The partial pressure of each component in equilibrium at 25 ∘ C with a solution of compositions χ p r o p = 0.25 , 0.50 , a n d 0.75 has to be calculated. The composition of the vapor phase has to be determined. Concept Introduction: According to Raoult’s law, in an ideal solution of two volatile liquids, the total vapor pressure of the solution is the sum of individual vapor pressure. P t o t a l = P A + P B .... ( 1 ) Where, P A = χ A P A ∘ a n d P B = χ B P B ∘ . The equation ( 1 ) can be further modified. P t o t a l = P A + P B P t o t a l = χ A P A ∘ + χ B P B ∘ P t o t a l = χ A P A ∘ + ( 1 − χ A ) P B ∘ [ ∵ χ A + χ B = 1 ] P t o t a l = P B ∘ + χ A ( P A ∘ − P B ∘ )

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Answer 1

Question

Chapter 16, Problem 16.54P

Interpretation Introduction

Interpretation:

The partial pressure of each component in equilibrium at 25∘C with a solution of compositions χprop=0.25, 0.50, and 0.75 has to be calculated. The composition of the vapor phase has to be determined.

Concept Introduction:

According to Raoult’s law, in an ideal solution of two volatile liquids, the total vapor pressure of the solution is the sum of individual vapor pressure.

Ptotal=PA+PB ....(1)

Where, PA=χAPA∘ and PB=χBPB∘ .

The equation (1) can be further modified.

Ptotal=PA+PBPtotal=χAPA∘+χBPB∘Ptotal=χAPA∘+(1−χA)PB∘ [∵ χA+χB=1]Ptotal=PB∘ +χA(PA∘−PB∘ )

Expert Solution & Answer

Answer to Problem 16.54P

The partial pressures of 1− propanol are 5.2 Torr, 10.45 Torr and 16 Torr. The partial pressures of isopropanol are 34 Torr, 23 Torr and 11 Torr.

The composition of vapor phase is χprop=0.13, 0.31, 0.59 and χiso=0.85, 0.70, and 0.41.

Explanation of Solution

Given that, the mole fraction of 1− propanol is 0.25, 0.50, and 0.75. Then the mole fraction of isopropanol is given below.

χiso=1−χprop=1−0.25=0.75χiso=1−χprop=1−0.50=0.50χiso=1−χprop=1−0.75=0.25

The partial pressure of each component can be calculated as given below.

Pprop=χpropPprop∘ =0.25×20.9 Torr=5.2 TorrPprop=χpropPprop∘ =0.50×20.9 Torr=10.45 TorrPprop=χpropPprop∘ =0.75×20.9 Torr=16 TorrPiso=χisoPiso∘ =0.75×45.2 Torr=34 TorrPiso=χisoPiso∘ =0.50×45.2 Torr=23 TorrPiso=χisoPiso∘ =0.25×45.2 Torr=11 Torr

The total vapor pressure can be calculated as given below.

When, χiso=0.75Ptotal=Pprop∘ +χiso(Piso∘−Pprop∘ )Ptotal=20.9 Torr +(0.75)(45.2−20.9)TorrPtotal=40 Torr.When, χiso=0.50Ptotal=20.9 Torr +(0.50)(45.2−20.9)TorrPtotal=33 Torr.When, χiso=0.25Ptotal=20.9 Torr +(0.25)(45.2−20.9)TorrPtotal=27 Torr.

The pressure of a gas is directly proportional to the number of moles of the gas. Hence, the mole fraction of each component in the vapor can be calculated as given below.

For 1-propanol:

When, Ptotal=40 Torryprop=npropnprop+niso=PpropPprop+Piso=PpropPtotal=5.2 Torr40 Torryprop=0.13.When, Ptotal=33 Torryprop=PpropPtotal=10.45 Torr33 Torryprop=0.31.When, Ptotal=27 Torryprop=PpropPtotal=16 Torr27 Torryprop=0.59.

