Essentials Of Materials Science And Engineering
Essentials Of Materials Science And Engineering
4th Edition
ISBN: 9781337385497
Author: WRIGHT, Wendelin J.
Publisher: Cengage,
Question
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Chapter 10, Problem 10.41P
Interpretation Introduction

(a)

Interpretation:

Degree of freedom for Tl-20 wt% Pb is to be determined at 325C and 400C.

Concept Introduction:

The Gibbs phase rule provides the relation between number of chemically independent component and the number of independent phases for a given system. When temperature and pressure of the system can be changed, it is defined as:

  2+C=F+P

Here, C is used for the number of chemically independent variables. This means that number of different elements or compounds which are required to specify the given system.

  F is the degree of freedom which are number of variables which are allowed to change in the given system without altering its number of phases. In this case, temperature and pressure are the two variables which are assumed to change which is represented in the equation by the constant "2". If pressure is constant for any system, then instead of " 2 ", use " 1 " is used in the above formula.

  P is the number of phases present in the given system which are in equilibrium which each other.

Expert Solution
Check Mark

Answer to Problem 10.41P

For temperature 325C, F=1.

For temperature 400C, F=2.

Explanation of Solution

The phase diagram given for the Tl-Pb solution is given as:

Essentials Of Materials Science And Engineering, Chapter 10, Problem 10.41P , additional homework tip  1

For Tl-20 wt% Pb, the straight line from 20 wt% is drawn until temperature 325C and 400C as shown. At the point 'a' representing temperature 325C, two phases are present which are solid solution and liquid. At point 'b', representing temperature 400C, only one phase is present which is liquid.

Essentials Of Materials Science And Engineering, Chapter 10, Problem 10.41P , additional homework tip  2

For the given system, the pressure is assumed to be constant. Thus, the formula used to calculate degree of freedom is:

  1+C=F+P ......... (1)

For temperature 325C ,

  C=2P=2

Apply equation (1) and calculate degree of freedom as:

  1+C=F+P1+2=F+2F=1

For temperature 400C ,

  C=2P=1

Apply equation (1) and calculate degree of freedom as:

  1+C=F+P1+2=F+1F=2

Interpretation Introduction

(b)

Interpretation:

Degree of freedom for Tl-40 wt% Pb is to be determined at 325C and 400C.

Concept Introduction:

The Gibbs phase rule provides the relation between number of chemically independent component and the number of independent phases for a given system. When temperature and pressure of the system can be changed, it is defined as:

  2+C=F+P

Here, C is used for the number of chemically independent variables. This means that number of different elements or compounds which are required to specify the given system.

  F is the degree of freedom which are number of variables which are allowed to change in the given system without altering its number of phases. In this case, temperature and pressure are the two variables which are assumed to change which is represented in the equation by the constant "2". If pressure is constant for any system, then instead of " 2 ", use " 1 " is used in the above formula.

  P is the number of phases present in the given system which are in equilibrium which each other.

Expert Solution
Check Mark

Answer to Problem 10.41P

For temperature 325C, F=2.

For temperature 400C, F=2.

Explanation of Solution

The phase diagram given for the Tl-Pb solution is given as:

Essentials Of Materials Science And Engineering, Chapter 10, Problem 10.41P , additional homework tip  3

For Tl-40 wt% Pb, the straight line from 40 wt% is drawn until temperature 325C and 400C as shown. At the point 'a' representing temperature 325C, only one phase is present which is solid. At point 'b', representing temperature 400C, only one phase is present which is liquid.

Essentials Of Materials Science And Engineering, Chapter 10, Problem 10.41P , additional homework tip  4

For the given system, the pressure is assumed to be constant. Thus, the formula used to calculate degree of freedom is:

  1+C=F+P ......... (1)

For temperature 325C ,

  C=2P=1

Apply equation (1) and calculate degree of freedom as:

  1+C=F+P1+2=F+1F=2

For temperature 400C ,

  C=2P=1

Apply equation (1) and calculate degree of freedom as:

  1+C=F+P1+2=F+1F=2

Interpretation Introduction

(c)

Interpretation:

Degree of freedom for Tl-90 wt% Pb is to be determined at 325C and 400C.

Concept Introduction:

The Gibbs phase rule provides the relation between number of chemically independent component and the number of independent phases for a given system. When temperature and pressure of the system can be changed, it is defined as:

  2+C=F+P

Here, C is used for the number of chemically independent variables. This means that number of different elements or compounds which are required to specify the given system.

  F is the degree of freedom which are number of variables which are allowed to change in the given system without altering its number of phases. In this case, temperature and pressure are the two variables which are assumed to change which is represented in the equation by the constant "2". If pressure is constant for any system, then instead of " 2 ", use " 1 " is used in the above formula.

