Chapter 4, Problem 4.19P

To determine

(a)

The plot of the logarithm of the rate of recrystallization versus the inverse of the absolute temperature and to check whether it is linear or not.

Answer to Problem 4.19P

The graph of logarithm of the rate of recrystallization versus the inverse of absolute temperature is shown in figure (1) and it is linear.

Explanation of Solution

Given:

The given table is,

S. No.	Temperature $\left(T\right)$ (Kelvin)	Time (Seconds)
$1$	$316$	$2.3\times {10}^6$
$2$	$361$	$33,000$
$3$	$375$	$10,000$
$4$	$385$	$7,000$
$5$	392	$4,000$
$6$	$408$	$1,500$

Calculation:

Draw the table to plot the graph of logarithm of the rate of recrystallization versus the inverse of absolute temperature.

S. No.	Temperature $\left(T\right)$ $\left(\text{K}\right)$	$\left(\frac{1}{T}\right)$ $\left({\text{K}}^{-1}\right)$	Time $\left(\text{s}\right)$	Rate $\left({\text{s}}^{-1}\right)$	$\ln \left(\text{Rate}\right)$
$1$	$316$	$0.003165$	$2.3\times {10}^6$	$4.35\times {10}^{-7}$	$-14.6484$
$2$	$361$	$0.00277$	$33,000$	$3.03\times {10}^{-5}$	$-10.4043$
$3$	$375$	$0.002667$	$10,000$	$1.00\times {10}^{-4}$	$-9.21034$
$4$	$385$	$0.002597$	$7,000$	$1.43\times {10}^{-4}$	$-8.85367$
$5$	392	$0.002551$	$4,000$	$2.50\times {10}^{-4}$	$-8.29405$
$6$	$408$	$0.002451$	$1,500$	$6.67\times {10}^{-4}$	$-7.31322$

The graph of logarithm of the rate of recrystallization versus the inverse of absolute temperature is shown below.

Figure (1)

Conclusion:

Therefore, the graph of logarithm of the rate of recrystallization versus the inverse of absolute temperature is shown in figure (1) and it is linear.

To determine

(b)

The activation enthalpy for the recrystallization of copper.

Answer to Problem 4.19P

The activation enthalpy for the recrystallization of copper is $0.89\text{eV}/\text{atom}$.

Explanation of Solution

Formula Used,

The activation enthalpy for the recrystallization of copper is given by,

$\Delta H_{\text{R}}=-\frac{\left[R\ln \left(\frac{R_1}{R_2}\right)\right]}{\left(\frac{1}{T_1}-\frac{1}{T_2}\right)}$   ....... (I)

Here, $\Delta H_{\text{R}}$ is activation enthalpy for the recrystallization of copper, $R_1$ is the rate of crystallization at the initial state, $R_2$ is the rate of crystallization at the final state, $T_1$ is the initial temperature, $T_2$ is the final temperature, and $R$ is the universal gas constant.

Calculation:

The activation enthalpy for the recrystallization of copper is calculated as,

Substitute $8.314\text{J}/\text{mole}\cdot \text{K}$ for $R$, $4.35\times {10}^{-7}{\text{s}}^{-1}$ for $R_1$, $6.67\times {10}^{-4}{\text{s}}^{-1}$ for $R_2$, $316\text{K}$ for $T_1$ and $408\text{K}$ for $T_2$ in equation (I).

$\begin{array}{c}\Delta H_{\text{R}}=-\frac{\left[\left(8.314\text{J}/\text{mole}\cdot \text{K}\right)\ln \left(\frac{4.35\times {10}^{-7}{\text{s}}^{-1}}{6.67\times {10}^{-4}{\text{s}}^{-1}}\right)\right]}{\left(\frac{1}{316\text{K}}-\frac{1}{408\text{K}}\right)}\\ =-\frac{\left[\left(8.314\text{J}/\text{mole}\cdot \text{K}\right)\left(-7.335\right)\right]}{\left(7.136\times {10}^{-4}\text{K}\right)}\\ =84,458\text{J}/\text{mole}\times \frac{1.04\times {10}^{-5}\text{eV}/\text{atom}}{1\text{J}/\text{mole}}\\ =0.89\text{eV}/\text{atom}\end{array}$

Conclusion:

Therefore, the activation enthalpy for the recrystallization of copper is $0.89\text{eV}/\text{atom}$.

To determine

(c)

The reason for activation enthalpy of recrystallization of copper being less than the activation enthalpy for vacancy diffusion.

Answer to Problem 4.19P

The activation enthalpy for the recrystallization is less than the activation enthalpy for vacancy diffusion because during the process of recrystallization, the size of grains increases and the vacancy site decreases.

Explanation of Solution

Introduction:

The Gibbs free energy is the energy that is available or free, to do work under conditions of constant pressure and temperature.

The expression for the Gibb’s free energy is given by,

$G=E+PV-TS$   ....... (II)

Here, $G$ is the Gibb’s free energy, $T$ is the temperature of the system, $S$ is the entropy of the system, $E$ is the total internal energy, $P$ is the pressure and $V$ is the volume.

The term $E+PV$ is termed as enthalpy. It is defined as the total heat content of a system.

$H=E+PV$   ....... (III)

Here, $H$ is the enthalpy.

During the process of recrystallization, the size of the grain changes to form a new crystal structure. By this process the grain size changes its size by diffusing into other grains, thereby increasing the size of the grains. When the size of the grain increases, the possibility for the vacancy site decreases; that’s why the activation enthalpy of recrystallization of copper is less than the activation enthalpy for vacancy diffusion.

Conclusion:

Therefore, the activation enthalpy for the recrystallization is less than the activation enthalpy for vacancy diffusion because during the process of recrystallization, the size of grains increases and the vacancy site decreases.

To determine

(d)

The temperature at which the copper completely crystallizes in $1$ hour.

Answer to Problem 4.19P

The temperature at which the copper completely crystallizes in $1$ hour is $394\text{K}$.

Explanation of Solution

Formula Used,

The logarithmic rate of crystallization is given by,

$\ln \left(\text{rate}\right)=\ln \left(\frac{1}{t}\right)$   ....... (IV)

Here, $\text{rate}$ is the rate of crystallization and $t$ is the time.

Calculation:

The logarithmic rate of crystallization is calculated as,

Substitute $3600\text{s}$ for $t$ in equation (IV).

$\begin{array}{c}\ln \left(\text{rate}\right)=\ln \left(\frac{1}{3600\text{s}}\right)\\ =-8.189{\text{s}}^{-1}\end{array}$

Drop a perpendicular from $-8.189$ in figure (1) and note down the corresponding inverse temperature. The perpendicular intersects the inverse temperature at $0.00254{\text{K}}^{-1}$.

$\begin{array}{c}\frac{1}{T}=0.00254{\text{K}}^{-1}\\ T=\frac{1}{0.00254{\text{K}}^{-1}}\\ =393.7\text{K}\\ \approx 394\text{K}\end{array}$

Conclusion:

Therefore, the temperature at which the copper completely crystallizes in $1$ hour is $394\text{K}$.