Question

The steady-state diffusion flux through a metal plate is 7.8 × 10-8 kg/m2-s at a temperature of 1220˚C ( 1493 K) and when the concentration gradient is -500 kg/m4. Calculate the diffusion flux at 1000˚C ( 1273 K) for the same concentration gradient and assuming an activation energy for diffusion of 145,000 J/mol.

Answer 1

Answer:

The diffusion flux is $7.3\times10^{-19}\ kg/m^2-s$.

Explanation:

Given that,

Diffusion flux $J =7.8\times10^{-8}\ kg/m^2-s$

Temperature = 1220°C =1493 K

Concentration gradient = -500 kg/m⁴

Diffusion energy = 145000 J/mol

We need to calculate the diffusion coefficient

Using formula of diffusion flux

$J = -D_{0}\dfrac{dc}{dx}$

$D_{0}=\dfrac{-J}{\dfrac{dc}{d}}$

Where, J = diffusion flux

$D_{0}$ = pre exponential factor

$\dfrac{dc}{dx}$ = Concentration gradient

Put the value into the formula

$D_{0}=\dfrac{-7.8\times10^{-8}}{-500 }$

$D_{0}=1.56\times10^{-10}\ m^2/s$

Using formula of diffusion coefficient formula

$D=D_{0}\ e^{\frac{-E}{RT}}$

Where, D = diffusion coefficient

R = gas constant

T = temperature

Put the value of D₀ into the formula

$D=1.56\times10^{-10}\ e^{\frac{-145000}{8.31\times1493}}$

$D=1.310\times10^{-15}\ m^2/s$

Now, we need to calculate the D at 1273 K

$D=1.310\times10^{-15}\ e^{\frac{-145000}{8.31\times1273}}$

$D=1.460\times10^{-21}\ m^2/s$

We need to calculate the diffusion flux

$J=-D\dfrac{dc}{dx}$

$J=-1.460\times10^{-21}\times(-500)$

$J=7.3\times10^{-19}\ kg/m^2-s$

Hence, The diffusion flux is $7.3\times10^{-19}\ kg/m^2-s$.