Problem 75 - University Physics with Modern Physics

Question

The speed of propagation of a sound wave in air at \(27^{\circ} \mathrm{C}\) is about 350 \(\mathrm{m} / \mathrm{s} .\) Calculate, for comparison, (a) \(v_{\mathrm{rms}}\) for nitrogen molecules and (b) the rms value of \(v_{x}\) at this temperature. The molar mass of nitrogen \(\left(\mathrm{N}_{2}\right)\) is 28.0 \(\mathrm{g} / \mathrm{mol} .\)

Step-by-Step Solution

Answer

(a) \( v_{\text{rms}} \approx 515 \text{ m/s} \), (b) \( v_{x,\text{rms}} \approx 297 \text{ m/s} \).

Step 1: Convert temperature to Kelvin

First, we need to convert the given temperature from degrees Celsius to Kelvin. The formula to convert Celsius to Kelvin is:\[ T_K = T_C + 273.15 \]Where \( T_C = 27^{\circ}C \), so:\[ T_K = 27 + 273.15 = 300.15 \ K \]

Step 2: Calculate the RMS speed of nitrogen molecules

The root mean square (RMS) speed \( v_{\text{rms}} \) for a gas is given by the formula:\[ v_{\text{rms}} = \sqrt{\frac{3kT}{m}} \]Where:- \( k = 1.38 \times 10^{-23} \text{ J/K} \) is the Boltzmann constant,- \( T = 300.15 \text{ K} \) is the temperature,- \( m \) is the mass of a nitrogen molecule.The molar mass of nitrogen \( N_2 \) is 28.0 g/mol, or \( 0.028 \text{ kg/mol} \). Since there are \( 6.022 \times 10^{23} \) molecules in a mole (Avogadro's number),\[ m = \frac{0.028}{6.022 \times 10^{23}} \text{ kg} \]Calculate \( v_{\text{rms}} \):\[ v_{\text{rms}} = \sqrt{\frac{3 \times 1.38 \times 10^{-23} \times 300.15}{4.65 \times 10^{-26}}} = \sqrt{\frac{1.2425 \times 10^{-20}}{4.65 \times 10^{-26}}} \text{ m/s} \]\[ v_{\text{rms}} \approx 515 \text{ m/s} \]

Step 3: Calculate the rms value of the x-component of velocity

The root mean square value of the x-component of the velocity, \( v_{x,\text{rms}} \), is related to the total RMS speed by:\[ v_{x,\text{rms}} = \frac{v_{\text{rms}}}{\sqrt{3}} \]Given \( v_{\text{rms}} \approx 515 \text{ m/s} \):\[ v_{x,\text{rms}} = \frac{515}{\sqrt{3}} \]\[ v_{x,\text{rms}} \approx 297 \text{ m/s} \]

Key Concepts

Sound Wave Propagation

When we talk about sound wave propagation, we refer to how sound travels through a medium, usually air. Sound is a mechanical wave, which means it requires a medium to travel. The speed of sound depends on factors such as the temperature and density of the medium.
At a standard room temperature of around 20°C, sound travels at approximately 343 meters per second in air. However, this speed increases with temperature because the molecules move faster, supporting quicker wave propagation.
For instance, at the given problem's temperature of 27°C, sound moves slightly faster, measured to be about 350 meters per second. This is essentially because warmer air leads to more energetic collisions among molecules, thus aiding in the transmission of sound energy across them.

Boltzmann Constant

The Boltzmann constant is a fundamental physical constant denoted by the symbol k. It bridges the macroscopic and microscopic worlds by linking the average kinetic energy of particles in a gas with the temperature of the gas.
Mathematically, it is used in equations like the formula for the root mean square (RMS) speed of gas molecules: \[ v_{\text{rms}} = \sqrt{\frac{3kT}{m}} \]where:

• \( k = 1.38 \times 10^{-23} \text{ J/K} \)
• \( T \) is the absolute temperature in Kelvin
• \( m \) is the mass of a gas molecule

This constant is crucial in statistical mechanics, helping us understand particle behavior in gases. It gives us a quantitative measure of the thermal agitation of particles.

Molar Mass

Molar mass is an essential concept in chemistry and physics, representing the mass of one mole of a substance in grams per mole (g/mol). It allows us to convert between the mass of a substance and the number of moles.
For molecules, the molar mass is the sum of the atomic masses of its constituent atoms. For nitrogen gas \((N_2)\), the molecule consists of 2 nitrogen atoms. If each nitrogen atom has an atomic mass of approximately 14 g/mol, then nitrogen gas has a molar mass of 28 g/mol.
In the context of gas motion, we use molar mass to find the mass of a single molecule, which is needed for calculating speeds like the root mean square speed. This is done by dividing the molar mass by Avogadro's number, giving us the mass of one molecule.

Avogadro's Number

Avogadro's number is a key constant in chemistry necessary for understanding the mole concept. It is the number of constituent particles (usually atoms or molecules) in one mole of a substance.
Its value is \( 6.022 \times 10^{23} \) particles per mole. This means if you have one mole of a substance, you have exactly \( 6.022 \times 10^{23} \) particles of that substance.
In calculations, Avogadro's number helps us transition between the macroscopic scale we observe and the microscopic world of atoms and molecules. For instance, in the problem, it aids in deriving the mass of a single nitrogen molecule by dividing the molar mass by this number, crucial for further calculations involving molecular speeds.