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Lecture_5b

# Lecture_5b - 5.5 Partial Molar Quantities Partial Molar...

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5.5. Partial Molar Quantities Partial Molar quantities are required to deal with open systems , i.e., systems that permit mass transfer between themselves and surroundings. Consider an open system with n 1 moles of component 1, n 2 moles of component 2, n 3 moles of component 3, etc. . We would write the free energy change dG for such a system as dG = æ è ¹ G ¹ P ö ø T , n 1 , n 2,. .. dP + æ è ¹ G ¹ T ö ø P , n 1 , n 2,. .. dT + æ è ç ¹ G ¹ n 1 ö ø ÷ P , T ,, n 2,. .. dn 1 + ... = VdP SdT + G 1 dn 1 + G 2 dn 2 + ... In the second equality, the quantities G 1 , G 2 , etc. . are called partial molar free energies . Similarly, we may define partial molar volumes, partial molar enthalpies, internal energies, and entropies: V 1 = æ è ç ¹ V ¹ n 1 ö ø ÷ P , T ,, n 2,. .. ; H 1 = æ è ç ¹ H ¹ n 1 ö ø ÷ P , T ,, n 2,. .. ;etc Because of their great importance in the thermodynamics of solutions, we discuss partial molar volumes and partial molar free energies further. Partial Molar Volume: The total volume of a solution of, say, two miscible liquids is given by (5.33) V = n 1 V 1 + n 2 V 2 . The units of partial molar volumes are the same as molar volumes. The relationship between the two, i.e., partial molar volume and the molar volume is a subtle but important one. ± In the case of ideal solutions, the partial molar volume of each component will be identical to the molar volume of the pure substance in the absence of the other component. ± However, in the case of non-ideal solutions, the presence of the second component has a measurable influence on the molar volume of the first component and vice versa. Therefore, in general, V 1 ! V 1 ± and V 2 ! V 2 ± . The standard state for defining partial molar quantities is a 1 molal solution, i.e., a solution that contains 1 mol of the substance in 1.0 kg of solvent.

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Physical Interpretation of partial molar quantities: It may appear that there is something &not quite right± about the following two equations: V = n 1 V 1 + n 2 V 2 , where V 1 = æ è ç ¹ V ¹ n 1 ö ø ÷ P , T , n 2 ,and V 2 = æ è ç ¹ V ¹ n 2 ö ø ÷ P , T , n 1 . Based on what we have seen so far, the first equation should be which is simply another application of the chain rule in partial dV = V 1 dn 1 + V 2 dn 2 , differentiation.
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Lecture_5b - 5.5 Partial Molar Quantities Partial Molar...

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