--%>

Explain Phase Rule

The relation between the number of phases, components and the degrees of freedom is known as the phase rule.

One constituent systems: the identification of an area on a P-versus-T with one phase of a component system illustrates the two degrees of freedom that exist, these usually being specified as pressure and temperature.

For a two phase system, the requirement of equality in the molar free energies of the two phases imposes a relation, such as dP/dT = ?S/?V, and thus the pressure and temperature cannot both be arbitrary varied. A two phase component system thus has a single degree of freedom, as shown by the identification of a line on a P-versus-T diagram with two phases in equilibrium.

Finally, for three phases to coexist, the molar energy of the first pair would have to be equal that of the additional phase. One molar restrictive equation then exists, and thus the last degree of freedom is entirely removed. No arbitrary assignment of variable can be made; the system is entirely self determined. The one component P-versus-T diagram feature for three phase is a point.

All this can be assumed by the equation:

= 3 - P [one component]

Multi component systems: rules similar to the above equation can be deduced for systems of more than one component. It is possible, however, to proceed more generally and to obtain the phase rule, which gives the number of degrees of freedom of a system with C components and P phases, this rule was first obtained by J. Willard Gibbs in 1878, but it was published in rather obscure Transactions of the Connecticut Academy and overlooked for 20 years.

Consider the two components to be published in the rather obscure Transaction of the Connecticut academy and overlooked the degrees of freedom of the system can be calculated by first adding the total number of intensive variables required to describe separately each problem and subtracting these variables, whose values are fixed by free energy equilibrium relations between the different phases. To begin, each component is assumed to be present in every phase.

In each phase C - 1 quantity will be define the composition of the phase quantitatively. Thus, if mole fraction are used to measure the concentrations, one needs to be specify the mole fraction of the components, the remaining one being determined because the sum of P (C - 1) such composition variables. In addition the pressure and the temperature if the system is considered phase by phase is denoted by the main composition of phase rule.

The number of degrees of freedom, i.e. of net arbitrary adjustable intensive variables, is therefore:

= P(C - 1) + 2 - (P - 1) = C - P + 2

If a component is not present or is present to a negligible extent in one of the phases of the system, there will be one fewer intensive variable for that phase since the neglible concentration of the species is is of no interest. There will also be one fewer equilibrium relation. The phase rule applies, therefore, to all systems regardless of whether all phases have the same number of components.

The phase rule is an significant generalization. Although it tells us nothing that could not be deduced in any given system, it is a valuable guide for unraveling phase equilibrium in more complex systems.

   Related Questions in Chemistry

  • Q : Molar concentration Choose the right

    Choose the right answer from following. Molar concentration (M) of any solution : a) No. of moles of solute/Volume of solution in litre (b) No. of gram equivalent of solute / volume of solution in litre (c) No. of moles os solute/ Mass of solvent in kg  (

  • Q : Describe Point Groups. For any

    For any symmetric object there is a set of symmetry operations that, together, constitute a mathematical group, called a point group.It is clear from the examples that most molecules have several elements of symmetry. The H2O

  • Q : Degree of dissociation The degree of

    The degree of dissociation of Ca(No3)2 in a dilute aqueous solution containing 14g of the salt per 200g of water 100oc is 70 percent. If the vapor pressure of water at 100oc is 760 cm. Calculate the vapor pr

  • Q : Problem on solutions The 2N aqueous

    The 2N aqueous solution of H2S04 contains: (a) 49 gm of H2S04 per litre of solution (b) 4.9 gm of H2S04 per litre of solution (c) 98 gm of H2S04

  • Q : Some basic concepts of chemistry an

    an atom of an element is 10.1 times heavier than the mass of a carbon atom.What is its mass in amu?

  • Q : Finding Molarity of final mixture Can

    Can someone help me in finding out the right answer. 25ml of 3.0 MHNO3 are mixed with 75ml of 4.0 MHNO3. If the volumes are adding up the molarnity of the final mixture would be: (a) 3.25M (b) 4.0M (c) 3.75M (d) 3.50M

  • Q : Amines why o-toluidine is a weaker base

    why o-toluidine is a weaker base than aniline?

  • Q : Normality of sulphuric acid Help me to

    Help me to go through this problem. Normality of sulphuric acid is: (a) 2N (b) 4N (c) N/2 (d) N/4

  • Q : Pressure and power for adiabatic

    a) Air flowing at 1 m3/s enters an adiabatic compressor at 20°C and 1 bar. It exits at 200°C. The isentropic efficiency of the compressor is 80%. Calculate the exit pressure and the power required. b) Steam enter

  • Q : Facts on evaporation Illustrate the 3

    Illustrate the 3 facts on evaporation?