Download Physcal Chemistry ERT 108 semester II 2010/2011

PRT 140 PHYSICAL CHEMISTRY
PROGRAMME INDUSTRIAL
CHEMICAL PROCESS
SEM 1 2013/2014
INTRODUCTION TO PHYSICAL CHEMISTRY
BY
PN ROZAINI ABDULLAH
SCHOOL OF BIOPROSES ENGINEERING
©RBA FTK RY 20 2013
LEARNING OUTCOME:
Ability to explain the phenomena, basic concepts, laws and
principles in physical chemistry.
In this chapter you may be able to:
©RBA FTK RY 20 2013
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.
©RBA FTK RY 20 2013
static phenomena
macroscopic
phenomena
equilibrium in macroscopic
systems
THERMODYNAMICS
ELECTROCHEMISTRY
dynamic phenomena
change of concentration as a
function of time
(macroscopic) KINETICS
(ELECTROCHEMISTRY)
STATISTICAL THEORY
OF MATTER
microscopic
phenomena
stationary states of particles
(atoms, molecules, electrons,
nuclei) e.g. during translation,
rotation, vibration
• bond breakage and
formation
• transitions between quantum
states
STRUCTURE OF MATTER
CHEMICAL BOND
STRUCTURE OF MATTER
(microscopic) KINETICS
CHEMICAL BOND
©RBA FTK RY 20 2013
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.
SURROUNDING
SYSTEM
Everything outside
surroundings.
a
system
is
called
The macroscopic part of the universe under
study in thermodynamics is called the system.
The boundary or wall separates a system
from it surroundings.
©RBA FTK RY 20 2013
Open System
Definition: An open system can exchange matter and energy with
its surroundings.
How does it work?
Example:
Surrounding
Matter
System
Energy
©RBA FTK RY 20 2013
Closed System
Definition: A closed system can exchange energy with its surroundings,
but it cannot exchange matter.
How does it work?
Example:
Surrounding
Matter
System
Energy
©RBA FTK RY 20 2013
Isolated System
Definition: An isolated system can exchange neither energy nor matter with
its surroundings.
How does it work?
Example:
Surrounding
Matter
System
Energy
©RBA FTK RY 20 2013
Walls
A system may be separated from its surroundings by various kinds of wall:
A
B
A
Wall
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.
©RBA FTK RY 20 2013
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
©RBA FTK RY 20 2013
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
©RBA FTK RY 20 2013
Equilibrium
Isolated system is in equilibrium when its macroscopic properties remain
constant with time.
(a) The system
macroscopic
properties
remain constant
with time
Non-isolated
system is in
equilibrium
when the
following
conditions
hold:
(b) Removal of
the system from
contact with its
surroundings
causes
no
change in the
properties of the
system
If (a) hold but (b) does not hold – the system is in a steady state.
©RBA FTK RY 20 2013
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.
©RBA FTK RY 20 2013
Thermodynamic Properties
Extensive
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.
Intensive
Variables
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
©RBA FTK RY 20 2013
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.
©RBA FTK RY 20 2013
Temperature
“2 system in thermal equilibrium with each other have the same temperature “
A
+
B
A
B
or
“2 system not in thermal equilibrium have different temperature”
A
+
B
A
B
The zeroth law of thermodynamics
Allows us to assert the existence of temperature as a state function
©RBA FTK RY 20 2013
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.
©RBA FTK RY 20 2013
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 ):
©RBA FTK RY 20 2013
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!).
©RBA FTK RY 20 2013
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
©RBA FTK RY 20 2013
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 1023 H
©RBA FTK RY 20 2013
Molar Mass
= the mass of substance i in a sample
,
= the number of moles of i in the sample
Mole fraction
©RBA FTK RY 20 2013
Ideal Gases
Boyle’s Law
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
©RBA FTK RY 20 2013
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
P2 =
=
V2
726 mmHg x 800 mL
= 1437.62 mmHg
404 mL
Convert 1437.62 mmHg to atm and pascal?
©RBA FTK RY 20 2013
Charles’ & Gay-Lussac’s Law
Measured the thermal expansion of gases and found a linear increase with
temperature.
VaT
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) + 273.15
©RBA FTK RY 20 2013
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 =507.15
K
V2 x T1
T2 =
=
V1
T2 = ?
2.0 L x 507.15 K
= 241.5 K
4.2 L
©RBA FTK RY 20 2013
Avogadro’s Law
V a number of moles (n)
V = constant x n
V1/n1 = V2/n2
Constant temperature
Constant pressure
©RBA FTK RY 20 2013
Ideal Gas Equation
Boyle’s law: V a
1
(at constant n and T)
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
nT
P
= R nT
P
R is the gas constant
= 8.314 47 JK-1mol-1
PV = nRT
©RBA FTK RY 20 2013
Ideal Gas Mixture
Partial Pressure
,
©RBA FTK RY 20 2013
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 22.414 L.
PV = nRT
R=
PV
nT
(1 atm)(22.414L)
=
(1 mol)(273.15 K)
R = 0.082057 L • atm / (mol • K)
©RBA FTK RY 20 2013
What is the volume (in liters) occupied by 64.9 g of HCl
at STP?
T = 0 0C = 273.15 K
P = 1 atm
PV = nRT
V=
nRT
P
1.78 mol x 0.0821
V=
1 mol HCl
n = 64.9 g x
= 1.78 mol
36.45 g HCl
L•atm
x 273.15 K
mol•K
1 atm
V = 39.9 L
©RBA FTK RY 20 2013
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)?
n, V and R are constant
PV = nRT
nR
P
=
T
V
P1
T1
= constant
P2
=
P1 = 1.20 atm
T1 = 305.15 K
P2 = ?
T2 = 368.15 K
T2
P2 = P1 x
T2
T1
= 1.20 atm x
368.15 K
= 1.45 atm
305.15 K
©RBA FTK RY 20 2013
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
©RBA FTK RY 20 2013
Consider a case in which two gases, A and B, are in a
container of volume V.
PA =
PB =
nART
V
nBRT
V
nA is the number of moles of A
nB is the number of moles of B
nB
nA
PT = PA + PB
PA = XA PT
XA =
nA + nB
XB =
nA + nB
PB = XB PT
Pi = X i PT
©RBA FTK RY 20 2013
A sample of natural gas contains 8.24 moles of CH4,
0.421 moles of C2H6, and 0.116 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 =
= 0.0132
8.24 + 0.421 + 0.116
Ppropane = 0.0132 x 1.37 atm
= 0.0181 atm
©RBA FTK RY 20 2013
The van der Waals Equation
R= gas constant
T = Temperature
P = pressure
b & a = van der waals coeeficient
Vm = V/n , volume
©RBA FTK RY 20 2013
Calculate the pressure (unit in atm) exerted by
2.0 mol C2H6 at 273.15 K in 22.414 dm3 behaving as
(i)
Perfect gas
(ii)
a van der Waals gas.
C2H6
C6H6
CH4
a
(atm.dm6.mol-2)
b
(10-2 dm3.mol-1)
5.507
18.57
2.273
6.51
11.93
4.31
©RBA FTK RY 20 2013
Thank you….
©RBA FTK RY 20 2013