# It took 4.5 minutes for 1.0 L helium to effuse through a porous barrier. How long will it take for 1.0 L Cl 2 gas to effuse under identical conditions?

### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

Chapter
Section

### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 8, Problem 114E
Textbook Problem
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## It took 4.5 minutes for 1.0 L helium to effuse through a porous barrier. How long will it take for 1.0 L Cl2 gas to effuse under identical conditions?

Interpretation Introduction

Interpretation: From the relative rates of effusion of two gases, the effusion time for a particular gas should be determined

Concept introduction:

Effusion is used to describe the passage of a gas through a tiny particle into an evacuated chamber.

The rate of effusion is the measure of speed at which the gas is transferred to the chamber.

According to Thomas Graham the rate of effusion of a gas is inversely proportional to the square root of the mass of its particles.

The relative rate of effusion of two gases at the same temperature and pressure are the inverse ratio of the square root of the masses of the gases particles. That is,

Rateofeffusionforgas1Rateofeffusionforgas2=M2M1orRate1Rate2=(M2M1)1/2

M1andM2arethemolarmassesofgas1andgas2

This equation is known as Graham’s law of effusion.

Effusion rate in some cases, equal to the volume of gas that effuses per unit volume.

### Explanation of Solution

Explanation

To determine: The effusion time for chlorine gas

The effusion time for chlorine gas, t =19min

According to Graham’s law of effusion,

Rate1Rate2=(M2M1)1/2 (1)

Here,Gas1=Heliumgas(He)Gas2=Chlorinegas(Cl2)M1=molarmassof He=4.003M2=molarmassofCl2=70.90Effusionrateinthisproblemareequaltothevolumeofgasperunittime.Thatis,inthecaseofhelium,Rarte1=Volumeofgas1(V1)Time(T1)Inthecasechlorine,Rate2=Volumeofgas2(V2)Time(T2)So,inthecaseofheliumgas,ittook4

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