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 5, Problem 5C.5P

(a)

Interpretation Introduction

Interpretation:

The phases present in the phase diagrams have to be stated.

Concept introduction:

The azeotropic mixture is defined as a mixture of two or more liquids which evaporates without altering the composition of the mixture.  The azeotropic mixture has a constant boiling point because of the same composition in liquid as well as vapor phase.  The components of azeotropic mixture are not separated by distillation processes.

The phase diagram represents the changes in the mixture on changing the parameters like temperature and pressure of the mixture.

(a)

Expert Solution
Check Mark

Answer to Problem 5C.5P

The phases have been stated in the phase diagrams as,

Atkins' Physical Chemistry, Chapter 5, Problem 5C.5P , additional homework tip  1

Atkins' Physical Chemistry, Chapter 5, Problem 5C.5P , additional homework tip  2

Explanation of Solution

The given phase diagrams are shown below as,

Atkins' Physical Chemistry, Chapter 5, Problem 5C.5P , additional homework tip  3

Figure 1

Atkins' Physical Chemistry, Chapter 5, Problem 5C.5P , additional homework tip  4

Figure 2

The region between the liquid phase curve and vapour phase curve has liquid and vapour phase in the equilibrium.  Thus, the number of phases present in that region is 2.

In the region above the curve, only vapour phase is present.

In the region below the curve, only liquid phase is present.

The regions are stated in the phase diagrams as,

Atkins' Physical Chemistry, Chapter 5, Problem 5C.5P , additional homework tip  5

Figure 3

Atkins' Physical Chemistry, Chapter 5, Problem 5C.5P , additional homework tip  6

Figure 4

(b)

Interpretation Introduction

Interpretation:

The vapor pressure of the mixture of heptane and hexane at 70°C under the given conditions has to be stated.

Concept introduction:

The azeotropic mixture is defined as a mixture of two or more liquids which evaporates without altering the composition of the mixture.  The azeotropic mixture has a constant boiling point because of the same composition in liquid as well as vapor phase.  The components of azeotropic mixture are not separated by distillation processes.

The phase diagram represents the changes in the mixture on changing the parameters like temperature and pressure of the mixture.

(b)

Expert Solution
Check Mark

Answer to Problem 5C.5P

The vapour pressure of the mixture of heptane and hexane at 70°C under the given conditions has been stated as 620Torr_.

Explanation of Solution

The number of moles of heptane and hexane present in the mixture is 1.00mol.

The mole fraction is calculated by the formula as,

    x=nntotal        (1)

The total number of moles of the mixture is calculated as,

    ntotal=nHexane+nHeptane=1+1=2

Substitute the values in the equation (1) to calculate the mole fraction of hexane as,

    x=nntotal=12=0.50

The pressure corresponding to xHexane=0.5 is equals to 620Torr which is taken by drawing the line on the y-axis as shown in phase diagram below.

Atkins' Physical Chemistry, Chapter 5, Problem 5C.5P , additional homework tip  7

Figure 5

Thus, the vapour pressure of the mixture of heptane and hexane at 70°C under the given conditions is 620Torr_.

(c)

Interpretation Introduction

Interpretation:

The vapor pressure of the mixture of heptane and hexane at 70°C when just one drop of the liquid remains has to be stated.

Concept introduction:

The azeotropic mixture is defined as a mixture of two or more liquids which evaporates without altering the composition of the mixture.  The azeotropic mixture has a constant boiling point because of the same composition in liquid as well as vapor phase.  The components of azeotropic mixture are not separated by distillation processes.

The phase diagram represents the changes in the mixture on changing the parameters like temperature and pressure of the mixture.

(c)

Expert Solution
Check Mark

Answer to Problem 5C.5P

The vapour pressure of the mixture of heptane and hexane at 70°C when just one drop of the liquid remains has been stated as more than 490Torr_.

Explanation of Solution

The number of moles of heptane and hexane present in the mixture is 1.00mol.

The mole fraction is calculated by the formula as,

    x=nntotal        (1)

The total number of moles of the mixture is calculated as,

    ntotal=nHexane+nHeptane=1+1=2

Substitute the values in the equation (1) to calculate the mole fraction of hexane as,

    x=nntotal=12=0.50

The pressure corresponding to xHexane=0.5 is equals to 490Torr which is taken by drawing the line on the y-axis as shown in phase diagram below.

