ChE 214 PHYSICAL CHEMISTRY Lecturer: Dr. Bedriye Ucpinar Durmaz Email: [email protected] Course 1 Books: 1- Ira N. Levine, Physical Chemistry, 6th Ed., McGraw Hill, Higher Education, 2002.Plotting V vs nB (measured values), slope of the curve; V?CONTENT OF PHYSICAL CHEMISTRY o Chapter 4: MATERIAL EQUILIBRIUM o Chemical Potential o Phase Equilibrium o Reaction Equilibrium o Chapter 9: SOLUTIONS o Solution Composition o Partial Molar Quantities o Ideal Solutions o Ideally Dilute Solutions o Chapter 10: NONIDEAL SOLUTIONS o Activities and Activity Coefficients o Excess Functions o Solutions of Electrolytes o Chapter 11: REACTION EQUILIBRIUM IN NONIDEAL SYSTEMS o Reaction Equilibrium in Nonelectrolyte Solutions o Reaction Equilibrium in Electrolyte Solutions o Reaction Equilibrium involving Pure Solids o Chapter 12: MULTICOMPONENT PHASE EQUILIBRIUM o Colligative Properties o Vapor Pressure Lowering o Freezing-Point Depression and Boiling-Point Elevation o Osmotic Pressure o Two Component Phase Diagrams o Three Component Systems CONTENT OF PHYSICAL CHEMISTRY o Solution Composition o Partial Molar Quantities o Ideal Solutions o Ideally Dilute Solutions Course 1- Ch. 9: SOLUTIONS ?( * ?V ?nj ) eqn; Vj * = Vm,j * ?Let's find an expression for the V of a solution; ? V depends on T, P, and the mole numbers. For fixed values of T, P, and solution mole fractions xi , the volume is directly proportional to the n V = n . f(T, P, x1 ,x2 ,.....) and n= ?i ni Relation between Solution Volume and Partial Molar Volumes dV = f(T, P, x1 ,x2 ,.....).dn dV = ?i V?V != V* 9.2 Partial Molar Quantities Vm,1= V n (single comp) Ex.: 50 mL water + 50 mL ethanol != 100 mL = 96.5 mL at 20?C The difference between V of the solution and V results from (a) Differences between intermolecular forces in the solution and those in the pure components (b) differences in sizes and shapes of the mixed molecules Partial molar volume ?We can write the above equation as: This equation gives the infinitesimal volume change dV that occurs when the T, P, ni of the solution are changed by dT, dP, dn1 , dn2 ,....(one by one) ?i ; (Vm : mean molar volume) Vm = V / n : n = ?i ni Problem 9.11 At 25 ?C and 1 atm, a 0.500 mol/kg solution of NaCl in water has V? B Soln.: Draw the tangent line at 0.1 mol of MgSO4 per kg of H2O Slope = 1001.9 -1001.7 0.3 -0.1 = 1 (cm3 /mol)

?V ?nB Hence V?MgSO4 = 1.0 (cm3 /mol), from curve V= 1001.7 cm3 Measurement of Partial Molar Volume Ex.:Use Fig.2- Robert G. Mortimer, , Physical Chemistry 3rd Ed., Elsevier 3- G. W. Castellan, Physical Chemistry, 3rd Ed., Addison Wesley - Benjamin Cummings Publishing Company, 1983.nB wA

nB nA MA ; Number of moles of B divided by the mass of the solvent (moles/kg) 9.1 Solution Composition ?i dn (T, P, xi = const.) f Relation between Solution Volume and Partial Molar Volumes ?H2O in MgSO4 (aq) at T=20?C, P=1atm, with molality 0.1 (mol/kg) Soln.: This solution has 0.1 mol of MgSO4 and 1000 g of H2O, which is 55.51 mol of H2O.ni V V: solutions volume For liquid solutions, it is called molarity (moles/l) 9.1 Solution Composition ?V of the solution depends on T & P ci = f (T, P) m : independent of T, P. 9.1 Solution Composition Partial molar volume ?We form a solution by mixing of <<1, 2, .dn + n dxi dni = xi .dn if we differentiate V at constant T, P, xi ; differentiate xi const.NaCl + nH2O .........?

