I Hope You all like it very much. I wish it is beneficial for all of you and you can get enough knowledge from it. Clear and appropriate objectives, in terms of what the audience ought to feel, think, and do as a result of seeing the presentation. Objectives are realistic – and may be intermediate parts of a wider plan.

1. Designed By: Dr. Sagir

2.  Gases are one of the most pervasive
aspects of our environment on the Earth.
We continually exist with constant
exposure to gases of all forms.
 The steam formed in the air during a hot
shower is a gas.
 The Helium used to fill a birthday balloon
is a gas.
 The oxygen in the air is an essential gas
for life.

3. A windy day or a still day is a result of the difference in pressure of gases
in two different locations. A fresh breeze on a mountain peak is a study in
basic gas laws.

4. Important Characteristics of Gases
1) Gases are highly compressible
An external force compresses the gas sample and decreases its
volume, removing the external force allows the gas volume to
increase.
2) Gases are thermally expandable
When a gas sample is heated, its volume increases, and when it is
cooled its volume decreases.
3) Gases have high viscosity
Gases flow much easier than liquids or solids.
4) Most Gases have low densities
Gas densities are on the order of grams per liter whereas liquids
and solids are grams per cubic cm, 1000 times greater.
5) Gases are infinitely miscible
Gases mix in any proportion such as in air, a mixture of many gases.

5.  Helium He 4.0
 Neon Ne 20.2
 Argon Ar 39.9
 Hydrogen H2 2.0
 Nitrogen N2 28.0
 Nitrogen Monoxide NO 30.0
 Oxygen O2 32.0
 Hydrogen Chloride HCl 36.5
 Ozone O3 48.0
 Ammonia NH3 17.0
 Methane CH4 16.0
Substances That Are Gases under
Normal Conditions
Substance Formula MM(g/mol)

6.  To fully understand the world around us
requires that we have a good
understanding of the behavior of gases.
The description of gases and their
behavior can be approached from
several perspectives.
 The Gas Laws are a mathematical
interpretation of the behavior of gases.
 However, before understanding the
mathematics of gases, a chemist must
have an understanding of the conceptual
description of gases. That is the purpose
of the Kinetic Molecular Theory.

7.  The Kinetic Molecular Theory is a single set of
descriptive characteristics of a substance known
as the Ideal Gas.
 All real gases require their own unique sets of
descriptive characteristics. Considering the large
number of known gases in the World, the task of
trying to describe each one of them individually
would be an awesome task.
 In order to simplify this task, the scientific
community has decided to create an imaginary
gas that approximates the behavior of all real
gases. In other words, the Ideal Gas is a
substance that does not exist.
 The Kinetic Molecular Theory describes that gas.
While the use of the Ideal Gas in describing all
real gases means that the descriptions of all real
gases will be wrong, the reality is that the
descriptions of real gases will be close enough to
correct that any errors can be overlooked.

8. Three basic assumptions of the kinetic
theory as it applies to gases:
1. Gas is composed of particles-
usually molecules or atoms
◦Small, hard spheres
◦Insignificant volume; relatively far
apart from each other
◦No attraction or repulsion between
particles

9. 2. Particles in a gas move rapidly
in constant random motion
◦Move in straight paths, changing
direction only when colliding with one
another or other objects
◦Average speed of O2 in air at 20 o
C is
an amazing 1660 km/h! (1.6km=1mile)

10. 3. Collisions are perfectly elastic-
meaning kinetic energy is transferred
without loss from one particle to
another- the total kinetic energy
remains constant
Newtonian Cradle-
Where the collisions between the balls elastic?
Yes, because kinetic energy was transferred with each
collision

11.  Why did the balls eventually stop swinging?
The collisions were not perfectly elastic, some
kinetic energy was lost as heat during each
collision.
 At constant temperatures and low to moderate
pressures, collisions between gas particles are
perfectly elastic

12.  Gas consists of large number of particles
(atoms or molecules)
 Particles make elastic collisions with each
other and with walls of container
 There exist no external forces (density
constant)
 Particles, on average, separated by
distances large compared to their diameters
 No forces between particles except when
they collide
Remember the assumptions

13. The potential energy
of the ball
Which is converted to
kinetic energy in the
ball
Which is converted
to potential energy
in the ball
Is converted to kinetic
energy in the ball
Which is converted into
the potential energy of
the ball…………..
…..but in reality the ball
loses height and
eventually stops bouncing
Why does this
happen?

