**Mixtures**

We have seen that the combined first- and second laws for an open system is written,

(1)Where the initial definitions of the

(2)An equivalent, and more easily understood, definition of the

(3)

(4)Equation 4 is an exact thermodynamic equation and contains no approximations.

Notice in Equation 3 that *μ _{i}*
is intensive, it is given by the extensive

(5)(We will use the bar over the symbol to indicate partial molar quantities. Some texts use a subscript, m, as in

Let us regard *V* as a function of temperature, pressure and composition:

(6a, b)If we hold

(7)which can be integrated (similar to the way we integrated Equation 1, above) to give,

(8a, b)The volumes given in Equations 7, 8a, and 8b are the partial molar volumes which we have already said are not necessarily equal to the molar volumes of the pure components. Thus, Equations 8a and b tell us that volumes may not be additive. That is, if we were to mix one liter of pure ethanol with one liter of pure water the final volume of the mixture would not likely be two liters. That is because water molecules interact with ethanol molecules differently than they interact with other water molecules.

As we have said above, we can define a partial molar quantity from any extensive variable. Thus we can define the partial molar entropy as

(9)or the partial molar enthalpy as,

(10)It is also true that

(11)and so on.

**How to measure partial molar volumes**

There are several ways that partial molar volumes can be measured. One way is to begin with one mole of a compound, call it component 1, add a small amount of component 2 and measure the volume, add a little more of component 2 and measure the volume again. Keep doing this until the desired concentration range has been covered. Then fit the volume data to a curve, for example, of the form,

(12)(Why would we not want to include that 1/2 power of

(13a, b)The partial molar volume of component 1 can be obtained from,

(14a, b)or

(15a, b) (The last step is because, in this case we have setn_{1}= 1.)

**Ideal Solutions**

We will define an ideal solution as a solution for which the chemical potential of each component is given by,

(16)where is the chemical potential of pure component

(Many texts define an ideal solution as a solution which obeys Raoult's law over the full range of composition,

(17)where is the vapor pressure of pure component

We will now prove that an ideal solution obeys Raoult's law (using our definition of an ideal solution).

Consider a solution of two components where the mole fraction of component
1 is *X*_{1}. We know that the chemical potential of component
1 must be the same in the solution as in the vapor in equilibrium with
the solution. That is,

(18)but the solution is ideal so,

(16)Also, we can approximate the vapor as an ideal gas so,

(19)where

(20)Equation 20 doesn't help us very much all by itself. However we have some more information. We know that for the pure component 1 we have

(21)Let us now subtract Equation 21 from Equation 20 to get

(22)from which we conclude that,

(23a, b)or

(17)which is Raoult's law.

**Example calculation using Raoult's law**

Benzene and toluene form a solution which is very nearly ideal. Consider
a mixture of benzene (Bz) and toluene (Tol) at 60^{o} C. At 60^{o}
C the vapor pressures of pure benzene and pure toluene are 385 Torr and
139 Torr, respectively. What are the vapor pressures of benzene and toluene
in a mixture with *X*_{Bz} = 0.400, and *X*_{Tol}
= 0.600, and what is the composition of the vapor in equilibrium with this
solution?

Use Raoult's law to find the vapor pressures of the two species,

The total pressure is the sum of these two individual pressures, 237 Torr.The composition of the vapor phase is obtained from the vapor pressures and Dalton's law of partial pressures,

Notice that the composition of the vapor is not the same as the composition of the liquid, the vapor phase is much richer in the more volatile compound, benzene. This fact will be important when we discuss vapor pressure diagrams and two-component phase diagrams.**Properties of ideal solutions**

Given the chemical potentials for the components of an ideal solution we can calculate a number of properties of ideal solutions. For example, the Gibbs free energy of mixing is the easiest to calculate,

(24a, b, c, d)To bring this equation into the usual form multiply and divide by the total number of moles,

(25)Notice that the Gibbs free energy of mixing is negative, as one would expect for a spontaneous process at constant temperature and pressure.

We can also calculate other properties of mixing ideal solutions. The entropy of mixing is,

(26)which is the same as the entropy of mixing for ideal gases.

The volume change on mixing can be found from,

(27)so that volumes are additive for an ideal solution. That is, if we were to mix one liter of benzene and one liter of toluene the final volume of the solution would be two liters.

We can also determine whether or not there is any heat of reaction.

(28)That is, there is no heat involved in the mixing of ideal solutions. If we mix several components and the mixture gets hot or cold we can be sure that the solution formed was not ideal.

WRS

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