Unit 1
(formerly Module 2)
Gases and Their Applications
Lesson 2-1
About Gases
2
R
 Gas is one of the three main states of matter
 Gas particles may be atoms or molecules,
depending on the type of substance (ie,
element or compound)
 Gas particles have much more space
between them than liquids or solids.
 Gases are said to be an expanded form of
matter, solids and liquids are condensed
forms of matter.
3
General Properties of a Gas
 Gases do have mass (although it is
sometimes difficult to measure).
 Gases have no definite volume,
 Gases have no definite shape.
 Gases are compressible, meaning they can
be squeezed into smaller containers, or can
expand to fill larger containers.
– Because gases compress, the density of gases
can only be compared under specific conditions.
4
Some Important Gases
 Oxygen (O2): clear, breathable, supports combustion.
 Ozone (O3): poisonous, unstable form of oxygen
 Nitrogen (N2): clear, low activity, most abundant gas in
the Earth’s atmosphere.
 Hydrogen (H2): clear, lighter than air,
flammable/explosive
 Carbon dioxide (CO2): clear, but turns limewater
cloudy. Does not support respiration but low toxicity.
Heavier than air. Largely responsible for the greenhouse
effect (global warming)
 Sulphur dioxide (SO2): smelly gas. When it combines
with oxygen and water vapour it can form H2SO4,
responsible for acid rain.
5
Some Important Gases
 Carbon monoxide (CO): clear, colourless, but very 
toxic. It destroys the ability of blood to carry oxygen. About
the same density as air.
 Ammonia (NH3): toxic, strong smell, refrigerant . Very
soluble in water, forms a basic solution called ammoniawater (NH4OH) which is found in some cleaners.
 Freon® or CFC: Non-toxic refrigerant used in airconditioners & freezers. Freon may catalyze
ozone breakdown. The original Freon formula is
now banned, but low chlorine versions are still in
use.
 Methane (CH4): flammable gas, slightly lighter
than air, produced by decomposition. Found in
natural gas. Methane is also a “greenhouse” gas.
 Helium (He): inert, lighter than air. Used in
balloons and in diver’s breathing mixtures.
6
 Acetylene (C2H2): AKA ethene, it is used as a fuel in
welding, lanterns and other devices.
 Propane (C3H8): used as a fuel in barbecues, stoves,
lanterns and other devices.
 Radon (Rn): A noble gas that is usually radioactive. It is
heavier than air, and sometimes found in poorly ventilated
basements.
 Neon (Ne) and Xenon (Xe): Noble gases found in
fluorescent light tubes, and as insulators inside windows.
They glow more brightly than other gases when electrons
pass through them. Neon is slightly lighter than air, Xenon
is quite a bit heavier.
 Compressed Air (78% N2, 21% O2): Not actually a pure
gas, but a gas mixture that acts much like a pure gas. It is
used by scuba divers (at shallow depths), and to run
pneumatic tools, and for producing foam materials.
7
Fun Gases
(of no real importance)
 Nitrous Oxide (N2O)
– AKA: Laughing gas, Happy gas, Nitro, NOS
– Once used as an anaesthetic in dentist offices, this
sweet-smelling gas reduces pain sensitivity and causes
euphoric sensations. It is an excellent oxidizer, reigniting
a glowing splint much like oxygen would. It is used in
racing where it is injected into the carburetor to
temporarily increase an engine’s horsepower.
 Sulfur Hexafluoride
– One of the densest gases in common use. Fun
with Sulfur hexafluoride
8
Match
the gas with the problem it causes
 Gas
Problem