14.  Through friction with the air (air
resistance)
 Through sound when it hits the floor
 Through deformation of the ball
 Through heat energy in the bounce

16.  The gas consists of objects with a defined m
 The gas particles travel randomly in straight-
 All collisions involving gas particles are elast
 The gas particles do not interact with each o
 The gas phase system will have an average k

19. Kinetic Molecular Theory (KMT) for an
ideal gas states that all gas particles:
 are in random, constant, straight-line
motion.
 are separated by great distances relative
to their size; the volume of the gas
particles is considered negligible.
 have no attractive forces between them.
 have collisions that may result in the
transfer of energy between gas particles,
but the total energy of the system remains
constant.

22.  Kinetic Theory Assumptions
◦ Point Mass
◦ No Forces Between Molecules
◦ Molecules Exert Pressure Via Elastic Collisions With
Walls
xx
(courtesy F. Remer)

23.  Non-Ideal Gas
◦ Violates Assumptions
 Volume of molecules
 Attractive forces of molecules
(courtesy F. Remer)

24.  A real gas is most like an ideal gas when
the real gas is at low pressure and high
temperature.
 At high pressures gas particles are close
therefore the volume of the gas particles is
considered.
 At low temperatures gas particles have low
kinetic energy therefore particles have some
attractive force
 Example
 Dry ice, liquid oxygen and nitrogen

25. Behave as described by the ideal gas
equation; no real gas is actually ideal
Within a few %, ideal gas equation
describes most real gases at room
temperature and pressures of 1 atm or
less
In real gases, particles attract each
other reducing the pressure
Real gases behave more like ideal gases

26.  Weight of column of air above your head.
 We can measure the density of the atmosphere by
measuring the pressure it exerts.

27. Effect of Atmospheric Pressure on
Objects at the Earth’s Surface

29. Pressure = Force per Unit Area
Atmospheric Pressure is the weight of
the column of air above a unit area. For
example, the atmospheric pressure felt
by a man is the weight of the column of
air above his body divided by the area
the air is resting on
P = (Weight of column)/(Area of base)
Standard Atmospheric Pressure:
1 atmosphere (atm)
14.7 lbs/in2
(psi)
760 Torr (mm Hg)
1013.25 KiloPascals or Millibars (kPa =
N/m2
)

30.  Torricelli determined from this
experiment that the pressure of the
atmosphere is approximately 30
inches or 76 centimeters (one
centimeter of mercury is equal to 13.3
millibars. He also noticed that height
of the mercury varied with changes in
outside weather conditions.
For climatological and meteorological purposes, standard sea-level pressure
is said to be 76.0 cm or 29.92 inches or 1013 millibars

31. Atmospheric pressure results from
the collisions of air molecules with
objects
◦Decreases as you climb a mountain
because the air layer thins out as
elevation increases
Barometer is the measuring
instrument for atmospheric
pressure; dependent upon weather

32. Common Units of Pressure
Unit Atmospheric Pressure Scientific Field
pascal (Pa); 1.01325 x 105
Pa SI unit; physics,
kilopascal(kPa) 101.325 kPa chemistry
atmosphere (atm) 1 atm* Chemistry
millimeters of mercury 760 mmHg* Chemistry, medicine,
( mm Hg ) biology
torr 760 torr* Chemistry
pounds per square inch 14.7 lb/in2
Engineering
( psi or lb/in2
)
bar 1.01325 bar Meteorology,
chemistry, physics

33. Converting Units of Pressure
Problem: A chemist collects a sample of carbon dioxide from the
decomposition of limestone (CaCO3) in a closed end manometer, the
height of the mercury is 341.6 mm Hg. Calculate the CO2 pressure in
torr, atmospheres, and kilopascals.
Plan: The pressure is in mmHg, so we use the conversion factors from
Table 5.2(p.178) to find the pressure in the other units.
Solution:
PCO2 (torr) = 341.6 mm Hg x = 341.6 torr
1 torr
1 mm Hg
converting from mmHg to torr:
converting from torr to atm:
PCO2( atm) = 341.6 torr x = 0.4495 atm
1 atm
760 torr
converting from atm to kPa:
PCO2(kPa) = 0.4495 atm x = 45.54 kPa101.325 kPa
1 atm

