Physical Chemistry Solutions s Partial Molar Quantities IT'+ P OT' P, n aP T, n (9.6) T,P,n1≠ T P The partial molar volume one- phase syst (9.7) j丿T,Pn+j the amount of a substance j ni:the amount of all other substances present are constant
Partial Molar Quantities Solutions r T P n r T P n P n T n dn n V dn n V dP P V dT T V dV i i r i i + + + + = , , 1 1 , , , , 1 (9.6) The partial molar volume . , , one phase syst n V V T P ni j j j − (9.7)* Physical Chemistry nj : the amount of a substance j nij : the amount of all other substances present are constant
Physical Chemistry Solutions Partial Molar Quantities The partial molar volume one- phase syst (9.7) T P Val e The partial molar volume is the slope of the graph of the total volume as the v(b) amount ofj is changed, P, T,amount of other components being constant Equation(9.6 )becomes x dT+ OT' P, n aP dP+∑vmn (9.8 T,n
Partial Molar Quantities Solutions The partial molar volume . , , one phase syst n V V T P ni j j j − (9.7)* + + = i i i P n T n dP V dn P V dT T V dV i i , , (9.8) Equation (9.6) becomes Physical Chemistry V(a) V(b) x V The partial molar volumeis the slope of the graph of the total volume as the amount of j is changed, P, T, amount of other components being constant
Physical Chemistry Solutions Partial Molar Quantities The partial molar volume one-phase syst (9.7) T, P,n (9.9) an T P. m,J (9.10) The partial molar volume of a pure substance is equal to its〃 olar volume HOwever, the partial molar volume of component j of a solution eis not necessarily equal to the molar volume of pure j
Partial Molar Quantities Solutions The partial molar volume . , , one phase syst n V V T P ni j j j − (9.7)* The partial molar volume of a pure substance is equal to its molar volume. ( , , , , ) (9.9) V i =V i T P x1 x2 However, the partial molar volume of component j of a solution is not necessarily equal to the molar volume of pure j. (9.10)* * , * V j =V m j * , , * , , * m j T P j j T P n j j V n V n V V i j = = Physical Chemistry
Physical Chemistry Solutions s Solution volume and partial molar volumes V=nf(,P,x, 142 (9.12 1 1 ls sonme function of T, P,x Differentiation of (9.12 )at constant T, P, x,x,,...,x gives dv=f(T, P,x,x2,.dn const. T, P,x:(. 13 Equation(9. 8)at constant Tand P becomes ∑ const. TP x or n;=x n dn,=d(x n)=x, dn+ndx At fixed xi , =0, and x d=∑ x Vidn const.T (9.15
Solution volume and partial molar volumes Solutions Differentiation of (9.12) at constant T, P, x1 , x2 , ……, xr gives ( , , , , ) (9.12) V = nf T P x1 x2 (9.15) = i i dV V idn const. T,P (9.13) i dV f (T, P, x , x , )dn const. T, P, x = 1 2 Equation (9.8) at constant T and P becomes or n x n n n x i i i i = = i i i dn ndxi dn = d(x n) = x + i i i i dV = x V dn const. T, P, x At fixed x dxi = 0, and dni = xi dn i , i i i n n , f is some function of T,P, x Physical Chemistry
Physical Chemistry Solutions s Solution volume and partial molar volumes dv=f(T, P,x,x2,.dn const. T, P,x,(9.13) ∑ xvidn const. T P (9.15 Comparison of the expressions(.13)and (9.15)for dv gives f=∑x V=nf(t, P,x, x2,... (9,12 Equation (9. 12) becomes V=nf=n〉x,V since x or n=Xn V=n,Vi one-phase syst (916)2
Solution volume and partial molar volumes Solutions Comparison of the expressions (9.13) and (9.15) for dV gives (9.13) i dV f (T, P, x , x , )dn const. T, P, x = 1 2 Equation (9.12) becomes i (9.15) i i i dV = x V dn const. T, P, x = i i f xi V = − i i i V n V one phase syst. (9.16)* ( , , , , ) (9.12) V = nf T P x1 x2 = = i i V nf n xi V since or n x n n n x i i i i = = Physical Chemistry