Ozone layer depletion
Global Warming
Toxic poisoning
Noxious smell
Acid Rain
Carbon Dioxide
CFCs
Methane
Carbon monoxide
Sulfur dioxide
Next slide: Summary
9
Some Gases Classified by
Relative Density
Low Density gases
Neutral Density Gases
High Density gases
“lighter than air”<25 g/mol
“similar to air” 29±4 g/mol
“Denser than air” (>34 g/mol)
Testable Property*:
Balloon will float in air
Balloon drops slowly
through air
Balloon drops quickly through
air
Examples:
Hydrogen (H2)
2
Helium (He)
4
Methane (CH4) 16
Ammonia (NH3) 17
Neon (Ne)
20
Hydrogen Fluoride (HF) 21
Examples:
“Cyanide“ (HCN)
Acetylene (C2H4)
Nitrogen (N2)
Carbon monoxide
Ethane (C2H6)
Oxygen (O2)
Examples:
Fluorine (F2)
Argon (Ar)
Carbon dioxide (CO2)
Propane (C3H8)
Butane (C4H10)
Sulphur Hexafluoride (SF6)
27
28
28
28
30
32
38
40
44
44
58
146
*balloon test: Fill a large, lightweight balloon with the gas, then
release it from a height of about 1.8 m in a room with still air. If
the gas is lighter than air the balloon will float upwards. If it is
close to air, the balloon will fall very slowly. If the gas is heavier
than air, the balloon will fall quickly.
Some Gases Classified by
Chemical Properties
Combustible gases
(combustion /explosion)
Reactive- oxidizing Gases Non-Reactive gases
(support combustion)
Testable property:
Burning splint produces
“pop”
Testable property:
Glowing splint reignites,
burning splint grows brighter
Testable property:
Burning splint is
extinguished, glowing
splint is dimmed
Other properties:
Useful as fuels
Other properties:
Cause metals and some
other materials to corrode or
oxidize. Can improve
combustion.
Other properties:
Can be used to preserve
foods by slowing
oxidation
Examples:
Hydrogen (H2)
Methane (CH4)
Propane (C3H8)
Acetylene (C2H4)
Examples:
Oxygen (O2)
Fluorine (F2)
Chlorine (Cl2)
Nitrous Oxide (NO2)
Examples:
Carbon dioxide (CO2)
Nitrogen (N2)
Argon (Ar)
Helium (He)
Textbook Assignments
 Read Chapter 1: pp. 37 to 50
 Do the exercises on pages 51 and 52
– Questions # 1 to 22
12
Summary:
• Know the properties of gases
• Know the features of some important
gases, esp:
• Oxygen
• Hydrogen
• Carbon dioxide
• Know the environmental problems
associated with some gases, eg.
• Carbon dioxide
• CFC’s
• Sulfur dioxide
13
Chapter 2
• The Kinetic Theory
“Moving, moving, moving,
Keep those atoms moving...”
14
Kinetic Theory
• Overview:
The kinetic theory of gases (AKA. kineticmolecular theory) tries to explain the behavior
of gases, and to a lesser extent liquids and
solids, based on the concept of moving
particles or molecules.
2.1
Page 54
The Kinetic Theory of Gases
(AKA: The Kinetic Molecular Theory)
• The Kinetic Theory of Gases tries to explain the
similar behaviours of different gases based on
the movement of the particles that compose
them.
• “Kinetic” refers to motion. The idea is that gas
particles* are in constant motion.
*
For simplicity, I usually call the gas particles “molecules”,
although in truth, they could include atoms or ions.
16
R
Not in text
The Particle Model
• The Kinetic Theory is part of the Particle
Model of matter, which includes the following
concepts:
– All matter is composed of particles (ions, atoms or
molecules) which are extremely small and have a
varying space between them, depending on their
state or phase.
– Particles of matter may attract or repel each other,
and the force of attraction or repulsion depends on
the distance that separates them.
+
+
– Particles of matter are always moving.
-
17
Kinetic Molecular Theory
And Temperature
• The absolute temperature of a gas (Kelvins) is
directly proportional to the average kinetic
energy of its molecules.
– In other words, when it is cold, molecules move
slowly and have lower kinetic energy.
– When the temperature increases, molecules speed
up and have more kinetic energy!
18
2.1.1
Page 54
R
Particle Motion
and Phases of Matter
• Recall that:
• In solids, the particles (molecules) are moving
relatively slowly. They have low kinetic energy
• In liquids, molecules move faster. They have
higher kinetic energy.
• In gases, the particles move fastest, and have high
kinetic energy.
• But, as we will find out later:
• Heavy particles moving slowly can have the same
kinetic energy as light particles moving faster.
19
Kinetic Theory Model of States
Solid
Particles vibrate
but don’t “flow”.
Strong molecular
attractions keep
them in place.
Liquid
Particles vibrate, move
and “flow”, but cohesion
(molecular attraction)
keeps them close
together.
Gas
Particles move freely
through container. The
wide spacing means
molecular attraction is
negligible.
20
2.1.1
Page 55
Kinetic Motion of Particles
• Particles (ie. Molecules) can have 3 types
of motion, giving them kinetic energy
– Vibrational kinetic energy (vibrating)
– Rotational kinetic energy (tumbling)
– Translational kinetic energy (moving)
21
2.1.1
Page 56
Kinetic Theory and Solids & Liquids
• When it is cold, molecules move slowly
• In solids, they move so slowly that they are held
in place and just vibrate (only vibrational energy)
• In liquids they move a bit faster, and can tumble
and flow, but they don’t escape from the
attraction of other molecules (more rotational
energy, along with a little bit of vibration & translation)
• In gases they move so fast that they go
everywhere in their container (more translational
energy, with a little bit of rotation & vibration).
22
Plasma, the “Fourth State”
(extension material)
• When strongly heated, or exposed to high
voltage or radiation, gas atoms may lose some
of their electrons. As they capture new
electrons, the atoms emit light—they glow. This
glowing, gas-like substance is called “plasma”
23
2.1.3
Page 61
Kinetic Theory and the Ideal Gas
• As scientists tried to understand how gas
particles relate to the properties of gases,
they saw mathematical relationships that
very closely, but not perfectly, described
the behaviour of many gases.
• They have developed theories and
mathematical laws that describe a
hypothetical gas, called “ideal gas.”
24
2.1.3
Page 61
• To make the physical laws (derived from
kinetic equations from physics) work, they
had to make five assumptions about how
molecules work.
• Four of these are listed on page 61 of your
textbook
• The fifth one is not.
1
2
kinetic equation : Ek  mv
2
25
2.1.3
Page 61
Kinetic Theory Hypotheses
about an Ideal Gas
1. The particles of an ideal gas are infinitely
small, so the size is negligible compared to
the volume of the container holding the gas.
2. The particles of an ideal gas are in constant
motion, and move in straight lines (until they
collide with other particles)
3. The particles of an ideal gas do not exert
any attraction or repulsion on each other.
4. The average kinetic energy of the particles
is proportional to the absolute temperature.
26
No Gas is Ideal
• Some of the assumptions on the previous
page are clearly not true.
• Molecules do have a size (albeit very tiny)
• Particles do exert forces on each other (slightly)
• As a result, there is no such thing as a
perfectly “ideal gas”
• However, the assumptions are very good
approximations of the real particle properties.
• Real gases behave in a manner very close to
“ideal gas”, in fact so close that we can usually
assume them to be ideal for the purposes of
calculations.
27
2.1.3
Page 61
Other “Imaginary Features”
of Ideal Gas
• An ideal gas would obey the gas
laws at all conditions of
temperature and pressure
• An ideal gas would never
condense into a liquid, nor
freeze into a solid.
• At absolute zero an ideal gas
would occupy no space at all.
28
Please Notice:
• Not all molecules move at exactly the same
speed. The kinetic theory is based on averages
of a great many molecules.
– Even if the molecules are identical and at a uniform
temperature, a FEW will be faster than the average,
and a FEW will be slower.
– If there are two different types of molecules, the
heavier ones will be slower than the light ones – ON
THE AVERAGE! – but there can still be variations.
That means SOME heavy molecules may be moving
as fast as the slowest of the light ones.
• Temperature is based on the average (mean)
kinetic energy of sextillions of individual
molecules.
29
The mean & mode
can help establish
“average” molecules
“Slow”
molecules
mean
Most molecules
mode
Increasing # molecules
The range of kinetic energies can be
represented as a sort of “bell curve.”
Maxwell’s Velocity Distribution Curve.
“Average”
molecules
“Fast”
Molecules
Increasing kinetic energy
Average
kinetic energy
30
So, Given two different gases at
the same temperature…
What is the same about them?
• The AVERAGE kinetic energy is the same.
• Not the velocity of individual molecules
• Not the mass of individual molecules.
• In fact, the lighter molecules will move faster
• Ek = mv2
kinetic energy of molecules
2
So, kinetic energy depends on both the speed
(v) and on the mass (m) of the molecules.
31
Faster
than
average molecules
Average kinetic energy of warmer molecules
Slower
than
average molecules
Average kinetic energy of molecules
Average kinetic energy of colder molecules
Number of molecules
Distribution of Particles Around
Average Kinetic Energies.
Kinetic Energy of molecules
(proportional to velocity of molecules)
32
Kinetic Theory Trivia
• The average speed of oxygen molecules at
20°C is 1656km/h.
• At that speed an oxygen molecule could travel from Montreal
to Vancouver in three hours…If it travelled in a straight line.
• Each air molecule has about 1010 (ten billion)
collisions per second
• 10 billion collisions every second means they bounce around
a lot!
• The number of oxygen molecules in a classroom
is about:
• 722 400 000 000 000 000 000 000 000
– that’s more than there are stars in the universe!
• The average distance air molecules travel
between collisions is about 60nm.
– 0.00000006m is about the width of a virus.
33
Videos
• Kinetic Molecular Basketball
– http://www.youtube.com/watch?v=t-Iz414g-ro&NR=1
• Average Kinetic Energies
– http://www.youtube.com/watch?v=UNn_trajMFo&NR=1
• Thermo-chemistry lecture on kinetics
34
Assignments
• Read pages 53 to 61
• Do Page 62 # 1-11
Chapter 2.2
• Behaviors of Gases
– Compressibility
– Expansion
– Diffusion and Effusion
– Graham’s Law
• 2.2.1 Compressibility:
– Because the distances between particles in a
gas is relatively large, gases can be squeezed
into a smaller volume.
– Compressibility makes it possible to store
large amounts of a gas compressed into small
tanks
• 2.2.2 Expansion:
– Gases will expand to fill any container they
occupy, due to the random motion of the
molecules.
37
2.2.3 Diffusion