34. Change in average
atmospheric pressure with
altitude.

35. Gas Pressure – defined as the
force exerted by a gas per unit
surface area of an object
◦Due to: a) force of collisions, and b)
number of collisions
◦No particles present? Then there
cannot be any collisions, and thus no
pressure – called a vacuum

37. Rules of thumb:
 When evaluating, start from the known
pressure end and work towards the
unknown end
 At equal elevations, pressure is
constant in the SAME fluid
 When moving down a manometer,
pressure increases
 When moving up a manometer,
pressure decreases
 Only include atmospheric pressure on
open ends
Manometers measure a pressure difference by balancing the
weight of a fluid column between the two pressures of interest

40. Find the pressure at
point A in this open u-
tube manometer with an
atmospheric pressure Po
PD = γH2O x hE-D + Po
Pc = PD
PB = PC - γHg x hC-B
PA = PB
Example 2
P = γ x h + PO

41.  What would Polly
Parcel look like if she
had no gas molecules
inside?
zero molecules = zero pressure inside
zero pressure inside = zero force on the
inside

42.  In order to describe gases,
mathematically, it is essential to be
familiar with the variables that are used.
There are four commonly accepted gas
law variables
 Temperature
 Pressure
 Volume
 Moles

43.  The temperature variable is always
symbolized as T.
 It is critical to remember that all temperature
values used for describing gases must be in
terms of absolute kinetic energy content for
the system.
 Consequently, T values must be converted to
the Kelvin Scale. To do so when having
temperatures given in the Celsius Scale
remember the conversion factor
 Kelvin = Celsius + 273
 According to the Kinetic Molecular Theory,
every particle in a gas phase system can have
its own kinetic energy. Therefore, when
measuring the temperature of the system, the
average kinetic energy of all the particles in

44.  The pressure variable is represented by
the symbol P.
 The pressure variable refers to the
pressure that the gas phase system
produces on the walls of the container that
it occupies.
 If the gas is not in a container, then the
pressure variable refers to the pressure it
could produce on the walls of a container
if it were in one.
 The phenomenon of pressure is really a
force applied over a surface area. It can
best be expressed by the equation

45.  Consider the Pressure equation and the impact
of variables on it.
 The force that is exerted is dependent upon the
kinetic energy of the particles in the system. If
the kinetic energy of the particles increases, for
example, then the force of the collisions with a
given surface area will increase. This would
cause the pressure to increase. Since the
kinetic energy of the particles is increased by
raising the temperature, then an increase in
temperature will cause an increase in pressure.
 If the walls of the container were reduced in
total surface area, there would be a change in
the pressure of the system. By allowing a given
quantity of gas to occupy a container with a
smaller surface area, the pressure of the system
would increase.

46.  As this container of gas is
heated, the temperature
increases. As a result, the
average kinetic energy of
the particles in the system
increases.
 With the increase in
kinetic energy, the force
on the available amount
of surface area increases.
As a result, the pressure
of the system increases.
 Eventually,......................
....Ka-Boom

47.  The Volume variable is represented by the
symbol V. It seems like this variable
should either be very easy to work with or
nonexistent.
 Remember, according to the Kinetic
Molecular Theory, the volume of the gas
particles is set at zero. Therefore, the
volume term V seems like it should be
zero.
 In this case, that is not true. The volume
being referred to here is the volume of the
container, not the volume of the gas
particles.
 The actual variable used to describe a gas
should be the amount of volume available

48.  Since the Kinetic Molecular Theory states
that the volume of the gas particles is
zero, then the equation simplifies.
 As a result, the amount of available space
for the gas particles to move around in is
approximately equal to the size of the
container.
 Thus, as stated before, the variable V is
the volume of the container.

