The change in entropy observed on mixing equal volumes of Ar gas in two different valves is to be stated. Whether the value of Δ S remains same if one of the volumes possesses Ar and another one possesses Ne or not is to be stated. Concept introduction: Mixing of two distinct gases generally cause a rise in system’s entropy. The corresponding reason is the higher microstates available for the gases in mixed system as compared to unmixed system. However, if the gases being mixed are same, then no entropy change is observed. The entropy change encountered on mixing two distinct gases is calculated by the formula shown. Δ S = − 2.303 R ( n A log x A + n B log x B )

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Answer 1

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Chapter 5, Problem 5.18CP

Interpretation Introduction

Interpretation: The change in entropy observed on mixing equal volumes of Ar gas in two different valves is to be stated. Whether the value of ΔS remains same if one of the volumes possesses Ar and another one possesses Ne or not is to be stated.

Concept introduction: Mixing of two distinct gases generally cause a rise in system’s entropy. The corresponding reason is the higher microstates available for the gases in mixed system as compared to unmixed system. However, if the gases being mixed are same, then no entropy change is observed. The entropy change encountered on mixing two distinct gases is calculated by the formula shown.

ΔS=−2.303R(nAlogxA+nBlogxB)

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Answer 2

Question

Chapter 5, Problem 5.18CP

Interpretation Introduction

Interpretation: The change in entropy observed on mixing equal volumes of Ar gas in two different valves is to be stated. Whether the value of ΔS remains same if one of the volumes possesses Ar and another one possesses Ne or not is to be stated.

Concept introduction: Mixing of two distinct gases generally cause a rise in system’s entropy. The corresponding reason is the higher microstates available for the gases in mixed system as compared to unmixed system. However, if the gases being mixed are same, then no entropy change is observed. The entropy change encountered on mixing two distinct gases is calculated by the formula shown.

ΔS=−2.303R(nAlogxA+nBlogxB)

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Answer 3

Question

Chapter 5, Problem 5.18CP

Interpretation Introduction

Interpretation: The change in entropy observed on mixing equal volumes of Ar gas in two different valves is to be stated. Whether the value of ΔS remains same if one of the volumes possesses Ar and another one possesses Ne or not is to be stated.

Concept introduction: Mixing of two distinct gases generally cause a rise in system’s entropy. The corresponding reason is the higher microstates available for the gases in mixed system as compared to unmixed system. However, if the gases being mixed are same, then no entropy change is observed. The entropy change encountered on mixing two distinct gases is calculated by the formula shown.

ΔS=−2.303R(nAlogxA+nBlogxB)

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Answer 4

Question

Chapter 5, Problem 5.18CP

Interpretation Introduction

Interpretation: The change in entropy observed on mixing equal volumes of Ar gas in two different valves is to be stated. Whether the value of ΔS remains same if one of the volumes possesses Ar and another one possesses Ne or not is to be stated.

Concept introduction: Mixing of two distinct gases generally cause a rise in system’s entropy. The corresponding reason is the higher microstates available for the gases in mixed system as compared to unmixed system. However, if the gases being mixed are same, then no entropy change is observed. The entropy change encountered on mixing two distinct gases is calculated by the formula shown.

ΔS=−2.303R(nAlogxA+nBlogxB)

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Answer 5

Chapter 5, Problem 5.18CP

Interpretation Introduction

Interpretation: The change in entropy observed on mixing equal volumes of Ar gas in two different valves is to be stated. Whether the value of ΔS remains same if one of the volumes possesses Ar and another one possesses Ne or not is to be stated.

Concept introduction: Mixing of two distinct gases generally cause a rise in system’s entropy. The corresponding reason is the higher microstates available for the gases in mixed system as compared to unmixed system. However, if the gases being mixed are same, then no entropy change is observed. The entropy change encountered on mixing two distinct gases is calculated by the formula shown.

ΔS=−2.303R(nAlogxA+nBlogxB)

Answer 6

Chapter 5, Problem 5.18CP

Interpretation Introduction

Interpretation: The change in entropy observed on mixing equal volumes of Ar gas in two different valves is to be stated. Whether the value of ΔS remains same if one of the volumes possesses Ar and another one possesses Ne or not is to be stated.

Concept introduction: Mixing of two distinct gases generally cause a rise in system’s entropy. The corresponding reason is the higher microstates available for the gases in mixed system as compared to unmixed system. However, if the gases being mixed are same, then no entropy change is observed. The entropy change encountered on mixing two distinct gases is calculated by the formula shown.

ΔS=−2.303R(nAlogxA+nBlogxB)

Answer 7

Chapter 5, Problem 5.18CP

Interpretation Introduction

Interpretation: The change in entropy observed on mixing equal volumes of Ar gas in two different valves is to be stated. Whether the value of ΔS remains same if one of the volumes possesses Ar and another one possesses Ne or not is to be stated.

Concept introduction: Mixing of two distinct gases generally cause a rise in system’s entropy. The corresponding reason is the higher microstates available for the gases in mixed system as compared to unmixed system. However, if the gases being mixed are same, then no entropy change is observed. The entropy change encountered on mixing two distinct gases is calculated by the formula shown.

ΔS=−2.303R(nAlogxA+nBlogxB)

Answer 8

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Answer 9

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Answer 10

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Answer 11

Ch. 5 - Prob. 5.1CP Ch. 5 - Prob. 5.2CP Ch. 5 - Prob. 5.3CP Ch. 5 - Prob. 5.4CP Ch. 5 - Prob. 5.5CP Ch. 5 - Prob. 5.6CP Ch. 5 - Prob. 5.7CP Ch. 5 - Prob. 5.8CP Ch. 5 - Prob. 5.9CP Ch. 5 - Prob. 5.10CP

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