Essentials Of Materials Science And Engineering
Essentials Of Materials Science And Engineering
4th Edition
ISBN: 9781337670845
Author: ASKELAND
Publisher: Cengage
Question
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Chapter 10, Problem 10.53P
Interpretation Introduction

(a)

Interpretation:

The composition of the solid phase and the liquid phase in mol% and wt% are to be calculated for NiO-20 mol% MgO at 2200C.

Concept Introduction:

On the temperature-composition graph of a ceramic, the curve above which the ceramic exist in the liquid phase is the liquidus curve. The temperature at this curve is the maximum temperature at which the crystals in the ceramic can coexist with its melt in the thermodynamic equilibrium.

Solidus curve is the locus of the temperature on the temperature composition graph of a ceramic, beyond which the ceramic is completely in solid phase.

Between the solidus and liquidus curve, the ceramic exits in a slurry form in which there is both crystals as well as ceramic melt.

Solidus temperature is always less than or equal to the liquidus temperature.

The formula to calculate the wt% from the mol% for a ceramic containing phases 1 and 2 is:

  (wt%)1=[( mol%)1×( M 1)][( mol%)1×( M 1)+( mol%)2×( M 2)]×100 ...... (1)

Here, (mol%)1 and (mol%)2 are the mole percent of the components present in the ceramic, and M1 and M2 are the molecular weights of the components.

Expert Solution
Check Mark

Answer to Problem 10.53P

Composition of the liquid phase in mol% is 15 mol% MgO.

Composition of the liquid phase in wt% is 8.69 mol% MgO.

Composition of the solid phase in mol% is 36 mol% MgO.

Composition of the solid phase in wt% is 23.28 mol% MgO.

Explanation of Solution

The phase diagram for NiO-MgO is given as:

Essentials Of Materials Science And Engineering, Chapter 10, Problem 10.53P , additional homework tip  1

Now, draw a straight line from temperature 2200C to 20 mol% MgO as:

Essentials Of Materials Science And Engineering, Chapter 10, Problem 10.53P , additional homework tip  2

Here, point 'a' represents NiO-20 mol% MgO at 2200C. Both the phases, solid and liquid are present at this condition. Point 'b' represents the liquid phase composition in mol% and point 'c' represents the solid phase composition in mol%. From the above phase diagram:

  L=15 mol% MgOS=36 mol% MgO

Molecular weight of NiO and MgO are 74.71 g/mol and 40.312 g/mol respectively.

Use equation (1) to convert mol% to wt% for liquid phase as:

  (wt%)MgO=[ ( mol% ) MgO×( M MgO )][ ( mol% ) MgO×( M MgO )+ ( mol% ) NiO×( M NiO )]×100=[15×40.312][15×40.312+85×74.71]×100=8.69 wt% MgO

Again, use equation (1) to convert mol% to wt% for solid phase as:

  (wt%)MgO=[ ( mol% ) MgO×( M MgO )][ ( mol% ) MgO×( M MgO )+ ( mol% ) NiO×( M NiO )]×100=[36×40.312][36×40.312+64×74.71]×100=23.28 wt% MgO

Interpretation Introduction

(b)

Interpretation:

The amount of each phase present in NiO-20 mol% MgO at 2200C in mol% and wt% are to be calculated.

Concept Introduction:

A matter can exist in different physical forms such as sold, liquid, gas, and plasma. These distinct physical forms are known as a Phase.

A phase has uniform physical and chemical properties and is bounded by a surface due to which two phases can be mechanically separated from each other.

The formula to calculate the wt% from the mol% for a ceramic containing phases 1 and 2 is:

  (wt%)1=[( mol%)1×( M 1)][( mol%)1×( M 1)+( mol%)2×( M 2)]×100 ...... (1)

Here, (mol%)1 and (mol%)2 are the mole percent of the components present in the ceramic, and M1 and M2 are the molecular weights of the components.