Diffusion is the tendency for molecules to
move from areas of high concentration to
areas of lower concentration, until the
concentration is uniform. They do this
because of the random motion of the
molecules.
 Effusion is the same process, but with the
molecules passing through a small hole or
barrier
Next slide:
38
Rate of Diffusion or Effusion
 It
has long been
known that lighter
molecules tend to
diffuse faster than
heavy ones, since
their average
velocity is higher,
but how much
faster?
 heavy particle
 light particle
39
Graham’s Law

Thomas Graham (c. 1840)
studied effusion (a type of
diffusion through a small hole)
and proposed the following law:
 “The rate of diffusion of a gas is
inversely proportional to the
square root of its molar mass.”


In other words, light gas particles
will diffuse faster than heavy gas
molecules, and there is a math
formula to calculate how much
faster.
Next slide: Example
Internet demo of effusion
M2
v1

v2
M1
Where: v1= rate of gas 1
v2= rate of gas 2
M1= molar mass of gas1
M2=molar mass of gas 2
40
Graham’s Law
Version #1, based on Effusion Rate
• The relationship between the rate of effusion or diffusion
and the molar masses is:
v1

v2
Where:
M2
M1
Note: See the inversion
of the 1 and 2 in the
2nd ratio!
v1 is the rate of diffusion of gas 1, in any appropriate rate units*
v2 is the rate of diffusion of gas 2, in the same units as gas 1
M1 is the molar mass of gas 1
M2 is the molar mass of gas 2
*Rate units must be an amount over a time for effusion (eg: mL/s or
L/min), or a distance over a time for diffusion (eg: cm/min or mm/s)
Thomas Graham
(1805-1869)
• Graham derived his law by treating
gases as ideal, and applying the kinetic
energy formula to them.
• Ek = ½ mv2
• All gases have the same kinetic energy
at the same temperature,
• Therefore, mv2 for the first gas = mv2
for the second gas: m1v12 = m2v22.
• A bit of algebra then gave him his
famous law.
And in my
spare time I
invented
dialysis, which
has saved the
lives of
thousands of
kidney patients
Graham’s Law
Version #2, Based on Effusion Time
• Sometimes it’s easier to measure the time it takes for a
gas to effuse completely, rather than the rate. Graham’s
law can be changed for this, but the relationship between
time and molar mass is direct as the square root:
t1

t2
Where:
M1
M2
Note: In this variant law,
the relationship is not
inverted!
t1 is the time it takes for the first gas to effuse completely.
t2 is the time it takes for an equal volume of the 2nd gas to effuse
M1 is the molar mass of the first gas
M2 is the molar mass of the second gas.
Example of Graham’s Law:





How much faster does He diffuse than N2?
MN =2x14.0=28 g/mol
Nitrogen (N2) has a molar
MHe=1x4.0=4 g/mol
mass of 28.0 g/mol
2
Helium (He) has a molar
mass of 4.0 g/mol
The difference between
their diffusion rates is:
Notice the reversal of
order!
So helium diffuses 2.6
times faster than nitrogen
Next slide: 2.3 Pressure of Gases
M N2
vHe

vN 2
M He
28 g / mol
5.3

 2.6
2
4 g / mol
44
Assignments
 Read
pages 63 to 67
 Do Questions 1 to 10 on
page 68
Outer Space (immeasurable)
100 km
Spaceship 1 (2006)
X15 (1963)
< 0.003 kPa
Edge of Space
Chapter 2.3
• Pressure of Gases
– What is Pressure
– Atmospheric Pressure
– Measuring Pressure
40 km
1 kPa
20 km
6 kPa
10 km
25 kPa
5 km
55 kPa
0 km
101 kPa
Highest Jet 4 kPa
Mt Everest 31 kPa
Mr. Smith
46
Pressure
• Pressure is the force exerted by a gas on
a surface.
• The surface that we measure the pressure on is
usually the inside of the gas’s container.
• Pressure and the Kinetic Theory
• Gas pressure is caused by billions of particles
moving randomly, and striking the sides of the
container.
F
• Pressure Formula: P 
A
Pressure = force divided by area
47
Atmospheric Pressure
• This is the force of a 100 km high
column of air pushing down on us.
• Standard atmospheric pressure is
•
•
•
•
1.00 atm (atmosphere), or
101.3 kPa (kilopascals), or
760 Torr (mmHg), or
14.7 psi (pounds per square inch)
• Pressure varies with:
• Altitude. (lower at high altitude)
• Weather conditions. (lower on cloudy days)
48
SP
Pressure conversions
P(units given)
P (units wanted )

SP(units given) SP(units wanted )
1.00 atm
760 mmHg
760 Torr
101.3 kPa
14.7 psi
1013 mB
29.9 inHg
Example 1: convert 540 mmHg to kilopascals
540 mmHg
P

760 mmHg 101.3 kPa
=72.0 kPa
Divide
Example 2: convert 155 kPa to atmospheres
155 kPa
P

