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) \(v_{\mathrm{ms}}\) 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) 517 m/s; (b) 298 m/s.

Step 1: Convert Celsius to Kelvin

First, convert the given temperature from Celsius to Kelvin. The conversion formula is: \[ T(K) = T(°C) + 273.15 \] For 27°C, it is:\[ T = 27 + 273.15 = 300.15 \] Thus, the temperature in Kelvin is 300.15 K.

Step 2: Calculate the molecular mass of nitrogen in kg

Convert the molar mass of nitrogen from grams per mole to kilograms per mole, which is needed for kinetic calculations. The conversion is:\[ M = 28.0 \text{ g/mol} = 28.0 \times 10^{-3} \text{ kg/mol} = 0.028 \text{ kg/mol} \] Thus, the molecular mass of nitrogen is 0.028 kg/mol.

Step 3: Use the formula for root mean square speed

The root mean square speed \(v_{rms}\) of a gas molecule is given by:\[ v_{rms} = \sqrt{\frac{3k_BT}{m}} \] Where \(k_B\) is the Boltzmann constant (\(1.38 \times 10^{-23} \text{ J/K}\)), \(T\) is the temperature in Kelvin, and \(m\) is the mass of the nitrogen molecule in kg.After calculating, \(m = \frac{M}{N_A} = \frac{0.028}{6.022 \times 10^{23}}\).

Step 4: Calculate the molecular mass per molecule

Determine the mass of a single nitrogen molecule using Avogadro's number:\[ m = \frac{0.028}{6.022 \times 10^{23}} \approx 4.65 \times 10^{-26} \text{ kg} \] This mass will be used to calculate the root mean square speed.

Step 5: Calculate RMS speed for nitrogen

Substitute the known values into the RMS speed formula:\[ v_{rms} = \sqrt{\frac{3 \times 1.38 \times 10^{-23} \times 300.15}{4.65 \times 10^{-26}}} \] Calculating this gives:\[ v_{rms} \approx 517 \text{ m/s} \] Thus, the root mean square speed of nitrogen molecules at 27°C is approximately 517 m/s.

Step 6: Calculate the RMS value of the x-component of velocity

For a completely isotropic gas, the RMS speed of any component (e.g., \(v_x\)) is given by dividing the total RMS speed by \(\sqrt{3}\):\[ v_{x_{rms}} = \frac{v_{rms}}{\sqrt{3}} \] Substitute\(v_{rms} = 517 \text{ m/s}\) into the formula:\[ v_{x_{rms}} = \frac{517}{\sqrt{3}} \approx 298 \text{ m/s} \] Thus, the RMS value of the x-component of the velocity is approximately 298 m/s.

Key Concepts

Root Mean Square Speed

The root mean square speed (RMS speed) is a useful concept in the kinetic theory of gases. It represents the average speed of particles in a gas. Mathematically, the RMS speed is given by \[ v_{rms} = \sqrt{\frac{3k_BT}{m}} \] where:

• \( v_{rms} \): Root mean square speed
• \( k_B \): Boltzmann constant (\(1.38 \times 10^{-23} \text{ J/K}\))
• \( T \): Absolute temperature in Kelvin
• \( m \): Mass of a single molecule

This formula shows how the speed of gas molecules increases with temperature and decreases with the mass of the molecules. At higher temperatures, particles have more kinetic energy, thus moving faster. In an example calculation for nitrogen at 27°C, we find \[ v_{rms} \approx 517 \text{ m/s} \].This means nitrogen molecules have an average speed that would allow them to cover 517 meters in one second if unobstructed.

Avogadro's Number

Avogadro's number is a basic tenet in chemistry and physics, reflecting the amount of particles found in one mole of a substance. Defined as \( 6.022 \times 10^{23} \) particles/mole, this number is crucial for conversions in solving chemistry problems. The use of Avogadro's number facilitates determining the mass of an individual molecule. Since most physical measurements occur on a macroscopic scale, using moles instead of individual molecules makes calculations manageable. In molecular simulation calculations, it is often necessary to convert macroscopic molar mass to the microscopic scale. For instance, knowing that the molar mass of nitrogen \(N_2\) is 28 g/mol, it can be converted into kilograms per molecule using Avogadro’s number \[ m = \frac{28 \times 10^{-3} \text{ kg/mol}}{6.022 \times 10^{23} \text{ molecules/mol}} \approx 4.65 \times 10^{-26} \text{ kg/molecule} \]. This small mass figure shows how much lower masses operate in quantum particle dimensions.

Boltzmann Constant

The Boltzmann constant (\( k_B \)) is a cornerstone in statistical mechanics and thermodynamics that relates the average kinetic energy of particles in a gas to the temperature of the gas. It is described by the equation:\[ k_B = 1.38 \times 10^{-23} \text{ J/K} \]This constant functions as a bridge between microscopic and macroscopic physical phenomena, enabling predictions about gas behaviors based on temperature. The Boltzmann constant comes into play in the root mean square speed formula as it allows for converting temperature (a macroscopic measure) into kinetic energy (a microscopic measure) for particles.In molecular physics problems, \( k_B \) helps determine how the energy scales of microscopic particles like gas molecules compare to human-scale observations. For example, in calculating \( v_{rms} \) of nitrogen molecules, \( k_B \) helps find how fast molecules are moving at a given temperature, translating thermal input into molecular motion.

Temperature Conversion

Temperature conversion is a fundamental step in thermodynamics calculations. Since many formulas, including those in gas dynamics, require temperatures in Kelvin, converting from Celsius (or any other temperature scale) is a standard practice. The conversion from Celsius to Kelvin is straightforward:\[ T(K) = T(°C) + 273.15 \]In any problem, ensuring temperatures are in Kelvin before performing calculations is crucial because Kelvin is the SI unit, directly relatable to physical laws without modifying the scale origin.Converting 27°C to Kelvin provides a temperature of 300.15 K. This step is essential as it establishes a baseline temperature compatible with constants like the Boltzmann constant (which operates naturally within Kelvin). Without this conversion, calculated properties such as the RMS speed of a gas would be inaccurate, leading to significant errors in scientific interpretations and applications.