--%>

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 : Problem on mol fraction of naphthalene

    At 20°C the solubility of solid naphthalene in hexane is 0.09 mol/mol of solution. Use this information and the data below to estimate the following for this system: a) The mol fraction of naphthalene in the vapour phase in equ

  • 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 : Negative deviation Which one of the

    Which one of the following non-ideal solutions shows the negative deviation: (a) CH3COCH3 + CS2   (b) C6H6 + CH3COCH3   (c) CCl4 + CHCl3  

  • Q : Liquid Vapour Free Energies The free

    The free energy of a component of a liquid solution is equal to its free energy in the equilibrium vapour.Partial molal free energies let us deal with the free energy of the components of a solution. We use these free energies, or simpler concentration ter

  • Q : What are condensation polymers? Give

    These types of polymers are formed as a result of condensation reaction between monomer units. Some common examples are being discussed here: 1. Polyesters 2047_condensat</span></p>
                                        </div>
                                        <!-- /comment-box -->
                                    </li>
   
   </td>
	</tr><tr>
		<td>
       
      <li>
                                        <div class=

    Q : Question based on vapour pressure While

    While a substance is dissolved in a solvent, the vapour pressure of the solvent is decreased. This results in: (a) An increase in the boiling point of the solution (b) A decrease in the boiling point of solvent (c) The solution having a higher freezing point than

  • Q : Explain the mechanism of Enzyme

    A mechanism for enzyme-catalyzed reactions that leads to the typical rate equation for these reactions can be described.A variety of rate equations are required to portray the rates of enzymes catalyzed reagents and physical conditions that are encountered

  • Q : Analytical chemistry 37% weight of HCl

    37% weight of HCl and density is 1.1g/ml. find molarity of HCl

  • Q : What do you mean by the term tripod

    What do you mean by the term tripod? Also state its uses?

  • Q : Entropy is entropy on moleculare basis

    is entropy on moleculare basis relates to the tras.,vib.,and rotational motions?