49.  The final gas law variable is the quantity of gas. This is
always expressed in terms of moles. The symbol that
represents the moles of gas is n. Notice that, unlike the
other variables, it is in lower case.
 Under most circumstances in chemistry, the quantity of a
substance is usually expressed in grams or some other unit
of mass. The mass units will not work in gas law
mathematics. Experience has shown that the number of
objects in a system is more descriptive than the mass of the
objects.
 Since each different gas will have its own unique mass for
the gas particles, this would create major difficulties when
working with gas law mathematics.
 The whole concept of the Ideal Gas says that all gases can
be approximated has being the same. Considering the large
difference in mass of the many different gases available,
using mass as a measurement of quantity would cause major
errors in the Kinetic Molecular Theory.
 Therefore, the mole will standardize the mathematics for all
gases and minimize the chances for errors.

50. There are four variables used mathematically for
describing a gas phase system. While the units used for
the variables may differ from problem to problem, the
conceptual aspects of the variables remain unchanged.
1. T, or Temperature, is a measure of the average kinetic
energy of the particles in the system and MUST be
expressed in the Kelvin Scale.
2. P, or Pressure, is the measure of the amount of force
per unit of surface area. If the gas is not in a container,
then P represents the pressure it could exert if it were in
a container.
3. V, or Volume, is a measure of the volume of the
container that the gas could occupy. It represents the
amount of space available for the gas particles to move
around in.
4. n, or Moles, is the measure of the quantity of gas. This
expresses the number of objects in the system and does
not directly indicate their masses.

51.  (1) When temperature is held constant, the
density of a gas is proportional to pressure, and
volume is inversely proportional to pressure.
Accordingly, an increase in pressure will cause
an increase in density of the gas and a decrease
in its volume. – Boyles’s Law
 (2) If volume is kept constant, the pressure of a
unit mass of gas is proportional to temperature. If
temperature increase so will pressure, assuming
no change in the volume of the gas.
 (3) Holding pressure constant, causes the
temperature of a gas to be proportional to
volume, and inversely proportional to density.
Thus, increasing temperature of a unit mass of

53.  Hyperbolic Relation Between Pressure and
Volume
p
V
p – V Diagramp – V Diagram
isotherms
T1 T2 T3 T3 >T2>T1
(courtesy F. Remer)

55.  Linear Relation Between Temperature and
Pressure
P
T (K)
0 100 200 300
P – T DiagramP – T Diagram
isochorsisochorsV1
V2
V3
V1 <V2 <V3
(courtesy F. Remer)

56. Real data must be
obtained above
liquefaction
temperature.
Experimental curves for
different gasses,
different masses,
different pressures all
extrapolate to a
common zero.

60.  What would Polly
Parcel look like if she
had a temperature of
absolute zero inside?
absolute zero = no molecular motion
no molecular motion = zero force on
the inside

61. The equality for the four variables
involved in Boyle’s Law, Charles’ Law,
Gay-Lussac’s Law and Avogadro’s law
can be written
PV = nRT
R = ideal gas constant

62. R is known as the universal gas constant
Using STP conditions
P V
R = PV = (1.00 atm)(22.4 L)
nT (1mol) (273K)
n T
= 0.0821 L-atm
mol-K

63. What is the value of R when the STP
value for P is 760 mmHg?

64. What is the value of R when the STP
value for P is 760 mmHg?
R = PV = (760 mm Hg) (22.4 L)
nT (1mol) (273K)
= 62.4 L-mm Hg
mol-K

65. Dinitrogen monoxide (N2O), laughing
gas, is used by dentists as an
anesthetic. If 2.86 mol of gas occupies
a 20.0 L tank at 23°C, what is the
pressure (mmHg) in the tank in the
dentist office?

66. Set up data for 3 of the 4 gas variables
Adjust to match the units of R
V = 20.0 L 20.0 L
T = 23°C + 273 296 K
n = 2.86 mol 2.86 mol
P = ? ?

