# What is Fick’s Law?

The **Fick ‘s Law** is a mathematical equation that relates the mass flow diffused in a medium having the concentration gradient or pressure. It was formulated in 1855 by the German physiologist and physician Adolf Fick, who inspired by the laws of Fourier (thermal conduction) and Ohm (electrical conduction), modeled the process of diffusion of oxygen towards the alveoli of the lungs.

Fick’s law has the particularity that it is not only applicable to chemical or biochemical diffusion phenomena, but to those of any type of nature. Therefore, it serves to model the diffusion of atoms between solids, being very useful in the physics of materials and engineering.

However, the central basis is the same for almost all diffusion phenomena, which is illustrated above. Purple particles, be they atoms or molecules, diffuse through a semipermeable membrane of thickness L and cross-sectional area A. On the left we have a higher concentration of C _{1} of particles than on the right, C _{2} .

Fick’s law states the following: the mass flow that diffuses through a surface is proportional to the concentration gradient (C _{2} -C _{1} / L) and to a constant D called the diffusion coefficient or diffusivity.

__Fick’s First Law__

__Fick’s First Law__

**Components and equation**

The thickness L of the semipermeable membrane represents the distance (x) that the particles must travel to reach the other side. As can be seen in the image, the more purple particles move away from the left compartment, where C _{1} is large, their concentration decreases to the value of C _{2} . That is, the concentration changes along the thickness of the membrane, being dependent on x.

This variation in concentration as a function of distance is known as the concentration gradient: (C _{2} -C _{1} ) / L or (C _{2} -C _{1} ) / x. Note that its value is negative (-1), because C _{2} > C _{1} .

On the other hand, we also have the speed with which the particles diffuse through the membrane or the space in question. This speed depends on the size and mass of the particles, as well as the nature of the medium and the temperature. The diffusion coefficient D represents this speed, and may or may not be constant during diffusion.

And finally, we have a mass flow ‘j’ that crosses the cross-sectional area of the membrane or channel through which the particles are diffused. Grouping these terms gives rise to the equation of Fick’s first law:

Where j is proportional to D and to (∂C / ∂x), the concentration gradient.

**Interpretation and units**

The negative symbol in the equation serves to neutralize the negative sign of the concentration gradient. Otherwise, j would have a negative value, which makes no physical sense. Likewise, the value of D is positive, so that when multiplied by the negative sign that precedes it, it gives a negative value.

Fick’s first law states the following: the greater the concentration gradient (∂C / ∂x), the greater the mass flow j. That is, the difference between C _{2} and C _{1} becomes larger and therefore more particles will diffuse through the membrane.

On the other hand, j also depends on D, which in turn is dependent on parameters such as temperature, viscosity, molecular weight , and cross-sectional area A:

D ∝ (A / L) (S / √M _{W} )

Where S is the solubility of the particle that diffuses with the medium, and M _{W} its molecular weight.

Regarding the units of the components or terms of the equation we have:

-C (kgm ^{-3} or molm ^{-3} )

-D (m ^{-2} s ^{-1} )

-j (kg · m ^{-2} · s ^{-1} or mol · m ^{-2} · s ^{-1} )

**Net mean square displacement**

During diffusion the particles collide with each other, and after short intervals of time they end up traveling enormous distances Δx. However, depending on the direction of these displacements, Δx can have negative or positive values (according to a point of origin). That is why the average of the Δx values for all molecules tends to 0.

On the other hand, the Δx values are very small compared to the distances that the particles travel. When they collide, they lose mobility in one direction, consequently having a limited net displacement; for example, they travel 2 cm in one direction after traveling hundreds of meters in collisions and rebounds.

Einstein in 1905 found a mathematical expression for the mean of the square of the displacement (therefore different from 0):

<(Δx) ^{2} > = 2Dt

Defining

(Δx) _{rms} ≡ <(Δx) ^{2} > ^{1/2} = (2Dt) ^{1/2}

(Δx) _{rms} is the net mean square displacement of the particles in question. (Δx) _{rms} tells us how much a particle moves on average (in a positive or negative direction) according to time. Some particles will travel distances farther or closer than (Δx) _{rms} , causing a Gaussian distribution.

__Fick’s second law__

__Fick’s second law__

**Equation**

Fick’s first law describes diffusion under stationary conditions, that is, the mass flow j does not vary with time. In real systems, however, we have non-stationary conditions, where the mass flow varies not only in space, but also with time. Therefore, it is interesting to determine (∂C / ∂t).

Below we have two equations that represent Fick’s second law:

The 2nd form is the most important of all, as it represents the general mathematical equation for any diffusion process; be it thermal, electrical, atomic, etc.

**Deduction**

Consider again the purple particles in a rectangular chamber. Between the distances x and x + Δx we have a flow j _{x} (incoming) and j _{x + Δx} (outgoing). The volume of the chamber between these distances is defined by:

ΔV = AΔX

Note that the graph C (x) vs x does not originate a straight line, so we have different values of j (j _{x} ≠ j _{x + Δx} ). We must determine ΔC / Δt.

