Question

Calculate the equilibrium concentration of vacancies per cubic meter in pure aluminium at 600°C. Energy of formation for Al is 0.76 eV, atomic weight is 2700 kg/m3 and the atomic weight of Al is 26.98 g/mol

Answer 1

Answer:

equilibrium concentration is 2.477 × $10^{24}$ vacancies/m³

Explanation:

given data

temperature = 600°C = 873 K

Energy of formation Al = 0.76 eV

atomic density = 2700 kg/m3

atomic weight of Al = 26.98 g/mol

to find out

equilibrium concentration of vacancies

solution

we have given atomic weight and density so we will apply formula that is

N = $\frac{Na*\rho}{atomic weight}$

here ρ is density and Na is avogadro number i.e 6.022 × $10^{23}$ atoms/mol and N is related to density

so

N = $\frac{6.022*10^{23}**2700}{26.98*10^{-3}}$

N = 6.027 ×$10^{28}$ atoms/m³

and

equilibrium concentration of vacancies is express as

n = $N*e^{-Q/Kt}$

here t is temperature given and Q is energy formation and K is 8.62 ×$10^{-5}$ eV/atom

so

n = $6.027*10^{28}*e^{(-0.76/8.62*10^{-5}*873)}$

n = 2.477 × $10^{24}$ vacancies/m³

so equilibrium concentration is 2.477 × $10^{24}$ vacancies/m³