101.3 kPa 1.00 atm
=1.53 atm
Measuring Pressure
• Barometer: measures atmospheric
pressure.
– Two types:
• Mercury Barometer
• Aneroid Barometer
• Manometer: measures pressure in a
container (AKA. Pressure guage)
• Dial Type: Similar to an aneroid barometer
• U-Tube: Similar to a mercury barometer
• Piston type: used in “tire guage”
50
the Mercury Barometer
• A tube at least 800 mm long is filled with
mercury (the densest liquid) and inverted
over a dish that contains mercury.
• The mercury column will fall until the air
pressure can support the mercury.
• On a sunny day at sea level, the air
pressure will support a column of mercury
760 mm high.
• The column will rise and fall slightly as the
weather changes.
• Mercury barometers are very accurate,
but have lost popularity due to the toxicity
of mercury.
51
The Aneroid Barometer
• In an aneroid barometer,
a chamber containing a
partial vacuum will
expand and contract in
response to changes in
air pressure
• A system of levers and
springs converts this into
the movement of a dial.
Manometers (Pressure Gauges)
• Manometers work much like
barometers, but instead of
measuring atmospheric
pressure, they measure the
pressure difference between
the inside and outside of a
container.
• Like barometers they come in
mercury and aneroid types.
There is also a cheaper
“piston” type used in tire
gauges, but not in science.
You Tube manometer
Tire gauge
(piston manometer)
U-tube manometer
(mercury manometer)
Pressure gauge
(aneroid)
Reading U-tube manometers
Atm.
pressure
Must be
in
mmHg,
not cm
or kPa!
After you finish, you can convert your
answer to kPa, or atm. Or whatever.
• When reading a mercury Utube manometer, you
measure the difference in
the heights of the two
columns of mercury.
• If the tube is “closed” then
the height (h) is the gas
pressure in mmHg.
P(mmHg)=h(mmHg)
• If the tube is “open” and h is
positive (the pressure you
are measuring is greater
than the atmosphere) then
you must add atmospheric
pressure in mmHg.
Pgas(mmHg) = Patm(mmHg)+h(mm)
Manometer Examples
on a day when the air pressure is 763mmHg (101.7 kPa)
4
cm
Closed tube: Pgas(mm Hg)=h (mm Hg)
Pgas = h = 4 cm = 40 mm Hg
Pgas = 40mm Hg 101.3kPa  5.3kPa
760mm Hg
6
Open: Pgas(mmHg)=P atm(mmHg) +h (mmHg)
Pgas = 763 + 60mm Hg =823 mm Hg
823mm Hg
Pgas =
101.3kPa  109.7kPa
760mm Hg
Open: Pgas(mmHg)=P atm(mmHg) -h (mmHg)
Pgas = 763 - 60mm Hg =703 mm Hg
Pgas = 703mmHg 101.3kPa  93.7kPa
760mmHg
9
Assignments
• Read pages 69 to 73.
• Do Page 74, Questions 1 to 4.
Chapter 2.4
• The Simple Gas Laws
–
–
–
–
Boyle’s Law
Charles’ Law
Gay-Lussac’s Law
Avogadro’s Law
Relates volume & pressure
Relates volume & temperature
Relates pressure & temperature
Relates to the number of moles
• Other Simple Laws that are a Gas:
– Cole’s Law
– Murphy’s Law
– Clarke’s Laws
Relates thinly sliced cabbage
to vinegar
Anything that can go wrong will.
Relates possible and impossible
57
Clarke’s Laws
of the impossible*
Clarke’s 1st Law: If an elderly and respected
science teacher (like me) tells you that
something is possible, he is probably right. If he
tells you something is impossible, he’s almost
certainly wrong.
 Clarke’s 2nd Law: The only way to find the limits
to what is possible is to go beyond them.
 Clarkes 3rd Law: Any sufficiently advanced
technology is indistinguishable from magic.

*these are slightly paraphrased, I quote them from memory. They were
developed by science fiction writer Arthur C. Clarke
Lesson 2.4.1
Boyle’s Law
Robert Boyle (1662)
For Pressure and Volume
“For a given mass of gas at a
constant temperature, the volume
varies inversely with pressure.”
1
P 
V
Next slide: Air in Syringe
59
Robert Boyle
 Born: 25 January 1627




Lismore, County Waterford, Ireland
Died 31 December 1691 (aged 64)
London, England
Fields: Physics, chemistry; Known for
Boyle's Law. Considered to be the
founder of modern chemistry
Influences: Robert Carew, Galileo
Galilei, Otto von Guericke, Francis
Bacon
Influenced: Dalton, Lavoisier, Charles,
Gay-Lussack, Avogadro.
Notable awards: Fellow of the Royal
Society
60
Pressure
 Gas pressure is the force placed on the sides of a
container by the gas it holds
 Pressure is caused by the collision of trillions of
gas particles against the sides of the container
 Pressure can be measured many ways
Atmospheres (atm)
Kilopascals (kPa)or(N/m2)
Millibars (mB)
Torr (torr) or mm mercury
Centimetres of mercury
Inches of mercury (inHg)
Pounds per sq. in (psi)
Standard Pressure
1 atm
101.3 kPa = 101.3 N/m2
1013 mB
760 torr = 760 mmHg
76 cmHg
29.9 inHg
(USA only)
14.7 psi
(USA only)
61
Example of Boyle’s Law:
Air trapped in a syringe
If some air is left in
a syringe, and the
needle removed
and sealed, you
can measure the
amount of force
needed to
compress the gas
to a smaller
volume.
Next slide: Inside syringe
62
Inside the syringe…

The harder you press, the smaller
the volume of air becomes.
Increasing the pressure makes the
volume smaller!
 The original pressure was low, the
volume was large. The new
pressure is higher, so the volume is
small.

Click Here for an internet demo using
psi (pounds per square inch) instead of
kilopascals (1kPa=0.145psi)
low
high
Next slide: PV
63
This means that:
 As
the volume of a contained gas
decreases, the pressure increases
 As the volume of a contained gas
increases, the pressure decreases
 This assumes that:


no more gas enters or leaves the container,
and
that the temperature remains constant.
 The
mathematical formula for this is given
on the next slide
Next slide: Example
64
Boyle’s Law
Relating Pressure and Volume of a Contained Gas
• By changing the shape of a gas container, such
as a piston cylinder, you can compress or
expand the gas. This will change the pressure
as follows:
P1V1  P2V2
Where:
P1 is the pressure* of the gas before the container changes shape.
P2 is the pressure after, in the same units as P1.
V1 is the volume of the gas before the container changes, in L or mL
V2 is the volume of the gas after, in the same units as V1
*appropriate pressure units include: kPa, mmHg, atm. Usable, but
inappropriate units include psi, inHg.
Example 1