For isopropanol:

When, Ptotal=40 Torryiso=nisoniso+nprop=PisoPprop+Piso=PisoPtotal=34 Torr40 Torryiso=0.85.When, Ptotal=33 Torryiso=nisoniso+nprop=PisoPtotal=23 Torr33 Torryiso=0.70.When, Ptotal=27 Torryiso=nisoniso+nprop=PisoPtotal=11 Torr27 Torryiso=0.41.

Therefore, the partial pressures of 1− propanol are 5.2 Torr, 10.45 Torr and 16 Torr. The partial pressures of isopropanol are 34 Torr, 23 Torr and 11 Torr. The composition of vapor phase is χprop=0.13, 0.31, 0.59 and χiso=0.85, 0.70, and 0.41.

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Answer 2

Question

Chapter 16, Problem 16.54P

Interpretation Introduction

Interpretation:

The partial pressure of each component in equilibrium at 25∘C with a solution of compositions χprop=0.25, 0.50, and 0.75 has to be calculated. The composition of the vapor phase has to be determined.

Concept Introduction:

According to Raoult’s law, in an ideal solution of two volatile liquids, the total vapor pressure of the solution is the sum of individual vapor pressure.

Ptotal=PA+PB ....(1)

Where, PA=χAPA∘ and PB=χBPB∘ .

The equation (1) can be further modified.

Ptotal=PA+PBPtotal=χAPA∘+χBPB∘Ptotal=χAPA∘+(1−χA)PB∘ [∵ χA+χB=1]Ptotal=PB∘ +χA(PA∘−PB∘ )

Expert Solution & Answer

Answer to Problem 16.54P

The partial pressures of 1− propanol are 5.2 Torr, 10.45 Torr and 16 Torr. The partial pressures of isopropanol are 34 Torr, 23 Torr and 11 Torr.

The composition of vapor phase is χprop=0.13, 0.31, 0.59 and χiso=0.85, 0.70, and 0.41.

Explanation of Solution

Given that, the mole fraction of 1− propanol is 0.25, 0.50, and 0.75. Then the mole fraction of isopropanol is given below.

χiso=1−χprop=1−0.25=0.75χiso=1−χprop=1−0.50=0.50χiso=1−χprop=1−0.75=0.25

The partial pressure of each component can be calculated as given below.

Pprop=χpropPprop∘ =0.25×20.9 Torr=5.2 TorrPprop=χpropPprop∘ =0.50×20.9 Torr=10.45 TorrPprop=χpropPprop∘ =0.75×20.9 Torr=16 TorrPiso=χisoPiso∘ =0.75×45.2 Torr=34 TorrPiso=χisoPiso∘ =0.50×45.2 Torr=23 TorrPiso=χisoPiso∘ =0.25×45.2 Torr=11 Torr

The total vapor pressure can be calculated as given below.

When, χiso=0.75Ptotal=Pprop∘ +χiso(Piso∘−Pprop∘ )Ptotal=20.9 Torr +(0.75)(45.2−20.9)TorrPtotal=40 Torr.When, χiso=0.50Ptotal=20.9 Torr +(0.50)(45.2−20.9)TorrPtotal=33 Torr.When, χiso=0.25Ptotal=20.9 Torr +(0.25)(45.2−20.9)TorrPtotal=27 Torr.

The pressure of a gas is directly proportional to the number of moles of the gas. Hence, the mole fraction of each component in the vapor can be calculated as given below.

For 1-propanol:

When, Ptotal=40 Torryprop=npropnprop+niso=PpropPprop+Piso=PpropPtotal=5.2 Torr40 Torryprop=0.13.When, Ptotal=33 Torryprop=PpropPtotal=10.45 Torr33 Torryprop=0.31.When, Ptotal=27 Torryprop=PpropPtotal=16 Torr27 Torryprop=0.59.

For isopropanol:

When, Ptotal=40 Torryiso=nisoniso+nprop=PisoPprop+Piso=PisoPtotal=34 Torr40 Torryiso=0.85.When, Ptotal=33 Torryiso=nisoniso+nprop=PisoPtotal=23 Torr33 Torryiso=0.70.When, Ptotal=27 Torryiso=nisoniso+nprop=PisoPtotal=11 Torr27 Torryiso=0.41.

