Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A 10g sample of neon gas has 1700J of thermal energy. Estimate the average speed of a neon atom.

Short Answer

Expert verified

The average spped of a neon atom isvavg=580m/s

Step by step solution

01

Step :1 Introduction 

TThe gravitational potential between atoms is 0 in the micro system, and the total energy of a monatomic gas equals translational kinetic energy.. As a result, a monatomic gas's thermal energy is Natoms is given by equation (20.27in the form

Eth=Nϵavg

Where ϵavgis the average energy. The molecule with mass mand velocity vhas a translational kinetic energy on average The average translational kinetic energy of a molecule is affected by its temperature, hence it is related to the temperature per molecule in the form.

ϵavg=12mvavg2

Where KBis Boltzmann's constant and in SIunit its value is

kB=1.38×1023J/K

Knowing the mass Mand the molar mass m, we can get the number of molecules by

N=Mm

02

Step : 2 Explanation 

Use this expressions of Nand ϵavg into equation (1) to get an expression for vavg2

Eth=Nϵavg

=Mm12mvavg2

=12Mvavg2

03

Step :3 Substitution 

Solve equation

(3) for vavgand use the values forMand Eth

vavg=2EthM

=2(1700J)0.01kg

=580m/s

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.

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

See all solutions

Recommended explanations on Physics Textbooks

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