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 <--- this answer is wrng. . plz help.
then somethings off..i got that ticked...it says 'The grader on this problem has been corrected, and the number of attempts has been increased to 10', so the answer has changed i guess..but why still i got the green tick??...
Me too got green tick only.
Luk in 3.091 final 2011. same question and same answer.. method given.
To determine the number of vacant sites per cm^3 at 550 K, we can use the relationship between temperature and the percentage of vacant sites. The relationship can be expressed as follows:
Vacancy percentage = A * exp(-Q/RT)
Where:
- A is a constant
- Q is the activation energy for vacancy formation
- R is the ideal gas constant
- T is the temperature in Kelvin
To solve for the number of vacancies per cm^3, we need to rearrange the equation and solve for A:
A = Vacancy percentage / exp(-Q/RT)
First, let's calculate the values of the constant A at 923 K and 757 K:
A_923 = 0.0700% / exp(-Q/(R * 923 K))
A_757 = 0.0090% / exp(-Q/(R * 757 K))
Now, let's calculate the value of the constant A at 550 K:
A_550 = A_757 * exp(-Q/(R * (757 K - 550 K)))
Finally, let's calculate the number of vacant sites per cm^3 at 550 K:
Vacancies/cm^3 = A_550 * exp(-Q/(R * 550 K))
Please note that we need the value of the activation energy (Q) to complete the calculation. If the value of Q is not provided, the equation cannot be solved accurately.