/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 2 The speed of sound. The speed of... [FREE SOLUTION] | 91影视

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Chapter 10: Problem 2

The speed of sound. The speed of sound in air is approximately the average velocity \(\left\langle v_{x}^{2}\right\rangle^{1 / 2}\) of the gas molecules. Compute this speed for \(T=0^{\circ} \mathrm{C}\), assuming that air is mostly nitrogen gas.

Short Answer

Expert verified
The speed of sound in air at 0掳C is approximately 493 m/s.

Step by step solution

01

- Find the necessary formula

The speed of sound in a gas can be approximated using the root-mean-square speed of gas molecules, which is given by the formula: \[ v_{rms} = \sqrt{\frac{3k_BT}{m}} \] where \(k_B\) is the Boltzmann constant, \(T\) is the temperature in Kelvin, and \(m\) is the mass of a gas molecule.
02

- Convert temperature to Kelvin

Convert the temperature from Celsius to Kelvin. Use the formula: \[ T(K) = T(\degree C) + 273.15 \] So for \( T = 0\degree C\): \[ T(K) = 0 + 273.15 = 273.15K \]
03

- Identify the constants

Identify the constants needed for the formula: \( k_B = 1.38 \times 10^{-23} \; J/K \) (Boltzmann constant)The mass of a nitrogen molecule, \( m \), is given by: \[ m = \frac{M}{N_A} \] where \( M \) is the molar mass of nitrogen (approximately \( 28 \, g/mol \) or \( 0.028 \, kg/mol \)) and \( N_A \) (Avogadro's number) is \( 6.022 \times 10^{23} \, molecules/mol \). Thus: \[ m = \frac{0.028}{6.022 \times 10^{23}} \approx 4.65 \times 10^{-26} \; kg \]
04

- Calculate the root-mean-square speed

Substitute the values into the root-mean-square speed formula: \[ v_{rms} = \sqrt{\frac{3 \times 1.38 \times 10^{-23} \times 273.15}{4.65 \times 10^{-26}}} \]
05

- Simplify and solve the equation

Perform the calculations step-by-step: Multiply the constants and the temperature in the numerator: \[ 3 \times 1.38 \times 10^{-23} \times 273.15 \approx 1.13 \times 10^{-20} \] Divide by the mass of a nitrogen molecule: \[ \frac{1.13 \times 10^{-20}}{4.65 \times 10^{-26}} \approx 2.43 \times 10^{5} \] Take the square root of the result: \[ v_{rms} \approx \sqrt{2.43 \times 10^{5}} \approx 493 \, m/s \]

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Key Concepts

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

root-mean-square speed
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Most popular questions from this chapter

The Boltzmann distribution of uniformly spaced energy levels. A system has energy levels uniformly spaced at \(3.2 \times 10^{-20} \mathrm{~J}\) apart. The populations of the energy levels are given by the Boltzmann distribution. What fraction of particles is in the ground state at \(T=300 \mathrm{~K}\) ?

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