Atkins' Physical Chemistry
Atkins' Physical Chemistry
11th Edition
ISBN: 9780198769866
Author: ATKINS, P. W. (peter William), De Paula, Julio, Keeler, JAMES
Publisher: Oxford University Press
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Chapter 16, Problem 16A.1ST
Interpretation Introduction

Interpretation:

The time taken by the 200mg of Cs atoms to effuse out of the oven is to be calculated.

Concept introduction:

Effusion is the motion of the gas particles inside a chamber through a hole. The rate of effusion according to Graham’s Law is inversely proportional to the mass of the gas particles raised to the power half or square root of the mass.

Expert Solution & Answer
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Answer to Problem 16A.1ST

The time taken by the 200mg of Cs atoms to effuse out of the oven is 8.77×103s.

Explanation of Solution

The formula to calculate the rate of effusion is shown below.

    Rateofeffusion=pA°NA2πMRT        (1)

Where,

  • p is the vapor pressure.
  • A° is the area of the hole.
  • M is the molar mass of gas in kilograms.
  • NA is the Avogadro’s number.
  • R is the gas constant.
  • T is the temperature.

The change in mass of atoms effuse out is given by the formula shown below.

    Δm=pA°M1/22πRT×Δt        (2)

Where,

  • Δt is the time taken by the Δm mass of atoms to effuse out.

The given data and some known values are stated below.

The value of the diameter of the hole is 1.0mm.

The value of T is given 500°C.

The value of R is 8.314JK 1mol1.

The molar mass (M) of cesium in kilograms is 132.91×103kgmol1.

The value of NA is 6.022×1023mol1.

The value of p is given 16Pa.

The value of Δm is given 200mg.

The area of the hole is given by the formula shown below.

    A°=πd24        (3)

Where,

  • d is the diameter of the hole.

Substitute the value of diameter in the equation (3) to find out area as shown below.

    A°=πd24=3.14×(1.0mm)24=3.144mm2=0.785×106m2(1mm=103m)

Substitute the value of Δm,M,T,p,R, and A° in equation (2) as shown below.

    200mg=16Pa(0.785×106m2)(132.91×103kgmol1)1/22π(8.314JK 1mol1)((500+273.15)K)×Δt200mg=16Pa(0.785×106m2)(3.65×101kg1/2mol1/2)2π(8.314JK 1mol1)(773.15K)×Δt

Rearrange above equation to calculate Δt as shown below.

    Δt=200mg2π(8.314JK 1mol1)(773.15K)16Pa(0.785×106m2)(3.65×101kg1/2mol1/2)=200mg40367.65Jmol145.84×107Pam2kg1/2mol1/2=200mg(200.92J1/2mol1/2)45.84×107Pam2kg1/2mol1/2=200×106kg(200.92J1/2mol1/2)45.84×107Pam2kg1/2mol1/2(1mg=106kg)

Simplify the above equation.

    Δt=200×106kg(200.92kg1/2ms1mol1/2)45.84×107kgm1s2m2kg1/2mol1/2(J=kgm2s2,Pa=kgm1s2)=4.02×10245.84×107s=0.0877×105s=8.77×103s

Therefore, the time taken by the 200mg of Cs atoms to effuse out of the oven is 8.77×103s.

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Chapter 16 Solutions

Atkins' Physical Chemistry

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