The explanation of the mass transport within crystals via the vacancy movement mechanism is interesting because the vacancies are present in equilibrium in all crystals. In the solid state transformations, the variations of pressure (P) and volume (V) are very small and this is why the energy to create a vacancy (HL) and the enthalpy (H)1 are practically the same i.e., H ≈ HL.
Given a crystal of N atomic positions and n vacancies, the increase in internal energy or enthalpy of the crystal can be considered a linear function of the number of vacancies
\[\triangle H = n H _ {L} \tag{2.15}\]
where HL is the needed energy to form a single vacancy.
The entropy of the “mixture” of full sites and vacancies in the crystalline network is calculated exactly the same way that configurational entropy of a binary bond. The configurational entropy SL is given by the Boltzmann equation:
\[S _ {L} = klogW \tag{2.16}\]
Where S is entropy, W is the possible number of configurations and k is the Boltzmann constant.
The number of possible arrangements of n vacancies in N atomic positions, W will be:
\[W = \frac{N!}{(N – n)!n!} \tag{2.17} \]
Substituting Eq. (2.17) into Eq. (2.16), we end up with:
\[S = k. \ln [\frac{N!}{(N – n)!n!}] \tag{2.18} \]
Substituting the equations (2.15) and (2.18) into the free energy equation \( \triangle G = H – T \triangle S \), we get
\[\triangle G = n \cdot H _ {L} – T \cdot k \cdot \ln [\frac{N!}{(N – n)!n!}] \tag{2.19} \]
Substituting the factorial logarithms of the Equation (2.19) by the Stirling approximation2 ( \(n ! \sim \sqrt{2 \pi n} (\frac{n}{e})^{n}\) ) for the factorial of a number, we have:
\[\triangle G = n \cdot H _ {L} – T \cdot k \ln [N \ln N – (N – n) \ln (N – n)-n \ln n] \tag{2.20} \]
The lowest point of this function (equation 2.20) is obtained when
\[\frac{\partial \triangle G}{\partial n} = 0 \tag{2.21}\]
Applying the first derivation of the equation 2.20 as a function of n, the result is:
\[\frac{n}{N-n} = exp (\frac {-H _ {L}}{kT}) \tag{2.22}\]
But, seeing as, the Equation (2.22) becomes:
\[\frac{n}{N} \sim exp (\frac {-H _ {L}}{kT}) \tag{2.23}\]
As an example, consider HL=20.000 cal/mol and T=1.000 K. With these conditions, a single position will be open per 10.000 (105).
1Enthalpy (H). The sum of internal energy (E) with the pruduct of pressure (P) and volume (V). H= E + PV
2James Stirling (1692-1770), English mathmatician accredited in the discovery of the formula (see text) to obtain, approximately, the factorial of a number.
