Physical Principles in BiologyBiology 3550

Fall 2021

Lecture 21:

Rates of Diffusion in Liquids and Gasses

Wednesday, 20 October 2021

©David P. GoldenbergUniversity of Utah

goldenberg@biology.utah.edu

A Quick Review from the Last Lecture

Kinetic energy of an object along the x-axis: Ek;x = mv 2x =2

RMS translational kinetic energy of a molecule: Ek;x = kT=2

RMS velocity of molecule: vx = ‹x=fi =pkT=m

The diffusion coefficient:

D =‹2x2fi

=vx‹x2

Calculating ‹x and fi from D and vx :

‹x =2D

vx=

2DpkT=m

fi =‹xvx

=‹2x2D

Random Walk of a Small Molecule, Bromophenol Blue,in Water

From diffusion experiment: D = 2× 10−10m2=s

Mass = 1.11×10−24 Kg

Velocity:

vx =pkT=m =

p1:38× 10−23 Kg ·m2s−2K−1 × 298K=1:1× 10−24 Kg

= 61m=s

RMS displacement of random-walk step:

‹x =2D

vx= 2× 2× 10−10m2=s÷ 61m=s

= 6:5× 10−12m

RMS time interval for random-walk step:

fi =‹xvx

= 6:5× 10−12m÷ 61m=s = 10−13 s

Clicker Question #1

A “typical” protein molecule: D = 1× 10−11m2=s and vx = 6m=s

What is the RMS step length, ‹x , as the molecule diffuses via a randomwalk?

A) 1×10−13 m

B) 3×10−13 m

C) 1×10−12 m

D) 3×10−12 m

E) 1×10−11 m

‹x =2D

vx=

2× 1× 10−11m2=s

6m=s= 3:3× 10−12m

Clicker Question #2

A “typical” protein molecule: D = 1× 10−11m2=s and vx = 6m=s.

What is the RMS average time between direction changes, fi , as themolecule diffuses via a random walk?

A) 5×10−14 s

B) 5×10−13 s

C) 5×10−12 s

D) 5×10−11 s

E) 5×10−10 m

fi =‹xvx

=3:3× 10−12m

6m=s= 5:5× 10−13 s

A Small Molecule Versus a Big One

Bromophenolblue

Protein

Molar mass 670 g/mol 60,000 g/mol

Velocity 60 m/s 6 m/s

D 2× 10−10m2=s 1× 10−11m2=s

‹x 6×10−12 m 3×10−12 m

fi 1×10−13 s 5×10−13 s

How do we think about the numbers on a molecular scale?

Have to think about liquids.

The Nature of Liquids

Molecules are very densely packed

Motions of molecules are limited bythe “energy barriers” surroundingthem.

Other molecules in a solution aresimilarly constrained.

Molecules move only very shortdistances before they changedirection.

The larger the molecule is, the lessfreedom it has and the more slowly itmoves.

From a simulation of liquid water.

RMS Distance of Diffusion

Random walk along one direction: 〈x2〉 = n‹2x

For diffusion: D =‹2x2fi

‹2x = 2Dfi

n = t=fi

〈x2〉 = n‹2x =t

fi2Dfi = 2Dt

RMS(x) =√2Dt

For bromophenol blue (and molecules of similar size):

RMS(x) =pt=s × 2× 10−5m

RMS Distance of Diffusion

RMS Distance of Diffusion

“Velocity” of Diffusion

Apparent velocity decreases with time and distance.

Time Required for Diffusion Over a Range of Distances

RMS(x) =√2Dt

2Dt = 〈x2〉

t = 〈x2〉=(2D)

Time required is inversely related to the diffusion coefficient.

Diffusion is effective over short distances, but not long.

Chemical Communication Between Neuronsand Between Neurons and Muscle Cells

Structure of a synapse

Synaptic cleft: ≈ 20 nm wide

Time for diffusion for a small molecule: ≈ 10−6 s = 1—s

Time to diffuse over the length of a sciatic axon (1 m): ≈ 2× 109 s = 80 yr

https://en.wikipedia.org/wiki/Chemical_synapse

Thomas Splettstoesser (www.scistyle.com)

Calculating Diffusion Coefficients

What determines diffusion coefficient?• Velocity of molecules (temperature)• Size and shape of molecules• How often molecules colide

Stokes-Einstein equation for spherical particles:

D =kT

6ı”r

r = sphere radius

” = viscosity

A key result from the 1905 Einstein paper on Brownian motion.

A testable prediction!

Viscosity: Property of a Liquid (or Gas)Resistance to Motion of Molecules Past One Another

A molecular property easily observed on the macroscopic scale.

From a 1960s TV ad for Prell Shampoohttps://www.youtube.com/watch?v=9lFsrjoLKq0

Reflects what we commonly think of as the “thickness” of a liquid.

Units are not straight forward! See the notes.

Some Calculated Diffusion Coefficients

The Stokes-Einstein equation for spherical particles:

D =kT

6ı”r

In water at 25◦C:• Small molecule (1 nm): 2×10−10 m2s−1

• Protein (10 nm): 2×10−11 m2s−1

• Bacterium (1 —m): 2×10−13 m2s−1

• 1 mm sphere: 2×10−16 m2s−1

Direction Change

Warning!

Diffusion of Gasses

Clicker Question #3

How Far Apart are Molecules in Air?(on average)

A) ≈ 1 nm

B) ≈ 1—m

C) ≈ 1mm

D) ≈ 1 cm

E) ≈ 1m

All answers count for now.

Clicker Question #4

Diffusion coefficients of gasses (at atmospheric pressure):≈ 2× 10−5m2=s

From a previous lecture: RMS velocity of N2: ≈ 300m=s.

Calculate the RMS distance an N2 molecule travels before changingdirection.

A) ≈ 1 nm

B) ≈ 10 nm

C) ≈ 100 nm

D) ≈ 1—m

E) ≈ 10—m

Diffusion in Gasses

Diffusion coefficients of gasses (at atmospheric pressure):≈ 2× 10−5m2=s

From a previous lecture: RMS velocity of N2: ≈ 300m=s.

D = ‹2x=(2fi), and v = ‹x=fi

‹x = 2D=v = 1:3× 10−7m = 130 nm

The “mean-free-path” distance.Is this the distance between molecules?

Time between collisions:

fi = ‹x=v ≈ 4:4× 10−10 s

Much longer random walk steps than in liquids, but still pretty short.