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

The dimension is
`D =  L ^(2) T^(-1)`

Step-by-step explanation:

From the question we are told that

`J  =  -D (dn)/(dx)`

Here
`[J] = (1)/(L^2 T)`

`[n] =(1)/(L^3)`

`[x] = L`

So

`(1)/(L^2 T) =  -D (d((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

`(1)/(L^2 T) =  -D (((1)/(L^3)))/([L])`

=>
`D =  (L^(-2) T^(-1) * L )/(L^(-3))`

=>
`D =  L ^(2) T^(-1)`