Question

A thin sheet of iron 0.5 mm thick has a nitrogen concentration on one side of 0.5 kg/m^3 and a nitrogen concentration on the other side of 3.5kg/m^3 . the diffusion coefficient is 3.44x10^-15 m^2/s. What is the flux of nitrogen through the iron?

Answer 1

2.064× 10⁻¹¹ kg/m²s

Step-by-step explanation:

The expression for the Fick's first law is shown below as:

`J=-D* \frac {dC}{dx}`

Where, J is the diffusion flux.

D is the diffusion coefficient =
`3.44* 10^(-15)\ m^2/s`

dC is the concentration change ( 3.5 - 0.5 kg/m³ = 3 kg/m³)

dx is the thickness = 0.5 mm = 0.0005 (As , 1 mm = 0.001 m )

Thus,

`J=-3.44* 10^(-15)* (3)/(0.0005)\ kg/m^2s`

Magnitude of the flux = 2.064× 10⁻¹¹ kg/m²s