--%>

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 : What are ion selective electrodes? Ion

    Ion Selective Electrodes An ion selective membrane can be used to form an electrochemical cell whose emf depends on the concentration of that ion. Before we proceed to an important application of emf measurements, brie

  • Q : Ionization Potential Second ionization

    Second ionization potential of Li, Be and B is in the order (a)Li>Be>B (b)Li>B>Be (c)Be>Li>B (d)B>Be>Li

  • Q : Wavelengths which the human eye can see

    Briefly state the wavelengths which the human eye can see?

  • Q : Question based on relative lowering of

    Give me answer of this question. When a non-volatile solute is dissolved in a solvent, the relative lowering of vapour pressure is equal to: (a) Mole fraction of solute (b) Mole fraction of solvent (c) Concentration of the solute in grams per litre (d) Concentratio

  • Q : Unit of molality Select the right

    Select the right answer of the question. The unit of molality is: (a) Mole per litre (b) Mole per kilogram (c) Per mole per litre (d) Mole litre

  • Q : Describe chemical properties of amines.

    Like ammonia, primary, secondary and tertiary amines have a single pair of electrons on N atom. Hence chemical behavior of amines is similar to ammonia. Amines are basic in nature, and in most of the reactions they act as nucleophiles.      1. Reaction wi

  • Q : What is Henry law constant and its

    1. The units of Henry Law constant are same as those of pressure, i.e. torr or h bar. 2. Different gases have dissimilar values of Henry law constant. The values of KH for some gases in water are given in tabl

  • Q : Neutralization of benzoic acid Choose

    Choose the right answer from following. How many grams of NaOH will be required to neutralize 12.2 grams of benzoic acid : (a) 40gms (b) 4gms (c)16gms (d)12.2gms

  • 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 : Relationship between Pressure and

    The pressure-temperature relation for solid-vapor or liquid vapor equilibrium is expressed by the Clausis-Clapeyron equation.We now obtain an expression for the pressure-temperature dependence of the state of equilibrium between two phases. To be specific,