The given equation is to be proved along with all the assumptions made. Concept Introduction: The Fick’s law equation for the diffusion of A through a binary mixture of A and B is written in the vector form as: N A = − c D A B ∇ y A + y A ( N A + N B ) .... (1) Here, N A is the molar flux vector of component A, N B is the molar flux vector of component B, y A is the mole fraction of A, c is the total molar concentration, and D A B is the diffusion coefficient of A through B. The term − c D A B ∇ y A is the molar flux due to the diffusion along the concentration gradient and the term y A ( N A + N B ) is the molar flux of component A due to its bulk movement in the mixture.

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Answer 1

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Chapter 24, Problem 24.1P

Interpretation Introduction

Interpretation:

The given equation is to be proved along with all the assumptions made.

Concept Introduction:

The Fick’s law equation for the diffusion of A through a binary mixture of A and B is written in the vector form as:

NA=−cDAB∇yA+yA(NA+NB) .... (1)

Here, NA is the molar flux vector of component A, NB is the molar flux vector of component B, yA is the mole fraction of A, c is the total molar concentration, and DAB is the diffusion coefficient of A through B.

The term −cDAB∇yA is the molar flux due to the diffusion along the concentration gradient and the term yA(NA+NB) is the molar flux of component A due to its bulk movement in the mixture.

Expert Solution & Answer

Explanation of Solution

Given information:

The Fick’s law equation for diffusion of A through the binary mixture of A and B is given as:

NA=−cDAB∇yA+yA(NA+NB)

The assumptions made for the mixture are:

1. The gas mixture is taken as ideal.

2. The density of the mixture is taken to be constant.

3. Total molar concentration is taken to be constant.

Let the molecular weight of A and B be MA and MB respectively. yB is the mole fraction of B. Now, the mass fraction of A in the gas mixture will be:

wA=yAMAyAMA+yBMB .... (2)

Total mass of the gas mixture will be:

M=yAMA+yBMB

Equation (2) is written as:

MwA=yAMA .... (3)

The mass flux of component A in vector form as:

nA=ρAvA

Here, ρA is the density of A and vA is the vector velocity of component A. vA is further written as:

nA=ρANAcA .... (4)

Here, cA is the molar concentration of A.

From equation (1), substitute the value of NA in equation (4) as:

nA=ρA(−cDAB∇yA+yA(NA+NB))cA

nA=MA(−cDAB∇yA+yA(NA+NB))

nA=−cDABMA∇yA+yAMA(NA+NB) ..... (5)

Use equation (3) in equation (5) and simplify as:

nA=−cDABMA∇yA+yAMA(NA+NB)

nA=−cDABMA∇cAc+wAM(nAM+nBM)

nA=−DAB∇ρA+wA(nA+nB)

Hence, the given statement is proved. Here,

yA=cAc

ρA=MAcA

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Answer 2

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Chapter 24, Problem 24.1P

Interpretation Introduction

Interpretation:

The given equation is to be proved along with all the assumptions made.

Concept Introduction:

The Fick’s law equation for the diffusion of A through a binary mixture of A and B is written in the vector form as:

NA=−cDAB∇yA+yA(NA+NB) .... (1)

Here, NA is the molar flux vector of component A, NB is the molar flux vector of component B, yA is the mole fraction of A, c is the total molar concentration, and DAB is the diffusion coefficient of A through B.

The term −cDAB∇yA is the molar flux due to the diffusion along the concentration gradient and the term yA(NA+NB) is the molar flux of component A due to its bulk movement in the mixture.

Expert Solution & Answer

Explanation of Solution

Given information:

The Fick’s law equation for diffusion of A through the binary mixture of A and B is given as:

NA=−cDAB∇yA+yA(NA+NB)

The assumptions made for the mixture are:

1. The gas mixture is taken as ideal.

2. The density of the mixture is taken to be constant.

3. Total molar concentration is taken to be constant.

Let the molecular weight of A and B be MA and MB respectively. yB is the mole fraction of B. Now, the mass fraction of A in the gas mixture will be:

wA=yAMAyAMA+yBMB .... (2)

Total mass of the gas mixture will be:

M=yAMA+yBMB

Equation (2) is written as:

MwA=yAMA .... (3)

The mass flux of component A in vector form as:

nA=ρAvA

Here, ρA is the density of A and vA is the vector velocity of component A. vA is further written as:

nA=ρANAcA .... (4)

Here, cA is the molar concentration of A.