67. Rearrange ideal gas law for unknown P
P = nRT
V
Substitute values of n, R, T and V and
solve for P
P = (2.86 mol)(62.4L-mmHg)(296
K)
(20.0 L) (K-mol)

68. A 5.0 L cylinder contains oxygen gas at
20.0°C and 735 mm Hg. How many
grams of oxygen are in the cylinder?

69. Solve ideal gas equation for n (moles)
n = PV
RT
= (735 mmHg)(5.0 L)(mol K)
(62.4 mmHg L)(293 K)
= 0. 20 mol O2 x 32.0 g O2 = 6.4 g
O2

70. What is the molar mass of a gas if 0.250
g of the gas occupy 215 mL at 0.813 atm
and 30.0°C?
n = PV = (0.813 atm) (0.215 L) =
0.00703 mol
RT (0.0821 L-atm/molK) (303K)
Molar mass = g = 0.250 g =
35.6 g/mol

71. Calculate the density in g/L of O2 gas at
STP.
From STP, we know the P and T.
P = 1.00 atm T = 273 K
Rearrange the ideal gas equation for
moles/L
PV = nRT PV = nRT P = n
RTV RTV RT

72. Substitute
(1.00 atm ) mol-K = 0.0446 mol
O2/L
(0.0821 L-atm) (273 K)
Change moles/L to g/L
0.0446 mol O2 x 32.0 g O2 =
1.43 g/L
1 L 1 mol O2
Therefore the density of O2 gas at STP is

73. A gas has a % composition by mass of
85.7% carbon and 14.3% hydrogen. At
STP the density of the gas is 2.50 g/L.
What is the molecular formula of the gas?

74. Calculate Empirical formula
85.7 g C x 1 mol C = 7.14 mol C/7.14
= 1 C
12.0 g C
14.3 g H x 1 mol H = 14.3 mol H/ 7.14
= 2 H
1.0 g H
Empirical formula = CH2

75. Using STP and density ( 1 L = 2.50 g)
2.50 g x 22.4 L = 56.0 g/mol
1 L 1 mol
n = EF/ mol = 56.0 g/mol = 4
14.0 g/EF
molecular formula
CH2 x 4 = C4H8

76. On December 1, 1783, Charles used 1.00 x
103
lb of iron filings to make the first ascent in
a balloon filled with hydrogen
Fe(s) + H2SO4(aq) → FeSO4(aq) + H2(g)
At STP, how many liters of hydrogen
gas were generated?

77. lb Fe → g Fe → mol Fe → mol H2 →
L H2
1.00 x 103
lb x 453.6 g x 1 mol Fe x 1
mol H2
1 lb 55.9 g 1 mol Fe
x 22.4 L H2 = 1.82 x 105
L H2
1 mol H2

78. How many L of O2 are need to react
28.0 g NH3 at 24°C and 0.950 atm?
4 NH3(g) + 5 O2(g) 4 NO(g) + 6
H2O(g)

79. Find mole of O2
28.0 g NH3 x 1 mol NH3 x 5 mol O2
17.0 g NH3 4 mol NH3
= 2.06 mol O2
V = nRT = (2.06 mol)(0.0821)(297K)
= 52.9 L

82. A.If the atmospheric pressure today is
745 mm Hg, what is the partial pressure
(mm Hg) of O2 in the air?
1) 35.6 2) 156 3) 760
B. At an atmospheric pressure of 714,
what is the partial pressure (mm Hg) N2
in the air?

83. A.If the atmospheric pressure today is
745 mm Hg, what is the partial pressure
(mm Hg) of O2 in the air?
2) 156
B. At an atmospheric pressure of 714,
what is the partial pressure (mm Hg) N2
in the air?

84. Partial Pressure
Pressure each gas in a mixture would
exert if it were the only gas in the
container
Dalton's Law of Partial Pressures
The total pressure exerted by a gas
mixture is the sum of the partial
pressures of the gases in that mixture.

85. The total pressure of a gas mixture
depends
on the total number of gas particles, not
on
the types of particles.
STP
P = 1.00 atm P =
1.00 atm
1.0 mol He
0.50 mol O2
+ 0.20 mol He
+ 0.30 mol N2

86. When a scuba diver is several hundred
feet
under water, the high pressures cause N2
from the tank air to dissolve in the blood.
If the diver rises too fast, the dissolved N2
will form bubbles in the blood, a
dangerous and painful condition called
"the bends". Helium, which is inert, less
dense, and does not dissolve in the blood,

87. A 5.00 L scuba tank contains 1.05 mole
of O2 and 0.418 mole He at 25°C.
What is the partial pressure of each
gas, and what is the total pressure in
the tank?

88. P = nRT PT = PO + PHe
V 2
PT = 1.47 mol x 0.0821 L-atm x
298 K
5.00 L (K mol)
= 7.19 atm