Therefore, the partial pressures of 1− propanol are 5.2 Torr, 10.45 Torr and 16 Torr. The partial pressures of isopropanol are 34 Torr, 23 Torr and 11 Torr. The composition of vapor phase is χprop=0.13, 0.31, 0.59 and χiso=0.85, 0.70, and 0.41.

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Answer 3

Question

Chapter 16, Problem 16.54P

Interpretation Introduction

Interpretation:

The partial pressure of each component in equilibrium at 25∘C with a solution of compositions χprop=0.25, 0.50, and 0.75 has to be calculated. The composition of the vapor phase has to be determined.

Concept Introduction:

According to Raoult’s law, in an ideal solution of two volatile liquids, the total vapor pressure of the solution is the sum of individual vapor pressure.

Ptotal=PA+PB ....(1)

Where, PA=χAPA∘ and PB=χBPB∘ .

The equation (1) can be further modified.

Ptotal=PA+PBPtotal=χAPA∘+χBPB∘Ptotal=χAPA∘+(1−χA)PB∘ [∵ χA+χB=1]Ptotal=PB∘ +χA(PA∘−PB∘ )

Expert Solution & Answer

Answer to Problem 16.54P

The partial pressures of 1− propanol are 5.2 Torr, 10.45 Torr and 16 Torr. The partial pressures of isopropanol are 34 Torr, 23 Torr and 11 Torr.

The composition of vapor phase is χprop=0.13, 0.31, 0.59 and χiso=0.85, 0.70, and 0.41.

Explanation of Solution

Given that, the mole fraction of 1− propanol is 0.25, 0.50, and 0.75. Then the mole fraction of isopropanol is given below.

χiso=1−χprop=1−0.25=0.75χiso=1−χprop=1−0.50=0.50χiso=1−χprop=1−0.75=0.25

The partial pressure of each component can be calculated as given below.

Pprop=χpropPprop∘ =0.25×20.9 Torr=5.2 TorrPprop=χpropPprop∘ =0.50×20.9 Torr=10.45 TorrPprop=χpropPprop∘ =0.75×20.9 Torr=16 TorrPiso=χisoPiso∘ =0.75×45.2 Torr=34 TorrPiso=χisoPiso∘ =0.50×45.2 Torr=23 TorrPiso=χisoPiso∘ =0.25×45.2 Torr=11 Torr

The total vapor pressure can be calculated as given below.

When, χiso=0.75Ptotal=Pprop∘ +χiso(Piso∘−Pprop∘ )Ptotal=20.9 Torr +(0.75)(45.2−20.9)TorrPtotal=40 Torr.When, χiso=0.50Ptotal=20.9 Torr +(0.50)(45.2−20.9)TorrPtotal=33 Torr.When, χiso=0.25Ptotal=20.9 Torr +(0.25)(45.2−20.9)TorrPtotal=27 Torr.

The pressure of a gas is directly proportional to the number of moles of the gas. Hence, the mole fraction of each component in the vapor can be calculated as given below.

For 1-propanol:

When, Ptotal=40 Torryprop=npropnprop+niso=PpropPprop+Piso=PpropPtotal=5.2 Torr40 Torryprop=0.13.When, Ptotal=33 Torryprop=PpropPtotal=10.45 Torr33 Torryprop=0.31.When, Ptotal=27 Torryprop=PpropPtotal=16 Torr27 Torryprop=0.59.

For isopropanol:

When, Ptotal=40 Torryiso=nisoniso+nprop=PisoPprop+Piso=PisoPtotal=34 Torr40 Torryiso=0.85.When, Ptotal=33 Torryiso=nisoniso+nprop=PisoPtotal=23 Torr33 Torryiso=0.70.When, Ptotal=27 Torryiso=nisoniso+nprop=PisoPtotal=11 Torr27 Torryiso=0.41.