You have 30 mL of air in a syringe at 100 kPa.
If you squeeze the syringe so that the air
occupies only 10 mL, what will the pressure
inside the syringe be?
P1 × V1 = P2 × V2, so..
100 kPa × 30 mL = ? kPa × 10 mL
3000 mL·kPa ÷ 10 mL = 300 kPa
The pressure inside the syringe will be 300 kPa
Next slide: Graph of Boyle’s Law
66
Graph of Boyle’s Law
The Pressure-Volume Relationship
8
Boyle’s Law produces
an inverse
relationship graph.
P(kpa) x V(L)
6
5
4
3
2 = 800
1.6 = 800
1.33 = 800
1.14 = 800
1 = 800
2
x
x
x
x
x
1
400
500
600
700
800
Volume (L) 
200 x 4 = 800
300 x 2.66 = 800
7
100 x 8 = 800
Next slide: Real Life Data
100
200 300 400 500 600 700 800
Pressure (kPa) 
67
Example 2: Real Life Data
30 35
2+4= 6kg : 20 mL
(120)
4+4=8kg : 15 mL
(120)
6+4=10kg: 12 mL
(120)
10
25
(116)
8+4=12kg: 10.5 mL
(126)
15
20
0+4= 4kg : 29 mL
5
At the beginning, Mr. Taylor
calculated the equivalent
of 4 kgf of atmospheric
pressure were exerted on
the syringe.
40
In an experiment Mr. Taylor
and Tracy put weights
onto a syringe of air.
2
4
6
8 10 12 14 16
Next slide: Boyle’s Law Experiment or skip to: Lesson 2.3 Charles’ Law:
18
68
Summary: Boyle’s law
1
P
V
• Formula: P1V1=P2V2
P1V1=P2V2
• Graph: Boyle’s law is usually
represented by an inverse
relationship graph (a curve)
Volume (L) 
• The volume of a gas is
inversely proportional to its
pressure
Pressure (kPa) 
69
70
Assignments on Boyle’s Law
• Read pages 75 to 79
• Do questions 1 to 10 on page 97
Boyle’s Law Lab Activity
• We will use the weight of a column of
mercury to compress and expand air (a
gas) sealed in a glass tube.
• Read the handout for details of the
procedure. (Note: You may shorten the
procedure section in your report by
including and referring to this handout as
part of a complete sentence.)
• You should still write all other report
sections (purpose, materials, diagram,
observations etc.) in full, as normal.
Diagram of Boyle’s Law Apparatus
#1. Horizontal
#2 Open end up
#3 Open end down
Collecting Data
• You will need to find the length of the mercury
column with the tube held horizontal:
(a) Position of “right”side of mercury
___
mm
(b) Position of “left” of mercury column
___
mm
(c) Height of mercury column (a) – (b)
(c)
mm
• You also need this atmospheric information:
(d) today’s temperature*
___
°C
(e) today’s barometric pressure (blackboard)
(e)
mmHg
*used to calibrate the barometer, not used in calculations
Collecting Data (continued)
Data set 1 - Horizontal Tube:
(f) Position of “left” side of column
mm
(g) Position of closure
mm
(h) “volume” of gas
(f) – (g)
(h)
Mm
Data set 2 - Open End Up:
(i) Position of bottom of column
mm
(j) Position of closure should be same as (g)
mm
(k) “volume” of gas
(i) – (j)
(k) mm
Collecting Data (continued)
Data set 3 - Open End Down:
l) Position of Top of column
m) Position of closure should be same as (g)
n) “volume” of gas
(l) – (m)
mm
mm
mm
This concludes the collection of data, now
we must process it and calculate the PV
(pressure x volume) values at each of the
three conditions.
Calculations
Barometric
pressure
Item (e)
Column
Height
Item (c)
“Pressure”
P
“Volume”
V
Horizontal
(e)
(c)
(e)
(h)
Open End
Up
(e)
(c)
(e)+(c)
(k)
Open End
Down
(e)
(c)
(e)- (c)
(n)
PV
PxV
Since we are using analogues for pressure & volume, the units don’t matter.
Conclusion and Discussion
• According to Boyle’s law, the PV values should
all be identical. In the real world they will not be
identical, but they should be very close.
• Analyze your results. While doing this you
should find the percentage similarity between
your largest and smallest result (smallest over
largest x 100%). This can help you conclude if
your results have supported Boyle’s Law or not.
• Discuss sources of error, and explain if they
were significant in your results.
• Discuss the meaning of Boyle’s law as it relates
to this activity.
Answers to Boyle’s Law Sheet
1. 1.00 L of a gas at standard temperature and
pressure (101 kPa) is compressed to 473 mL.
What is the new pressure of the gas?
1 mark
1 mark
formula
P1 • V1 =P2 • V2
Known
P1= 101 kPa
V1= 1.00x103 mL 101kPa • 1000 mL = P kPa • 473 mL
2
P2= unknown
1 mark
V2= 473 mL
P2 = 101•1000 kPa•mL = 213.53 kPa
473 mL
1 mark
Answer: the pressure will be about 214 kilopascals
2.
In a thermonuclear device the pressure of 0.050 L of gas
reaches 4.0x108kPa. When the bomb casing explodes,
the gas is released into the atmosphere where it reaches
a pressure of 1.00x102kPa. What is the volume of the
gas after the explosion?
formula
P1 • V1 =P2 • V2
1 mark
Known 1 mark
P1= 4.0x108kPa
V1= 0.050 L
4.0x108kPa • 0.050L = 1x102kPa • V2L
P2= 1x102kPa
1 mark
V2=unknown
V2 = 4x108•0.05 kPa•L = 2.00x105 L
1x102kPa
1 mark
Answer: there will be 2.00x105Litres (or 200 000L) of gas
3. synthetic diamonds can be manufactured at
pressures of 6.00x104 atm. If we took 2.00L of
gas at 1.00 atm and compressed it to 6.00x104
atm, what would the volume be?
Known
1 mark
P1= 1.00 atm
V1= 2.00 L
P2= 6.0x104 atm
V2= unknown
or
P1=1.01x102kPa,
P2=6.06x106kPa.
Formula
P1V1=P2V2
1 mark
1.00•2.00 = 6.0•104 • V2
V2 = 2.00 ÷ 6.0x104
V2 = 3.33 x10-5 L
The volume would be 3.33x10-5 Litres
1 mark
1 mark
4.
Divers get the bends if they come up too fast because
gas in their blood expands, forming bubbles in their
blood. If a diver has 0.0500L of gas in his blood at a
depth of 50m where the pressure is 5.00x103 kPa, then
rises to the surface where the pressure is
1.00x102kPa, what will the volume of gas in his blood
be? Do you think this will harm the diver?
1 mark
Known
P1=5.00x103 kPa
V1=0.0500 L
P2= 1.00x102 kPa
V2= Unknown
Formula
P1V1=P2V2
1 mark
5.0x103kPa • 0.0500L = 1x102kPa • V2L
1 mark
V2 = 5x103•0.05 kPa•L = 2.50 L
1x102kPa
The sudden appearance of 2½ litres of gas in the
diver’s bloodstream could be quite deadly.
1 mark
Lesson 2.4.2
Charles’ Law
The Relationship between Temperature
and Volume.
“Volume varies directly with Temperature”
V T
Next slide: Jacques Charles
83
Jacques Charles (1787)
“The volume of a fixed mass
of gas is directly proportional
to its temperature (in kelvins)
if the pressure on the gas is
kept constant”
This assumes that the
container can expand, so that
the pressure of the gas will
not rise.
Next slide: The Mathematical formula for this law
Born: November 12, 1746
(1746-11-12) Beaugency,
Orléanais
Died: April 7, 1823
(1823-04-08) (aged 76),
Paris
Nationality: France
Fields: physics,
mathematics, hot air
ballooning
Institutions: Conservatoire
des Arts et Métiers
Charles’ Law
Relating Volume and Temperature of a Gas
• If you place a gas in an expandable container,
such as a piston or balloon, as you heat the gas
its volume will increase, as you cool it the
volume will decrease.
V1 V2

