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ERT 108/3
PHYSICAL CHEMISTRY
INTRODUCTION
Prepared by:
Pn. Hairul Nazirah Abdul Halim
Chemistry is the study of matter and the changes it
undergoes
What is Physical Chemistry?
• is the study of macroscopic, atomic, subatomic, and
particulate phenomena in chemical systems in
terms of physical laws and concepts.
• It applies the principles, practices and concepts of
physics such as motion, energy, force, time,
thermodynamics, quantum chemistry, statistical
chemistry and dynamic.
Subtopic
• The properties of gases
a) The perfect gas
b) Real Gases
THE PERFECT GAS
• The general form of an equation of state:
p  f (T , V , n)
• If we know the values of T, V and n for a particular
substance, then the pressure has a fixed value.
• Equation of state of a ‘perfect gas’:
nRT
p
V
a) PRESSURE
•
Definition: A force divided by the area to which the
force is applied.
•
p = F/A
•
SI unit of pressure: Pascal (Pa) = Nm-2 = kgm-1s-2
•
Pressure is measured with a barometer
Units of Pressure
•
•
•
•
•
•
Pascal (Pa)
Bar
Atmosphere
Torr
mmHg
Psi
1Pa = 1 Nm-2
1 bar = 105 Pa
1 atm = 1.01325 bar
760 Torr = 1 atm
1mmHg = 133.322 Pa
1 psi = 6.894757 kPa
• When a region of high pressure is separated from a
low pressure by a movable wall, the wall will be
pushed into one region.
• If the two pressure is identical, the wall will not
move.
Example 1.1
Calculating pressure
Suppose Isaac Newton weighed 65 kg. Calculate
the pressure he exerted on the ground when
wearing boots with soles of total area 250 cm2 in
contact with the ground.
Solution
(65 kg)(9.81 ms -2 )
4
-2
4
p
 2.6 x 10 Nm  2.6 x 10 Pa
-2
2
2.50 x 10 m
b) Temperature
•
•
Is the property that indicates the direction of the
flow of energy through a thermally conducting,
rigid wall.
Two types of boundary:
a) Diathermic – if a change of state is observed
when two objects at different temp. are bought
into contact. Example: metal container
b) Adiabatic – if no change occurs even though
the two objects have different temperatures.
• Thermal equilibrium:
no change of state
occurs when two
objects are in contact
through a diathermic
boundary.
• Zeroth Law
Thermodynamics:
If A is in thermal equilibrium with B,
and B is in thermal equilibrium with C,
then C is also in thermal equilibrium with A.
The Gas Laws
a)
b)
c)
d)
e)
f)
Boyle’s law
Charles’s Law
Avogadro’s Principle
The perfect gas law
Mixture of gases
Mole fractions and partial pressures
a) Boyle’s Law
•
At constant temp., the pressure of a sample gas
is inversely proportional to its volume.
pV  constant
•
The volume it occupies is inversely proportional to
its pressure:
1
p
V
1
V
p
Fig 1.5
The pressure-volume
dependence of a
fixed amount of
perfect gas at
different
temperatures.
Each curve is a
hyperbola
(pV = constant) and is
called an isotherm.
• Straight lines are obtained when the pressure is
plotted against 1/V at constant temp.
b) Charles’s Law
V  constant x T
p  constant x T
at constant pressure
at constant volume
c) Avogadro’s principle
V  constant x n
At constant pressure and temp.
d) The perfect gas law
• perfect gas = ideal gas
pV  nRT
R = gas constant
PV
R
nT
Molar volume (Vm) of a perfect gas under Standard
Ambient Temperature Pressure (SATP):
• Temp. at 298.15 K
• Pressure at 1 bar (105 Pa)
RT
Vm 
p
2
-1
-1
(8.31447 x10 L bar K mol )( 298.15 K)
Vm 
(1 bar)
Vm  24.789 L mol
-1
• To calculate the change in conditions when
constant amount of gas is subjected to different
temp., pressures and volume:
P1V1
 nR
T1
P2V2
 nR
T2
• Combined gas equation:
P1V1 P2V2

T1
T2
Example 1.3
Using combined gas equation
In an industrial process, nitrogen is heated to 500K
in a vessel of constant volume. If it enters the
vessel at 100 atm and 300K, what pressure would it
exert at the working temp., if it behaved as a
perfect gas?
Solution
P1 P2

