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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 \(\mathrm{g} / \mathrm{mol} .\)

Short Answer

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

Step by step solution

01

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:\[ T_K = 27 + 273.15 = 300.15 \ K \]
02

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 \( v_{\text{rms}} \):\[ v_{\text{rms}} = \sqrt{\frac{3 \times 1.38 \times 10^{-23} \times 300.15}{4.65 \times 10^{-26}}} = \sqrt{\frac{1.2425 \times 10^{-20}}{4.65 \times 10^{-26}}} \text{ m/s} \]\[ v_{\text{rms}} \approx 515 \text{ m/s} \]
03

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 \( v_{\text{rms}} \approx 515 \text{ m/s} \):\[ v_{x,\text{rms}} = \frac{515}{\sqrt{3}} \]\[ v_{x,\text{rms}} \approx 297 \text{ m/s} \]

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Key Concepts

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

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.

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Most popular questions from this chapter

Helium gas with a volume of \(2.60 \mathrm{L},\) under a pressure of 0.180 atm and at a temperature of \(41.0^{\circ} \mathrm{C},\) is warmed until both pressure and volume are doubled. (a) What is the final temperature? (b) How many grams of helium are there? The molar mass of helium is 4.00 \(\mathrm{g} / \mathrm{mol} .\)

The size of an oxygen molecule is about 2.0 \(\times 10^{-10} \mathrm{m}\) Make a rough estimate of the pressure at which the finite volume of the molecules should cause noticeable deviations from ideal-gas behavior at ordinary temperatures \((T=300 \mathrm{K})\) .

A cylindrical tank has a tight-fitting piston that allows the volume of the tank to be changed. The tank originally contains 0.110 \(\mathrm{m}^{3}\) of air at a pressure of 0.355 atm. The piston is slowly pulled out until the volume of the gas is increased to 0.390 \(\mathrm{m}^{3} .\) If the temperature remains constant, what is the final value of the pressure?

(a) Oxygen (O) has a molar mass of 32.0 \(\mathrm{g} / \mathrm{mol} .\) What is the average translational kinetic energy of an oxygen molecule at a temperature of 300 \(\mathrm{K}\) ? (b) What is the average value of the square of its speed? (c) What is the root-mean-square speed? (d) What is the momentum of an oxygen molecule traveling at this speed? (e) Suppose an oxygen molecule traveling at this speed bounces back and forth between opposite sides of a cubical vessel 0.10 \(\mathrm{m}\) on a side. What is the average force the molecule exerts on one of the walls of the container? (Assume that the molecule's velocity is perpendicular to the two sides that it strikes.) (f) What is the average force per unit area? (g) How many oxygen molecules traveling at this speed are necessary to produce an average pressure of 1 atm? (h) Compute the number of oxygen molecules that are actually contained in a vessel of this size at 300 \(\mathrm{K}\) and atmospheric pressure. (i) Your answer for part (h) should be three times as large as the answer for part (g). Where does this discrepancy arise?

Modern vacuum pumps make it easy to attain pressures of the order of \(10^{-13}\) atm in the laboratory. Consider a volume of air and treat the air as an ideal gas. (a) At a pressure of \(9.00 \times 10^{-14}\) atm and an ordinary temperature of \(300.0 \mathrm{K},\) how many molecules are present in a volume of 1.00 \(\mathrm{cm}^{3} ?\) (b) How many molecules would be present at the same temperature but at 1.00 atm instead?

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