Chapter 4, Problem 4.18P

To determine

(a)

The validity of the given statement for copper by comparing the vacancy concentration when there is one vacancy in the atom positions surrounding each atom with the equilibrium vacancy.

Answer to Problem 4.18P

The statement that the FCC metals melt when there is on the average one vacancy in the atom positions surrounding each atom is wrong.

Explanation of Solution

Given:

The vacancy formation enthalpy in copper is $1.0\text{eV}$ per vacancy.

The melting temperature of copper is $1356\text{K}$.

Formula Used:

The formula to calculate the concentration of vacancies in silicon is given by,

$\frac{n_{\text{v}}}{N}=\exp \left(-\frac{\Delta H_{\text{v}}}{kT}\right)$   ........ (I)

Here, $\frac{n_{\text{v}}}{N}$ is the concentration of vacancies in silicon, $k$ is Boltzmann’s constant, $\Delta H_{\text{v}}$ is enthalpy of vacancy formation, and $T$ is temperature at which the vacancy sites are formed.

Calculations:

The concentration of vacancies is calculated as,

Substitute $1.0\text{eV}/\text{vacancy}$ for $\Delta H_{\text{v}}$, $8.62\times {10}^{-5}\text{eV}/\text{atom}\cdot \text{K}$ for $k$ and $1356\text{K}$ for $T$ in equation (I).

$\begin{array}{c}\frac{n_{\text{v}}}{N}=\exp \left(-\frac{1.0\text{eV}/\text{vacancy}}{\left(8.62\times {10}^{-5}\text{eV}/\text{atom}\cdot \text{K}\right)\left(1356\text{K}\right)}\right)\\ =\exp \left(-8.555\right)\\ =1.9\times {10}^{-4}\text{vacancies}/\text{atom}\end{array}$

In FCC structure every atom is surrounded by $12$ atoms, so if there exists one vacancy out these 12 atoms. Then,

The vacancy concentration for this condition is given by,

$\begin{array}{c}V=\frac{1}{12}\\ =8.33\times {10}^{-2}\end{array}$

Thus, the vacancy concentration is less at the melting temperature, when compared to the given statement.

Conclusion:

Therefore, the statement that the FCC metals melt when there is on the average one vacancy in the atom positions surrounding each atom is wrong.

To determine

(b)

The entropy of mixing the equilibrium number of vacancies into a copper crystal at the melting temperature.

Answer to Problem 4.18P

The entropy of mixing the equilibrium number of vacancies into a copper crystal at the melting temperature is $943\times {10}^{14}\text{eV}/\text{mole}\cdot \text{K}$.

Explanation of Solution

Given:

The total number of crystal sites is equal to $1\text{mole}$.

Formula Used:

The concentration of copper for mixing is given by,

$C_{\text{Cu}}=1-C_{\text{v}}$   ........ (II)

Here, $C_{\text{Cu}}$ is concentration of copper, and $C_{\text{v}}$ is the equilibrium vacancy concentration.

The entropy of mixing equilibrium vacancy concentration into copper at the melting temperature is given by,

$\Delta S_{\text{m}}=-N_{\text{A}}k\left(C_{\text{Cu}}\ln C_{\text{Cu}}+C_{\text{v}}\ln C_{\text{v}}\right)$   ........ (III)

Here, $\Delta S_{\text{m}}$ is entropy of mixing, and $N_{\text{A}}$ is the number of vacancy sites in one mole of copper.

Calculations:

The entropy of mixing the equilibrium number of vacancies into a copper crystal at the melting temperature is calculated as,

Substitute $1.9\times {10}^{-4}$ for $C_{\text{v}}$ in equation (II).

$C_{\text{Cu}}=\left(1-1.9\times {10}^{-4}\right)\text{Cu}\text{atoms}/\text{total}\text{sites}$

Substitute $6.02\times {10}^{23}\text{Total}\text{sites}/\text{mole}$ for $N_{\text{A}}$, $8.62\times {10}^{-5}\text{eV}/\text{atom}\cdot \text{K}$ for $k$, $1.9\times {10}^{-4}\text{vacancies}/\text{Total}\text{site}$ for $C_{\text{v}}$, and $\left(1-1.9\times {10}^{-4}\right)\text{Cu}\text{atoms}/\text{total}\text{sites}$ for $C_{\text{Cu}}$ in equation (III).

