However there are cases where such estimations are

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Unformatted text preview: ,nB (n + n )2 (n A + nB ) (n A + nB ) A B Therefore the partial molar volume of A is given by: € # 1− x & # ∂v & A( % VA = (n A + nB )% ( % n +n (+ v $ ∂x A 'P,T,nB $ ( A B )' # ∂v & VA = (1− x A )% +v ( ∂x A 'P,T,n $ B # dv & VA = (1− x A )% (+ v $ dx A ' € Marand’s Notes: Chapter 5 - The Properties of Simple Mixtures 159 The partial molar volume of A is therefore equal to the average molar volume plus the quantity (1- xA)(dv/dxA). Consider the graph above and you should see that the quantity (1- xA)(dv/dxA) is exactly equal to the length T’H’. That this is so, arises from the fact that the tangent TT’ of the curve v = f(xA) at point P (of abcissa xA and ordinate v) has a slope equal to (dv/dXA) = T’H’/PH’. Also we know that SS’ = PH’ = 1 - xA Therefore the sum of the distances T’H’ and H’S’ is exactly equal to (1- xA)(dv/dxA) + v. Similarly, you can show that the partial molar volume of substance B at the point P is given by the intercept of the tangent at P to the curve v = f(xA) with the ordinate axis for xA = 0 (point T on the graph). Partial molar volume of B = OT = OH - TH = v - xA(dv/dxA) So the final results are: # dv & VA = v + (1− x A )% ( $ dx A ' # dv & VB = v − x A % ( $ dx A ' € Note the general result: V = n A VA + nB VB or v = x A VA + xB VB € Marand’s Notes: Chapter 5 - The Properties of Simple Mixtures 160 The same kind of approach may be used to calculate any partial molar quantity. You just have to plot the thermodynamic property for the whole system (on...
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