From equation (1), substitute the value of NA in equation (4) as:

nA=ρA(−cDAB∇yA+yA(NA+NB))cA

nA=MA(−cDAB∇yA+yA(NA+NB))

nA=−cDABMA∇yA+yAMA(NA+NB) ..... (5)

Use equation (3) in equation (5) and simplify as:

nA=−cDABMA∇yA+yAMA(NA+NB)

nA=−cDABMA∇cAc+wAM(nAM+nBM)

nA=−DAB∇ρA+wA(nA+nB)

Hence, the given statement is proved. Here,

yA=cAc

ρA=MAcA

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Answer 3

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Chapter 24, Problem 24.1P

Interpretation Introduction

Interpretation:

The given equation is to be proved along with all the assumptions made.

Concept Introduction:

The Fick’s law equation for the diffusion of A through a binary mixture of A and B is written in the vector form as:

NA=−cDAB∇yA+yA(NA+NB) .... (1)

Here, NA is the molar flux vector of component A, NB is the molar flux vector of component B, yA is the mole fraction of A, c is the total molar concentration, and DAB is the diffusion coefficient of A through B.

The term −cDAB∇yA is the molar flux due to the diffusion along the concentration gradient and the term yA(NA+NB) is the molar flux of component A due to its bulk movement in the mixture.

Expert Solution & Answer

Explanation of Solution

Given information:

The Fick’s law equation for diffusion of A through the binary mixture of A and B is given as:

NA=−cDAB∇yA+yA(NA+NB)

The assumptions made for the mixture are:

1. The gas mixture is taken as ideal.

2. The density of the mixture is taken to be constant.

3. Total molar concentration is taken to be constant.

Let the molecular weight of A and B be MA and MB respectively. yB is the mole fraction of B. Now, the mass fraction of A in the gas mixture will be:

wA=yAMAyAMA+yBMB .... (2)

Total mass of the gas mixture will be:

M=yAMA+yBMB

Equation (2) is written as:

MwA=yAMA .... (3)

The mass flux of component A in vector form as:

nA=ρAvA

Here, ρA is the density of A and vA is the vector velocity of component A. vA is further written as:

nA=ρANAcA .... (4)

Here, cA is the molar concentration of A.

From equation (1), substitute the value of NA in equation (4) as:

nA=ρA(−cDAB∇yA+yA(NA+NB))cA

nA=MA(−cDAB∇yA+yA(NA+NB))

nA=−cDABMA∇yA+yAMA(NA+NB) ..... (5)

Use equation (3) in equation (5) and simplify as:

nA=−cDABMA∇yA+yAMA(NA+NB)

nA=−cDABMA∇cAc+wAM(nAM+nBM)

nA=−DAB∇ρA+wA(nA+nB)

Hence, the given statement is proved. Here,

yA=cAc

ρA=MAcA

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Answer 4

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Chapter 24, Problem 24.1P

Interpretation Introduction

Interpretation:

The given equation is to be proved along with all the assumptions made.

Concept Introduction:

The Fick’s law equation for the diffusion of A through a binary mixture of A and B is written in the vector form as:

NA=−cDAB∇yA+yA(NA+NB) .... (1)

Here, NA is the molar flux vector of component A, NB is the molar flux vector of component B, yA is the mole fraction of A, c is the total molar concentration, and DAB is the diffusion coefficient of A through B.

The term −cDAB∇yA is the molar flux due to the diffusion along the concentration gradient and the term yA(NA+NB) is the molar flux of component A due to its bulk movement in the mixture.