Therefore, the partial pressures of 1− propanol are 5.2 Torr, 10.45 Torr and 16 Torr. The partial pressures of isopropanol are 34 Torr, 23 Torr and 11 Torr. The composition of vapor phase is χprop=0.13, 0.31, 0.59 and χiso=0.85, 0.70, and 0.41.

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Answer 4

Question

Chapter 16, Problem 16.54P

Interpretation Introduction

Interpretation:

The partial pressure of each component in equilibrium at 25∘C with a solution of compositions χprop=0.25, 0.50, and 0.75 has to be calculated. The composition of the vapor phase has to be determined.

Concept Introduction:

According to Raoult’s law, in an ideal solution of two volatile liquids, the total vapor pressure of the solution is the sum of individual vapor pressure.

Ptotal=PA+PB ....(1)

Where, PA=χAPA∘ and PB=χBPB∘ .

The equation (1) can be further modified.

Ptotal=PA+PBPtotal=χAPA∘+χBPB∘Ptotal=χAPA∘+(1−χA)PB∘ [∵ χA+χB=1]Ptotal=PB∘ +χA(PA∘−PB∘ )

Expert Solution & Answer

Answer to Problem 16.54P

The partial pressures of 1− propanol are 5.2 Torr, 10.45 Torr and 16 Torr. The partial pressures of isopropanol are 34 Torr, 23 Torr and 11 Torr.

The composition of vapor phase is χprop=0.13, 0.31, 0.59 and χiso=0.85, 0.70, and 0.41.

Explanation of Solution

Given that, the mole fraction of 1− propanol is 0.25, 0.50, and 0.75. Then the mole fraction of isopropanol is given below.

χiso=1−χprop=1−0.25=0.75χiso=1−χprop=1−0.50=0.50χiso=1−χprop=1−0.75=0.25

The partial pressure of each component can be calculated as given below.

Pprop=χpropPprop∘ =0.25×20.9 Torr=5.2 TorrPprop=χpropPprop∘ =0.50×20.9 Torr=10.45 TorrPprop=χpropPprop∘ =0.75×20.9 Torr=16 TorrPiso=χisoPiso∘ =0.75×45.2 Torr=34 TorrPiso=χisoPiso∘ =0.50×45.2 Torr=23 TorrPiso=χisoPiso∘ =0.25×45.2 Torr=11 Torr

The total vapor pressure can be calculated as given below.

When, χiso=0.75Ptotal=Pprop∘ +χiso(Piso∘−Pprop∘ )Ptotal=20.9 Torr +(0.75)(45.2−20.9)TorrPtotal=40 Torr.When, χiso=0.50Ptotal=20.9 Torr +(0.50)(45.2−20.9)TorrPtotal=33 Torr.When, χiso=0.25Ptotal=20.9 Torr +(0.25)(45.2−20.9)TorrPtotal=27 Torr.

The pressure of a gas is directly proportional to the number of moles of the gas. Hence, the mole fraction of each component in the vapor can be calculated as given below.

For 1-propanol:

When, Ptotal=40 Torryprop=npropnprop+niso=PpropPprop+Piso=PpropPtotal=5.2 Torr40 Torryprop=0.13.When, Ptotal=33 Torryprop=PpropPtotal=10.45 Torr33 Torryprop=0.31.When, Ptotal=27 Torryprop=PpropPtotal=16 Torr27 Torryprop=0.59.

For isopropanol:

When, Ptotal=40 Torryiso=nisoniso+nprop=PisoPprop+Piso=PisoPtotal=34 Torr40 Torryiso=0.85.When, Ptotal=33 Torryiso=nisoniso+nprop=PisoPtotal=23 Torr33 Torryiso=0.70.When, Ptotal=27 Torryiso=nisoniso+nprop=PisoPtotal=11 Torr27 Torryiso=0.41.

Therefore, the partial pressures of 1− propanol are 5.2 Torr, 10.45 Torr and 16 Torr. The partial pressures of isopropanol are 34 Torr, 23 Torr and 11 Torr. The composition of vapor phase is χprop=0.13, 0.31, 0.59 and χiso=0.85, 0.70, and 0.41.

