1
$\begingroup$

What is the meaning of minus sign in the first Fick's law, $$ J=-D\nabla\varphi, $$ where $J$ is the diffusion flux, $D$ the diffusivity and $\varphi$ the concentration?

I read that the flux goes from region of high concentration to region of low concentration and that the flux is larger where the concentration gradient is largest. Why?

$\endgroup$
0
1
$\begingroup$

You want the flux in a given direction to be larger when the change in concentration is greater in that direction.

However as you point out the direction of the flux is from high concentration to low concentration so the concentration gradient (related to final concentration - initial concentration) needs to be more negative for the flux to be increased.

So if the concentration gradient becomes more negative (smaller) then minus the concentration gradient becomes more positive (larger) and it is that quantity which then mirrors the increase (more positive ) in the flux.

$\endgroup$
1
$\begingroup$

The gradient is measured as the change in concentration over the change in position. Molecules tend to move from high a concentration to a low concentration, which is a negative change. Since flux is usually used to measure the amount of molecules moving across an area to a low concentration, the negative sign is added so flux ends up positive.

$\endgroup$
-1
$\begingroup$

Matter and energy move only in response to a gradient. The steeper the gradient, the more entropy is produced when matter/energy shifts to eliminate it. Just as matter spontaneously moves down a gradient in the chemical potential (which is often approximated by the concentration), thermal energy moves down a temperature gradient, charge moves down an electric field gradient, boundaries move down a pressure gradient, and so on. All to increase entropy in accordance with the Second Law.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.