Consider the data below
Temperature(0C) Diffusivity(m2/s)
736 2 x 10-13
782 5 x 10-13
835 1.3 x 10-12
Given this data, determine the activation energy for the diffusion process (units of Joules per atom or molecule) in this material and the pre-exponential factor (use units of m2/s).
To determine the activation energy and pre-exponential factor, we can use the Arrhenius equation, which relates the diffusivity (D) to the temperature (T), activation energy (Ea), and pre-exponential factor (D0) as follows:
D = D0 * exp(-Ea / (RT))
Where:
D = diffusivity
D0 = pre-exponential factor
Ea = activation energy
R = gas constant (8.314 J/(mol·K))
T = temperature in Kelvin (K)
To use this equation, we need to transform the temperature from Celsius to Kelvin. The temperature in Kelvin can be obtained by adding 273.15 to the Celsius value.
Now, let's calculate the activation energy and pre-exponential factor using the given data:
1. Convert the temperatures to Kelvin:
736°C + 273.15 = 1009.15 K
782°C + 273.15 = 1055.15 K
835°C + 273.15 = 1108.15 K
2. Apply the Arrhenius equation to each temperature and diffusivity pair to get three equations:
Equation 1: 2 x 10^(-13) = D0 * exp(-Ea / (8.314 * 1009.15))
Equation 2: 5 x 10^(-13) = D0 * exp(-Ea / (8.314 * 1055.15))
Equation 3: 1.3 x 10^(-12) = D0 * exp(-Ea / (8.314 * 1108.15))
3. Take the natural logarithm (ln) of both sides of each equation to linearize our system of equations:
ln(2 x 10^(-13)) = ln(D0) - (Ea / (8.314 * 1009.15))
ln(5 x 10^(-13)) = ln(D0) - (Ea / (8.314 * 1055.15))
ln(1.3 x 10^(-12)) = ln(D0) - (Ea / (8.314 * 1108.15))
4. Let's denote ln(D0) as B, and solve for the activation energy, Ea:
ln(D0) = B
Ea = -slope * 8.314 * average_temperature
To obtain an accurate average temperature, sum up all the temperatures in Kelvin and divide by the number of data points.
5. Finally, we can find the pre-exponential factor (D0) by taking the exponential function (e) of B.
Now, plug in these calculations and solve for the activation energy and pre-exponential factor.