--%>

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 : Particles of quartz Particles of quartz

    Particles of quartz are packed by:(i) Electrical attraction forces  (ii) Vander Waal's forces  (iii) Covalent bond forces  (iv) Strong electrostatic force of attraction Answer: (iii)

  • Q : Amines arrange in decreasing order of

    arrange in decreasing order of basicity pi pyridine,pyridine,pyrrole, morphine

  • Q : Molarity A solution has volume 200ml

    A solution has volume 200ml and molarity 0.1.if it is diluted 5times then calculate the molarity of reasulying solution and the amount of water added to it.

  • Q : Molecular Properties Symmetry Molecular

    Molecular orbitals and molecular motions belong to certain symmetry species of the point group of the molecule.Examples of the special ways in which vectors or functions can be affected by symmetry operations are illustrated here. All wave functions soluti

  • Q : Question based on strength of solution

    Help me to go through this problem. On dissolving 1 mole of each of the following acids in 1 litre water, the acid which does not give a solution of strength 1N is: (a) HCl (b) Perchloric acid (c) HNO3 (d) Phosphoric acid

  • Q : Volume of solution containing solute

    What volume of solution contains 0.1 mole of the solute: (a) 100ml (b) 125ml  (c) 500ml (d) 62.5ml Choose the right answer from above.

  • Q : Neutralisation of phosphorous acids

    Provide solution of this question. To neutralise completely 20 mL of 0.1 M aqueous solution of phosphorous acid (H3 PO3) the volume of 0.1 M aqueous KOH solution required is: (a) 40 mL (b) 20 mL (c) 10 mL (d) 60 mL

  • Q : What are aliphatic amines and its

    In common system, the aliphatic amines are named by using prefix for alkyl group followed by the word amine.In case of mixed amines, the name of alkyl groups are arranged in alphabetical order. This is followed by the word amine. However, for simple secondary or tertiary amines anothe

  • Q : Which is largest planet in our solar

    which is largest planet in our solar system

  • Q : Application of colligative properties

    Choose the right answer from following. Colligative properties are used for the determination of: (a) Molar Mass (b) Equivalent weight (c) Arrangement of molecules (d) Melting point and boiling point (d) Both (a) and (b)