The rate of increase in entropy should be determined for the mixture of nitrogen and hydrogen gases flowing at given rates. The gases should be assumed to be ideal and they mix adiabatically in a steady-flow process. Concept introduction: Molar rate of flow is equal to the ratio of mass flow rate to the molar mass of the compound. Now, entropy of mixing of molar flows should be calculated with the formula as: Δ S m i x = − R × t o t a l m o l a r f l o w [ y 1 ln ( y 1 ) + y 2 ln ( y 2 ) ] Where, y 1 and y 2 are molar flow fractions of the two gases in respective vessels.

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Answer 1

Question

Chapter 10, Problem 10.3P

Interpretation Introduction

Interpretation:

The rate of increase in entropy should be determined for the mixture of nitrogen and hydrogen gases flowing at given rates. The gases should be assumed to be ideal and they mix adiabatically in a steady-flow process.

Concept introduction:

Molar rate of flow is equal to the ratio of mass flow rate to the molar mass of the compound.

Now, entropy of mixing of molar flows should be calculated with the formula as:

ΔSmix=−R×total molar flow[y1ln(y1)+y2ln(y2)]

Where, y₁ and y₂are molar flow fractions of the two gases in respective vessels.

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Answer 2

Question

Chapter 10, Problem 10.3P

Interpretation Introduction

Interpretation:

The rate of increase in entropy should be determined for the mixture of nitrogen and hydrogen gases flowing at given rates. The gases should be assumed to be ideal and they mix adiabatically in a steady-flow process.

Concept introduction:

Molar rate of flow is equal to the ratio of mass flow rate to the molar mass of the compound.

Now, entropy of mixing of molar flows should be calculated with the formula as:

ΔSmix=−R×total molar flow[y1ln(y1)+y2ln(y2)]

Where, y₁ and y₂are molar flow fractions of the two gases in respective vessels.

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Answer 3

Question

Chapter 10, Problem 10.3P

Interpretation Introduction

Interpretation:

The rate of increase in entropy should be determined for the mixture of nitrogen and hydrogen gases flowing at given rates. The gases should be assumed to be ideal and they mix adiabatically in a steady-flow process.

Concept introduction:

Molar rate of flow is equal to the ratio of mass flow rate to the molar mass of the compound.

Now, entropy of mixing of molar flows should be calculated with the formula as:

ΔSmix=−R×total molar flow[y1ln(y1)+y2ln(y2)]

Where, y₁ and y₂are molar flow fractions of the two gases in respective vessels.

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Answer 4

Question

Chapter 10, Problem 10.3P

Interpretation Introduction

Interpretation:

The rate of increase in entropy should be determined for the mixture of nitrogen and hydrogen gases flowing at given rates. The gases should be assumed to be ideal and they mix adiabatically in a steady-flow process.

Concept introduction:

Molar rate of flow is equal to the ratio of mass flow rate to the molar mass of the compound.

Now, entropy of mixing of molar flows should be calculated with the formula as:

ΔSmix=−R×total molar flow[y1ln(y1)+y2ln(y2)]

Where, y₁ and y₂are molar flow fractions of the two gases in respective vessels.

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Answer 5

Chapter 10, Problem 10.3P

Interpretation Introduction

Interpretation:

The rate of increase in entropy should be determined for the mixture of nitrogen and hydrogen gases flowing at given rates. The gases should be assumed to be ideal and they mix adiabatically in a steady-flow process.

Concept introduction:

Molar rate of flow is equal to the ratio of mass flow rate to the molar mass of the compound.

Now, entropy of mixing of molar flows should be calculated with the formula as:

ΔSmix=−R×total molar flow[y1ln(y1)+y2ln(y2)]

Where, y₁ and y₂are molar flow fractions of the two gases in respective vessels.

Answer 6

Chapter 10, Problem 10.3P

Interpretation Introduction

Interpretation:

The rate of increase in entropy should be determined for the mixture of nitrogen and hydrogen gases flowing at given rates. The gases should be assumed to be ideal and they mix adiabatically in a steady-flow process.

Concept introduction:

Molar rate of flow is equal to the ratio of mass flow rate to the molar mass of the compound.

Now, entropy of mixing of molar flows should be calculated with the formula as:

ΔSmix=−R×total molar flow[y1ln(y1)+y2ln(y2)]

Where, y₁ and y₂are molar flow fractions of the two gases in respective vessels.

Answer 7

Chapter 10, Problem 10.3P

Interpretation Introduction

Interpretation:

The rate of increase in entropy should be determined for the mixture of nitrogen and hydrogen gases flowing at given rates. The gases should be assumed to be ideal and they mix adiabatically in a steady-flow process.

Concept introduction:

Molar rate of flow is equal to the ratio of mass flow rate to the molar mass of the compound.

Now, entropy of mixing of molar flows should be calculated with the formula as:

ΔSmix=−R×total molar flow[y1ln(y1)+y2ln(y2)]

Where, y₁ and y₂are molar flow fractions of the two gases in respective vessels.

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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Chapter 10 Solutions