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Answer 5

Chapter 16, Problem 16.54P

Interpretation Introduction

Interpretation:

The partial pressure of each component in equilibrium at 25∘C with a solution of compositions χprop=0.25, 0.50, and 0.75 has to be calculated. The composition of the vapor phase has to be determined.

Concept Introduction:

According to Raoult’s law, in an ideal solution of two volatile liquids, the total vapor pressure of the solution is the sum of individual vapor pressure.

Ptotal=PA+PB ....(1)

Where, PA=χAPA∘ and PB=χBPB∘ .

The equation (1) can be further modified.

Ptotal=PA+PBPtotal=χAPA∘+χBPB∘Ptotal=χAPA∘+(1−χA)PB∘ [∵ χA+χB=1]Ptotal=PB∘ +χA(PA∘−PB∘ )

Expert Solution & Answer

Answer to Problem 16.54P

The partial pressures of 1− propanol are 5.2 Torr, 10.45 Torr and 16 Torr. The partial pressures of isopropanol are 34 Torr, 23 Torr and 11 Torr.

The composition of vapor phase is χprop=0.13, 0.31, 0.59 and χiso=0.85, 0.70, and 0.41.

Explanation of Solution

Given that, the mole fraction of 1− propanol is 0.25, 0.50, and 0.75. Then the mole fraction of isopropanol is given below.

χiso=1−χprop=1−0.25=0.75χiso=1−χprop=1−0.50=0.50χiso=1−χprop=1−0.75=0.25

The partial pressure of each component can be calculated as given below.

Pprop=χpropPprop∘ =0.25×20.9 Torr=5.2 TorrPprop=χpropPprop∘ =0.50×20.9 Torr=10.45 TorrPprop=χpropPprop∘ =0.75×20.9 Torr=16 TorrPiso=χisoPiso∘ =0.75×45.2 Torr=34 TorrPiso=χisoPiso∘ =0.50×45.2 Torr=23 TorrPiso=χisoPiso∘ =0.25×45.2 Torr=11 Torr

The total vapor pressure can be calculated as given below.

When, χiso=0.75Ptotal=Pprop∘ +χiso(Piso∘−Pprop∘ )Ptotal=20.9 Torr +(0.75)(45.2−20.9)TorrPtotal=40 Torr.When, χiso=0.50Ptotal=20.9 Torr +(0.50)(45.2−20.9)TorrPtotal=33 Torr.When, χiso=0.25Ptotal=20.9 Torr +(0.25)(45.2−20.9)TorrPtotal=27 Torr.

The pressure of a gas is directly proportional to the number of moles of the gas. Hence, the mole fraction of each component in the vapor can be calculated as given below.

For 1-propanol:

When, Ptotal=40 Torryprop=npropnprop+niso=PpropPprop+Piso=PpropPtotal=5.2 Torr40 Torryprop=0.13.When, Ptotal=33 Torryprop=PpropPtotal=10.45 Torr33 Torryprop=0.31.When, Ptotal=27 Torryprop=PpropPtotal=16 Torr27 Torryprop=0.59.

For isopropanol:

When, Ptotal=40 Torryiso=nisoniso+nprop=PisoPprop+Piso=PisoPtotal=34 Torr40 Torryiso=0.85.When, Ptotal=33 Torryiso=nisoniso+nprop=PisoPtotal=23 Torr33 Torryiso=0.70.When, Ptotal=27 Torryiso=nisoniso+nprop=PisoPtotal=11 Torr27 Torryiso=0.41.

Therefore, the partial pressures of 1− propanol are 5.2 Torr, 10.45 Torr and 16 Torr. The partial pressures of isopropanol are 34 Torr, 23 Torr and 11 Torr. The composition of vapor phase is χprop=0.13, 0.31, 0.59 and χiso=0.85, 0.70, and 0.41.

