厦门大学：《物理化学 Physical Chemistry》课程电子教案（PPT教学课件，英文版）chapter09-1 Solutions

Chapter 9 Solutions

Chapter 9 Solutions

Physical Chemistry Solutions Solution Composition A solution is a homogeneous mixture: that is a solution is a one-phase system with more than one component e The mole fraction x; of species i is defined by x (1.6 ot The(molar) concentration (or volume concentration)c of species i is defined by (9.1) The mass concentration p, of species i in a solution of → volume v is defined by (92)

Solution Composition Solutions V n c i i  (9.1)* A solution is a homogeneous mixture; that is, a solution is a one-phase system with more than one component. The mole fraction xi of species i is defined by tot i i n n x  (1.6) The (molar) concentration (or volume concentration) ci of species i is defined by The mass concentration i of species i in a solution of volume V is defined by V mi i  (9.2)* Physical Chemistry

Physical Chemistry Solutions Solution Composition The molarity is the molar concentration of a species for liquid solutions in moles per liter(dm Solvent(A) Solution Solute(B The molarity m, of species i in a solution is defined by The solute molarity mB IS M4=nAM4(1.4) B B B (9,3) aA

Solution Composition Solutions A i i w n m  The molarity is the molar concentration of a species for liquid solutions in moles per liter (dm3 ). The molarity mi of species i in a solution is defined by The solute molarity mB is A A B A B B n M n w n m  = (9.3)* Solution Solvent (A) Solute (B) wA = nA MA (1.4) Physical Chemistry

Physical Chemistry Solutions Partial Molar Quantities Partial molar volumes mn+nym2+…+nm=∑nm (9.4) V(H,O)/cm cm 7550250 The star indicates a property 1001 of a pure substance or a collection of pure substances 99 Addition of 50.0 cm of water to 98 50.0 cm of ethanol at 20 oC and 97 I atm gives a solution of 96.5 cm(Fig. 9.1, left) 96 0255075100 (ethanol)cm ≠卩

Partial Molar Quantities Solutions The star indicates a property of a pure substance or a collection of pure substances. = + + + = i V nVm n Vm nr Vm r ni Vm i * , * , * 2 ,2 * 1 ,1 *  (9.4) Partial Molar Volumes 0 25 50 75 100 V (ethanol)/cm3 100 99 98 97 96 75 50 25 0 V (H2O)/cm3 V /cm3 Addition of 50.0 cm3 of water to 50.0 cm3 of ethanol at 20 oC and 1 atm gives a solution of 96.5 cm3 (Fig. 9.1, left). * V  V Physical Chemistry

Physical Chemistry Solutions s Partial Molar Quantities (1)different intermolecular forces ≠ (2)different packing of molecules V/cm V(H,O)/cm (different sizes and shapes of the 7550250 100 molecules being mixed 99 V=V(T,P,n12…,n) U=U(7,P,n1 (9.5) 98 The total differential of v 97 o dt+ T aP P. T 96 (9.6) 0255075100 o o dn,+…+ V(ethanol)/cm r/T P, nier

Partial Molar Quantities Solutions 0 25 50 75 100 V (ethanol)/cm3 100 99 98 97 96 75 50 25 0 V (H2O)/cm3 V /cm3 * V  V (1) different intermolecular forces (2) different packing of molecules (different sizes and shapes of the molecules being mixed) ( , , , , ) ( , , , , ), 1 1 r r U U T P n n V V T P n n   = = (9.5) The total differential of V 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) Physical Chemistry

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 nij : 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