molecule as a rotating dumbbell and find the rms angular velocity at this temperature of a nitrogen molecule around the z-axis, as shown in Figure 20.10.

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Understand the problem: The nitrogen molecule is modeled as a rotating dumbbell, and we are tasked with finding the root-mean-square (rms) angular velocity (ω_rms) at a given temperature. This involves concepts of rotational kinetic energy and the equipartition theorem.
Apply the equipartition theorem: For a rotating molecule, the rotational kinetic energy is distributed equally among its degrees of freedom. A diatomic molecule like nitrogen has two rotational degrees of freedom (around the x-axis and y-axis, assuming the z-axis is the bond axis). The average rotational kinetic energy per degree of freedom is given by \( \frac{1}{2} k_B T \), where \( k_B \) is the Boltzmann constant and \( T \) is the temperature.
Write the total rotational kinetic energy: The total rotational kinetic energy for the molecule is \( E_{rot} = 2 \cdot \frac{1}{2} k_B T = k_B T \). This energy is also expressed as \( \frac{1}{2} I \omega_{rms}^2 \), where \( I \) is the moment of inertia of the molecule and \( \omega_{rms} \) is the root-mean-square angular velocity.
Calculate the moment of inertia: The moment of inertia for a diatomic molecule rotating about an axis perpendicular to the bond is \( I = \mu r^2 \), where \( \mu = \frac{m_1 m_2}{m_1 + m_2} \) is the reduced mass of the two nitrogen atoms, and \( r \) is the bond length (0.11 nm). Use the mass of a nitrogen atom (\( m_N \)) to compute \( \mu \).
Solve for \( \omega_{rms} \): Equate the expressions for rotational kinetic energy: \( k_B T = \frac{1}{2} I \omega_{rms}^2 \). Rearrange to solve for \( \omega_{rms} \): \( \omega_{rms} = \sqrt{\frac{2 k_B T}{I}} \). Substitute \( I = \mu r^2 \) into the equation to get \( \omega_{rms} = \sqrt{\frac{2 k_B T}{\mu r^2}} \). Finally, substitute the known values for \( k_B \), \( T \), \( \mu \), and \( r \) to compute \( \omega_{rms} \).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Bond Length

Bond length is the distance between the nuclei of two bonded atoms, which in this case is 0.11 nm for a nitrogen molecule. This distance is crucial for understanding the molecular structure and the forces acting between the atoms. It influences the molecule's rotational and vibrational properties, which are essential for calculating angular velocity.
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RMS Angular Velocity

Root Mean Square (RMS) angular velocity is a statistical measure of the average angular velocity of a rotating object. It provides a way to quantify the rotational motion of the nitrogen molecule around the z-axis. This concept is important for understanding how temperature affects molecular motion, as higher temperatures generally lead to increased angular velocities.
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Rotational Motion

Rotational motion refers to the movement of an object around an axis. In the context of the nitrogen molecule, it can be modeled as a rotating dumbbell, where the two nitrogen atoms rotate about their center of mass. This concept is fundamental for analyzing the dynamics of molecules, especially in relation to temperature and energy distribution in gases.
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