T1 T2
T2
P2 
x P1
T1
500 K
P2 
x 100 atm  167 atm
300 K
e) Mixture of Gases
•Dalton’s Law: the pressure exerted by a mixture of
gases is the sum of the partial pressures of the
gases.
p  pA  pB  ...
•Where
p = total pressure
pA = partial pressure of perfect gas A
pB = partial pressure of perfect gas B
Example 1.4
Using Dalton’s Law
A container of volume 10.0L holds 1.00 mol N2 and
3.0 mol H2 at 298 K. What is the total pressure in
atmospheres if each component behaves as a
perfect gas?
Solution
RT
p  p A  p B  (n A  n B )
V
(8.206 x 10 -2 L atm K -1 mol -1 )( 298K)
p  (1.00 mol  3.00 mol) x
10.0 L
p  9.78atm
f) Mole fractions and partial pressures
nJ
XJ 
n
n  nA  nB  ...
Where;
XJ = mole fraction; amount of J expressed as a
fraction of the total amount of molecules, n in the
sample.
x A  xB  ...  1
Example
Mole fractions.
A mixture of 1.0 mol N2 and 3.0 mol H2 consist of:
Mol fraction of N2 = 1.0 mol / (1.0 + 3.0 mol) = 0.25
Mol fraction of H2= 3.0 mol / (1.0 + 3.0 mol) = 0.75
Partial Pressure, pJ of a gas J in a mixture
pJ  xJ p
• Where
p = total pressure
The sum of partial pressures is equal to the total
pressure:
p A  pB  ...  ( x A  xB  ...) p  p
Example 1.5
Calculating Partial Pressures
The mass percentage composition of dry air at sea
level is approximately N2 = 75.5; O2 = 23.2; Ar = 1.3.
What is the partial pressure of each component
when the total pressure is 1.00 atm?
Solution
Assume; total mass of the sample = 100g.
100 g
n( N 2 )  0.755 x
 2.69 mol
-1
28.02 gmol
100 g
n(O 2 )  0.232 x
 0.725 mol
-1
32.00 gmol
100 g
n( Ar )  0.013 x
 0.033 mol
-1
39.95 gmol
pJ
xJ 
p
Total mol  3.45 mol
N2
O2
Ar
Mole fraction
0.780
0.210
0.0096
Partial pressure (atm)
0.780
0.210
0.0096
Real Gases
• Real gases do not obey the perfect gas law exactly.
1.3
Molecular Interactions
• Real gases show deviations from the perfect gas
law because molecules interact with each other.
• Repulsive forces between molecules assist
expansion.
• Attractive forces assist compression.
a) The compression factor
Compression factor, Z is the ratio of its molar
o
V
volume, Vm to the molar volume of a perfect gas, m
at the same pressure and temp.
Vm
Z o
Vm
•
•
For perfect gas, Z = 1
Deviation of Z from 1 is a measure of departure
from perfect behavior.
Fig 1.13
The variation of
compression factor, Z, with
pressure for several gases
at 00C.
• At very low pressures, all gases have Z≈1 and
behave nearly perfectly.
• At high pressure, all the gases have Z > 1,
signifying they have a larger molar volume than a
perfect gas. Repulsive forces are dominant.
• At intermediate pressure, most gases have Z < 1.
Attractive forces are reducing the molar volume.
b) Virial
Coefficients
Fig 1.14
Experimental
isotherms of
carbon dioxide at
several
temperatures.
b) Virial Coefficients
•
•
Refer to Figure 1.14
At large molar volumes and high temp., the realgas isotherms do not differ greatly from perfect gas
isotherms.
•
The small different suggest that the perfect gas
law is in fact the first term in an expression of the
form:
pVm  RT (1  B p  C p  ...)
'
•
'
Known as ‘Virial equation of state’
2
• A more convenient expansion for many applications is:


B
C
pVm  RT 1 
 2  ...
 Vm Vm

• The term in parentheses can be identified with the
compression factor, Z.
• Known as ‘Virial equation of state’
• First virial coefficient = 1
• Second virial coefficient = B
• Third virial coefficient = C
1.4 The van der Waals equation
nRT
n
p
 a 
V  nb
V 
2
Eq. 1.25a
The equation is often written in terms of the molar
volume Vm = V/n.
 a 
RT
p
  2 
Vm  b  Vm 
Eq. 1.25b
Constant a and b = van der Waals coefficients.
Example 1.6
Using the van der Waals equation
to estimate a molar volume
Estimate the molar volume of CO2 at 500K and 100
atm by treating it as a van der Waals gas.
Solution
1. Arrange equation 1.25b to become:

RT  2  a 
ab
Vm   Vm 
V   b 
0
p 
p

 p
3
m
 a
RT
p
  2
Vm  b  Vm



RTV m2  aVm  ab
p
3
2
Vm  bVm


p Vm3  bVm2  RTV m2  aVm  ab
 RT
V  bV  
 p
3
m
2
m
 2 a
ab
Vm   Vm 
p

 p

RT  2  a 
ab
Vm   Vm 
V   b 
0
p 
p

 p
3
m
• Refer to Table 1.5,
For CO2, a = 3.610 L2 atm mol-2
b = 4.29 x 10-2 L mol-1


RT
8.2057 x 10 L atm K mol 500K 

 0.410 L mol -1
p
100 atm
RT
2
-1
b
 4.29 x 10  0.410  0.453L mol
p
a 3.61

 3.61x10  2 (L mol -1 ) 2
p 100
2
-1
-1
ab (3.61x10  2 )( 4.29 x 10  2 )

 1.55x 10 3 (L mol -1 ) 3
p
100
• Thefore, on writing x = Vm, the equation to solve is:
x 3  0.453x 2  (3.61E  2) x  (1.55E  3)  0
• The acceptable root is x = 0.366
• Hence Vm = 0.366 L mol-1.