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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

This document was uploaded on 02/28/2014 for the course CHEM 311 at LA Tech.

Ask a homework question - tutors are online