--%>

What is Ideal Mixtures

Ideal mixing properties can be recognized in the formation of an ideal gas mixture from ideal gases.

Consider the formation of a mixture of gases i.e. a gaseous solution, from two mixtures of pure gases. A useful characterization of an ideal mixture, or solution, can be obtained by beginning with Dalton's law of partial pressures. That law, as seen in the pressure needed to confine a mixture of gases to a container is equal to the sum of the pressures that would be needed to confine the gas components separately to the same container.

The formation of Dalton's law binary mixture can be pictured by the process suggested in the fig. we begin with the gas sample containing of the separate components, each at pressure P. the mixing process consists of the expansion of each component to fill the entire container.

Suppose there are two containers nA mol of A and nB mol of B. the gas sample, both before and after mixing, has a volume V, and pressure to confine the gas to this volume is P. before mixing, the components are both occupy the total volume, and the pressures, or partial pressures, needed to confine them are also proportion to the number of moles. The relations that are implied are shown in fig.

The exponent of each component in this ideal gas mixture process occurs without regard to the presence of the other component. The change that occurs in the mixing is the sum of the changes experienced by each component.

From the relation between free energy and pressure for an ideal gas so that we have:

GA (in mixture) - GA (pure) = nRT in xB

G(in mixture) - GB (pure) = nRT in xB  


Ideal solutions: the free energy result of the above equation was developed by piecing together features of ideal behavior. In a more elegant procedure, adherence to the equation and to the consequences of this equation is used as the definition of ideal solution behavior. The entropy and free energy changes for the formation of 1 mol of an ideal gas solution are shown in the fig. and along with enthalpy it is accurate. Gas mixtures, except a high pressures or low temperatures, confirm to these ideal mixture characteristics. In what follows we treat gas mixtures as ideal.

Liquid mixtures, i.e. solutions, generally do not; behave according to these ideal mixing expressions. The volume of the solution is not always equal to the sum of the volume of the separate components. In the formation of a solution energy must often be exchanged with the thermal surroundings to maintain a constant temperature. Only for a few solutions are the free energy and entropy changes given by the ideal solution expressions. 

Entropy and free energy change at 25°C for formation of 1 mol of an ideal binary solution:

Mole fraction (xA) Mole fraction (xB) xA R In xA, Jk-1mol-1 xB R In xB, Jk-1mol-1 ΔSmix, JK-1mol-1 T ΔSmix, J mol-1 ΔGmix, J mol-1
1 0 0 0 0 0 -0
0.9 0.1 -0.79 -1.91 2.70 805 -805
0.8 0.2 -1.48 -2.68 4.16 1240 -1240
0.7 0.3 -2.08 -3.00 5.08 1510 -1510
0.6 0.4 -2.55 -3.05 5.60 1670 -1670
0.5 0.5 -2.88 -2.88 5.76 1720 -1720
0.4 0.6 -3.05 -2.55 5.60 1670 -1670
0.3 0.7 -3.00 -2.08 5.08 1510 -1510
0.2 0.8 -2.68 -1.48 4.16 1240 -1240
0.1 0.9 -1.91 -0.79 2.70 805 -805
0 0 0 0 0 0 -0

   Related Questions in Chemistry

  • Q : Define tripod and its use Illustrate a

    Illustrate a tripod? And how it’s used?

  • Q : Law of vapour pressure Select the right

    Select the right answer of the question. "The relative lowering of the vapour pressure is equal to the mole fraction of the solute." This law is called: (a) Henry's law (b) Raoult's law (c) Ostwald's law (d) Arrhenius's law

  • Q : Concentration of an aqueous solution

    Give me answer of this question. The concentration of an aqueous solution of 0.01M CH3OH solution is very nearly equal to which of the following : (a) 0.01%CH3OH (b) 0.1%CH3OH (c) xCH3OH= 0.01 (d) 0.99MH2O (

  • Q : Moles of HCl present in .70 L of a .33

    Detail the moles of HCl which are present in .70 L of a .33 M HCl solution?

  • Q : Group IV Cations Chromium(III)

    Chromium(III) hydroxide is highly insoluble in distilled water but dissolves readily in either acidic or basic solution. Briefly explain why the compound can dissolve in acidic or in basic but not in neutral solution. Write appropriate equations to support your answer.

  • Q : Problem on decomposition reaction

    Nitrogen tetroxide (melting point: -11.2°C, normal boiling point 21.15°C) decomposes into nitrogen dioxide according to the following reaction: N2O4(g) ↔ 2 NO2(g)<

  • Q : Strength of Nacl in solution To 5.85gm

    To 5.85gm of Nacl one kg of water is added to prepare of solution. What is the strength of Nacl in this solution (mol. wt. of nacl = 58.5)? (a) 0.1 Normal (b) 0.1 Molal (c) 0.1 Molar (d) 0.1 FormalAnswer:

  • Q : Explain reactions of carbonyl oxygen

    In these reaction oxygen atom of carbonyl group is replaced by either one divalent group or two monovalent groups. Reaction by ammonia derivatives: aldehydes and ketones react with a number of ammonia derivatives such as hydroxylaminem hydrazine, semicarbazide etc. in weak acidic medium.

  • Q : Acid Solutions Choose the right answer

    Choose the right answer from following. Volume of water needed to mix with 10 ml 10N NHO3 to get 0.1 N HNO3: (a) 1000 ml (b) 990 ml (c) 1010 ml (d) 10 ml

  • Q : Mole fraction of Carbon dioxide Choose

    Choose the right answer from following. If we take 44g of CO2 and 14g of N2 what will be mole fraction of CO2 in the mixture: (a) 1/5 (b) 1/3 (c) 2/3 (d) 1/4