vol. = X3V
Fig. 7.1 Two methods of mixing xA moles of pure A with xB moles of pure B to form 1 mole of an ideal gas mixture. The temperature is constant p
Fig. 7.1 Two methods of mixing xA moles of pure A with xB moles of pure B to form 1 mole of an ideal gas mixture. The temperature is constant
The initial states of A and B at pressure p obey the ideal gas law:
(the mole fractions appear on both sides of these equations to facilitate subsequent explanations). The two pure gases are mixed in two ways.
The upper path in Fig. 7.1, called the Amagat route, simply mixes the two pure components at constant pressure p. The lower path, called the Dalton route, is more circuitous but produces the same end result. In this route, the pure components are first isothermally expanded from their initial volumes to the final volume v of the mixture.
When expanded to volume v, the pressures of the pure components are reduced according to the ideal gas law:
Comparing the left hand sides of the above pairs of equations shows that the partial pressures are:
When the two expanded volumes of A and B are mixed as shown in Fig. 7.1, the pressures pA and pB are retained in the mixture. This is the origin of the term "partial" pressures. Adding the two equations in (7.3) yields:
The partial pressures play duel roles as measures of mixture composition and as contributions to the total gas pressure. Equations (7.3) and (7.4) are collectively known as Dalton's rule.
The components of an ideal gas mixture individually obey the ideal gas law. Each constituent species independently occupies the same volume at the same temperature as the others. The equation of state for species i in a general multicomponent gas mixture is:
The mixture obeys the ideal gas law at the specified total pressure:
7.2.2 Entropy of mixing
In isothermal, isobaric mixing of ideal substances, all thermodynamic properties remain unchanged:
except for the entropy. Despite the absence of intermolecular interactions in ideal gases, mixing of pure components increases the entropy because the mixture is more random state than the separated pure components; mixing is an irreversible process that requires work to undo. The irreversible nature of mixing resides in the Amagat route in Fig. 7.1, which entails an entropy increase ASmix. The utility of the Dalton route in this figure is that it provides a way of calculating ASmix. Figure 7.2 shows the entropy changes for the various steps in the two routes. Because the change in a thermodynamic property is path-independent, these entropy changes are related by:
The first two terms on the right represent isothermal pressure changes for which the entropy changes are given by Eq (3.10):
The entropy change on mixing by the Dalton path, AS', is zero.
Pure components Xa moles A XBm°les B 1 mole Mixture
Pure components Xa moles A XBm°les B 1 mole Mixture
Proof of this assertion is based on the cylinder-piston apparatus shown in Fig. 7.3. The pure components at their reduced pressures are initially contained in compartments separated by a fixed membrane that is permeable to component A but that will not pass component B (top view). These two compartments are equivalent to the two lower boxes in Fig. 7.2. They both have volume v.
The apparatus in Fig. 7.3 also contains moveable pistons connected by a rigid rod. The piston on the left permits component B to pass but is a barrier to component A. The piston on the right is impermeable to both gases. In the initial position of the rod, A is kept in the right-hand chamber by the left piston and B is confined to the left chamber by the fixed central membrane.
The pistons are frictionless and can move from right to left without requiring work to be done on them; the net force on the connected pistons is zero because pB acts equally on both faces of the B-permeable membrane and pA acting on the right hand face of the B-permeable membrane is balanced by the same pressure acting on the left hand face of the impermeable membrane. The lower plus sign in Fig. 7.2 represents the mixing process depicted in the middle view of Fig. 7.3.
As the moveable pair of connected pistons moves to the left, component B flows without resistance into the mixture region and component A does likewise through the fixed central membrane.There is no internal energy change during mixing of the gases since they are ideal. Since W = 0 and AU = 0, the First law requires that Q = 0. Since the process is reversible, the Second law yields AS' = Q/T = 0.
Using this result along with Eq (7.9) in Eq (7.8) and expressing the pressure ratios by Dalton's rule, Eq (7.3), yields:
Membrane permeable to A only(fixed)
D permeable to B only
I impermeable to both
Fig. 7.3 Apparatus for mixing ideal gases without an entropy change
Eq (7.10) applies to one mole of a binary gas mixture. For gas mixtures containing ni moles of each components, the general form of the entropy of mixing formula is:
For nonideal gases or mixing processes that are neither isothermal nor isobaric, or for initial states that are mixtures rather than pure components, Eq (7.11) is but one component of the entropy change. These effects are illustrated in the following examples.
Example: Calculate the changes in enthalpy and entropy when two moles of helium (species 1) at Ti = 100oC and pi = 1 atm are mixed with one mole of nitrogen (species 2) at T2 = 200oC and p2 = 0.5 atm. Mixing takes place adiabatically and the mixture volume is the same as the sum of the volumes of the initial pure species. This process represents opening of a valve between two insulated tanks of the gases and allowing their contents sufficient time to mix to a uniform state.
First, the temperature, pressure, volume, and composition of the mixture are calculated. These properties are designated without a subscript. The composition is characterized by the helium mole fraction xHe = nHe/n = 0.667. The nitrogen mole fraction is 1- xHe = 0.333. The mixture volume is the sum of the volumes of the two tanks:
To calculate the mixture temperature, the First law is applied. Since the tanks are insulated (Q = 0) and do no work because they do not change volume (W = 0), the First law requires the internal energy of the mixture to equal the sum of the internal energies of the pure gases. For ideal gases the internal energy is dependent on temperature only and can be expressed in terms of the heat capacities from Eq (7.2d):
Cv(T - Tref) = xHeCvHe(Ti - Tref) + (1 - xHe)CvN2(T2 - Tref) Tref is an arbitrary reference temperature and Cvi = (3/2)R and Cv2 = (5/2)R are the specific heats at constant volume of helium and nitrogen, respectively. Using Eq (7.2d), the specific heat of the mixture is the mole-fraction average of the two pure species: Cy = (11/6)R. Solving the above equation for T yields:
The final pressure is obtained by applying the ideal gas law to the mixture:
V 0.141 Pa
To obtain the enthalpy change upon mixing the pure gases, the specific heats at constant pressure are needed. These are equal to the specific heats at constant volume plus the gas constant: CPi = CVi + R = (5/2)R and CP2 = CV2 + R = (7/2)R. The mixture CP is (17/6)R
The enthalpy change for the mixing process is:
AH = n [Cp(T -Tref) - xCH(T1 - Tref) - (1-x)CP2(T2 - Tref)] = 305 J
(the terms involving Tref cancel).
To calculate the entropy change, the pure components must first be brought to the final p and T of the mixture. Using Eq (3.10), this step incurs the following changes in entropy:
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