The reason for straight line in the µ versus T segments for the three phases compare to curvature needs to be explained. Also, the curvature upward or downward on a µ versus T plot needs to be explained. Concept Introduction: All substances can mainly exist in three phases; solid, liquid and gases. These three phases can convert into each other by the application of temperature and pressure such as by heating of solid, it converts to liquid and then gaseous state of matter . Similarly, the gases can be compressed at high pressure to liquefy. The interconversion of three phases of substance can be show on the phase diagram.

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Answer 1

Question

Chapter 8, Problem 8.1CP

Interpretation Introduction

Interpretation:The reason for straight line in the µ versus T segments for the three phases compare to curvature needs to be explained. Also, the curvature upward or downward on a µ versus T plot needs to be explained.

Concept Introduction: All substances can mainly exist in three phases; solid, liquid and gases. These three phases can convert into each other by the application of temperature and pressure such as by heating of solid, it converts to liquid and then gaseous state of matter.

Similarly, the gases can be compressed at high pressure to liquefy. The interconversion of three phases of substance can be show on the phase diagram.

Expert Solution & Answer

Explanation of Solution

The relation between chemical potential and T with entropy at constant pressure can be written as:

$( ∂μ ∂T)P= - S$

Since the entropy change with increase in T is very less therefore it is taken as constant in normal experimental conditions and this is the reason to plot the lines as a straight lines in the three phases segaments.

$( ∂ 2 μ ∂ T 2 )P= - (∂S∂T)$

Since $(∂S∂T)>0$ and $( ∂ 2 μ ∂ T 2 )P<0$ therefore the curvature must be downwards.

Conclusion

Since the entropy change with increase in T is very less therefore it is taken as constant in normal experimental conditions and this is the reason to plot the lines as a straight lines in the three phases segaments and the curvature must be downwards.

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Answer 2

Question

Chapter 8, Problem 8.1CP

Interpretation Introduction

Interpretation:The reason for straight line in the µ versus T segments for the three phases compare to curvature needs to be explained. Also, the curvature upward or downward on a µ versus T plot needs to be explained.

Concept Introduction: All substances can mainly exist in three phases; solid, liquid and gases. These three phases can convert into each other by the application of temperature and pressure such as by heating of solid, it converts to liquid and then gaseous state of matter.

Similarly, the gases can be compressed at high pressure to liquefy. The interconversion of three phases of substance can be show on the phase diagram.

Expert Solution & Answer

Explanation of Solution

The relation between chemical potential and T with entropy at constant pressure can be written as:

$( ∂μ ∂T)P= - S$

Since the entropy change with increase in T is very less therefore it is taken as constant in normal experimental conditions and this is the reason to plot the lines as a straight lines in the three phases segaments.

$( ∂ 2 μ ∂ T 2 )P= - (∂S∂T)$

Since $(∂S∂T)>0$ and $( ∂ 2 μ ∂ T 2 )P<0$ therefore the curvature must be downwards.

Conclusion

Since the entropy change with increase in T is very less therefore it is taken as constant in normal experimental conditions and this is the reason to plot the lines as a straight lines in the three phases segaments and the curvature must be downwards.

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Answer 3

Question

Chapter 8, Problem 8.1CP

Interpretation Introduction

Interpretation:The reason for straight line in the µ versus T segments for the three phases compare to curvature needs to be explained. Also, the curvature upward or downward on a µ versus T plot needs to be explained.

Concept Introduction: All substances can mainly exist in three phases; solid, liquid and gases. These three phases can convert into each other by the application of temperature and pressure such as by heating of solid, it converts to liquid and then gaseous state of matter.

Similarly, the gases can be compressed at high pressure to liquefy. The interconversion of three phases of substance can be show on the phase diagram.

Expert Solution & Answer

Explanation of Solution

The relation between chemical potential and T with entropy at constant pressure can be written as:

$( ∂μ ∂T)P= - S$

Since the entropy change with increase in T is very less therefore it is taken as constant in normal experimental conditions and this is the reason to plot the lines as a straight lines in the three phases segaments.

$( ∂ 2 μ ∂ T 2 )P= - (∂S∂T)$

Since $(∂S∂T)>0$ and $( ∂ 2 μ ∂ T 2 )P<0$ therefore the curvature must be downwards.

Conclusion

Since the entropy change with increase in T is very less therefore it is taken as constant in normal experimental conditions and this is the reason to plot the lines as a straight lines in the three phases segaments and the curvature must be downwards.

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Answer 4

Question

Chapter 8, Problem 8.1CP

Interpretation Introduction

Interpretation:The reason for straight line in the µ versus T segments for the three phases compare to curvature needs to be explained. Also, the curvature upward or downward on a µ versus T plot needs to be explained.

Concept Introduction: All substances can mainly exist in three phases; solid, liquid and gases. These three phases can convert into each other by the application of temperature and pressure such as by heating of solid, it converts to liquid and then gaseous state of matter.

Similarly, the gases can be compressed at high pressure to liquefy. The interconversion of three phases of substance can be show on the phase diagram.

