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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.

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okpalawalter8

Answer:

The  dimension is  [tex]D =  L ^{2} T^{-1}[/tex]

Explanation:

From the question we are told that

     [tex]J  =  -D \frac{dn}{dx}[/tex]

Here  [tex][J] = \frac{1}{L^2 T}[/tex]

       [tex][n] =\frac{1}{L^3}[/tex]

        [tex][x] = L[/tex]

So

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

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

So

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

=>   [tex]D =  \frac{L^{-2} T^{-1} * L }{L^{-3}}[/tex]

=>   [tex]D =  L ^{2} T^{-1}[/tex]

   

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