1. INTRODUCTION TO PHYSICAL CHEMISTRY SCHOOL OF BIOPROSES ENGINEERING
PRT 140 PHYSICAL CHEMISTRY PROGRAMME INDUSTRIAL CHEMICAL PROCESS SEM /2014
INTRODUCTION TO PHYSICAL CHEMISTRY BY
PN ROZAINI ABDULLAH
SCHOOL OF BIOPROSES ENGINEERING
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2. LEARNING OUTCOME: In this chapter you may be able to:
Ability to explain the phenomena, basic concepts, laws and principles in physical chemistry.
In this chapter you may be able to:
Able to classify, define and explain terms in thermodynamics and laws in thermodynamics
2.Able to differentiate and calculate the laws in thermodynamics.
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3. What is Physical Chemistry?
Physical chemistry is a study of the physical basis of phenomena related to the chemical composition and structure of substances. OR
“It is the science of explaining physical
phenomenon in chemical terms”
What is Chemical Systems?
A chemical system can be studied from either a microscopic or a
macroscopic viewpoint.
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4. (macroscopic) KINETICS (microscopic) KINETICS
static phenomena
dynamic phenomena
equilibrium in macroscopic systems
THERMODYNAMICS
ELECTROCHEMISTRY
change of concentration as a function of time
(macroscopic) KINETICS
(ELECTROCHEMISTRY)
macroscopic
phenomena
STATISTICAL THEORY
OF MATTER
stationary states of particles (atoms, molecules, electrons, nuclei) e.g. during translation, rotation, vibration
STRUCTURE OF MATTER
CHEMICAL BOND
bond breakage and formation
transitions between quantum states
STRUCTURE OF MATTER
(microscopic) KINETICS
CHEMICAL BOND
microscopic
phenomena
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5. What is Thermodynamic?
Thermodynamic is a microscopic science that studies the interrelationships of the various equilibrium properties of a system and the changes in equilibrium properties.
Thermodynamic is the study of heat, work, energy and the changes they produce in the state of system.
Everything outside a system is called surroundings.
SYSTEM
SURROUNDING
The macroscopic part of the universe under study in thermodynamics is called the system.
The boundary or wall separates a system from it surroundings.
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6. Open System
Definition: An open system can exchange matter and energy with its surroundings.
How does it work?
Example:
Surrounding
System
Matter
Energy
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7. Closed System
Definition: A closed system can exchange energy with its surroundings, but it cannot exchange matter.
How does it work?
Example:
Surrounding
System
Matter
Energy
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8. Isolated System
Definition: An isolated system can exchange neither energy nor matter with its surroundings.
Example:
How does it work?
Surrounding
System
Matter
Energy
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9. Walls
A system may be separated from its surroundings by various kinds of wall:
Wall A B A B
1. A wall can be rigid or non rigid (movable).
2. A wall may be permeable or impermeable (allows no matter to pass through it.)
3. A wall may be adiabatic or nonadiabatic.
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10. * Does not conduct heat at all
An adiabatic (isolated) system is one that does not permit the passage of energy as heat through its boundary even if there is a temperature difference between the system and its surroundings. It has adiabatic walls.
Adiabatic Wall
Energy
* Does not conduct heat at all
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11. A nonadiabatic or diathermic (closed) system is one that allows energy to escape as heat through its boundary if there is a difference in temperature between the system and its surroundings. It has diathermic walls.
Diathermic Wall
Energy
* Does conduct heat
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12. Equilibrium
Isolated system is in equilibrium when its macroscopic properties remain constant with time.
(b) Removal of the system from contact with its surroundings causes no change in the properties of the system
Non-isolated system is in equilibrium when the following conditions hold:
(a) The system macroscopic properties remain constant with time
If (a) hold but (b) does not hold – the system is in a steady state.
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13. Types of Equilibrium 1. Mechanical equilibrium
• No unbalanced forces act on or within the system; hence the system undergoes no acceleration, and there is no turbulence within the system.