Expert Solution & Answer

Explanation of Solution

The relation between chemical potential and T with entropy at constant pressure can be written as:

$( ∂μ ∂T)P= - S$

Since the entropy change with increase in T is very less therefore it is taken as constant in normal experimental conditions and this is the reason to plot the lines as a straight lines in the three phases segaments.

$( ∂ 2 μ ∂ T 2 )P= - (∂S∂T)$

Since $(∂S∂T)>0$ and $( ∂ 2 μ ∂ T 2 )P<0$ therefore the curvature must be downwards.

Conclusion

Since the entropy change with increase in T is very less therefore it is taken as constant in normal experimental conditions and this is the reason to plot the lines as a straight lines in the three phases segaments and the curvature must be downwards.

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Answer 5

Chapter 8, Problem 8.1CP

Interpretation Introduction

Interpretation:The reason for straight line in the µ versus T segments for the three phases compare to curvature needs to be explained. Also, the curvature upward or downward on a µ versus T plot needs to be explained.

Concept Introduction: All substances can mainly exist in three phases; solid, liquid and gases. These three phases can convert into each other by the application of temperature and pressure such as by heating of solid, it converts to liquid and then gaseous state of matter.

Similarly, the gases can be compressed at high pressure to liquefy. The interconversion of three phases of substance can be show on the phase diagram.

Expert Solution & Answer

Explanation of Solution

The relation between chemical potential and T with entropy at constant pressure can be written as:

$( ∂μ ∂T)P= - S$

Since the entropy change with increase in T is very less therefore it is taken as constant in normal experimental conditions and this is the reason to plot the lines as a straight lines in the three phases segaments.

$( ∂ 2 μ ∂ T 2 )P= - (∂S∂T)$

Since $(∂S∂T)>0$ and $( ∂ 2 μ ∂ T 2 )P<0$ therefore the curvature must be downwards.

Conclusion

Since the entropy change with increase in T is very less therefore it is taken as constant in normal experimental conditions and this is the reason to plot the lines as a straight lines in the three phases segaments and the curvature must be downwards.

Answer 6

Chapter 8, Problem 8.1CP

Interpretation Introduction

Interpretation:The reason for straight line in the µ versus T segments for the three phases compare to curvature needs to be explained. Also, the curvature upward or downward on a µ versus T plot needs to be explained.

Concept Introduction: All substances can mainly exist in three phases; solid, liquid and gases. These three phases can convert into each other by the application of temperature and pressure such as by heating of solid, it converts to liquid and then gaseous state of matter.

Similarly, the gases can be compressed at high pressure to liquefy. The interconversion of three phases of substance can be show on the phase diagram.

Expert Solution & Answer

Explanation of Solution

The relation between chemical potential and T with entropy at constant pressure can be written as:

$( ∂μ ∂T)P= - S$

Since the entropy change with increase in T is very less therefore it is taken as constant in normal experimental conditions and this is the reason to plot the lines as a straight lines in the three phases segaments.

$( ∂ 2 μ ∂ T 2 )P= - (∂S∂T)$

Since $(∂S∂T)>0$ and $( ∂ 2 μ ∂ T 2 )P<0$ therefore the curvature must be downwards.

Conclusion

Since the entropy change with increase in T is very less therefore it is taken as constant in normal experimental conditions and this is the reason to plot the lines as a straight lines in the three phases segaments and the curvature must be downwards.

Answer 7

Chapter 8, Problem 8.1CP

Interpretation Introduction

Interpretation:The reason for straight line in the µ versus T segments for the three phases compare to curvature needs to be explained. Also, the curvature upward or downward on a µ versus T plot needs to be explained.

Concept Introduction: All substances can mainly exist in three phases; solid, liquid and gases. These three phases can convert into each other by the application of temperature and pressure such as by heating of solid, it converts to liquid and then gaseous state of matter.

Similarly, the gases can be compressed at high pressure to liquefy. The interconversion of three phases of substance can be show on the phase diagram.

Answer 8

Expert Solution & Answer

Explanation of Solution

The relation between chemical potential and T with entropy at constant pressure can be written as:

$( ∂μ ∂T)P= - S$

Since the entropy change with increase in T is very less therefore it is taken as constant in normal experimental conditions and this is the reason to plot the lines as a straight lines in the three phases segaments.

$( ∂ 2 μ ∂ T 2 )P= - (∂S∂T)$

Since $(∂S∂T)>0$ and $( ∂ 2 μ ∂ T 2 )P<0$ therefore the curvature must be downwards.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Conclusion

Since the entropy change with increase in T is very less therefore it is taken as constant in normal experimental conditions and this is the reason to plot the lines as a straight lines in the three phases segaments and the curvature must be downwards.

Answer 12

Ch. 8 - Prob. 8.1CP Ch. 8 - Prob. 8.2CP Ch. 8 - Prob. 8.3CP Ch. 8 - Prob. 8.4CP Ch. 8 - Prob. 8.5CP Ch. 8 - Prob. 8.6CP Ch. 8 - Prob. 8.7CP Ch. 8 - Prob. 8.8CP Ch. 8 - Prob. 8.9CP Ch. 8 - Prob. 8.10CP

Explanation of Solution

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Chapter 8 Solutions