Answer 6

Chapter 16, Problem 16.54P

Interpretation Introduction

Interpretation:

The partial pressure of each component in equilibrium at 25∘C with a solution of compositions χprop=0.25, 0.50, and 0.75 has to be calculated. The composition of the vapor phase has to be determined.

Concept Introduction:

According to Raoult’s law, in an ideal solution of two volatile liquids, the total vapor pressure of the solution is the sum of individual vapor pressure.

Ptotal=PA+PB ....(1)

Where, PA=χAPA∘ and PB=χBPB∘ .

The equation (1) can be further modified.

Ptotal=PA+PBPtotal=χAPA∘+χBPB∘Ptotal=χAPA∘+(1−χA)PB∘ [∵ χA+χB=1]Ptotal=PB∘ +χA(PA∘−PB∘ )

Expert Solution & Answer

Answer to Problem 16.54P

The partial pressures of 1− propanol are 5.2 Torr, 10.45 Torr and 16 Torr. The partial pressures of isopropanol are 34 Torr, 23 Torr and 11 Torr.

The composition of vapor phase is χprop=0.13, 0.31, 0.59 and χiso=0.85, 0.70, and 0.41.

Explanation of Solution

Given that, the mole fraction of 1− propanol is 0.25, 0.50, and 0.75. Then the mole fraction of isopropanol is given below.

χiso=1−χprop=1−0.25=0.75χiso=1−χprop=1−0.50=0.50χiso=1−χprop=1−0.75=0.25

The partial pressure of each component can be calculated as given below.

Pprop=χpropPprop∘ =0.25×20.9 Torr=5.2 TorrPprop=χpropPprop∘ =0.50×20.9 Torr=10.45 TorrPprop=χpropPprop∘ =0.75×20.9 Torr=16 TorrPiso=χisoPiso∘ =0.75×45.2 Torr=34 TorrPiso=χisoPiso∘ =0.50×45.2 Torr=23 TorrPiso=χisoPiso∘ =0.25×45.2 Torr=11 Torr

The total vapor pressure can be calculated as given below.

When, χiso=0.75Ptotal=Pprop∘ +χiso(Piso∘−Pprop∘ )Ptotal=20.9 Torr +(0.75)(45.2−20.9)TorrPtotal=40 Torr.When, χiso=0.50Ptotal=20.9 Torr +(0.50)(45.2−20.9)TorrPtotal=33 Torr.When, χiso=0.25Ptotal=20.9 Torr +(0.25)(45.2−20.9)TorrPtotal=27 Torr.

The pressure of a gas is directly proportional to the number of moles of the gas. Hence, the mole fraction of each component in the vapor can be calculated as given below.

For 1-propanol:

When, Ptotal=40 Torryprop=npropnprop+niso=PpropPprop+Piso=PpropPtotal=5.2 Torr40 Torryprop=0.13.When, Ptotal=33 Torryprop=PpropPtotal=10.45 Torr33 Torryprop=0.31.When, Ptotal=27 Torryprop=PpropPtotal=16 Torr27 Torryprop=0.59.

For isopropanol:

When, Ptotal=40 Torryiso=nisoniso+nprop=PisoPprop+Piso=PisoPtotal=34 Torr40 Torryiso=0.85.When, Ptotal=33 Torryiso=nisoniso+nprop=PisoPtotal=23 Torr33 Torryiso=0.70.When, Ptotal=27 Torryiso=nisoniso+nprop=PisoPtotal=11 Torr27 Torryiso=0.41.

Therefore, the partial pressures of 1− propanol are 5.2 Torr, 10.45 Torr and 16 Torr. The partial pressures of isopropanol are 34 Torr, 23 Torr and 11 Torr. The composition of vapor phase is χprop=0.13, 0.31, 0.59 and χiso=0.85, 0.70, and 0.41.

Answer 7

Chapter 16, Problem 16.54P

Interpretation Introduction

Interpretation:

The partial pressure of each component in equilibrium at 25∘C with a solution of compositions χprop=0.25, 0.50, and 0.75 has to be calculated. The composition of the vapor phase has to be determined.

