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
Question
Book Icon
Chapter 16, Problem 16A.2AE

(i)

Interpretation Introduction

Interpretation:

The diffusion constant of argon at 20 °C and 1.00 Pa is to be calculated.  The flow of gas due to diffusion at 1.00 Pa has to be calculated.

Concept introduction:

Diffusion of gas is a process in which the gas follows from higher pressure area to lower pressure area.  The process of diffusion requires a potential gradient.  The diffusion coefficient of a gas is given by the expression as shown below.

    D=13(kTσp)(8RTπM)1/2

(i)

Expert Solution
Check Mark

Answer to Problem 16A.2AE

The diffusion constant of argon at 20 °C and 1.00 Pa is 147.69 m2 s1_.  The flow of gas due to diffusion at 1.00 Pa is 6062.8 mol m2 s1_.

Explanation of Solution

The diffusion coefficient of a gas is given by the expression as shown below.

    D=13(kTσp)(8RTπM)1/2        (1)

Where,

  • R is the gas constant (8.314J K1 mol1).
  • T is the temperature.
  • k is the Boltzmann constant (1.381×1023J K1).
  • σ is the collision cross section area.
  • M is the molar mass.
  • p is the pressure.

The temperature of argon gas is 20 °C.

The conversion of temperature in Kelvin is shown below.

    T=(273+20.0 °C)K=293 K

The pressure of argon gas at 20 °C is 1.00 Pa.

The conversion of pressure in kg m1 s2 is shown below.

    p=(1.00 Pa)(1 kg m1 s21 Pa)=1.00 kg m1 s2

The value of σ for argon has is 0.36 nm2.

The conversion of σ in m2 is shown below.

    σ=(0.36 nm2)(1 m21020 nm)=0.36×1020 m2

The molar mass of argon is 39.948×103 kg mol1.

Substitute the value of R, T, k, σ, M, and p in the equation (1).

    D=(13((1.381×1023J K1)(293 K)(0.36×1020 m2)(1.00 kg m1 s2))(8(8.314J K1 mol1)(1kg m2 s21J)(293 K)(3.14)(39.948×103 kg mol1))1/2)=(0.3747)(155361.3328)1/2 m2 s1=(0.3747)(394.159) m2 s1=147.69 m2 s1_

Therefore, the diffusion constant of argon at 20 °C and 1.00 Pa is 147.69 m2 s1_

The pressure gradient of the pipe is 1.0 bar m1.

The flow of gas due to diffusion (JzNA) is given by the expression as shown below.

    JzNA=(DRT)dpdz

Where,

  • dpdz is the pressure gradient.

Substitute the values of dpdz, D, R, and T in the above equation.

    JzNA=(147.69m2 s1(8.314J K1 mol1)(1kg m2 s21J)(293 K))(1.0 bar m1)=(0.0606 mol s kg1)(1.0 bar m1)(105 kg m1 s21 bar)=6062.8 mol m2 s1_

Therefore, the flow of gas due to diffusion at 1.00 Pa is 6062.8 mol m2 s1_.

(b)

Interpretation Introduction

Interpretation:

The diffusion constant of argon at 20 °C and 100 kPa is to be calculated.  The flow of gas due to diffusion at 100 kPa has to be calculated.

Concept introduction:

As mentioned in the concept introduction in part (a).

(b)

Expert Solution
Check Mark

Answer to Problem 16A.2AE

The diffusion constant of argon at 20 °C and 100 kPa is 1.4769×103 m2 s1_.  The flow of gas due to diffusion at 100 kPa is 6.06289×10-7 mol m2 s1_.

Explanation of Solution

The diffusion coefficient of a gas is given by the expression as shown below.

    D=13(kTσp)(8RTπM)1/2        (1)

Where,

  • R is the gas constant (8.314J K1 mol1).
  • T is the temperature.
  • k is the Boltzmann constant (1.381×1023J K1).
  • σ is the collision cross section area.
  • M is the molar mass.
  • p is the pressure.

The temperature of argon gas is 20 °C.

The conversion of temperature in Kelvin is shown below.

    T=(273+20.0 °C)K=293 K

The pressure of argon gas at 20 °C is 100 kPa.

The conversion of pressure in kg m1 s2 is shown below.

    p=(100 kPa)(103 kg m1 s21 kPa)=1.00×105 kg m1 s2

The value of σ for argon has is 0.36 nm2.

The conversion of σ in m2 is shown below.

    σ=(0.36 nm2)(1 m21020 nm)=0.36×1020 m2

The molar mass of argon is 39.948×103 kg mol1.

Substitute the value of R, T, k, σ, M, and p in the equation (1).

