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

equilibrium concentration is 2.477 ×
`10^(24)` vacancies/m³

Step-by-step 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 =
`(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 =
`(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