--%>

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 : Mole fraction of water and ethanol Give

    Give me answer of this question. A solution contains 1 mole of water and 4 mole of ethanol. The mole fraction of water and ethanol will be: (a) 0.2 water + 0.8 ethanol (b) 0.4 water + 0.6 ethanol (c) 0.6 water + 0.8 ethanol (d) 0.8 water + 0.2 ethanol

  • Q : Problem based on molality of glucose

    Select the right answer of the question. If 18 gm of glucose (C6H12O6) is present in 1000 gm of an aqueous solution of glucose, it is said to be: (a)1 molal (b)1.1 molal (c)0.5 molal (d)0.1 molal

  • Q : Normality of solution containing

    Can someone please help me in getting through this problem. Determine the normality of a solution having 4.9 gm H3PO4 dissolved in 500 ml water: (a) 0.3  (b) 1.0  (c) 3.0   (d) 0.1

  • Q : Describe Transformation Matrices. Each

    Each symmetry operation can be represented by a transformation matrix.You have seen what happens when a molecule is subjected to the symmetry operation that corresponds to any of the symmetry elements of the point group to which the molecule belongs. The m

  • Q : Chem Silicon has three naturally

    Silicon has three naturally occurring isotopes. 28Si, mass = 27.976927; 29Si, mass = 28.976495; 30Si, mass = 29.973770 and 3.10% abundance. What is the abundance of 28Si?

  • Q : Linde liquefaction process Liquefied

    Liquefied natural gas (LNG) is produced using a Linde liquefaction process from pure methane gas at 3 bar and 280 K (conditions at point 1 in figure below). A three-stage compressor with interceding is used to compress the methane to 100 bar (point 2). The first stage

  • Q : Preparation of normal solution Give me

    Give me answer of this question. What weight of ferrous ammonium sulphate is requiored to prepare 100 ml of 0.1 normal solution (mol. wt. 392): (a) 39.2 gm (b) 3.92 gm (c)1.96 gm (d)19.6 gm

  • Q : Product of HCl Zn Illustrate  the

    Illustrate  the product of HCl Zn?

  • Q : How haloalkanes are prepared from

    This is the common method for preparing haloalkanes in laboratory. Alcohols can be converted to haloalkanes by substitution of - OH group with a halogen atom. Different reagents can be used to get haloa

  • Q : Problem on making solution Select the

    Select the right answer of the question. The weight of H2C2O42H2O required to prepare 500ml of 0.2N solution is : (a) 126g (b) 12.6g (c) 63g (d) 6.3g