We have seen that the combined first- and second laws for an open system is written,
(1)Where the initial definitions of the μi were given by,
(2)An equivalent, and more easily understood, definition of the μi was given by,
(3)μi is called the "chemical potential" and it can be described in words as the Gibbs free energy per mole of substance. Another name for this quantity is "partial molar Gibbs free energy." We found, using the integrated form of Equation 1 and the definition of the Gibbs free energy, that the Gibbs free energy could be written as,
(4)Equation 4 is an exact thermodynamic equation and contains no approximations.
Notice in Equation 3 that μi is intensive, it is given by the extensive G divided by the extensive ni. Also, the variables that are being held constant are temperature and pressure along with the number of moles of the other components of the system. We can define an equation similar to Equation 3 for any extensive property. The partial molar Gibbs free energy is the most important of these because it provides a measure of the "driving force" for chemical processes. The next most important of these quantities is the partial molar volume,
(5)(We will use the bar over the symbol to indicate partial molar quantities. Some texts use a subscript, m, as in Vmi to indicate them.) The partial molar volume in Equation 5 can be thought of in several ways. It is the incremental volume obtained by adding a small amount of component i to the mixture while holding the temperature, pressure, and the number of moles of all the other components constant divided by the number of moles of component i. Another way to look at it is to say that it is the incremental volume obtained by adding one mole of component i to an infinite sample of the mixture. The partial molar volume is not necessarily the same as the volume of one mole of the pure component. (The intermolecular interaction between molecules of one component and molecules of other components may be different than the interaction of molecules of a component with other molecules of the same component.)
Let us regard V as a function of temperature, pressure and composition:
(6a, b)If we hold T and p constant we get
(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 n2 in Equation 12? People use whatever powers of n2 they need fit the data.) The constants, a, b, c, etc are obtained from the curve fitting and the first term is the molar volume of pure component 1. Then the partial molar volume of component 2 can be obtained by direct differentiation,
(13a, b)The partial molar volume of component 1 can be obtained from,
(15a, b) (The last step is because, in this case we have set n1 = 1.)
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 i, and Xi is the mole fraction of component i in the solution.
(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 i.)
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 X1. 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 p1 is the vapor pressure (partial pressure) of component 1 above the solution. Combining Equations 16, 18, and 19 we get,
(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 X1 = 1, and we know that the pressure of component 1 vapor in equilibrium with the liquid is just the vapor pressure of the pure liquid, p1*, so that,
(21)Let us now subtract Equation 21 from Equation 20 to get
(22)from which we conclude that,
(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 60o C. At 60o 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 XBz = 0.400, and XTol = 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, n, and bring the 1/n inside the parentheses to convert the number of moles of each component into a mole fraction,
(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.
Copyright 2004, W. R. Salzman
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