Expert Solution & Answer

Explanation of Solution

Given information:

The Fick’s law equation for diffusion of A through the binary mixture of A and B is given as:

NA=−cDAB∇yA+yA(NA+NB)

The assumptions made for the mixture are:

1. The gas mixture is taken as ideal.

2. The density of the mixture is taken to be constant.

3. Total molar concentration is taken to be constant.

Let the molecular weight of A and B be MA and MB respectively. yB is the mole fraction of B. Now, the mass fraction of A in the gas mixture will be:

wA=yAMAyAMA+yBMB .... (2)

Total mass of the gas mixture will be:

M=yAMA+yBMB

Equation (2) is written as:

MwA=yAMA .... (3)

The mass flux of component A in vector form as:

nA=ρAvA

Here, ρA is the density of A and vA is the vector velocity of component A. vA is further written as:

nA=ρANAcA .... (4)

Here, cA is the molar concentration of A.

From equation (1), substitute the value of NA in equation (4) as:

nA=ρA(−cDAB∇yA+yA(NA+NB))cA

nA=MA(−cDAB∇yA+yA(NA+NB))

nA=−cDABMA∇yA+yAMA(NA+NB) ..... (5)

Use equation (3) in equation (5) and simplify as:

nA=−cDABMA∇yA+yAMA(NA+NB)

nA=−cDABMA∇cAc+wAM(nAM+nBM)

nA=−DAB∇ρA+wA(nA+nB)

Hence, the given statement is proved. Here,

yA=cAc

ρA=MAcA

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Answer 5

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Answer 6

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Answer 7

Chapter 24, Problem 24.1P

Interpretation Introduction

Interpretation:

The given equation is to be proved along with all the assumptions made.

Concept Introduction:

The Fick’s law equation for the diffusion of A through a binary mixture of A and B is written in the vector form as:

NA=−cDAB∇yA+yA(NA+NB) .... (1)

Here, NA is the molar flux vector of component A, NB is the molar flux vector of component B, yA is the mole fraction of A, c is the total molar concentration, and DAB is the diffusion coefficient of A through B.

The term −cDAB∇yA is the molar flux due to the diffusion along the concentration gradient and the term yA(NA+NB) is the molar flux of component A due to its bulk movement in the mixture.

Expert Solution & Answer

Explanation of Solution

Given information:

The Fick’s law equation for diffusion of A through the binary mixture of A and B is given as:

NA=−cDAB∇yA+yA(NA+NB)

The assumptions made for the mixture are:

1. The gas mixture is taken as ideal.

2. The density of the mixture is taken to be constant.

3. Total molar concentration is taken to be constant.

Let the molecular weight of A and B be MA and MB respectively. yB is the mole fraction of B. Now, the mass fraction of A in the gas mixture will be:

wA=yAMAyAMA+yBMB .... (2)

Total mass of the gas mixture will be:

M=yAMA+yBMB

Equation (2) is written as:

MwA=yAMA .... (3)

The mass flux of component A in vector form as:

nA=ρAvA

Here, ρA is the density of A and vA is the vector velocity of component A. vA is further written as:

nA=ρANAcA .... (4)

Here, cA is the molar concentration of A.

From equation (1), substitute the value of NA in equation (4) as:

nA=ρA(−cDAB∇yA+yA(NA+NB))cA

nA=MA(−cDAB∇yA+yA(NA+NB))

nA=−cDABMA∇yA+yAMA(NA+NB) ..... (5)

Use equation (3) in equation (5) and simplify as:

nA=−cDABMA∇yA+yAMA(NA+NB)

nA=−cDABMA∇cAc+wAM(nAM+nBM)

nA=−DAB∇ρA+wA(nA+nB)

Hence, the given statement is proved. Here,

yA=cAc

ρA=MAcA

Answer 8

Chapter 24, Problem 24.1P

Interpretation Introduction

Interpretation:

The given equation is to be proved along with all the assumptions made.

Concept Introduction:

The Fick’s law equation for the diffusion of A through a binary mixture of A and B is written in the vector form as:

NA=−cDAB∇yA+yA(NA+NB) .... (1)

Here, NA is the molar flux vector of component A, NB is the molar flux vector of component B, yA is the mole fraction of A, c is the total molar concentration, and DAB is the diffusion coefficient of A through B.

The term −cDAB∇yA is the molar flux due to the diffusion along the concentration gradient and the term yA(NA+NB) is the molar flux of component A due to its bulk movement in the mixture.