ChE 214
PHYSICAL CHEMISTRY
Lecturer: Dr. Bedriye Üçpınar Durmaz
Email: [email protected]
Course 1
Books: 1- Ira N. Levine, Physical Chemistry, 6th Ed., McGraw Hill, Higher Education, 2002.
2- Robert G. Mortimer, , Physical Chemistry 3rd Ed., Elsevier
3- G. W. Castellan, Physical Chemistry, 3rd Ed., Addison Wesley – Benjamin Cummings Publishing Company, 1983.
CONTENT OF PHYSICAL CHEMISTRY
• Chapter 4: MATERIAL EQUILIBRIUM
• Chemical Potential
• Phase Equilibrium
• Reaction Equilibrium
• Chapter 9: SOLUTIONS
• Solution Composition
• Partial Molar Quantities
• Ideal Solutions
• Ideally Dilute Solutions
• Chapter 10: NONIDEAL SOLUTIONS
• Activities and Activity Coefficients
• Excess Functions
• Solutions of Electrolytes
• Chapter 11: REACTION EQUILIBRIUM IN NONIDEAL SYSTEMS
• Reaction Equilibrium in Nonelectrolyte Solutions
• Reaction Equilibrium in Electrolyte Solutions
• Reaction Equilibrium involving Pure Solids
• Chapter 12: MULTICOMPONENT PHASE EQUILIBRIUM
• Colligative Properties
• Vapor Pressure Lowering
• Freezing-Point Depression and Boiling-Point Elevation
• Osmotic Pressure
• Two Component Phase Diagrams
• Three Component Systems
CONTENT OF PHYSICAL CHEMISTRY
• Solution Composition
• Partial Molar Quantities
• Ideal Solutions
• Ideally Dilute Solutions
Course 1- Ch. 9: SOLUTIONS
➢ The majority of chemical and biochemical processes occur in solution.
Solution; a homogeneous mixture of two or more substances in any ratio. Solvent + Solute
✓The phases of solvent, solute and solution may be solid, liquid, or gas.
➢How can we specify the quantities of these components? ; ‘Solution Compositions’
➢What is the the thermodynamics of solutions? ; ‘Partial molar properties.’
SOLUTIONS
The composition of a solution can be specified in several ways:
▪ The mole fraction of component i : xi xi ≡
𝒏𝒊
𝒏𝒕𝒐𝒕
ni
: number of moles of i
ntot : total number of moles of all species in solution
▪ The (molar) concentration (or volume concentration) of component i : ci
ci ≡
𝒏𝒊
𝑽
V: solutions volume
For liquid solutions, it is called molarity (moles/l)
9.1 Solution Composition
▪ The mass concentration of i in a solution: i
i ≡
𝒎𝒊
𝑽
(g/l) , mi
: mass of i present, V: solutions volume
For liquid and solid solutions, it is convenient to treat solvent differently from the solutes.
We denote : solvent; A , solute; B
▪ The molality of species B in a solution: mB
mB ≡
𝒏𝑩
𝒘𝑨

𝒏𝑩
𝒏𝑨 𝑴𝑨
; Number of moles of B divided by the mass of the solvent
(moles/kg)
9.1 Solution Composition
▪ The weight percent of species B (solute) in a solution:
wt% ≡
𝒘𝑩
𝒘
x 100 wB
: mass of B w: mass of the solution
❑ V of the solution depends on T & P ci = f (T, P)
m : independent of T, P.
9.1 Solution Composition
Partial molar volume
➢We form a solution by mixing of «1, 2, . . . , r» components at const. T and P

• moles of substances; n1
  , n2
  ,. . . , nr
  • Vm,1, . . . , Vm,r ; be the molar volumes of pure substances 1, 2, . . . , r
  • V* ; the total volume of the unmixed (pure) components
  V* = n1 Vm,1 + n2 Vm,2 +….+ nr Vm,r = σ𝒊 ni V
  ∗
  m,i
  After mixing of these components that the volume of the solution is not (in general)
  equal to the unmixed volume!
  V ≠ V
  9.2 Partial Molar Quantities
  V∗m,1=
  𝑽
  𝒏
  (single comp)
  Ex.: 50 mL water + 50 mL ethanol ≠ 100 mL
  = 96.5 mL at 20°C
  The difference between V of the solution and V* results from
  (a) Differences between intermolecular forces in the solution and
  those in the pure components
  (b) differences in sizes and shapes of the mixed molecules
  Partial molar volume
  ▪ Similar equations can be written for any extensive property, such as U, H, Cp
  , S, G, at const. T & P.
  ▪ Each of these properties generally changes on mixing the pure components
  V = V(T, P, n1
  , ……,nr
  ) ; U = U(T, P, n1
  , ……,nr
  ) ; S; H; G; … etc.
  ➢The total differential of V can be written as:
  We define partial molar volume 𝑽ഥ
  𝒋
  of substance j in solution:
  𝑽ഥ
  𝒋 ≡ ( 𝝏𝑽
  𝝏𝒏𝒋
  )
  Partial molar volume
  ▪ We can write the above equation as:
  This equation gives the infinitesimal volume change dV that occurs when the T, P, ni of the solution are changed
  by dT, dP, dn1
  , dn2
  ,….(one by one)
  ▪ Partial molar volume 𝑽ഥ
  𝒋
  = ratio of infinitesimal changes in two extensive properties intensive property
  𝑽ഥ
  𝒊 = 𝑽ഥ
  𝒊
  (T, P, x1
  , x2
  ,…)
  ➢If P, T, and ni≠j
  are hold const. and dnj moles of j is added to the solution;
  Partial molar volume
  𝑽ഥ
  𝒋 ≡ (
  𝒅𝑽
  𝒅𝒏𝒋
  )
  Ex.:
  Partial molar volume
  ❖ Addition of dnj moles of j to a solution at
  const. T, P produces a change dV in the
  solutions volume
  dV = 𝑽ഥ
  𝒋
  . dnj
  ➢ The volume of pure substance j; Vj
  Vj = nj
  . Vm,j (T, P) (Vm,j : molar volume of pure j)
  ➢ If a pure substance is taken as a special case of a solution;
  𝑽𝒋

