Question

When we do dimensional analysis, we do something analogous to stoichiometry, but with multiplying instead of adding. Consider the diffusion constant that appears in Fick's first law:
J= -Ddn/dx

In this expression, J represents a flow of particles: number of particles per unit area per second, n represents a concentration of particles: number of particles per unit volume; and x represents a distance. We can assume that they have the following dimensionalities:

[J] = 1/L2T
[n] = 1/L3
[x] = L

Required:
From this, determine the dimensionality of D.

Answer 1

Answer:

The  dimension is  $D = L ^{2} T^{-1}$

Explanation:

From the question we are told that

     $J = -D \frac{dn}{dx}$

Here  $[J] = \frac{1}{L^2 T}$

       $[n] =\frac{1}{L^3}$

        $[x] = L$

So

    $\frac{1}{L^2 T} = -D \frac{d(\frac{1}{L^3})}{d[L]}$

Given that the dimension represent the unites of  n and  x then the differential  will not effect on them

So

$\frac{1}{L^2 T} = -D \frac{(\frac{1}{L^3})}{[L]}$

=>   $D = \frac{L^{-2} T^{-1} * L }{L^{-3}}$

=>   $D = L ^{2} T^{-1}$