2. Material equilibrium
• No net chemical reactions are occurring in the system, nor is there any net transfer of matter from one part of the system to another or between the system and its surroundings; the concentrations of the chemical species in the various parts of the system are constant in time.
3. Thermal equilibrium between a system and its surroundings
• There must be no change in the properties of the system or surroundings when they are separated by a thermally conducting wall.
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14. Thermodynamic Properties
Extensive Variables
Intensive Variables
Is one whose value is equal to the
sum of its values for the parts of the system. Thus, if we divide a system into parts, the mass of the system is the sum of the masses of the parts; mass is an extensive property.
Is one whose value does not depend on the size of the system, provided the system remains of macroscopic.
Examples: pressure, density
Examples: mass, volume, energy
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15. A system compose of 2 or more phases.
Homogeneous System:
Each of intensive macroscopic property is constant throughout a system.
Heterogeneous System:
A system compose of 2 or more phases.
Phase:
Homogeneous part of a (possibly) heterogeneous system.
Equilibrium condition:
 The macroscopic properties do not change without external influence.
 The system returns to equilibrium after a transient perturbation.
 In general exists only a single true equilibrium state.
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16. The zeroth law of thermodynamics
Temperature
“2 system in thermal equilibrium with each other have the same temperature “ A B + or
“2 system not in thermal equilibrium have different temperature” A B +
The zeroth law of thermodynamics
Allows us to assert the existence of temperature as a state function
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17. The Zeroth Law of thermodynamics:
If A is in thermal equilibrium with B, and B is in thermal equilibrium with C, than C is also in thermal equilibrium with A. All these systems have a common property: the same temperature.
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18. The Mole
The ratio of the average mass of an atom of an element to the mass of some chosen standard.
The Relative Atomic Mass of a chemical element gives us an idea of how heavy it feels (the force it makes when gravity pulls on it).
The relative masses of atoms are measured using an instrument called a mass spectrometer.
Look at the periodic table, the number at the bottom of the symbol is the Relative Atomic Mass (Ar ):
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19. Relative Molecular Mass, Mr
Most atoms exist in molecules.
To work out the Relative Molecular Mass, simply add up the Relative Atomic Masses of each atom in the molecule:
A relative molecular mass can be calculated easily by adding together the relative atomic masses of the constituent atoms. For example, Nitrate, NO3, has a Mr of 62 g/mol (Try it!).
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20. Molecular mass of O2 = 32Gram
Gram Molecular Mass
Molecular mass expressed in grams is numerically equal to gram molecular mass of the substance.
Molecular mass of O2 = 32Gram
Calculation of Molecular Mass
Molecular mass is equal to sum of the atomic masses of all atoms present in one molecule of the substance.
Example:
– H2O Mass of H atom = 18g
– NaCl = 58.44g
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21. “The number of 12C atoms in exactly 12 g of 12C”
Avogadro’s Number
“The number of 12C atoms in exactly 12 g of 12C”
Avogadro's number = 6.02 x 1023
Atomic Mass or Molecular Mass
The average mass of an atom or molecule
Mole
A mole of some substances is define as an amount of that substance which contains Avogadro’s Number of elementary entities.
E.g: 1 mole of hydrogen atoms contain 6.02 x H
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22. = the mass of substance i in a sample
Molar Mass
= the mass of substance i in a sample ,
= the number of moles of i in the sample
Mole fraction
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23. Ideal Gases Boyle’s Law P1 x V1 = P2 x V2
Boyle investigated the relationship between pressure & volume of gases.
P a 1/V
P x V = constant
P1 x V1 = P2 x V2
Constant temperature
Constant amount of gas
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24. A sample of nitrogen gas occupies a volume of 800 mL at a pressure of 726 mmHg. What is the pressure of the gas (in mmHg) if the volume is reduced at constant temperature to 404 mL?
P1 x V1 = P2 x V2
P1 = 726 mmHg
P2 = ?
V1 = 800 mL
V2 = 404 mL
P1 x V1 V2
726 mmHg x 800 mL
404 mL =
P2 =
= mmHg
Convert mmHg to atm and pascal?