Atkins' Physical Chemistry, Chapter 5, Problem 5C.5P , additional homework tip  8

Figure 6

When just one drop of the liquid remains, then the vapour pressure of the mixture is more than 490Torr as till 490Torr, the amount of liquid phase is high in the mixture.

Thus, the vapour pressure of the mixture of heptane and hexane at 70°C when just one drop of the liquid remains is more than 490Torr_.

(d)

Interpretation Introduction

Interpretation:

The mole fraction of hexane in the liquid and vapour phase for the conditions of part b has to be calculated.

Concept introduction:

The azeotropic mixture is defined as a mixture of two or more liquids which evaporates without altering the composition of the mixture.  The azeotropic mixture has a constant boiling point because of the same composition in liquid as well as vapor phase.  The components of azeotropic mixture are not separated by distillation processes.

The phase diagram represents the changes in the mixture on changing the parameters like temperature and pressure of the mixture.

(d)

Expert Solution
Check Mark

Answer to Problem 5C.5P

The mole fraction of hexane in the liquid and vapour phase for the conditions of part b has been calculated as 0.5_ each.

Explanation of Solution

The number of moles of hexane present in the mixture is given as 1.00mol.

The mole fraction is calculated by the formula as,

    x=nntotal        (1)

The total number of moles of the mixture is calculated as,

    ntotal=nHexane+nHeptane=1+1=2

Substitute the values in the equation (1) to calculate the mole fraction of hexane as,

    x=nntotal=12=0.50_

The mole fraction of hexane in the liquid phase is taken by drawing the line on the liquid curve of the phase diagram which is shown by the line AB’ in phase diagram below.

Atkins' Physical Chemistry, Chapter 5, Problem 5C.5P , additional homework tip  9

Figure 7

Thus, mole fraction of hexane in the liquid and vapour phase for the conditions of part b is 0.5_ each.

(e)

Interpretation Introduction

Interpretation:

The mole fraction of hexane in the liquid and vapour phase for the conditions of part c has to be calculated.

Concept introduction:

The azeotropic mixture is defined as a mixture of two or more liquids which evaporates without altering the composition of the mixture.  The azeotropic mixture has a constant boiling point because of the same composition in liquid as well as vapor phase.  The components of azeotropic mixture are not separated by distillation processes.

The phase diagram represents the changes in the mixture on changing the parameters like temperature and pressure of the mixture.

(e)

Expert Solution
Check Mark

Answer to Problem 5C.5P

The mole fraction of hexane in the liquid and vapour phase for the conditions of part c has been calculated as 0.8_ and 0.5_ respectively.

Explanation of Solution

The number of moles of hexane present in the mixture is given as 1.00mol.

The mole fraction is calculated by the formula as,

    x=nntotal        (1)

The total number of moles of the mixture is calculated as,

    ntotal=nHexane+nHeptane=1+1=2

Substitute the values in the equation (1) to calculate the mole fraction of hexane as,

    x=nntotal=12=0.50_

The mole fraction of hexane in the liquid phase is taken by drawing a parallel line on the liquid curve of the phase diagram which is shown by the line F’G’ in phase diagram below.

Atkins' Physical Chemistry, Chapter 5, Problem 5C.5P , additional homework tip  10

Figure 8

The mole fraction of hexane in the liquid phase is found out to be 0.8 from the phase diagram.

Thus, mole fraction of hexane in the liquid and vapour phase for the conditions of part c is 0.8_ and 0.5_ respectively.

(f)

Interpretation Introduction

Interpretation:

The amounts of substances in the liquid and vapour phase at 85°C and 760Torr having zheptane=0.40 have to be calculated.

Concept introduction:

The azeotropic mixture is defined as a mixture of two or more liquids which evaporates without altering the composition of the mixture.  The azeotropic mixture has a constant boiling point because of the same composition in liquid as well as vapor phase.  The components of azeotropic mixture are not separated by distillation processes.

The phase diagram represents the changes in the mixture on changing the parameters like temperature and pressure of the mixture.

