Physical Chemistry 2.pdf - OER@AVU - African Virtual University

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African Virtual University After the barrier is removed, the gases mix so that the total pressure is the sum of their partial pressure P = P A + P B . Gibbs energy is then given by G F inal = n A ⎡ ° ⎛ PA ⎞ ⎤ ⎢µ A + RT ln ⎝ ⎜ P° ⎠ ⎟ ⎥ ⎣ ⎦ + nB µ ⎡ ° ⎛ PB ⎞ ⎤ ⎢ B + RT ln ⎝ ⎜ P° ⎠ ⎟ ⎥ ⎣ ⎦ (1.13) The change in Gibbs free energy on mixing is the difference between the initial free energy, G initial , and the final G final i.e. ∆ mixG = nA RT ln P ⎛ A ⎞ ⎝ ⎜ P° ⎠ ⎟ + nB RT ln P ⎛ B ⎞ ⎝ ⎜ P° ⎠ ⎟ (1.14) The may also be written in terms of mole fractions as since n i = x i n and P i /P = x i . 1.7 What about the entropy of mix? (1.15) The entropy of mixing is derived from the thermodynamic expression that, − S = (∂G/∂T) P,n . It then follows from the equation (1.15) that the entropy of mixing is given by the expression ∆ mix S = − ∂∆ mixG ⎛ ⎝ ⎜ ∂T ⎞ ⎠ ⎟ P ,T , nA , nB = − nR ( xA ln xA + xB ln xB ) (1.16) From previous discussions in thermodynamics you will recall that Gibbs energy is related to enthalpy through the expression ΔG = ΔH – TΔS. For the mixing of two gases we can write that Δ mix G = Δ mix H – TΔ mix S, which gives Δ mix H= Δ mix G + TΔ mix S. (1.17)
African Virtual University 0 If the gases under consideration are perfect gases then at constant temperature ΔH mix = 0 from equation (1.15) and (1.16). The significance of this result is that it confirms our expectations that there are no interactions in molecules of a perfect gas. Chemical potential of liquid mixture The formation of a liquid-liquid solution can be viewed as involving three main steps. The first step involves the introduction of the solute liquid into the solvent. Molecules of one liquid have to be moved apart to make way for the other liquid molecules. Forces of attraction have to be overcome during this process. This process thus requires energy (endothermic) and leads to an increase in the potential energy of the system. The second step involves expansion of the solvent molecules to make way for the solute liquid. Energy is also required and similarly leads to an increase in the energy of the system. In the final step the molecules need to mix. The molecules experience mutual attraction giving off energy and decreasing the systems energy. From the preceding discussion, the net energy change is the balance when the three steps are considered. The energy diagram to illustrate the changes in energy is shown in Figure 1.5. If the energy change (input) in steps 1 and 2 is equal to energy change (output) in step 3, the heat of solution, ΔH solution = 0. This signifies a condition where the intermolecular forces in both the solute and solvent are the same. A typical example is benzene and carbon tetrachloride. The solution formed from the mixing of benzene and carbon tetrachloride approximates to an ideal solution. Figure 1.5 Diagram showing energy changes during the mixing of two liquids
Page 1 and 2: Physical Chemistry 2 Prepared by On
Page 3 and 4: Table of ConTenTs African Virtual U
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Page 7 and 8: Unit IV Electrochemistry (20 hours)
Page 9 and 10: Unit Learning objective(s) African
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Page 17 and 18: List of required readings • Solut
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Page 21 and 22: Practice problem 1 Which of the fol
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Page 25 and 26: Visual demonstration: Forces of coh
Page 27 and 28: Simulation: Solution of salt 1.4 Th
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Page 73 and 74: Practice problem 1 What is X in the
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(ii) Photoelectric absorption Afric
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Reading # 9 INTRODUCTION TO RADIOAC
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African Virtual University Rational
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Useful Link #2 Title : Chem1 Virtua
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Useful Link # 5 Title : Chemistry U
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XVI. summative evaluation African V
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Physical Chemistry 2.pdf - OER@AVU - African Virtual University

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