We have to prove excess properties of binary gas mixtures in given terms. Concept introduction: The gas mixture fugacity coefficients are partial molar properties of the excess Gibbs free energy, hence partial molar properties of the total fugacity can be written as follows: ln ϕ = [ ∂ ( G E / R T ) ∂ n i ] P , T , n j Excess properties are the difference between real gas property to ideal gas property. Equation (3.36) and (10.62) mentioned in question are: Z = 1 + B P R T B = ∑ i ∑ j y i y j B i , j All excess properties in terms of excess Gibbs free energy given in Table 10.1 in the book are: V E = ( ∂ ( G E ) ∂ P ) T , y S E = − ( ∂ ( G E ) ∂ T ) P , y H E = G E + T S E H E = G E + T [ − ( ∂ ( G E ) ∂ T ) P , y ] H E = − R T 2 ( ∂ ( G E / R T ) ∂ T ) P , y C P E = ( ∂ H E ∂ T ) P , y C P E = − T ( ∂ 2 G E ∂ T 2 ) P , y

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Answer 1

Question

Chapter 10, Problem 10.32P

Interpretation Introduction

Interpretation:

We have to prove excess properties of binary gas mixtures in given terms.

Concept introduction:

The gas mixture fugacity coefficients are partial molar properties of the excess Gibbs free energy, hence partial molar properties of the total fugacity can be written as follows:

lnϕ=[∂( G E /RT)∂ni]P,T,nj

Excess properties are the difference between real gas property to ideal gas property.

Equation (3.36) and (10.62) mentioned in question are:

Z=1+BPRTB=∑i∑j y i y j B i,j

All excess properties in terms of excess Gibbs free energy given in Table 10.1 in the book are:

VE=( ∂( GE ) ∂P)T,ySE=−( ∂( GE ) ∂T)P,yHE=GE+TSEHE=GE+T[−( ∂( G E ) ∂T )P,y]HE=−RT2( ∂( GE/RT ) ∂T)P,yCPE=( ∂ H E ∂T)P,yCPE=−T( ∂ 2 G E ∂ T 2 )P,y

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Answer 2

Question

Chapter 10, Problem 10.32P

Interpretation Introduction

Interpretation:

We have to prove excess properties of binary gas mixtures in given terms.

Concept introduction:

The gas mixture fugacity coefficients are partial molar properties of the excess Gibbs free energy, hence partial molar properties of the total fugacity can be written as follows:

lnϕ=[∂( G E /RT)∂ni]P,T,nj

Excess properties are the difference between real gas property to ideal gas property.

Equation (3.36) and (10.62) mentioned in question are:

Z=1+BPRTB=∑i∑j y i y j B i,j

All excess properties in terms of excess Gibbs free energy given in Table 10.1 in the book are:

VE=( ∂( GE ) ∂P)T,ySE=−( ∂( GE ) ∂T)P,yHE=GE+TSEHE=GE+T[−( ∂( G E ) ∂T )P,y]HE=−RT2( ∂( GE/RT ) ∂T)P,yCPE=( ∂ H E ∂T)P,yCPE=−T( ∂ 2 G E ∂ T 2 )P,y

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Answer 3

Question

Chapter 10, Problem 10.32P

Interpretation Introduction

Interpretation:

We have to prove excess properties of binary gas mixtures in given terms.

Concept introduction:

The gas mixture fugacity coefficients are partial molar properties of the excess Gibbs free energy, hence partial molar properties of the total fugacity can be written as follows:

lnϕ=[∂( G E /RT)∂ni]P,T,nj

Excess properties are the difference between real gas property to ideal gas property.

Equation (3.36) and (10.62) mentioned in question are:

Z=1+BPRTB=∑i∑j y i y j B i,j

All excess properties in terms of excess Gibbs free energy given in Table 10.1 in the book are:

VE=( ∂( GE ) ∂P)T,ySE=−( ∂( GE ) ∂T)P,yHE=GE+TSEHE=GE+T[−( ∂( G E ) ∂T )P,y]HE=−RT2( ∂( GE/RT ) ∂T)P,yCPE=( ∂ H E ∂T)P,yCPE=−T( ∂ 2 G E ∂ T 2 )P,y

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Answer 4

Question

Chapter 10, Problem 10.32P

Interpretation Introduction

Interpretation:

We have to prove excess properties of binary gas mixtures in given terms.

