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 $\text{NiO-20 mol\% MgO}$ at ${2200}^{\circ }\text{C}$.

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 $1\text{ and }2$ is:

  ${\left(\text{wt\%}\right)}_1=\frac{\left[{\left(\text{mol\%}\right)}_1\times \left({\text{M}}_1\right)\right]}{\left[{\left(\text{mol\%}\right)}_1\times \left({\text{M}}_1\right)+{\left(\text{mol\%}\right)}_2\times \left({\text{M}}_2\right)\right]}\times 100$ ...... (1)

Here, ${\left(\text{mol\%}\right)}_1\text{ and }{\left(\text{mol\%}\right)}_2$ are the mole percent of the components present in the ceramic, and ${\text{M}}_1{\text{ and M}}_2$ are the molecular weights of the components.

Answer to Problem 10.53P

Composition of the liquid phase in mol% is $15\text{ mol\% MgO}$.

Composition of the liquid phase in wt% is $8.69\text{ mol\% MgO}$.

Composition of the solid phase in mol% is $36\text{ mol\% MgO}$.

Composition of the solid phase in wt% is $23.28\text{ mol\% MgO}$.

Explanation of Solution

The phase diagram for NiO-MgO is given as:

Now, draw a straight line from temperature ${2200}^{\circ }\text{C}$ to $20\text{ mol\% MgO}$ as:

Here, point 'a' represents $\text{NiO-}20\text{ mol\% MgO}$ at ${2200}^{\circ }\text{C}$. 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:

  $\begin{array}{c}L=15\text{ mol\% MgO}\\ S=36\text{ mol\% MgO}\end{array}$

Molecular weight of NiO and MgO are $74.71\text{ g/mol and 40}\text{.312 g/mol}$ respectively.

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

  $\begin{array}{c}{\left(\text{wt\%}\right)}_{\text{MgO}}=\frac{\left[{\left(\text{mol\%}\right)}_{\text{MgO}}\times \left({\text{M}}_{\text{MgO}}\right)\right]}{\left[{\left(\text{mol\%}\right)}_{\text{MgO}}\times \left({\text{M}}_{\text{MgO}}\right)+{\left(\text{mol\%}\right)}_{\text{NiO}}\times \left({\text{M}}_{\text{NiO}}\right)\right]}\times 100\\ =\frac{\left[15\times 40.312\right]}{\left[15\times 40.312+85\times 74.71\right]}\times 100\\ =8.69\text{ wt\% MgO}\end{array}$

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

  $\begin{array}{c}{\left(\text{wt\%}\right)}_{\text{MgO}}=\frac{\left[{\left(\text{mol\%}\right)}_{\text{MgO}}\times \left({\text{M}}_{\text{MgO}}\right)\right]}{\left[{\left(\text{mol\%}\right)}_{\text{MgO}}\times \left({\text{M}}_{\text{MgO}}\right)+{\left(\text{mol\%}\right)}_{\text{NiO}}\times \left({\text{M}}_{\text{NiO}}\right)\right]}\times 100\\ =\frac{\left[36\times 40.312\right]}{\left[36\times 40.312+64\times 74.71\right]}\times 100\\ =23.28\text{ wt\% MgO}\end{array}$

Interpretation Introduction

(b)

Interpretation:

The amount of each phase present in $\text{NiO-20 mol\% MgO}$ at ${2200}^{\circ }\text{C}$ 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 $1\text{ and }2$ is:

  ${\left(\text{wt\%}\right)}_1=\frac{\left[{\left(\text{mol\%}\right)}_1\times \left({\text{M}}_1\right)\right]}{\left[{\left(\text{mol\%}\right)}_1\times \left({\text{M}}_1\right)+{\left(\text{mol\%}\right)}_2\times \left({\text{M}}_2\right)\right]}\times 100$ ...... (1)

Here, ${\left(\text{mol\%}\right)}_1\text{ and }{\left(\text{mol\%}\right)}_2$ are the mole percent of the components present in the ceramic, and ${\text{M}}_1{\text{ and M}}_2$ 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:

  $\text{Phase mol\%}=\frac{\text{opposite arm of lever}}{\text{total length of the tie line}}\times 100$ ...... (2)

Answer to Problem 10.53P

Amount of liquid phase in mol% is $76.19\text{ mol\%}$.

Amount of liquid phase in wt% is $78.07\text{ wt\%}$.

Amount of solid phase in mol% is $23.81\text{ mol\%}$.

Amount of solid phase in wt% is $21.93\text{ wt\%}$.

