ChE 214
PHYSICAL CHEMISTRY
Lecturer: Dr. Bedriye Ucpinar Durmaz
Email: bedriyeucpinar@hotmail.com
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
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
?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