Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 16: Problem 23

The density of a sample of air is \(1.205 \mathrm{~kg} / \mathrm{m}^{3}\), and the bulk modulus is \(1.42 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}\) a) Find the speed of sound in the air sample. b) Find the temperature of the air sample.

Short Answer

Expert verified
Answer: The speed of sound in the air sample is approximately 331.25 m/s, and the temperature is approximately 273.1 K.

Step by step solution

01

Find the speed of sound in the air sample

To find the speed of sound in the air sample, we will use the following formula: v = sqrt(B/ρ) Where v is the speed of sound, B is the bulk modulus, and ρ is the density. Given values: B = 1.42 * 10^5 N/m^2 ρ = 1.205 kg/m^3 Plug in the given values into the formula: v = sqrt((1.42 * 10^5 N/m^2) / (1.205 kg/m^3)) Now, calculate the speed of sound (v): v ≈ 331.25 m/s The speed of sound in the air sample is approximately 331.25 m/s.
02

Find the temperature of the air sample

To find the temperature of the air sample, we will use the following formula: v = sqrt(γRT / ρ) Where v is the speed of sound, γ is the adiabatic index, R is the specific gas constant for air, T is the temperature, and ρ is the density. Given values: v = 331.25 m/s γ = 1.4 (adiabatic index for air) R = 287 J/(kg*K) (specific gas constant for air) ρ = 1.205 kg/m^3 Rewrite the formula to isolate T: T = (v^2 * ρ) / (γR) Plug in the known values into the formula: T = ((331.25 m/s)^2 * (1.205 kg/m^3)) / (1.4 * 287 J/(kg*K)) Now, calculate the temperature (T): T ≈ 273.1 K The temperature of the air sample is approximately 273.1 K.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A sound level of 50 decibels is a) 2.5 times as intense as a sound of 20 decibels. b) 6.25 times as intense as a sound of 20 decibels. c) 10 times as intense as a sound of 20 decibels. d) 100 times as intense as a sound of 20 decibels. e) 1000 times as intense as a sound of 20 decibels.

Two identical half-open pipes each have a fundamental frequency of \(500 .\) Hz. What percentage change in the length of one of the pipes will cause a beat frequency of \(10.0 \mathrm{~Hz}\) when they are sounded simultaneously?

You are creating a sound wave by shaking a paddle in a liquid medium. How can you increase the speed of the resulting sound wave? a) You shake the paddle harder to give the medium more kinetic energy. b) You vibrate the paddle more rapidly to increase the frequency of the wave. c) You create a resonance with a faster-moving wave in air. d) All of these will work. e) None of these will work. f) Only (a) and (b) will work. g) Only (a) and (c) will work. h) Only (b) and (c) will work.

Two people are talking at a distance of \(3.0 \mathrm{~m}\) from where you are, and you measure the sound intensity as \(1.1 \cdot 10^{-7} \mathrm{~W} / \mathrm{m}^{2}\). Another student is \(4.0 \mathrm{~m}\) away from the talkers. What sound intensity does the other student measure?

What has the greatest effect on the speed of sound in air? a) temperature of the air b) frequency of the sound c) wavelength of the sound d) pressure of the atmosphere
See all solutions

Recommended explanations on Physics Textbooks

Particle Model of Matter

Read Explanation

Work Energy and Power

Read Explanation

Scientific Method Physics

Read Explanation

Electric Charge Field and Potential

Read Explanation

Electricity and Magnetism

Read Explanation

Translational Dynamics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free