Concept Introduction:

According to Raoult’s law, in an ideal solution of two volatile liquids, the total vapor pressure of the solution is the sum of individual vapor pressure.

Ptotal=PA+PB ....(1)

Where, PA=χAPA∘ and PB=χBPB∘ .

The equation (1) can be further modified.

Ptotal=PA+PBPtotal=χAPA∘+χBPB∘Ptotal=χAPA∘+(1−χA)PB∘ [∵ χA+χB=1]Ptotal=PB∘ +χA(PA∘−PB∘ )

Answer 8

Expert Solution & Answer

Answer to Problem 16.54P

The partial pressures of 1− propanol are 5.2 Torr, 10.45 Torr and 16 Torr. The partial pressures of isopropanol are 34 Torr, 23 Torr and 11 Torr.

The composition of vapor phase is χprop=0.13, 0.31, 0.59 and χiso=0.85, 0.70, and 0.41.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given that, the mole fraction of 1− propanol is 0.25, 0.50, and 0.75. Then the mole fraction of isopropanol is given below.

χiso=1−χprop=1−0.25=0.75χiso=1−χprop=1−0.50=0.50χiso=1−χprop=1−0.75=0.25

The partial pressure of each component can be calculated as given below.

Pprop=χpropPprop∘ =0.25×20.9 Torr=5.2 TorrPprop=χpropPprop∘ =0.50×20.9 Torr=10.45 TorrPprop=χpropPprop∘ =0.75×20.9 Torr=16 TorrPiso=χisoPiso∘ =0.75×45.2 Torr=34 TorrPiso=χisoPiso∘ =0.50×45.2 Torr=23 TorrPiso=χisoPiso∘ =0.25×45.2 Torr=11 Torr

The total vapor pressure can be calculated as given below.

When, χiso=0.75Ptotal=Pprop∘ +χiso(Piso∘−Pprop∘ )Ptotal=20.9 Torr +(0.75)(45.2−20.9)TorrPtotal=40 Torr.When, χiso=0.50Ptotal=20.9 Torr +(0.50)(45.2−20.9)TorrPtotal=33 Torr.When, χiso=0.25Ptotal=20.9 Torr +(0.25)(45.2−20.9)TorrPtotal=27 Torr.

The pressure of a gas is directly proportional to the number of moles of the gas. Hence, the mole fraction of each component in the vapor can be calculated as given below.

For 1-propanol:

When, Ptotal=40 Torryprop=npropnprop+niso=PpropPprop+Piso=PpropPtotal=5.2 Torr40 Torryprop=0.13.When, Ptotal=33 Torryprop=PpropPtotal=10.45 Torr33 Torryprop=0.31.When, Ptotal=27 Torryprop=PpropPtotal=16 Torr27 Torryprop=0.59.

For isopropanol:

When, Ptotal=40 Torryiso=nisoniso+nprop=PisoPprop+Piso=PisoPtotal=34 Torr40 Torryiso=0.85.When, Ptotal=33 Torryiso=nisoniso+nprop=PisoPtotal=23 Torr33 Torryiso=0.70.When, Ptotal=27 Torryiso=nisoniso+nprop=PisoPtotal=11 Torr27 Torryiso=0.41.

Therefore, the partial pressures of 1− propanol are 5.2 Torr, 10.45 Torr and 16 Torr. The partial pressures of isopropanol are 34 Torr, 23 Torr and 11 Torr. The composition of vapor phase is χprop=0.13, 0.31, 0.59 and χiso=0.85, 0.70, and 0.41.

Answer 12

Ch. 16 - Prob. 16.1P Ch. 16 - Prob. 16.2P Ch. 16 - Prob. 16.3P Ch. 16 - Prob. 16.4P Ch. 16 - Prob. 16.5P Ch. 16 - Prob. 16.6P Ch. 16 - Prob. 16.7P Ch. 16 - Prob. 16.8P Ch. 16 - Prob. 16.9P Ch. 16 - Prob. 16.10P

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