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
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Chapter 14, Problem 14.8IA

(a)

Interpretation Introduction

Interpretation:

An expression for the osmotic virial coefficient, B, has to be deduced for a freely jointed chain.  The value of B has to be evaluated.

Concept introduction:

The osmotic pressure is generated in the case when there is osmosis of solvent molecules through a semi-permeable membrane to a high concentration of solute molecules.

It is given by the formula shown below.

  Π=cRT

In the above expression, Π is the osmotic pressure, c is the concentration, T is temperature and R is the gas constant.

(a)

Expert Solution
Check Mark

Answer to Problem 14.8IA

The expression for the osmotic virial coefficient, B, for a freely jointed chain is shown below.

    B=163NAπγ3l3(N6)3/2

The value of B is 0.39m3mol1_.

Explanation of Solution

The expression for the osmotic pressure (Π) in terms of osmotic virial coefficient B is given below.

    Π=RT{cM+B×(cM)2+......}        (1)

Where,

  • R is the gas constant.
  • T is the temperature.
  • c is the concentration.
  • M is the molar mass.

The osmotic virial coefficient B, arises from the effect of excluded volume.  B is the excluded volume per mole of the molecules.

The volume of a molecule (V) is given by the expression below.

    V=43πa3

Where,

  • a is the effective radius of a molecule.

The formula to determine B is shown below.

    B=4NAV        (2)

Where,

  • NA is Avogadro’s number. (6.022×1023mol1)

Substitute the value of V in equation (2).

    B=4NA(43πa3)=163NAπa3        (3)

The formula to determine a is shown below.

    a=γRg

Where,

  • Rg is the radius of gyration.
  • γ is a constant.

Substitute the value of a in equation (3).

    B=163NAπ(γRg)3=163NAπγ3Rg3        (4)

For a freely jointed chain, the formula for radius of gyration (Rg) is shown below.

    Rg=(N6)1/2l

Where,

  • N is the number of units.
  • l is the length of the coil.

Substitute the value of Rg in equation (4).

    B=163NAπγ3((N6)1/2l)3=163NAπγ3l3(N6)3/2        (5)

Therefore, the expression for osmotic virial coefficient, B, for a freely jointed chain is shown below.

    B=163NAπγ3l3(N6)3/2

The value of l is 154pm.

The conversion of pm to m is shown below.

    1pm=1012m

Therefore, the conversion of 154pm to m is shown below.

    154pm=154×1012m

The value of N is 4000.

The value of γ is 0.85.

Substitute the value of N, γ, l and NA in equation (5).

    B=163×6.022×1023mol1×3.14×(0.85)3×(154×1012m)3×(40006)3/2=(163×6.022×1023mol1×3.14×0.614125×3.652264×1030m3×17213.25932)=0.39m3mol1_

Therefore, the value of B is 0.39m3mol1_.

(b)

Interpretation Introduction

Interpretation:

An expression for the osmotic virial coefficient, B, has to be deduced for a chain with tetrahedral bond angles.  The value of B has to be evaluated.

Concept introduction:

The osmotic pressure is generated in the case when there is osmosis of solvent molecules through a semi-permeable membrane to a high concentration of solute molecules.

It is given by the formula shown below.

  Π=cRT

In the above expression, Π is the osmotic pressure, c is the concentration, T is temperature and R is the gas constant.

(b)

Expert Solution
Check Mark

Answer to Problem 14.8IA

The expression for the osmotic virial coefficient, B, for a chain with tetrahedral bond angles is shown below.

    B=163NAπγ3l3(N3)3/2

The value of B is 1.1m3mol1_.

Explanation of Solution

The expression for the osmotic pressure (Π) in terms of osmotic virial coefficient B is given below.

    Π=RT{cM+B×(cM)2+......}        (1)

Where,

  • R is the gas constant.
  • T is the temperature.
  • c is the concentration.
  • M is the molar mass.

The osmotic virial coefficient B, arises from the effect of excluded volume.  B is the excluded volume per mole of the molecules.

The volume of a molecule (V) is given by the expression below.

    V=43πa3

Where,

  • a is the effective radius of a molecule.

The formula to determine B is shown below.

    B=4NAV        (2)

Where,

  • NA is Avogadro’s number. (6.022×1023mol1)

Substitute the value of V in equation (2).

    B=4NA(43πa3)=163NAπa3        (3)

The formula to determine a is shown below.

    a=γRg

Where,

  • Rg is the radius of gyration.
  • γ is a constant.

Substitute the value of a in equation (3).

    B=163NAπ(γRg)3=163NAπγ3Rg3        (4)

For a chain tetrahedral bond angles, the formula for radius of gyration (Rg) is shown below.

    Rg=(N3)1/2l

Where,

  • N is the number of units.
  • l is the length of the coil.

Substitute the value of Rg in equation (4).

    B=163NAπγ3((N3)1/2l)3=163NAπγ3l3(N3)3/2        (5)

Therefore, the expression for osmotic virial coefficient, B, for a freely jointed chain is shown below.

    B=163NAπγ3l3(N3)3/2

The value of l is 154pm.

The conversion of pm to m is shown below.

    1pm=1012m

Therefore, the conversion of 154pm to m is shown below.

