Question

Calculate the activation energy for vacancy formation in aluminum, given that the equilibrium number of vacancies at 500°C (773 K) is 7.57 × 1023 m−3 . The atomic weight and density (at 500°C) for aluminum are, respectively, 26.98 g/mol and 2.62 g/cm3 .

Answer 1

The answer & explanation for this question is given in the attachment below.

Answer 2

The required activation energy is 0.75 ev/atm.

Given that,

M=26.98 g/mol

T=$500^{\circ}C$(773K)

$N_v=n=7.57\times 10^{23} /m^3$

Density=2.62 g/$cm^3$

We know that,

$N=\frac{N_A\rho _{Ad}}{M}\\=\frac{6.02\times10^{23}\times 2.62}{26.98}\\=5.848\times 10^{22} atm/m^3$

Then,

$N=\frac{N_A\rho _{Ad}}{M}\\=\frac{6.02\times10^{23}\times 2.62}{26.98}\\=5.848\times 10^{22} atm/m^3\\Q_v =-KTln(\frac{N_v}{N})\\=(-8.62\times10^{-5} (773)ln(\frac{7.57\times 10^{23}}{5.843\times10^{28}}\\=0.7499 ev/atm\\\approx 0.75 ev/atm$

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