Ptnb a b v va 1 x a v x a ptn b

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Unformatted text preview: the volume occupied by one mole of A, in a mixture of A and B, is the partial molar volume of A and generally depends Marand’s Notes: Chapter 5 - The Properties of Simple Mixtures 157 on the composition of the mixture. To describe composition, we use the mole fraction (defined by the symbol x). The mole fraction of a substance J is simply: xJ = € nJ nA which for a binary (A, B) mixture leads to: x A = n A + nB nJ ∑ Experimentally, we can measure the total volume of a mixture as a function € of the composition. Mix nA moles of A with nB moles of B, measure the volume and divide by the total number of moles (nA + nB). This gives us v, the average molar volume of the A/B solution. Repeat this procedure for different solutions made up with different numbers of moles of A and B and we can plot v vs xA (see figure below for an arbitrary solution) v T’ H H’ P VA T v = f(xA) VB S’ O 0 S 1 Marand’s Notes: Chapter 5 - The Properties of Simple Mixtures xA 158 We want to express the partial molar volume of A and B in terms of v and xA. First, we start with the definition of partial molar volumes: # ∂ (n + n )v # ∂V & A B VA = % =% ( ∂n A $ ∂n A 'P,T,nB % $ [ & ( ( 'P,T,nB # ∂[n + n ] & # ∂v & A B ( VA = (n A + nB )% + v% ( % ∂n ( ∂n A 'P,T,n $ A $ 'P,T,nB B # ∂v & VA = (n A + nB )% +v ( $ ∂n A 'P,T,nB # ∂v & # ∂v & # ∂x A & =% % ( ( % ( $ ∂n A 'P,T,nB $ ∂x A 'P,T,nB $ ∂n A 'P,T,nB € The partial derivative of xA with respect to nA at constant nB is given by: # ∂x A & (1− x A ) nB xB = = = % ( $ ∂n A 'P,T...
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This note was uploaded on 01/26/2014 for the course CHEM 3615 taught by Professor Aresker during the Spring '07 term at Virginia Tech.

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