A comparison of density values obtained by X-ray diffraction reveals that the percentages of vacant sites in aluminum (Al) are 0.0700% at 923 K and 0.0090% at 757 K. Given this data, determine the number of vacant sites per cm3 at 550 K. Express your answer as vacancies/cm3.
7.4*10^18
Thanks
thank u
thanks
No problem.....:)
wrong
To determine the number of vacant sites per cm3 at 550 K, we can use the relationship between vacancy concentration and temperature known as the "Equation of Vacancies," which is given by:
Nv / N = exp(-Q / RT)
Where Nv is the number of vacancies, N is the total number of atomic sites, Q is the activation energy for vacancies, R is the ideal gas constant, and T is the temperature in Kelvin.
In this case, we have the following data:
At 923 K: % of vacant sites = 0.0700%
At 757 K: % of vacant sites = 0.0090%
To find the number of vacancies per cm3 at 550 K, we need to find the activation energy Q for vacancies in aluminum. Unfortunately, this information is not given in the question.
The activation energy Q can vary depending on the material, so we need additional information or a reference source to determine its value for aluminum.
Once we have the activation energy Q, we can use the equation above to calculate the number of vacancies per cm3 at 550 K.