T1
T2
Where:
T1 is Temperature of the gas before it is heated, in kelvins.
T2 is Temperature of the gas after it is heated, in kelvins
V1 is the volume of the gas before it was heated, in L or mL
V2 is the volume of the gas after it was heated, in the same units.
Charles Law Evidence
Charles used cylinders and pistons to
study and graph the expansion of
gases in response to heat.
See the next two slides for diagrams
of his apparatus and graphs.
Lord Kelvin (William Thompson) used
one of Charles’ graphs to discover
the value of absolute zero.
Next slide: Diagram of Cylinder & Piston
86
Charles Law Example
Piston
Cylinder
Trapped Gas
Click Here for a simulated
internet experiment
Next slide: Graph of Charles’ Law
87
Graph of Charles Law
6L
Lord Kelvin
traced it back
to absolute
zero.
Charles discovered
the direct
relationship
5L
4L
3L
Liquid state
Solid state
-250°C -200°C -150°C
1L
Expansion
of most
real gases
-100°C -50°C
-273.15°C
0°C
50°C
100°C
273°C
2L
150°C 200°C 250°C
Next slide: Example
Example




If 2 Litres of gas at 27°C are heated in a cylinder,
and the piston is allowed to rise so that pressure is
kept constant, how much space will the gas take up
at 327°C?
Convert temperatures to kelvins: 27°C =300k,
327°C = 600k
Use Charles’ Law: (see below)
Answer: 4 Litres
V1 V2
2 Litres x Litres

, so :

T1 T2
300 K
600 K
Next slide: Lesson 2.4 Gay Lussac’s Law
Standard Temperature & Pressure
(STP)
• Since the volume of a gas can change with
pressure and temperature, gases must be
compared at a specific temperature and
pressure. The long-standing standard for
comparing gases is called Standard
Temperature and Pressure (STP)
• Standard Temperature =0°C = 273 K
• Standard Pressure
=101.3 kPa
Ambient Temperature
• Some chemists prefer to compare gases at 25°C rather
than 0°C. At zero it is freezing, a temperature difficult to
maintain inside the lab. This alternate set of conditions
is known as Standard Ambient Temperature and
Pressure (SATP). Although not widely used, you should
be aware of it, and always watch carefully in case a
question uses AMBIENT temperature instead of
STANDARD temperature.
• Ambient Temperature
• Standard Pressure
= 25°C = 298 K
= 101.3 kPa
Comparison
Standard and Ambient Conditions
Pressure
Temperature °C
Temperature K
Molar Volume
Standard Temperature &
Pressure
Ambient Temperature &
Pressure
(STP)
(SATP)
101.3 kPa
0 °C
273.15 K
22.4 L/mol
101.3 kPa
25 °C
298.15 K
24.5 L/mol
Summary: Charles’ law
• The volume of a gas is
directly proportional to its
temperature
V1 V2

T1 T2
Volume (L) 
• Formula:
• Graph: Charles’ law is
usually represented by a
direct relationship graph
(straight line)
• Video1
V T
Absolute zero
0°C=273K
Temp
Charles’ Law Worksheet
1. The temperature inside my fridge is about 4˚C, If I place a
balloon in my fridge that initially has a temperature of 22˚C
and a volume of 0.50 litres, what will be the volume of the
balloon when it is fully cooled? (for simplicity, we will
assume the pressure in the balloon remains the same)
Temperatures must be converted to kelvin
Data:
T1=22˚C =295K
So:
T2=4˚C =277K
1
2 V2=V1 x T2 ÷ T1
V1=0.50 L
V2=0.5L x 277K
divide
1
2
To find:
295K
V2= unknown
V =0.469 L
V V

T T
2
The balloon will have a volume of 0.47 litres
94
2.
A man heats a balloon in the oven. If the balloon has
an initial volume of 0.40 L and a temperature of
20.0°C, what will the volume of the balloon be if he
heats it to 250°C.
Convert temperatures to kelvin
Data
20+273= 293K, 250+273=523k
V1= 0.40L
T1= 20°C =293 K Use Charles’ Law
T2= 250°C =523 K
V V
0. 4 L
V
V2= ?0.7139L
1
T1

2
T2
...
293K

2
523K
0.40L x 523 K ÷ 293 K = 0.7139L
Answer: The balloon’s volume will be 0.71 litres
95
3. On hot days you may have noticed that potato chip bags
seem to inflate. If I have a 250 mL bag at a temperature
of 19.0°C and I leave it in my car at a temperature of
60.0°C, what will the new volume of the bag be?
Convert temperatures to kelvin
19+273= 292K, 60+273=333K
Data:
V1=250 mL
Use Charles’ Law
T1= 19.0°C=292 K
T2=60.0°C =333 K V
V2
250mL
1

...