Concept introduction:

The gas mixture fugacity coefficients are partial molar properties of the excess Gibbs free energy, hence partial molar properties of the total fugacity can be written as follows:

lnϕ=[∂( G E /RT)∂ni]P,T,nj

Excess properties are the difference between real gas property to ideal gas property.

Equation (3.36) and (10.62) mentioned in question are:

Z=1+BPRTB=∑i∑j y i y j B i,j

All excess properties in terms of excess Gibbs free energy given in Table 10.1 in the book are:

VE=( ∂( GE ) ∂P)T,ySE=−( ∂( GE ) ∂T)P,yHE=GE+TSEHE=GE+T[−( ∂( G E ) ∂T )P,y]HE=−RT2( ∂( GE/RT ) ∂T)P,yCPE=( ∂ H E ∂T)P,yCPE=−T( ∂ 2 G E ∂ T 2 )P,y

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Answer 5

Chapter 10, Problem 10.32P

Interpretation Introduction

Interpretation:

We have to prove excess properties of binary gas mixtures in given terms.

Concept introduction:

The gas mixture fugacity coefficients are partial molar properties of the excess Gibbs free energy, hence partial molar properties of the total fugacity can be written as follows:

lnϕ=[∂( G E /RT)∂ni]P,T,nj

Excess properties are the difference between real gas property to ideal gas property.

Equation (3.36) and (10.62) mentioned in question are:

Z=1+BPRTB=∑i∑j y i y j B i,j

All excess properties in terms of excess Gibbs free energy given in Table 10.1 in the book are:

VE=( ∂( GE ) ∂P)T,ySE=−( ∂( GE ) ∂T)P,yHE=GE+TSEHE=GE+T[−( ∂( G E ) ∂T )P,y]HE=−RT2( ∂( GE/RT ) ∂T)P,yCPE=( ∂ H E ∂T)P,yCPE=−T( ∂ 2 G E ∂ T 2 )P,y

Answer 6

Chapter 10, Problem 10.32P

Interpretation Introduction

Interpretation:

We have to prove excess properties of binary gas mixtures in given terms.

Concept introduction:

The gas mixture fugacity coefficients are partial molar properties of the excess Gibbs free energy, hence partial molar properties of the total fugacity can be written as follows:

lnϕ=[∂( G E /RT)∂ni]P,T,nj

Excess properties are the difference between real gas property to ideal gas property.

Equation (3.36) and (10.62) mentioned in question are:

Z=1+BPRTB=∑i∑j y i y j B i,j

All excess properties in terms of excess Gibbs free energy given in Table 10.1 in the book are:

VE=( ∂( GE ) ∂P)T,ySE=−( ∂( GE ) ∂T)P,yHE=GE+TSEHE=GE+T[−( ∂( G E ) ∂T )P,y]HE=−RT2( ∂( GE/RT ) ∂T)P,yCPE=( ∂ H E ∂T)P,yCPE=−T( ∂ 2 G E ∂ T 2 )P,y

Answer 7

Chapter 10, Problem 10.32P

Interpretation Introduction

Interpretation:

We have to prove excess properties of binary gas mixtures in given terms.

Concept introduction:

The gas mixture fugacity coefficients are partial molar properties of the excess Gibbs free energy, hence partial molar properties of the total fugacity can be written as follows:

lnϕ=[∂( G E /RT)∂ni]P,T,nj

Excess properties are the difference between real gas property to ideal gas property.

Equation (3.36) and (10.62) mentioned in question are:

Z=1+BPRTB=∑i∑j y i y j B i,j

All excess properties in terms of excess Gibbs free energy given in Table 10.1 in the book are:

VE=( ∂( GE ) ∂P)T,ySE=−( ∂( GE ) ∂T)P,yHE=GE+TSEHE=GE+T[−( ∂( G E ) ∂T )P,y]HE=−RT2( ∂( GE/RT ) ∂T)P,yCPE=( ∂ H E ∂T)P,yCPE=−T( ∂ 2 G E ∂ T 2 )P,y

Answer 8

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Answer 9

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Answer 10

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Answer 11

Ch. 10 - Prob. 10.1P Ch. 10 - Prob. 10.2P Ch. 10 - Prob. 10.3P Ch. 10 - What is the ideal work for the separation of an...Ch. 10 - Prob. 10.5P Ch. 10 - Prob. 10.6P Ch. 10 - Prob. 10.7P Ch. 10 - Prob. 10.8P Ch. 10 - Prob. 10.9P Ch. 10 - Prob. 10.10P

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