    154pm=154×1012m

The value of N is 4000.

The value of γ is 0.85.

Substitute the value of N, γ, l and NA in equation (5).

    B=163×6.022×1023mol1×3.14×(0.85)3×(154×1012m)3×(40003)3/2=(163×6.022×1023mol1×3.14×0.614125×3.652264×1030m3×48686.45)=1.1m3mol1_

Therefore, the value of B is 1.1m3mol1_.

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Chapter 14 Solutions

Atkins' Physical Chemistry

Ch. 14 - Prob. 14A.3DQCh. 14 - Prob. 14A.1AECh. 14 - Prob. 14A.1BECh. 14 - Prob. 14A.2AECh. 14 - Prob. 14A.2BECh. 14 - Prob. 14A.3AECh. 14 - Prob. 14A.3BECh. 14 - Prob. 14A.4AECh. 14 - Prob. 14A.4BECh. 14 - Prob. 14A.5AECh. 14 - Prob. 14A.5BECh. 14 - Prob. 14A.6AECh. 14 - Prob. 14A.6BECh. 14 - Prob. 14A.7AECh. 14 - Prob. 14A.7BECh. 14 - Prob. 14A.8AECh. 14 - Prob. 14A.8BECh. 14 - Prob. 14A.9AECh. 14 - Prob. 14A.9BECh. 14 - Prob. 14A.1PCh. 14 - Prob. 14A.2PCh. 14 - Prob. 14A.3PCh. 14 - Prob. 14A.4PCh. 14 - Prob. 14A.5PCh. 14 - Prob. 14A.6PCh. 14 - Prob. 14A.7PCh. 14 - Prob. 14A.8PCh. 14 - Prob. 14A.10PCh. 14 - Prob. 14A.12PCh. 14 - Prob. 14A.13PCh. 14 - Prob. 14B.1DQCh. 14 - Prob. 14B.2DQCh. 14 - Prob. 14B.3DQCh. 14 - Prob. 14B.4DQCh. 14 - Prob. 14B.5DQCh. 14 - Prob. 14B.1AECh. 14 - Prob. 14B.1BECh. 14 - Prob. 14B.2AECh. 14 - Prob. 14B.2BECh. 14 - Prob. 14B.3AECh. 14 - Prob. 14B.3BECh. 14 - Prob. 14B.4AECh. 14 - Prob. 14B.4BECh. 14 - Prob. 14B.5AECh. 14 - Prob. 14B.5BECh. 14 - Prob. 14B.6AECh. 14 - Prob. 14B.6BECh. 14 - Prob. 14B.1PCh. 14 - Prob. 14B.2PCh. 14 - Prob. 14B.3PCh. 14 - Prob. 14B.4PCh. 14 - Prob. 14B.5PCh. 14 - Prob. 14B.6PCh. 14 - Prob. 14B.7PCh. 14 - Prob. 14B.8PCh. 14 - Prob. 14B.10PCh. 14 - Prob. 14C.1DQCh. 14 - Prob. 14C.2DQCh. 14 - Prob. 14C.1AECh. 14 - Prob. 14C.1BECh. 14 - Prob. 14C.2AECh. 14 - Prob. 14C.2BECh. 14 - Prob. 14C.3AECh. 14 - Prob. 14C.3BECh. 14 - Prob. 14C.4AECh. 14 - Prob. 14C.4BECh. 14 - Prob. 14C.1PCh. 14 - Prob. 14C.2PCh. 14 - Prob. 14D.1DQCh. 14 - Prob. 14D.2DQCh. 14 - Prob. 14D.3DQCh. 14 - Prob. 14D.4DQCh. 14 - Prob. 14D.5DQCh. 14 - Prob. 14D.1AECh. 14 - Prob. 14D.1BECh. 14 - Prob. 14D.2AECh. 14 - Prob. 14D.2BECh. 14 - Prob. 14D.3AECh. 14 - Prob. 14D.3BECh. 14 - Prob. 14D.4AECh. 14 - Prob. 14D.4BECh. 14 - Prob. 14D.5AECh. 14 - Prob. 14D.5BECh. 14 - Prob. 14D.6AECh. 14 - Prob. 14D.6BECh. 14 - Prob. 14D.8AECh. 14 - Prob. 14D.8BECh. 14 - Prob. 14D.9AECh. 14 - Prob. 14D.9BECh. 14 - Prob. 14D.2PCh. 14 - Prob. 14D.3PCh. 14 - Prob. 14D.4PCh. 14 - Prob. 14D.6PCh. 14 - Prob. 14D.7PCh. 14 - Prob. 14D.8PCh. 14 - Prob. 14D.9PCh. 14 - Prob. 14D.10PCh. 14 - Prob. 14E.1DQCh. 14 - Prob. 14E.2DQCh. 14 - Prob. 14E.3DQCh. 14 - Prob. 14E.4DQCh. 14 - Prob. 14E.5DQCh. 14 - Prob. 14E.1AECh. 14 - Prob. 14E.1BECh. 14 - Prob. 14E.1PCh. 14 - Prob. 14E.3PCh. 14 - Prob. 14.1IACh. 14 - Prob. 14.2IACh. 14 - Prob. 14.6IACh. 14 - Prob. 14.8IA
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