V2= ?285.10 mL
T1
T2
292 K
V2
333K
250mL x 333 K ÷ 292 K = 285.10mL
Answer: The bag will have a volume of 285mL
Although only the answers are shown here, in order to get
full marks you need to show all steps of the solution!
4. The volume of air in my lungs will be 2.35
litres
Be sure to show your known information
Change the temperature to Kelvins and show them.
Show the formula you used and your calculations
State the answer clearly.
5.
6. The temperature is 279.7 K, which corresponds to 6.70 C.
jacket or sweater would be appropriate clothing for this
weather.
A
Charles’ Law Assignments
• Read pages 80 to 84
• Do questions 11 to 21 on pages 97 and 98
Lesson 2.4.3
Gay-Lussac’s Law
For Temperature-Pressure changes.
“Pressure varies directly with Temperature”
P T
Next slide:’
99
Joseph Gay-Lussac (1802)
“The pressure of a gas is
directly proportional to
the temperature (in
kelvins) if the volume is
kept constant.”
Born 6 December 1778
Saint-Léonard-de-Noblat
Died 9 May 1850 @
Saint-Léonard-de-Noblat
Nationality: French
Fields: Chemistry
Known for Gay-Lussac's law
Next slide:’
10
Gay-Lussac’s Law
Relating Pressure and Temperature of a Gas
P1 P2

T1 T2
Where:
P1 is the pressure* of the gas before the temperature change.
P2 is the pressure after the temperature change, in the same units.
T1 is the temperature of the gas before it changes, in kelvins.
T2 is the temperature of the gas after it changes, in kelvins.
*appropriate pressure units include: kPa, mmHg, atm.
Gay-Lussac’s Law
 As
the gas in a sealed
container that cannot
expand is heated, the
pressure increases.
pressure
 For
calculations, you
must use Kelvin
temperatures:

K=°C+273
10
Example

A sealed can contains 310 mL of air at
room temperature (20°C) and an internal
pressure of 100 kPa. If the can is heated
to 606 °C what will the internal pressure
be? Remove irrelevant fact
Data:
P1= 100kPa
V1=310 mL
T1=20˚C
P2=unknown
T2=606˚C
Next slide: T vs P graph
˚Celsius must be converted to kelvins
20˚C ==293K
293 K
606˚C ==879K
879 K
Formula:
P1 P2

T1 T2
100kPa
x

divideK
293
879 K
x = 87900 ÷ 293
x = 300
Answer: the pressure
will be 300 kPa
10
Temperature & Pressure Graph
 The
graph of temperature in Kelvin vs.
pressure in kilopascals is a straight line.
Like the temperature vs. volume graph, it
can be used to find the value of absolute
zero.
10
Pressure (kPa) 
Graph of Pressure-Temperature Relationship
(Gay-Lussac’s Law)
Temperature (K) 
Next slide:’
273K
10
Summary: Gay-Lussac’s law
• The pressure of a gas is
directly proportional to its
temperature
• Formula:
• Graph: Gay-Lussac’s law is
usually represented by an
direct relationship graph
(straight line)
Pressure 
P1 P2

T1 T2
Absolute zero
0°C=273K
Temp
Assignment on Gay-Lussac’s Law
• Read pages 85 to 87
• Answer questions #22 to 30 on page 98
Lesson 2.4.4
Avogadro’s Law
For amount of gas.
“The volume of a gas is directly related to the
number of moles of gas”
V n
Next slide: Lorenzo Romano Amedeo Carlo Avogadro di Quaregna
10
Lorenzo Romano Amedeo Carlo
Avogadro di Quaregna
“Equal volumes of gas at
the same temperature
and pressure contain
the same number of
moles of particles.”
Amedeo Avogadro
Born: August 9, 1776
Turin, Italy
Died: July 9, 1856
Field: Physics
University of Turin
Known for Avogadro’s
hypothesis,
Avogadro’s number.
 You already know most of the facts that
relate to Avogadro’s Law:
– That a mole contains a certain number of
particles (6.02 x 1023)
– That a mole of gas at standard temperature and
pressure will occupy 22.4 Litres (24.5 at SATP)
 The only new thing here, is how changing
the amount of gas present will affect
pressure or volume.
– Increasing the amount of gas present will
increase the volume of a gas (if it can expand),
– Increasing the amount of gas present will
increase the pressure of a gas (if it is unable to
expand).
110
It’s mostly common sense…
 If you pump more gas into a
balloon, and allow it to expand
freely, the volume of the balloon
will increase.
 If you pump more gas into a
container that can’t expand, then
the pressure inside the container
will increase.
111
Avogadro’s Laws
Relating Moles of Gas to Volume or Pressure
V1 V2

n1 n2
Where:
or
P1 P2

n1 n2
V1 = volume before, in appropriate volume units.
V2 = volume after, in the same volume units
P1=pressure before, in appropriate pressure units.
P2=pressure after, in the same pressure units.
n1 = #moles before
n2 = #moles after
112
Assignments on Avogadro’s Law
• Read pages 92 to 96
• Do Questions 31 to 36 on page 98
113
Lesson 2.5
The General Gas Law and the
Ideal Gas Law
Next slide:
114
The Combined or General
Gas Law
• The general (or combined) gas law replaces
the four simple gas laws. It puts together:
•
•
•
•
Boyle’s Law
Charles’ Law
Gay-Lussac’s Law
Avogadro’s Law
= General Gas Law
• Advantages of the Combined Gas Law:
• It is easier to remember one law than four.
• It can handle changing more than one variable at a
time (eg. Changing both temperature and
pressure)
115
The General Gas Law
Relating all the Simple Laws Together
P1V1 P2V2

n1T1 n2T2
Where:
P1 P2 are the pressure of the gas before and after changes.
V1, V2 are the volume of the gas before and after changes.
T1 T2 are the temperatures, in kelvins
n 1, n2 is the number of moles of the gas.
The neat thing about the General gas law is that it
can replace the three original gas laws.
Just cross out or cover the parts that don’t
change, and you have the other laws:
If the temperature is
constant, then you have
Boyle’s law.
If, instead, pressure
remains constant, you
have Charles’ Law
And
the the
Mostfinally,
of the iftime,
volume
constant,
number stays
of moles
stays
then
you have
Gaythe same,
so you
can
Lussac’s
Law from the
remove moles
equation.
P1V1 P2V2

n1T1 n2T2
117
The Ideal Gas Law
The Ideal Gas Law is derived from the
General Gas Law in several mathematical
steps.
First, start with the general gas law,
including P, V, T, and the amount of gas in
moles (n) .
P1V1 P2V2

n1T1 n2T2
Next slide:
Remember Standard Temperature & Pressure
(STP)
 Standard
Temperature is 0°C or more to
the point, 273K (@SATP = 25°C = 298K)
 Standard Pressure is 101.3 kPa (one
atmospheric pressure at sea level)
 At STP one mole of an ideal gas occupies
exactly 22.4 Litres (@SATP = 24.5 L)
The Ideal Gas Law: Calculating the
Ideal Gas Constant.