Amount of each phase in mol% is calculated using lever rule. At a particular temperature and ceramic composition, a tie line is drawn on the phase diagram of the ceramic between the solidus and liquidus curve. Then the portion of the lever opposite to the phase whose amount is to be calculated is considered in the formula used as:

  Phase mol%=opposite arm of levertotal length of the tie line×100 ...... (2)

Expert Solution
Check Mark

Answer to Problem 10.53P

Amount of liquid phase in mol% is 76.19 mol%.

Amount of liquid phase in wt% is 78.07 wt%.

Amount of solid phase in mol% is 23.81 mol%.

Amount of solid phase in wt% is 21.93 wt%.

Explanation of Solution

The phase diagram for NiO-MgO is given as:

Essentials Of Materials Science And Engineering, Chapter 10, Problem 10.53P , additional homework tip  3

Now, draw a straight line from temperature 2200C to 20 mol% MgO as:

Essentials Of Materials Science And Engineering, Chapter 10, Problem 10.53P , additional homework tip  4

Here, point 'a' represents NiO-20 mol% MgO at 2200C. Both the phases, solid and liquid are present at this condition. Point 'b' represents the liquid phase composition in mol% and point 'c' represents the solid phase composition in mol%. From the above phase diagram:

  L=15 mol% MgOS=36 mol% MgO

To calculate amount of liquid phase, lever 'ac' will be used and to calculate amount of solid phase, lever 'ba' will be used. Use equation (2) to calculate the amount of each phase as:

  Liquid mol%=length of 'ac'length of 'bc'×100=36203615×100=76.19 mol%Solid mol%=length of 'ba'length of 'bc'×100=20153615×100=23.81 mol%

To calculate the amount of liquid and solid phases in wt%, first convert the original mol% of MgO in wt% using equation (1) and molecular weights of NiO and MgO as:

  (wt%)MgO=[ ( mol% ) MgO×( M MgO )][ ( mol% ) MgO×( M MgO )+ ( mol% ) NiO×( M NiO )]×100=[20×40.312][20×40.312+80×74.71]×100=11.89 wt% MgO

To apply the lever rule, use the corresponding wt% for the liquid and solid phases as calculated in part (a) as:

  L=8.69 wt% MgOS=23.28 wt% MgO

Apply lever rule as:

  Liquid wt%=length of 'ac'length of 'bc'×100=23.2811.8923.288.69×100=78.07 wt%Solid wt%=length of 'ba'length of 'bc'×100=11.898.6923.288.69×100=21.93 wt%

Interpretation Introduction

(c)

Interpretation:

The amount of each phase is to be calculated in vol%.

Concept Introduction:

The formula to convert wt% to vol% using density (ρ) is:

  (Vol%)1=[ ( wt% ) 1 ρ 1][ ( wt% ) 1 ρ 1+ ( wt% ) 2 ρ 2]×100 ...... (3)

Expert Solution
Check Mark

Answer to Problem 10.53P

The amount of liquid phase in vol% is 75.91 Vol%.

The amount of solid phase in vol% is 24.09 Vol%.

Explanation of Solution

Given information:

A ceramic containing NiO-20 mol% MgO is heated to 2200C. The solids present in it at this point has density 6.32 g/cm3 and the density of the liquid is 7.14 g/cm3.

From part (b), the amount of liquid and solid phases in wt% is calculated as:

  L=78.07 wt%S=21.93 wt%

Use equation (3) along with the given densities of the phases to calculate the vol% as:

  (Vol%)L=[ ( wt% ) L ρ L ][ ( wt% ) L ρ L + ( wt% ) S ρ S ]×100=[ 78.07 7.14][ 78.07 7.14+ 21.93 6.32]×100=75.91 Vol%(Vol%)S=[ ( wt% ) S ρ S ][ ( wt% ) L ρ L + ( wt% ) S ρ S ]×100=[ 21.93 6.32][ 78.07 7.14+ 21.93 6.32]×100=24.09 Vol%

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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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