Explanation of Solution

The phase diagram for NiO-MgO is given as:

Now, draw a straight line from temperature ${2200}^{\circ }\text{C}$ to $20\text{ mol\% MgO}$ as:

Here, point 'a' represents $\text{NiO-}20\text{ mol\% MgO}$ at ${2200}^{\circ }\text{C}$. 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:

  $\begin{array}{c}L=15\text{ mol\% MgO}\\ S=36\text{ mol\% MgO}\end{array}$

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:

  $\begin{array}{c}\text{Liquid mol\%}=\frac{\text{length of 'ac'}}{\text{length of 'bc'}}\times 100\\ =\frac{36-20}{36-15}\times 100\\ =76.19\text{ mol}\%\\ \text{Solid mol\%}=\frac{\text{length of 'ba'}}{\text{length of 'bc'}}\times 100\\ =\frac{20-15}{36-15}\times 100\\ =23.81\text{ mol}\%\end{array}$

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:

  $\begin{array}{c}{\left(\text{wt\%}\right)}_{\text{MgO}}=\frac{\left[{\left(\text{mol\%}\right)}_{\text{MgO}}\times \left({\text{M}}_{\text{MgO}}\right)\right]}{\left[{\left(\text{mol\%}\right)}_{\text{MgO}}\times \left({\text{M}}_{\text{MgO}}\right)+{\left(\text{mol\%}\right)}_{\text{NiO}}\times \left({\text{M}}_{\text{NiO}}\right)\right]}\times 100\\ =\frac{\left[20\times 40.312\right]}{\left[20\times 40.312+80\times 74.71\right]}\times 100\\ =11.89\text{ wt\% MgO}\end{array}$

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

  $\begin{array}{c}L=8.69\text{ wt\% MgO}\\ S=23.28\text{ wt\% MgO}\end{array}$

Apply lever rule as:

  $\begin{array}{c}\text{Liquid wt\%}=\frac{\text{length of 'ac'}}{\text{length of 'bc'}}\times 100\\ =\frac{23.28-11.89}{23.28-8.69}\times 100\\ =78.07\text{ wt}\%\\ \text{Solid wt\%}=\frac{\text{length of 'ba'}}{\text{length of 'bc'}}\times 100\\ =\frac{11.89-8.69}{23.28-8.69}\times 100\\ =21.93\text{ wt}\%\end{array}$

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 $\left(\rho \right)$ is:

  ${\left(\text{Vol\%}\right)}_1=\frac{\left[\frac{{\left(\text{wt\%}\right)}_1}{{\rho }_1}\right]}{\left[\frac{{\left(\text{wt\%}\right)}_1}{{\rho }_1}+\frac{{\left(\text{wt\%}\right)}_2}{{\rho }_2}\right]}\times 100$ ...... (3)

Answer to Problem 10.53P

The amount of liquid phase in vol% is $75.91\text{ Vol\%}$.

The amount of solid phase in vol% is $24.09\text{ Vol\%}$.

Explanation of Solution

Given information:

A ceramic containing $\text{NiO-20 mol\% MgO}$ is heated to ${2200}^{\circ }\text{C}$. The solids present in it at this point has density $6.32{\text{ g/cm}}^3$ and the density of the liquid is $7.14{\text{ g/cm}}^3$.

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

  $\begin{array}{c}L=78.07\text{ wt\%}\\ S=21.93\text{ wt\%}\end{array}$

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

  $\begin{array}{c}{\left(\text{Vol\%}\right)}_L=\frac{\left[\frac{{\left(\text{wt\%}\right)}_L}{{\rho }_L}\right]}{\left[\frac{{\left(\text{wt\%}\right)}_L}{{\rho }_L}+\frac{{\left(\text{wt\%}\right)}_S}{{\rho }_S}\right]}\times 100\\ =\frac{\left[\frac{78.07}{7.14}\right]}{\left[\frac{78.07}{7.14}+\frac{21.93}{6.32}\right]}\times 100\\ =75.91\text{ Vol\%}\\ {\left(\text{Vol\%}\right)}_S=\frac{\left[\frac{{\left(\text{wt\%}\right)}_S}{{\rho }_S}\right]}{\left[\frac{{\left(\text{wt\%}\right)}_L}{{\rho }_L}+\frac{{\left(\text{wt\%}\right)}_S}{{\rho }_S}\right]}\times 100\\ =\frac{\left[\frac{21.93}{6.32}\right]}{\left[\frac{78.07}{7.14}+\frac{21.93}{6.32}\right]}\times 100\\ =24.09\text{ Vol\%}\end{array}$