The diffusion coefficient of silver (Ag) in copper (Cu) has the following values:
T (K) D (m2s-1)
923 5.42 x 10-16
1173 1.34 x 10-13
(a) Calculate the activation energy for diffusion of silver in copper. Express your answer in units of J/mol.
(b) You have two specimens of copper, both of them polycrystalline. Specimen A has an average grain size of 3.091 μm, while Specimen B has an average grain size of 7.777 μm. Which specimen will silver diffuse through more quickly? Choose the best explanation.
I)Assuming both specimens are at the same temperature, the diffusivity should be the same through both samples as it is a materials property.
II)Specimen A should allow silver to diffuse more quickly due to its smaller grain size and increased grain boundary area.
III)Specimen A should allow silver to diffuse more quickly due to its smaller grain size and decreased grain boundary area.
IV)Specimen B should allow silver to diffuse more quickly due to its larger grain size and decreased grain boundary area.
V)Specimen B should allow silver to diffuse more quickly due to its larger grain size and increased grain boundary area.
(b) II
Someone please answer this ;(
1. 193000 J/mol
tuhin your answer is wrong
tuhin is a red X!
Answer is: 193000
Thank you anonymous
how did u get it
To answer part (a), we can use the Arrhenius equation, which relates the diffusion coefficient (D) to the activation energy (Ea), temperature (T), and a pre-exponential factor (D0):
D = D0 * exp(-Ea / (R * T))
where R is the gas constant (8.314 J/mol*K).
We can rearrange the equation to solve for the activation energy:
Ea = -R * T * ln(D / D0)
Now, let's plug in the given values:
For T = 923 K, D = 5.42 x 10^-16 m^2/s
For T = 1173 K, D = 1.34 x 10^-13 m^2/s
Using the formula, we can calculate the activation energy (Ea) at each temperature and then take the average:
For T = 923 K:
Ea1 = - (8.314 J/mol*K) * (923 K) * ln(5.42 x 10^-16 / D0)
For T = 1173 K:
Ea2 = - (8.314 J/mol*K) * (1173 K) * ln(1.34 x 10^-13 / D0)
Average activation energy:
Ea_avg = (Ea1 + Ea2) / 2
This will give us the activation energy for diffusion of silver in copper in units of J/mol.
To answer part (b), we need to consider the effect of grain size on diffusivity. Generally, smaller grain sizes result in faster diffusion due to the increased grain boundary area, which provides more paths for diffusion. Therefore, we can eliminate options IV and V.
Now, we consider the effect of grain size on diffusivity. Smaller grain sizes also imply more grain boundaries. Since it is known that grain boundaries hinder diffusion, smaller grain sizes (more grain boundaries) tend to decrease the diffusivity. So, we can eliminate option III.
Therefore, the correct answer is option II:
Specimen A should allow silver to diffuse more quickly due to its smaller grain size and increased grain boundary area.