  P is the number of phases present in the given system which are in equilibrium which each other.

Expert Solution
Check Mark

Answer to Problem 10.41P

For temperature 325C, F=1.

For temperature 400C, F=2.

Explanation of Solution

The phase diagram given for the Tl-Pb solution is given as:

Essentials Of Materials Science And Engineering, Chapter 10, Problem 10.41P , additional homework tip  5

For Tl-90 wt% Pb, the straight line from 90 wt% is drawn until temperature 325C and 400C as shown. At the point 'a' representing temperature 325C, two phases are present which are solid solution and liquid. At point 'b', representing temperature 400C, only one phase is present which is liquid.

Essentials Of Materials Science And Engineering, Chapter 10, Problem 10.41P , additional homework tip  6

For the given system, the pressure is assumed to be constant. Thus, the formula used to calculate degree of freedom is:

  1+C=F+P ......... (1)

For temperature 325C ,

  C=2P=2

Apply equation (1) and calculate degree of freedom as:

  1+C=F+P1+2=F+2F=1

For temperature 400C ,

  C=2P=1

Apply equation (1) and calculate degree of freedom as:

  1+C=F+P1+2=F+1F=2

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Chapter 10 Solutions

Essentials Of Materials Science And Engineering

Ch. 10 - Prob. 10.11PCh. 10 - Prob. 10.12PCh. 10 - Prob. 10.13PCh. 10 - Prob. 10.14PCh. 10 - Prob. 10.15PCh. 10 - Prob. 10.16PCh. 10 - Prob. 10.17PCh. 10 - Prob. 10.18PCh. 10 - Prob. 10.19PCh. 10 - Prob. 10.20PCh. 10 - Prob. 10.21PCh. 10 - Prob. 10.22PCh. 10 - Prob. 10.23PCh. 10 - Prob. 10.24PCh. 10 - Prob. 10.25PCh. 10 - Prob. 10.26PCh. 10 - Prob. 10.27PCh. 10 - Prob. 10.28PCh. 10 - Prob. 10.29PCh. 10 - Prob. 10.30PCh. 10 - Prob. 10.31PCh. 10 - Prob. 10.32PCh. 10 - Prob. 10.33PCh. 10 - Prob. 10.34PCh. 10 - Prob. 10.35PCh. 10 - Prob. 10.36PCh. 10 - Prob. 10.37PCh. 10 - Prob. 10.38PCh. 10 - Prob. 10.39PCh. 10 - Prob. 10.40PCh. 10 - Prob. 10.41PCh. 10 - Prob. 10.42PCh. 10 - Prob. 10.43PCh. 10 - Prob. 10.44PCh. 10 - Prob. 10.45PCh. 10 - Prob. 10.46PCh. 10 - Prob. 10.47PCh. 10 - Prob. 10.48PCh. 10 - Prob. 10.49PCh. 10 - Prob. 10.50PCh. 10 - Prob. 10.51PCh. 10 - Prob. 10.52PCh. 10 - Prob. 10.53PCh. 10 - Prob. 10.54PCh. 10 - Prob. 10.55PCh. 10 - Prob. 10.56PCh. 10 - Prob. 10.57PCh. 10 - Prob. 10.58PCh. 10 - Prob. 10.59PCh. 10 - Prob. 10.60PCh. 10 - Prob. 10.61PCh. 10 - Prob. 10.62PCh. 10 - Prob. 10.63PCh. 10 - Prob. 10.64PCh. 10 - Prob. 10.65PCh. 10 - Prob. 10.66PCh. 10 - Prob. 10.67PCh. 10 - Prob. 10.68PCh. 10 - Prob. 10.69PCh. 10 - Prob. 10.70PCh. 10 - Prob. 10.71PCh. 10 - Prob. 10.72PCh. 10 - Prob. 10.73PCh. 10 - Prob. 10.74PCh. 10 - Prob. 10.75PCh. 10 - Prob. 10.76PCh. 10 - Prob. 10.77PCh. 10 - Prob. 10.78PCh. 10 - Prob. 10.79PCh. 10 - Prob. 10.80PCh. 10 - Prob. 10.81PCh. 10 - Prob. 10.82PCh. 10 - Prob. 10.83PCh. 10 - Prob. 10.84PCh. 10 - Prob. 10.85PCh. 10 - Prob. 10.86PCh. 10 - Prob. 10.87PCh. 10 - Prob. 10.88DPCh. 10 - Prob. 10.89DPCh. 10 - Prob. 10.90DPCh. 10 - Prob. 10.91DPCh. 10 - Prob. 10.92CPCh. 10 - Prob. 10.93CPCh. 10 - Prob. 10.94CPCh. 10 - Prob. K10.1KP
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