Lecture_05

# V 1 v n n 1 p t n 2 2 p t n 1 based on what

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Unformatted text preview: es is a 1 molal solution, i.e., a solution that contains 1 mol of the substance in 1.0 kg of solvent. 7 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 , and V 2 = æ ¹V ö . V 1 = æ ¹V ö ç ¹n ÷ ç ¹n ÷ è 1 ø P, T , n 2 è 2 ø P, T , n 1 Based on what we have seen so far, the first equation should be dV = V 1 dn 1 + V 2 dn 2 ,which is simply another application of the chain rule in partial differentiation. However, Eq. (5.33) can indeed be justified on physical grounds as follows. Consider a large volume of solution containing ethanol (E) and water (W). We now add a small amount of water, say, ∆nW moles of water, to this solution. We would want to express the new volume of the solution as V new = V old + Dn W V  ,m , W where V  ,m is the molar volume of pure water. However, this will give us the final W volume only in the case of an ideal solution. In the ethanol-water solution, the effective molar volumes of both substances are different from their molar volumes in the absence of the other substance. Designating...
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## This document was uploaded on 02/28/2014 for the course CHEM 311 at LA Tech.

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