--%>

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 : Problem based on normality Choose the

    Choose the right answer from following. NaClO solution reacts with H2SO3 as,. NaClO + H2SO3→NaCl+ H2SO4. A solution of NaClO utilized in the above reaction contained 15g of NaClO per litre. The

  • Q : Mass percent Help me to go through this

    Help me to go through this problem. 10 grams of a solute is dissolved in 90 grams of a solvent. Its mass percent in solution is : (a) 0.01 (b) 11.1 (c)10 (d) 9

  • Q : Hydrocarbons list and identify

    list and identify differences between the major classes of hydrocarbons

  • Q : Problem on partial pressure i) Show

    i) Show that the equilibrium constant Kp for the reaction CaCo3(s) ↔ CaO(s) +CO2(g)is about unity (i.e. = 1.0) at 895 °C.ii) If two grams of calcium carbonate are pl

  • Q : Question based on vapour pressure and

    Give me answer of this question. The vapour pressure of water at 20degreeC is 17.54 mm. When 20g of a non-ionic, substance is dissolved in 100g of water, the vapour pressure is lowered by 0.30 mm. What is the molecular weight of the substances: (a) 210.2 (b) 206.88

  • Q : Calculating molarity of a solution

    Select the right answer of the question .The molarity of a 0.2 N N2Co3 solution will be: (a) 0.05 M (b) 0.2 M (c) 0.1 M (d)0.4 M

  • Q : Question associated to vapour pressure

    Choose the right answer from following. The vapour pressure lowering caused by the addition of 100 g of sucrose(molecular mass = 342) to 1000 g of water if the vapour pressure of pure water at 25degree C is 23.8 mm Hg: (a)1.25 mm Hg (b) 0.125 mm Hg (c) 1.15 mm H

  • Q : Death cap musrooms the death cap

    the death cap mushroom based on your knowledge of the biochemistry of dna and rna

  • Q : Problem associated to vapour pressure

    Provide solution of this question. 60 gm of Urea (Mol. wt 60) was dissolved in 9.9 moles, of water. If the vapour pressure of pure water is P0 , the vapour pressure of solution is:(a) 0.10P0 (b) 1.10P0 (c) 0.90P0 (d) 0.99P0

  • 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