Respuesta :
Answer:
The diffusion coefficients for iron in nickel are given at two temperatures:
T (K) 1273 1473
D ([tex]m^{2}[/tex]/s) 9.4 × [tex]10^{-16}[/tex] 2.4 × [tex]10^{-14}[/tex]
(a) Determine the values of Do and the activation energy Qd.
(b) What is the magnitude of D at 1100°C (1373 K)?
A
The pre-exponential factor Do = 2.1 x [tex]10^{-16}[/tex]
The activation energy Qd = 252,609 J/mol
B
The diffusion coefficient D= 5.14 x [tex]10^{-15}[/tex]
Explanation:
The full explanation is contained in the attached images;
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a) The value of D0 is 2.18 * 10^-5 m^2 / s while the value of the activation energy is Qd = 252.609 kj/mol
b) The coefficient of diffusion at 100 C is 5.4 * 10^-15 m^2 / s
What is Diffusion?
Diffusion is the movement of a gas from a region of high concentration to that of low cooncetration.
From coefficient of diffusion equation
[tex]D = D_{0} exp(-\frac{Q_{d} }{RT} )[/tex]
D = coefficient of diffusiion
D0 = pre- exponential factor
Qd = activation energy of diffusion
R = gas constant
T = thermodynamic temperature
Given
T1 = 273
D1 = 9.4 * 10^`-16
T2 = 1473
D2 = 2.4 * 10^-14
R = 8.314
D1 = D2 exp ( Qd / R T )
D1 - D2 = exp ( Qd / R T )
Qd = { in ( 2.4 * 10^-16 ) - in ( 9.4 * 10^`-16 ) } * 8.314 ) / ( 1 / 1473 - 1 / 273 )
Qd = 252.609 kj/mol
Pre- exponential factor, D0
D0 = D1 / exp ( - Qd / R T )
D0 = 9.4 * 10^`-16 / exp ( - 252609 / ( 8.314 * 1273)
D0 = 2.18 * 10^-5 m^2 / s
b) magnitude of D at 1100 C
convert to Kelvin: 1100 + 273 = 1373
D3 = D2 / exp ( - Qd / R T )
D3 = 2.4 * 10^-16 exp ( - 252609 / ( 8.314 * 1373 )
D3 = D at 100 = 5.4 * 10^-15 m^2 / s
Read more on Diffusion coefficient here: https://brainly.com/question/13843639