$\begin{array}{c}\Delta S_{\text{m}}=\left[-\left\{\begin{array}{l}\left(6.02\times {10}^{23}\frac{\text{Total}\text{sites}}{\text{mole}}\right)\\ \left(8.62\times {10}^{-5}\text{eV}/\text{atom}\cdot \text{K}\right)\end{array}\right\}\left(\begin{array}{l}\left\{\begin{array}{l}\left(\left(1-1.9\times {10}^{-4}\right)\frac{\text{Cu}\text{atoms}}{\text{total}\text{sites}}\right)\\ \ln \left(\left(1-1.9\times {10}^{-4}\right)\frac{\text{Cu}\text{atoms}}{\text{total}\text{sites}}\right)\end{array}\right\}+\\ \left\{\begin{array}{l}\left(1.9\times {10}^{-4}\frac{\text{vacancies}}{\text{Total}\text{site}}\right)\\ \ln \left(1.9\times {10}^{-4}\frac{\text{vacancies}}{\text{Total}\text{site}}\right)\end{array}\right\}\end{array}\right)\right]\\ =-\left(6.02\times {10}^{23}\frac{\text{Total}\text{sites}}{\text{mole}}\right)\left(8.62\times {10}^{-5}\text{eV}/\text{atom}\cdot \text{K}\right)\left(-1.9\times {10}^{-4}-1.628\times {10}^{-3}\right)\\ =943\times {10}^{14}\text{eV}/\text{mole}\cdot \text{K}\end{array}$

Conclusion:

Therefore, the entropy of mixing the equilibrium number of vacancies into a copper crystal at the melting temperature is $943\times {10}^{14}\text{eV}/\text{mole}\cdot \text{K}$.

To determine

(c)

The change in the Gibbs free energy.

Answer to Problem 4.18P

The change in Gibbs free energy is $-1.35\times {10}^{19}\text{eV}/\text{mole}$.

Explanation of Solution

Formula Used:

The change in Gibbs free energy when the equilibrium vacancy concentration forms relative to that of copper are given by,

$\Delta G_{\text{v}}=C_{\text{v}}{N}_{\text{A}}\Delta H_{\text{v}}-T\Delta S_{\text{m}}$   ........ (IV)

Here, $\Delta G_{\text{v}}$ is the change in Gibbs free energy, $T$ is the temperature at which mixing takes place.

Calculations:

The concentration of vacancies is calculated as,

Substitute $6.02\times {10}^{23}\frac{\text{total}\text{sites}}{\text{mole}}$ for $N_A$, $1.9\times {10}^{-4}\frac{\text{vacancies}}{\text{total}\text{sites}}$ for $C_V$, $\text{1356}\text{K}$ for $T$, $943\times {10}^{14}\text{eV}/\text{mole}\cdot \text{K}$ for $S_{\text{m}}$ and $1.0\text{eV}/\text{vacancy}$ for $\Delta H_V$ in equation (IV).

$\begin{array}{c}\Delta G_{\text{v}}=\left[\begin{array}{l}\left(1.9\times {10}^{-4}\frac{\text{vacancies}}{\text{total}\text{sites}}\right)\left(6.02\times {10}^{23}\frac{\text{total}\text{sites}}{\text{mole}}\right)\left(1.0\text{eV}/\text{vacancy}\right)\\ -\left(\text{1356}\text{K}\right)\left(943\times {10}^{14}\text{eV}/\text{mole}\cdot \text{K}\right)\end{array}\right]\\ =1.144\times {10}^{20}\text{eV}/\text{mole}-1.279\times {10}^{20}\text{eV}/\text{mole}\\ =-1.35\times {10}^{19}\text{eV}/\text{mole}\end{array}$

Thus, the change in Gibbs free energy when the equilibrium vacancy concentration forms relative to that of copper is $-1.35\times {10}^{19}\text{eV}/\text{mole}$. As, the change in Gibbs free energy is negative, the change is spontaneous.

Conclusion:

Therefore, the change in Gibbs free energy is $-1.35\times {10}^{19}\text{eV}/\text{mole}$.