   Chapter 10, Problem 43PS

Chapter
Section
Textbook Problem

If the rms speed of an oxygen molecule is 4.28 × 104 cm/s at a given temperature, what is the rms speed of a CO2 molecule at the same temperature?

Interpretation Introduction

Interpretation: For the given rms speed value of oxygen molecule at particular temperature the rms speed for the CO2 molecule at same temperature should be determined.

Concept introduction:

The root mean square velocity μ is defined as the measure of velocity of particle in gas. It is the method to determine the single velocity value for particles.

Root mean square velocity can be determined,

μrms=(3RTM)1/2 (1)

(gas constant)R=8.314JKmolM=Molarmass

Molar mass: The molar mass of a substance is determined by dividing the given mass of substance by the amount of the substance.

Explanation

The expression used to calculate the rms speed is as follows,

μrms=(3RTM)1/2

Both the given gases differ only in the molar mass values since the temperature are same.

Now, equating the rms expression for both the given gases

μrmsof O=(3RTMO)1/2μrmsof CO2=(3RTMCO2)1/2whereMO= molarmass of oxygenMCO2=molarmass ofcarbondioxide

While equating both the expressions the term 3RT gets cancelled since it is same for both the molecules whereas the rms speed value and the molecular value remain since they are different for both the given gases

Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started 