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25. V1/T1 = V2/T2 Charles’ & Gay-Lussac’s Law T (K) = t (0C) + 273.15
Measured the thermal expansion of gases and found a linear increase with temperature.
V a T
V = constant x T, at constant n, p
p= constant x T, at constant n, V
V1/T1 = V2/T2
Temperature must be
in Kelvin
T (K) = t (0C)
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26. A sample of carbon monoxide gas occupies 4. 2 L at 234 0C
A sample of carbon monoxide gas occupies 4.2 L at 234 0C. At what temperature will the gas occupy a volume of 2.0 L if the pressure remains constant?
V1/T1 = V2/T2
V1 = 4.20 L
V2 = 2.0 L
T1 = K
T2 = ?
V2 x T1 V1
2.0 L x K
4.2 L =
T2 =
= K
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27. Avogadro’s Law V a number of moles (n) V = constant x n V1/n1 = V2/n2
Constant temperature
Constant pressure
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28. Ideal Gas Equation PV = nRT Boyle’s law: V a (at constant n and T) 1 P
Charles’ law: V a T (at constant n and P)
Avogadro’s law: V a n (at constant P and T)
V a nT P
V = constant x = R nT P
R is the gas constant
= JK-1mol-1
PV = nRT
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29. Ideal Gas Mixture
Partial Pressure ,
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30. The conditions 0 0C and 1 atm are called standard temperature and pressure (STP).
Experiments show that at STP, 1 mole of an ideal gas occupies L.
PV = nRT
R = PV nT =
(1 atm)(22.414L)
(1 mol)( K)
R = L • atm / (mol • K)
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31. What is the volume (in liters) occupied by 64.9 g of HCl at STP?
T = 0 0C = K
P = 1 atm
PV = nRT
n = 64.9 g x
1 mol HCl
36.45 g HCl
= 1.78 mol
V =
nRT P
V =
1 atm
1.78 mol x x K
L•atm
mol•K
V = 39.9 L
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32. Argon is an inert gas used in lightbulbs to retard the vaporization of the filament. A certain lightbulb containing argon at 6.20 atm and 32 0C is heated to 95 0C at constant volume. What is the final pressure of argon in the lightbulb (in atm)?
PV = nRT
n, V and R are constant nR V = P T
= constant
P1 = 1.20 atm
T1 = K
P2 = ?
T2 = K P1 T1 P2 T2 =
P2 = P1 x T2 T1
= 1.20 atm x K K
= 1.45 atm
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33. Dalton’s Law of Partial Pressures
The pressure exerted by a mixture of gasses is the sum of the pressure that each one would exert if it occupied the container alone.
V and T are constant P1 P2
Ptotal = P1 + P2
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34. nA is the number of moles of A
Consider a case in which two gases, A and B, are in a container of volume V.
PA =
nART V
nA is the number of moles of A
PB =
nBRT V
nB is the number of moles of B
XA = nA
nA + nB
XB = nB
nA + nB
PT = PA + PB
PA = XA PT
PB = XB PT
Pi = Xi PT
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35. A sample of natural gas contains 8. 24 moles of CH4, 0
A sample of natural gas contains 8.24 moles of CH4, moles of C2H6, and moles of C3H8. If the total pressure of the gases is 1.37 atm, what is the partial pressure of propane (C3H8)?
Pi = Xi PT
PT = 1.37 atm
0.116
Xpropane = =
Ppropane = x 1.37 atm
= atm
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36. The van der Waals Equation
R= gas constant
T = Temperature
P = pressure
b & a = van der waals coeeficient
Vm = V/n , volume
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37. Calculate the pressure (unit in atm) exerted by
2.0 mol C2H6 at K in dm3 behaving as
(i) Perfect gas
(ii) a van der Waals gas. a
(atm.dm6.mol-2) b
(10-2 dm3.mol-1)
C2H6
5.507
6.51
C6H6
18.57
11.93
CH4
2.273
4.31
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38. Thank you….
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