(f)

Expert Solution
Check Mark

Answer to Problem 5C.5P

The mole fraction of hexane in the liquid and vapour phase at 85°C and 760Torr is 0.8_ and 0.3_ respectively and the mole fraction of heptane in the liquid and vapour phase at 85°C and 760Torr is 0.20_ and 0.70_ respectively.

Explanation of Solution

The overall mole fraction of heptane present in the mixture is given as 0.40.

Thus, the overall mole fraction of hexane present in the mixture is calculated as,

    xHexane=1xHeptane=10.40=0.60

The mole fraction of hexane in the liquid phase is taken by drawing a line passing through xHexane=0.60.  This line is then extrapolated onto the liquid curve and vapour curve (RS) as shown in the phase diagram below.

Atkins' Physical Chemistry, Chapter 5, Problem 5C.5P , additional homework tip  11

Figure 9

The mole fraction of hexane in the liquid phase is found out to be approximately 0.8_ from the phase diagram.

Thus, the mole fraction of heptane in the liquid phase is calculated as,

    xHeptane=1xHexane=10.80=0.20_

The mole fraction of hexane in the vapour phase is found out to be 0.3_ from the phase diagram.

Thus, the mole fraction of heptane in the vapour phase is calculated as,

    yHeptane=1yHexane=10.30=0.70_

Thus, mole fraction of hexane in the liquid and vapour phase at 85°C and 760Torr is 0.8_ and 0.3_ respectively and the mole fraction of heptane in the liquid and vapour phase at 85°C and 760Torr is 0.20_ and 0.70_ respectively.