Expert Solution & Answer

Explanation of Solution

Given information:

The Fick’s law equation for diffusion of A through the binary mixture of A and B is given as:

NA=−cDAB∇yA+yA(NA+NB)

The assumptions made for the mixture are:

1. The gas mixture is taken as ideal.

2. The density of the mixture is taken to be constant.

3. Total molar concentration is taken to be constant.

Let the molecular weight of A and B be MA and MB respectively. yB is the mole fraction of B. Now, the mass fraction of A in the gas mixture will be:

wA=yAMAyAMA+yBMB .... (2)

Total mass of the gas mixture will be:

M=yAMA+yBMB

Equation (2) is written as:

MwA=yAMA .... (3)

The mass flux of component A in vector form as:

nA=ρAvA

Here, ρA is the density of A and vA is the vector velocity of component A. vA is further written as:

nA=ρANAcA .... (4)

Here, cA is the molar concentration of A.

From equation (1), substitute the value of NA in equation (4) as:

nA=ρA(−cDAB∇yA+yA(NA+NB))cA

nA=MA(−cDAB∇yA+yA(NA+NB))

nA=−cDABMA∇yA+yAMA(NA+NB) ..... (5)

Use equation (3) in equation (5) and simplify as:

nA=−cDABMA∇yA+yAMA(NA+NB)

nA=−cDABMA∇cAc+wAM(nAM+nBM)

nA=−DAB∇ρA+wA(nA+nB)

Hence, the given statement is proved. Here,

yA=cAc

ρA=MAcA

Answer 9

Chapter 24, Problem 24.1P

Interpretation Introduction

Interpretation:

The given equation is to be proved along with all the assumptions made.

Concept Introduction:

The Fick’s law equation for the diffusion of A through a binary mixture of A and B is written in the vector form as:

NA=−cDAB∇yA+yA(NA+NB) .... (1)

Here, NA is the molar flux vector of component A, NB is the molar flux vector of component B, yA is the mole fraction of A, c is the total molar concentration, and DAB is the diffusion coefficient of A through B.

The term −cDAB∇yA is the molar flux due to the diffusion along the concentration gradient and the term yA(NA+NB) is the molar flux of component A due to its bulk movement in the mixture.

Answer 10

Expert Solution & Answer

Explanation of Solution

Given information:

The Fick’s law equation for diffusion of A through the binary mixture of A and B is given as:

NA=−cDAB∇yA+yA(NA+NB)

The assumptions made for the mixture are:

1. The gas mixture is taken as ideal.

2. The density of the mixture is taken to be constant.

3. Total molar concentration is taken to be constant.

Let the molecular weight of A and B be MA and MB respectively. yB is the mole fraction of B. Now, the mass fraction of A in the gas mixture will be:

wA=yAMAyAMA+yBMB .... (2)

Total mass of the gas mixture will be:

M=yAMA+yBMB

Equation (2) is written as:

MwA=yAMA .... (3)

The mass flux of component A in vector form as:

nA=ρAvA

Here, ρA is the density of A and vA is the vector velocity of component A. vA is further written as:

nA=ρANAcA .... (4)

Here, cA is the molar concentration of A.

From equation (1), substitute the value of NA in equation (4) as:

nA=ρA(−cDAB∇yA+yA(NA+NB))cA

nA=MA(−cDAB∇yA+yA(NA+NB))

nA=−cDABMA∇yA+yAMA(NA+NB) ..... (5)

Use equation (3) in equation (5) and simplify as:

nA=−cDABMA∇yA+yAMA(NA+NB)

nA=−cDABMA∇cAc+wAM(nAM+nBM)

nA=−DAB∇ρA+wA(nA+nB)

Hence, the given statement is proved. Here,

yA=cAc

ρA=MAcA

Answer 11

Expert Solution & Answer

Answer 12

Expert Solution & Answer

Answer 13

Ch. 24 - Prob. 24.1P Ch. 24 - Starting with Ficks rate equation for the...Ch. 24 - Prob. 24.18P Ch. 24 - Prob. 24.19P Ch. 24 - Prob. 24.23P

Explanation of Solution

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