Chapter 12, Problem 12.3P

Interpretation Introduction

Interpretation:

The value of constants n and c for the given equation needs to be calculated.

Concept Introduction:

At constant temperature, the determination of solid phase transformation is done by using the Avarmi equation. Generally, it is used to describe the rate of kinetic crystallization.

This equation is also known as the JMAK equation as it was founded by Johnson-Mehl-Avrami-Kolmology. This equation was first found in the year $1937$ by Kolmogorov. And the first paper was published by Melvin Avrami during the years $1939$ and $1941$ in the Journal of chemical equations.

Answer to Problem 12.3P

The constant value of $\text{c = 8}{\text{.9×10}}^{\text{-6}}$ and $\text{n = 2}\text{.81}$.

Explanation of Solution

Given:

Eqauation: ${\text{f=1-exp(-ct}}^{\text{n}}\text{)}$

Temperature = $\text{T =140 K}$

Assume:

Elapsed time − $\text{t}$

Temperature constants - $\text{c }$ and $\text{n}$

From the given equations of curves, the values of f and time are:

$\text{f}$	Time (min)
$0.1$	$28$
$0.2$	$37$
$0.3$	$44$
$0.4$	$50$
$0.5$	$55$
$0.6$	$60$
$0.7$	$67$

Table $1$

To calculate the constants value of c and n by using the below formula,

  ${\text{f=1-exp(-ct}}^{\text{n}}\text{)}$

  $\begin{array}{l}\text{1-f = exp}\left({\text{-ct}}^{\text{n}}\right)\\ \text{ln}\left(\text{1-f}\right)\text{= -ct}\\ \end{array}$

Taking natural logarithm on both sides,

  $\ln \left[\text{-ln}\left(\text{1-f}\right)\right]\text{= ln}\left(\text{c}\right)\text{+n×ln}\left(\text{t}\right)$

$\text{f}$	$\text{t}$ (min)	$\text{ln}\left(\text{t}\right)$	$\ln \left[\text{-ln}\left(\text{1-f}\right)\right]$
$0.1$	$28$	$3.33$	$-2.2504$
$0.2$	$37$	$3.61$	$-1.4999$
$0.3$	$44$	$3.78$	$-1.0309$
$0.4$	$50$	$3.91$	$-0.6717$
$0.5$	$55$	$4.01$	$-0.3665$
$0.6$	$60$	$4.09$	$-0.0874$
$0.7$	$67$	$4.2$	$0.1856$
$0.8$	$73$	$4.29$	$0.4759$
$0.9$	$86$	$4.45$	$0.8340$

Table $1$

From the above table, the solpe of the graph is plotted as:

Fig. $2$ Slope graph

From graph ,

  $\text{n}$ - temperature constant

  $\text{n = 2}\text{.81}$

Then to find value of $\text{c}$, we can use −

  $\ln \left[\text{-ln}\left(\text{1-f}\right)\right]\text{= ln}\left(\text{c}\right)\text{+n×ln}\left(\text{t}\right)$

  $\begin{array}{l}\text{ln}\left[\text{-ln}\left(\text{1-0}\text{.2}\right)\right]\text{= ln}\left(\text{c}\right)\text{+2}\text{.81×ln}\left(\text{37}\right)\\ \text{ln}\left(\text{c}\right)\text{=-11}\text{.64}\\ {\text{c = e}}^{\text{-11}\text{.64}}\\ \text{c}=8.9\times {10}^{-9}\end{array}$

Conclusion

Thus, the value of $\text{c = 8}{\text{.9×10}}^{\text{-6}}$ and $\text{n = 2}\text{.81}$ which are obtained by using the slope of graph.