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Chapter 5 Solutions

Atkins' Physical Chemistry

Ch. 5 - Prob. 5A.4DQCh. 5 - Prob. 5A.5DQCh. 5 - Prob. 5A.1AECh. 5 - Prob. 5A.1BECh. 5 - Prob. 5A.2AECh. 5 - Prob. 5A.2BECh. 5 - Prob. 5A.3AECh. 5 - Prob. 5A.3BECh. 5 - Prob. 5A.4AECh. 5 - Prob. 5A.4BECh. 5 - Prob. 5A.5AECh. 5 - Prob. 5A.5BECh. 5 - Prob. 5A.6AECh. 5 - Prob. 5A.6BECh. 5 - Prob. 5A.7AECh. 5 - Prob. 5A.7BECh. 5 - Prob. 5A.8AECh. 5 - Prob. 5A.8BECh. 5 - Prob. 5A.9AECh. 5 - Prob. 5A.9BECh. 5 - Prob. 5A.10AECh. 5 - Prob. 5A.10BECh. 5 - Prob. 5A.11AECh. 5 - Prob. 5A.11BECh. 5 - Prob. 5A.1PCh. 5 - Prob. 5A.3PCh. 5 - Prob. 5A.4PCh. 5 - Prob. 5A.5PCh. 5 - Prob. 5A.6PCh. 5 - Prob. 5A.7PCh. 5 - Prob. 5B.1DQCh. 5 - Prob. 5B.2DQCh. 5 - Prob. 5B.3DQCh. 5 - Prob. 5B.4DQCh. 5 - Prob. 5B.5DQCh. 5 - Prob. 5B.6DQCh. 5 - Prob. 5B.7DQCh. 5 - Prob. 5B.1AECh. 5 - Prob. 5B.1BECh. 5 - Prob. 5B.2AECh. 5 - Prob. 5B.2BECh. 5 - Prob. 5B.3AECh. 5 - Prob. 5B.3BECh. 5 - Prob. 5B.4AECh. 5 - Prob. 5B.4BECh. 5 - Prob. 5B.5AECh. 5 - Prob. 5B.5BECh. 5 - Prob. 5B.6AECh. 5 - Prob. 5B.6BECh. 5 - Prob. 5B.7AECh. 5 - Prob. 5B.7BECh. 5 - Prob. 5B.8AECh. 5 - Prob. 5B.8BECh. 5 - Prob. 5B.9AECh. 5 - Prob. 5B.9BECh. 5 - Prob. 5B.10AECh. 5 - Prob. 5B.10BECh. 5 - Prob. 5B.11AECh. 5 - Prob. 5B.11BECh. 5 - Prob. 5B.12AECh. 5 - Prob. 5B.12BECh. 5 - Prob. 5B.1PCh. 5 - Prob. 5B.2PCh. 5 - Prob. 5B.3PCh. 5 - Prob. 5B.4PCh. 5 - Prob. 5B.5PCh. 5 - Prob. 5B.6PCh. 5 - Prob. 5B.9PCh. 5 - Prob. 5B.11PCh. 5 - Prob. 5B.13PCh. 5 - Prob. 5C.1DQCh. 5 - Prob. 5C.2DQCh. 5 - Prob. 5C.3DQCh. 5 - Prob. 5C.1AECh. 5 - Prob. 5C.1BECh. 5 - Prob. 5C.2AECh. 5 - Prob. 5C.2BECh. 5 - Prob. 5C.3AECh. 5 - Prob. 5C.3BECh. 5 - Prob. 5C.4AECh. 5 - Prob. 5C.4BECh. 5 - Prob. 5C.1PCh. 5 - Prob. 5C.2PCh. 5 - Prob. 5C.3PCh. 5 - Prob. 5C.4PCh. 5 - Prob. 5C.5PCh. 5 - Prob. 5C.6PCh. 5 - Prob. 5C.7PCh. 5 - Prob. 5C.8PCh. 5 - Prob. 5C.9PCh. 5 - Prob. 5C.10PCh. 5 - Prob. 5D.1DQCh. 5 - Prob. 5D.2DQCh. 5 - Prob. 5D.1AECh. 5 - Prob. 5D.1BECh. 5 - Prob. 5D.2AECh. 5 - Prob. 5D.2BECh. 5 - Prob. 5D.3AECh. 5 - Prob. 5D.3BECh. 5 - Prob. 5D.4AECh. 5 - Prob. 5D.4BECh. 5 - Prob. 5D.5AECh. 5 - Prob. 5D.5BECh. 5 - Prob. 5D.6AECh. 5 - Prob. 5D.1PCh. 5 - Prob. 5D.2PCh. 5 - Prob. 5D.3PCh. 5 - Prob. 5D.4PCh. 5 - Prob. 5D.5PCh. 5 - Prob. 5D.6PCh. 5 - Prob. 5D.7PCh. 5 - Prob. 5E.1DQCh. 5 - Prob. 5E.2DQCh. 5 - Prob. 5E.3DQCh. 5 - Prob. 5E.4DQCh. 5 - Prob. 5E.1AECh. 5 - Prob. 5E.1BECh. 5 - Prob. 5E.2AECh. 5 - Prob. 5E.2BECh. 5 - Prob. 5E.3AECh. 5 - Prob. 5E.3BECh. 5 - Prob. 5E.4AECh. 5 - Prob. 5E.4BECh. 5 - Prob. 5E.5AECh. 5 - Prob. 5E.5BECh. 5 - Prob. 5E.1PCh. 5 - Prob. 5E.2PCh. 5 - Prob. 5E.3PCh. 5 - Prob. 5F.1DQCh. 5 - Prob. 5F.2DQCh. 5 - Prob. 5F.3DQCh. 5 - Prob. 5F.4DQCh. 5 - Prob. 5F.5DQCh. 5 - Prob. 5F.1AECh. 5 - Prob. 5F.1BECh. 5 - Prob. 5F.2AECh. 5 - Prob. 5F.2BECh. 5 - Prob. 5F.3AECh. 5 - Prob. 5F.3BECh. 5 - Prob. 5F.4AECh. 5 - Prob. 5F.4BECh. 5 - Prob. 5F.5AECh. 5 - Prob. 5F.5BECh. 5 - Prob. 5F.6AECh. 5 - Prob. 5F.6BECh. 5 - Prob. 5F.7AECh. 5 - Prob. 5F.7BECh. 5 - Prob. 5F.8AECh. 5 - Prob. 5F.8BECh. 5 - Prob. 5F.1PCh. 5 - Prob. 5F.2PCh. 5 - Prob. 5F.3PCh. 5 - Prob. 5F.4PCh. 5 - Prob. 5.1IACh. 5 - Prob. 5.2IACh. 5 - Prob. 5.3IACh. 5 - Prob. 5.4IACh. 5 - Prob. 5.5IACh. 5 - Prob. 5.6IACh. 5 - Prob. 5.8IACh. 5 - Prob. 5.9IACh. 5 - Prob. 5.10IA
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