    D=(13((1.381×1023J K1)(293 K)(0.36×1020 m2)(1.00×105 kg m1 s2))(8(8.314J K1 mol1)(1kg m2 s21J)(293 K)(3.14)(39.948×103 kg mol1))1/2)=(3.747×106)(155361.3328)1/2 m2 s1=(3.747×106)(394.159) m2 s1=1.4769×103 m2 s1_

Therefore, the diffusion constant of argon at 20 °C and 100 kPa is 1.4769×103 m2 s1_

The pressure gradient of the pipe is 1.0 bar m1.

The flow of gas due to diffusion (JzNA) is given by the expression as shown below.

    JzNA=(DRT)dpdz

Where,

  • dpdz is the pressure gradient.

Substitute the values of dpdz, D, R, and T in the above equation.

    JzNA=(1.4769×103m2 s1(8.314J K1 mol1)(1kg m2 s21J)(293 K))(1.0 bar m1)=(6.062 mol s kg1)(1.0 bar m1)(105 kg m1 s21 bar)=6.06289×10-7 mol m2 s1_

Therefore, the flow of gas due to diffusion at 100 kPa is 6.06289×10-7 mol m2 s1_.

(c)

Interpretation Introduction

Interpretation:

The diffusion constant of argon at 20 °C and 10.0 MPa is to be calculated.  The flow of gas due to diffusion at 10.0 MPa has to be calculated.

Concept introduction:

As mentioned in the concept introduction in part (a).

(c)

Expert Solution
Check Mark

Answer to Problem 16A.2AE

The diffusion constant of argon at 20 °C and 10.0 MPa is 1.4769×105 m2 s1_.  The flow of gas due to diffusion at 10.0 MPa is 6.0628×109 mol m2 s1_.

Explanation of Solution

The diffusion coefficient of a gas is given by the expression as shown below.

    D=13(kTσp)(8RTπM)1/2        (1)

Where,

  • R is the gas constant (8.314J K1 mol1).
  • T is the temperature.
  • k is the Boltzmann constant (1.381×1023J K1).
  • σ is the collision cross section area.
  • M is the molar mass.
  • p is the pressure.

The temperature of argon gas is 20 °C.

The conversion of temperature in Kelvin is shown below.

    T=(273+20.0 °C)K=293 K

The pressure of argon gas at 20 °C is 10.0 MPa.

The conversion of pressure in kg m1 s2 is shown below.

    p=(10.0 MPa)(106 kg m1 s21 MPa)=1.00×107 kg m1 s2

The value of σ for argon has is 0.36 nm2.

The conversion of σ in m2 is shown below.

    σ=(0.36 nm2)(1 m21020 nm)=0.36×1020 m2

The molar mass of argon is 39.948×103 kg mol1.

Substitute the value of R, T, k, σ, M, and p in the equation (1).

    D=(13((1.381×1023J K1)(293 K)(0.36×1020 m2)(1.00×107 kg m1 s2))(8(8.314J K1 mol1)(1kg m2 s21J)(293 K)(3.14)(39.948×103 kg mol1))1/2)=(3.747×108)(155361.3328)1/2 m2 s1=(3.747×108)(394.159) m2 s1=1.4769×105 m2 s1_

Therefore, the diffusion constant of argon at 20 °C and 10.0 MPa is 1.4769×105 m2 s1_.

The pressure gradient of the pipe is 1.0 bar m1.

The flow of gas due to diffusion (JzNA) is given by the expression as shown below.

    JzNA=(DRT)dpdz

Where,

  • dpdz is the pressure gradient.

Substitute the values of dpdz, D, R, and T in the above equation.

    JzNA=(1.4769×105m2 s1(8.314J K1 mol1)(1kg m2 s21J)(293 K))(1.0 bar m1)=(0.0606 mol s kg1)(1.0 bar m1)(105 kg m1 s21 bar)=6.0628×109 mol m2 s1_

Therefore, the flow of gas due to diffusion at 10.0 MPa is 6.0628×109 mol m2 s1_.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Chapter 16 Solutions

Atkins' Physical Chemistry

Knowledge Booster
Background pattern image
Recommended textbooks for you
Text book image
Chemistry
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Text book image
Chemistry
Chemistry
ISBN:9781259911156
Author:Raymond Chang Dr., Jason Overby Professor
Publisher:McGraw-Hill Education
Text book image
Principles of Instrumental Analysis
Chemistry
ISBN:9781305577213
Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:Cengage Learning
Text book image
Organic Chemistry
Chemistry
ISBN:9780078021558
Author:Janice Gorzynski Smith Dr.
Publisher:McGraw-Hill Education
Text book image
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Text book image
Elementary Principles of Chemical Processes, Bind...
Chemistry
ISBN:9781118431221
Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:WILEY