Problem 82 - University Physics with Modern Physics

Question

(a) Calculate the total rotational kinetic energy of the molecules in 1.00 mol of a diatomic gas at 300 \(\mathrm{K}\) . (b) Calculate the moment of inertia of an oxygen molecule \(\left(\mathrm{O}_{2}\right)\) for rotation about either the \(y\) - or \(z\) -axis shown in Fig. 18.18 . Treat the molecule as two massive points (representing the oxygen atoms) separated by a distance of \(1.21 \times 10^{-10} \mathrm{m}\) . The molar mass of oxygen atoms is \(16.0 \mathrm{g} / \mathrm{mol} .\) (c) Find the rms angular velocity of rotation of an oxygen molecule about either the \(y\) - or \(z\) -axis shown in Fig. 18.15. How does your answer compare to the angular velocity of a typical piece of rapidly rotating machinery \((10,000 \mathrm{rev} / \mathrm{min}) ?\)

Step-by-Step Solution

Answer

(a) 2494.2 J; (b) 1.17168 x 10^-30 kg·m²; (c) 5.256 x 10¹¹ rad/s; much higher than 1047.2 rad/s.

Step 1: Kinetic Energy of Diatomic Gas

The total rotational kinetic energy for a diatomic gas is given by \( KE_{rot} = \frac{f}{2} nRT \), where \( f = 2 \) (degrees of freedom for rotation, since it's diatomic), \( n \) is the number of moles, \( R = 8.314 \frac{J}{mol \cdot K} \) is the gas constant, and \( T = 300 \ K \) is the temperature. Therefore, \( KE_{rot} = \frac{2}{2} \times 1.00 \ mol \times 8.314 \frac{J}{mol \cdot K} \times 300 \ K = 2494.2 \ Joules \).

Step 2: Moment of Inertia Calculation

For a diatomic molecule, such as \( O_2 \), treated as two point masses separated by a distance \( d \), the moment of inertia about the center of mass is \( I = \mu d^2 \). Here, \( \mu \) is the reduced mass. The reduced mass \( \mu = \frac{m}{2} \) where \( m = \frac{16.0 \ g/mol}{1000} = 0.016 \ kg/mol \) (mass of one \( O \) atom). So, \( \mu = \frac{0.016 \ kg/mol}{2} = 0.008 \ kg/mol \). Using \( d = 1.21 \times 10^{-10} \ m \), \( I = 0.008 \ kg/mol \times (1.21 \times 10^{-10} \ m)^2 = 1.17168 \times 10^{-30} \ kg \cdot m^2 \).

Step 3: Root Mean Square Angular Velocity

The root mean square (rms) angular velocity \( \omega_{rms} \) is given by \( \omega_{rms} = \sqrt{\frac{2 \times KE_{rot}}{N \times I}} \), where \( N = N_A \) (Avogadro's number, \( 6.022 \times 10^{23} \)). Thus, \( \omega_{rms} = \sqrt{\frac{2 \times 2494.2}{1.17168 \times 10^{-30} \times 6.022 \times 10^{23}}} = 5.256 \times 10^{11} \ rad/s \).

Step 4: Compare with Machinery Angular Velocity

Convert the machinery angular velocity: \( 10,000 \ rev/min \) ➔ \( \frac{10,000 \ times 2\pi}{60} \ rad/s = 1047.2 \ rad/s \). Thus, \( \omega_{rms} \) of an oxygen molecule is much higher than typical rotating machinery.

Key Concepts

Diatomic Gas

Diatomic gases, such as oxygen (\(O_2\)) and nitrogen (\(N_2\)), are molecules composed of two atoms bonded together. The rotational kinetic energy of diatomic gases is significant because it involves both rotational and translational degrees of freedom. Here, we focus on rotational kinetic energy, which arises from the molecule's rotation about its center of mass.

In the context of rotational kinetic energy, each molecule has certain degrees of freedom that contribute to its energy. Diatomic molecules, being quite simple, have two rotational degrees of freedom due to their linear shape. This means they can rotate about two axes perpendicular to the bond axis.

The general formula for calculating the rotational kinetic energy of a diatomic gas is:

• \( KE_{rot} = \frac{f}{2} nRT \)

Where:

• \( f = 2 \): degrees of freedom for rotation.
• \( n \): number of moles.
• \( R \): universal gas constant \( (8.314 \, \frac{J}{mol\cdot K}) \).
• \( T \): temperature in Kelvin.

Through this formula, we can calculate how much of the gas's energy is stored in its rotational motion, especially relevant at higher temperatures.

Moment of Inertia

The moment of inertia is a critical concept that measures how the mass of an object is distributed relative to its rotational axis. For a diatomic molecule like oxygen, this can be calculated by treating the two atoms as point masses a certain distance apart.

The formula to find the moment of inertia \( I \) for a diatomic molecule is:

• \( I = \mu d^2 \)

where \( \mu \) is the reduced mass given by \( \frac{m}{2} \), with \( m \) being the mass of a single atom and \( d \) the distance between the two atoms.

For oxygen, the reduced mass \( \mu \) is calculated using its molar mass, converted from grams to kilograms. With the separation distance between the two oxygen atoms being \( 1.21 \times 10^{-10} \, m \), the moment of inertia is derived, helping us understand the resistance of an \( O_2 \) molecule to changes in its rotational state.

Understanding moment of inertia helps in visualizing how different molecular structures respond to rotational forces, which ultimately influences their kinetic energy and behavior in various conditions.

Root Mean Square Angular Velocity

Root Mean Square (RMS) Angular Velocity \( \omega_{rms} \) provides a measure of the average angular velocity of molecular rotation. It's essential when comparing the molecular motion to more familiar macroscopic motions, such as rotating machinery.

To find \( \omega_{rms} \) of a diatomic gas, we use the formula:

• \( \omega_{rms} = \sqrt{\frac{2 \times KE_{rot}}{N \times I}} \)

where \( KE_{rot} \) is the rotational kinetic energy we computed earlier, \( N \) is Avogadro's number (\( 6.022 \times 10^{23} \)), and \( I \) is the moment of inertia.

This calculation gives us insight into how fast, on average, an oxygen molecule spins under given conditions. Compared to a high-speed machine operating at \( 10,000 \) revolutions per minute (converted to \( 1047.2 \, rad/s \)), the \( \omega_{rms} \) of an \( O_2 \) molecule at room temperature can be in the range of \( 5.256 \times 10^{11} \ rad/s \), showcasing the incredibly high-speed dynamics at the molecular level. Understanding \( \omega_{rms} \) helps appreciate molecular kinetics compared to everyday mechanical systems.