• = 𝝏𝑽𝒋
  ∗
  𝝏𝒏𝒋 T,P = Vm,j plug V*j in 𝑽ഥ
  𝒋 ≡ ( *
  𝝏𝑽
  𝝏𝒏𝒋
  ) eqn; 𝑽𝒋 * = Vm,j

➢Let’s find an expression for the V of a solution;
▪ V depends on T, P, and the mole numbers. For fixed values of T, P, and solution mole fractions xi
, the volume
is directly proportional to the n
V = n . f(T, P, x1
,x2
,…..) and n= σ𝑖 𝑛𝑖
Relation between Solution Volume and Partial Molar Volumes
dV = f(T, P, x1
,x2
,…..).dn
dV = σ𝑖 𝑉ഥ
𝑖
. 𝑑𝑛𝑖
; T, P = const.
xi =
𝑛𝑖
𝑛
ni = xi
.n dni = xi
. dn + n dxi dni = xi
.dn
if we differentiate V at constant T, P, xi
;
differentiate xi
const.
dxi = 0
dV = σ𝑖 𝑉ഥ
𝑖
xi
.dn
dV = σ𝑖 𝑥𝑖 𝑉ഥ
𝑖 dn (T, P, xi = const.)
f
Relation between Solution Volume and Partial Molar Volumes
▪ Remember; dV = f(T, P, x1
,x2
,…..).dn and V=n. f(T, P, x1
,x2
,…..)
V = n.f = n . σ𝑖 𝑥𝑖 𝑉ഥ
𝑖
; xi =
𝑛𝑖
𝑛 V = σ𝒊 𝒏𝒊 𝑽ഥ
𝒊
Solution’s volume V ; desired expression
▪ The change in volume on mixing the solution from its pure components at const. T,P is given by
Vmix = V – V* = σ𝑖 𝑛𝑖(𝑉ഥ
𝑖 − 𝑉𝑚,𝑖
∗
) (where mix stands for mixing (and not for mixture))
Vm = σ𝑖 𝑥𝑖 𝑉ഥ
𝑖
; (Vm : mean molar volume)
Vm = V / n : n = σ𝑖 𝑛𝑖
Problem 9.11
At 25 °C and 1 atm, a 0.500 mol/kg solution of NaCl in water has 𝑉ത
NaCl = 18.63 cm3
/mol
and 𝑉ത
H2O = 18.062 cm3
/mol. Find the volume at 25°C and 1 atm of a solution prepared by
dissolving 0.5 mol NaCl in 1000.0 g of water. (MA-H2O = 18 g/mol)
Soln:
V = σ𝑖 𝑛𝑖 𝑉ഥ
𝑖 = nNaCl . 𝑉ത
NaCl + nH2O
. 𝑉ത
H2O
= 0.5 mol.(18.63) cm3
/mol +
1000.0 𝑔
18 𝑔/𝑚𝑜𝑙
.(18.062) cm3
/mol
V = 1012.76 cm3
➢Consider a solution composed of A and B.
➢To measure 𝑉ത
B we should prepare solutions at desired T, P with a fixed nA but varying nB
.
➢ Plotting V vs nB
(measured values), slope of the curve; 𝑽ഥ
B
Soln.: Draw the tangent line at 0.1 mol of MgSO4 per kg of H2O
Slope =
1001.9 −1001.7
0.3 −0.1
= 1 (cm3
/mol)

∆𝑽
∆𝒏𝑩
Hence 𝑉ത
MgSO4 = 1.0 (cm3
/mol), from curve V= 1001.7 cm3
Measurement of Partial Molar Volume
Ex.:Use Fig. to find 𝑉ത
MgSO4 and 𝑉ത
H2O in MgSO4
(aq) at T=20°C, P=1atm, with molality 0.1 (mol/kg)
Soln.: This solution has 0.1 mol of MgSO4
and 1000 g of H2O, which is 55.51 mol of H2O.
V = nA 𝑉ത
A + nB 𝑉ത
B
V = 1001.7 cm3 = (55.51 mol) 𝑉ത
H2O + (0.10 mol) (1.0 cm3
/mol), 𝑉ത
H2O = 18.04 cm3
/mol
Question; Find 𝑉ത
MgSO4 and 𝑉ത
H2O in 0.20 mol/kg MgSO4
(aq) at 20°C and 1 atm