The mass m _{x} will be equal to:

m _{x} = j _{x} AΔt

A dimensional analysis helps to understand why:

kg = (kg · m ^{-2} · s ^{-1} ) (m ^{2} ) (s)

In the same way we calculate m _{x + Δx} :

m _{x + Δx} = j _{x + Δx} AΔt

Being the mass that accumulates in said region equal to Δm:

Δm = m _{x} – m _{x + Δx}

= (j _{x} – j _{x + Δx} ) AΔt

= – (j _{x + Δx} – j _{x} ) AΔt

= -ΔjAΔt

And knowing that ΔC = Δm / ΔV

ΔC = -ΔjAΔt / ΔV

= -ΔjAΔt / AΔx

= -ΔjΔt / Δx

We solve for ΔC / Δt

ΔC / Δt = -Δj / Δx

This expression indicates that the variation of the concentration over time is equal to the variation of the flow j with respect to its displacement. Applying the limits for Δt and Δx tending to 0 we obtain the same expression as a partial derivative:

∂C / ∂t = – (∂j / ∂x) (1st form)

Finally, the 2nd form is obtained by substituting j with Fick’s first law:

∂C / ∂t = -∂ / ∂x (-D∂C / ∂x)

= D (∂ ^{2} C / ∂x ^{2} ) (2nd form)

__Solved exercises__

__Solved exercises__

In the following exercises we will consider very simple systems whose unknowns can be solved using Fick’s first law.

**Exercise 1**

**Statement**

In a pipe 15 meters long and 21 centimeters wide, and which is also saturated with nitrogen, a stream of oxygen is diffused from one end to the other at a temperature of 0 ºC. Knowing that the pressure on the left side (P _{1} ) is 20 kg / m ^{3} , and that the pressure on the right side (P _{2} ) is 10 kg, determine:

a) the mass flow that is diffused

b) How many kilograms of O _{2} will diffuse through the pipe in 17 minutes?

c) the concentration or pressure gradient

d) the pressure of O _{2} at a distance of 7 meters from the inlet to the pipe

e) How long will it take 80 kg of O _{2} to diffuse through this pipe?

Consider that D _{O2-N2} is equal to 1.8 · 10 ^{-5} m ^{2} · s ^{-1} .

**Resolution**

From Fick’s first law we have to solve part a):

j = -D (P _{2} -P _{1} ) / L

= – (1.8 10 ^{-5} m ^{2} s ^{-1} ) (10-20) (kg / m ^{3} ) / (15m)

= 1.2 10 ^{-5} kg m ^{-2} s ^{-1}

For b) we need the area of the pipe:

A = π (0.21 m) ^{2}

= 0.14 m ^{2}

And we multiply j by A and time t to determine the mass of diffused O _{2} :

m _{O2} = (1.2 · 10 ^{-5} kg · m ^{-2} · s ^{-1} ) (17 s) (0.14 m ^{2} )

= 3.57 10 ^{-5} kg

Now, for part c) we have that the gradient is equal to:

Gradient = (P _{2} -P _{1} ) / L

= (10-20) (kg / m ^{3} ) / 15 m

= -2/3 (kg / m ^{3} ) m ^{-1}

But we take the positive value, which makes physical sense:

2/3 (kg / m ^{3} ) m ^{-1}

This value will then help us to solve part d) if the gradient is interpreted correctly: every meter the pressure of O _{2} will fall 2/3 kg / m ^{3} . When spreading 7 meters we will therefore have:

2/3 (kg / m ^{3} ) m ^{-1} (7 m) = 14/3 or 4.7 kg / m ^{3}

That is, the pressure at that distance will be:

(20-4.7) (kg / m ^{3} ) = 15.3 kg / m ^{3}

And finally, part e) is similar to b), only now we clear the time and not the mass:

m _{O2} = jAt

t = m _{O2} / jA

= (80 kg) / (1.2 · 10 ^{-5} kg · m ^{-2} · s ^{-1} ) (0.14 m ^{2} )

= 47619.04 s or 0.55 day

**Exercise 2**

**Statement**

Determine (Δx) _{rms} for sucrose in water at t = 1 min, 1 h and 1 day. The diffusion coefficient of sucrose in water is 0.52 · 10 ^{-5} cm ^{2} · s ^{-1} .

**Resolution**

We apply the equation:

(Δx) _{rms} ≡ <(Δx) ^{2} > ^{1/2} = (2Dt) ^{1/2}

We evaluate (Δx) _{rms} with the times expressed in seconds. For t = 1 min or 60 s:

(Δx) _{rms} = ((2 (0.52 · 10 ^{-5} cm ^{2} · s ^{-1} ) (60s)) ^{1/2}

= 0.025 cm

For t = 1 h or 3600 s:

(Δx) _{rms} = ((2 (0.52 · 10 ^{-5} cm ^{2} · s ^{-1} ) (3600s)) ^{1/2}

= 0.19 cm

And finally for t = 1 day or 86400 s:

(Δx) _{rms} = ((2 (0.52 · 10 ^{-5} cm ^{2} · s ^{-1} ) (86400s)) ^{1/2}

= 0.95 cm

Note that as time passes the sucrose molecules have not even been able to move 1 cm in either direction.