--%>

Diffusion Molecular View

When the diffusion process is treated as the movement of particles through a solvent the diffusion coefficient can be related to the effective size of diffusing particles and the viscosity of the medium.

To see how the experimental coefficients can be treated to properties of the system and particularly of the solute macromolecules we take a molecular view of the diffusion process. Consider across a distance interval dx over which the concentration changes from c to c-dc. The force that drives the molecules to the ore dilute region can be related to the difference in the, molar free energy of the solute at concentration c and at concentration c-dc. If deal behaviour is assumed, the free energy differences per molecule is

Gc - dc - Gc = RT/N In (c -dc)/c 

Or

dG = RT/N In (1 - dc/c) - RT/N dc/c  where the relation In (1 - y) = -y for small y has been used.

This free energy difference corresponds to the mechanical energy needed to transfer one macromolecule across the distance dx. This energy can therefore be written as a force times the distance dx. Thud dG = driving force × dx, or

Driving force = dG/dx = RT/N 1/c dc/dx

A frictional force sets in and balances this diffusion force when some constant velocity is reached. The frictional force exerted by a viscous solvent fluid of viscosity η has been derived for a macroscopic sphere of radius r by G. G strokes as 

Frictional force = 6∏rη dx/dt

It appears suitable to apply this expression to the motion of reasonably spherical macromolecules. The diffusion velocity increases, therefore, until the force balances that equation. Then

6∏rη dx/dt = - RT/N 1/c dc/dx 

Or

cdx/dt = - RT/(6∏rη) dc/dx

Since c implies a mass per unit volume measure of concentrations, the product c dx/dt can be interrupted as the rate with which the diffusing substance moves through a unit cross section at x. this follows suggests, from the fact that dx/dt, the average diffusion velocity in the x direction, is the distance the diffusing molecules travel per unit time. Thus all the molecules within a distance dx/dt of a cross section will pass cross section in unit time. These molecules are in a volume equal to dx/dt times the cross section area. The mass of these molecules is the product of this volume and the concentration expressed as mass per unit volume. Thus c dx/dt is the amount per unit time, i.e. the rate with which the solute passes through the cross section. We can write now

D ∂c/∂x = - RT/(6∏rη) ∂c/∂x

This leads to the identification

D = RT/(6∏rη) 

And 6∏rη = RT/DN

Measurements of D and η could therefore lead to a value of the radius r for the macromolecule. Such a procedure is a little unsatisfactory. Molecules do not necessarily obey Strokes' law, even if they are spherical. Furthermore, macromolecules will generally be solvated and in moving through the solution will to some extent vary along this salvation layer. Equation is important however, in that it provides a way of determining the effective value of the group of terms 6∏rη for a solute characterized by molecules with radius r and a solvent characterized by viscosity η

   Related Questions in Chemistry

  • Q : Electrochemistry ( electrolysis of

    1. Define Faraday's first law of electrolysis 2. define Faraday's second law of electrolysis

  • Q : Explain alcohols and phenols in organic

    Alcohols and phenols are the compounds

  • Q : Explain the polymers and its types.

    Polymers are the chief products of modern chemical industry which form the backbone of present society. Daily life without the discovery and varied applications of polymers would not have been easier and colourful. The materials made of polymers find multifarious uses and applications in all walk

  • Q : Macromolecules what are condensation

    what are condensation polymerization give in with 2 examples

  • Q : Schrodinger equation with particle in a

    Three dimensional applications of the Schrodinger equation are introduced by the particle-in-a-box problem.So far only a one-dimensional problem has been solved by application of the Schrodinger equation. Now the allowed energies and the probability functi

  • Q : Molar concentration of hydrogen 20 g of

    20 g of hydrogen is present in 5 litre of vessel. Determine he molar concentration of hydrogen: (a) 4  (b) 1 (c) 3 (d) 2 Choose the right answer from above.

  • Q : Vander Waals forces Wax is an example

    Wax is an example of: (a) Ionic crystal  (b) Covalent crystal  (c) Metallic crystal  (d) Molecular crystalAnswer: (d) Iodine crystals are molecular crystals, in which constituent particles are molecules having inter particle

  • Q : Problem on Redlich-Kwong equation i)

    i) Welcome to Beaver Gas Co.! Your first task is to calculate the annual gross sales of our superpure-grade nitrogen and oxygen gases. a) The total gross sales of N2 is 30,000 units. Take the volume of the cylinder to be

  • Q : Net charge of a non-ionized atom

    Describe the net charge of a non-ionized atom?

  • Q : Vapour pressure of water Give me answer

    Give me answer of this question. 5cm3 of acetone is added to 100cm3 of water, the vapour pressure of water over the solution: (a) It will be equal to the vapour pressure of pure water (b) It will be less than the vapour pressure of pure water