We are going to
calculate a new constant
by substituting in values
for P2, V2, T2 and n2
 At STP we know all the
conditions of the gas.
 Substitute and solve to
give us a constant
P1V1 P2V2

n1T1 n2T2
P1V1 101.3 kPa  22.4 L

n1T1
1 mol  273 K
P1V1
 8.31 L  kPa / K  mol
n1T1
Next slide: R-- The Ideal Gas Constant
The Ideal Gas Constant
is the proportionality constant that makes the ideal gas law work
 The
Ideal Gas Constant has the symbol R
R=8.31
L· kPa /
K·mol
 The Ideal Gas constant is 8.31 litrekilopascals per kelvin-mole.
Next slide: Ideal Gas Formula


So, if
P1V1
R
n1T1
Then, by a bit of algebra: P1V1=n1RT1
 Since we are only using one set of
subscripts here, we might as well remove
them: PV=nRT
The Ideal Gas Law
Relating Conditions to the Ideal Gas Constant
PV  nRT
Where:
P=Pressure, in kPa
V=Volume, in Litres
n= number of moles, in mol
R= Ideal Gas constant, 8.31
T = Temperature, in kelvins
LkPa/
Kmol
 The
Ideal gas law is best to use when you
don’t need a “before and after” situation.
 Just one set of data (one volume, one
pressure, one temperature, one amount of
gas)
 If you know three of the data, you can find
the missing one.
Sample Problem









8.0 g of oxygen gas is at a pressure of 2.0x102
kPa (ie: 200 Kpa w. 2 sig fig) and a temperature of
15°C. How many litres of oxygen are there?
Formula:
PV = nRT
Variables:
P=200 kPa
V=? (our unknown)= x
n= 8.0g ÷ 32 g/mol =0.25 mol
R=8.31 L·kPa/K·mol (ideal gas constant)
T= 15°C + 273 = 288K
200 x = (0.25)(8.31)(288) , therefore
x= (0.25)(8.31)(288) ÷ 200=2.99 L
There are 3.0 L of oxygen (rounded to 2 S.D.)
Sample problem

8.0 g of oxygen gas is at a pressure of 2.0x102 kPa (ie:
200KPa) and a temperature of 15°C. How many litres
of oxygen are there? (assume 2 significant digits)
Data:
P=200 kPa
V=unknown = X
n= not
given
0.25
mol
R=8.31 L·kPa/K·mol
T= 15°C + 273 = 288K
--m (O2) = 8g
M (O2) = 32.0 g/mol
Temperature has been converted to
kelvins
Calculate the value of n using the mole
formula:
m
8g
n

 0.25mol
M 32 g / mol
PV  nRT
200 x = (0.25)(8.31)(288) , therefore
x= (0.25)(8.31)(288) ÷ 200=2.99 L
There are 3.0 L of oxygen
(rounded to 2 S.D.)
Sample Problem
• 8.0 g of oxygen gas is at a pressure of 2.0x102 kPa (ie:
200KPa) and a temperature of 15°C. How many litres
of oxygen are there? (give answer to 2 significant digits)
Data:
P = 200 kPa
R = 8.31 L·kPa/K·mol
T = 15+273 = 288K
m(O2)= 8.0 g
M(O2)= 32.0 g/mol
8g
n=
 0.25 mol
32 g / mol
To find:
V
Next slide: Ideal vs. Real
Formula:
PV  nRT
Work:
200kPa  V  0.25 mol  8.31
L  kPa
 288K
K  mol
L  kPa
0.25 mol  8.31
 288K
K  mol
V
200kPa
V  2.99 L
Ideal vs. Real Gases
The gas laws were worked out by assuming that gases are ideal, that
is, that they obey the gas laws at all temperatures and pressures. In
reality gases will condense or solidify at low temperatures and/or high
pressures, at which point they stop behaving like gases. Also,
attraction forces between molecules may cause a gas’ behavior to vary
slightly from ideal.
A gas is ideal if its particles are extremely small (true for most gases),
the distance between particles is relatively large (true for most gases
near room temperature) and there are no forces of attraction between
the particles (not always true)
At the temperatures where a substance is a gas, it follows
the gas laws closely, but not always perfectly.
For our calculations, unless we are told otherwise, we will
assume that a gas is behaving ideally. The results will be
accurate enough for our purposes!
Next slide: Summary
Testing if a gas is ideal

If you know all the important properties of a
gas (its volume, pressure, temperature in
kelvin, and the number of moles) substitute
them into the ideal gas law, but don’t put in the
value of R. Instead, calculate to see if the
value of R is close to 8.31, if so, the gas is
ideal, or very nearly so. If the calculated value
of R is quite different from 8.31 then the gas is
far from ideal.
Example







A sample of gas contains 1 mole of particles and
occupies 25L., its pressure 100 kPa is and its
temperature is 27°C. Is the gas ideal?
Convert to kelvins: 27°C+273=300K
PV=nRT
(ideal gas law formula)
100kPa25L=1molR300K, so…
R=100kPa25L÷(300K1mol)
R=8.33 kPaL /Kmol expected value: 8.31 kPaL /Kmol
So the gas is not ideal, but it is fairly close to an
ideal gas,
(8.33  8.31)
100%  0.24%

It varies from ideal by only 0.24%
8.31
Gas Laws Overview
• When using gas laws, remember that
temperatures are given in Kelvins (K)
– Based on absolute zero: –273°C
• The three original gas laws can be combined,
and also merged with Avogadro’s mole concept
to give us the Combined Gas Law.
• Rearranging the Combined Gas Law and doing a
bit of algebra produces the Ideal Gas Law.
• Substituting in the STP conditions we can find
the Ideal Gas Constant.
• “Ideal gases” are gases that obey the gas laws
at all temperatures and pressures. In reality, no
gas is perfectly ideal, but most are very close.
Gas Laws: Summary
Simple gas laws
– Boyle’s Law:
– Charles’ Law:
– Gay-Lussac’s Law:
– Combined gas law:
1
P
V
P1V1  P2V2
V T P  T
V1 V2

T1 T2
P1 P2

T1 T2
P1V1 P2V2

n1T1 n2T2
– Ideal gas law:
PV  nRT
– The ideal gas constant:
R=8.31 Lkpa/Kmol
Video
• Simple gas laws
Assignments on the Simple Gas
Laws
• Finish Exercises p. 99 #37 to 52
Extra Assignments
• Old text References:
– Textbook Chapter 10: pp. 221 to 240
– Student Study Guide pp. 2-4 to 2-11
• Old Textbook: page 241 # 25 to 30
– Do these in your assignments folder.
Extra practice:
• Study guide: pp 2.12 to 2.17 # 1 to 22
– There is an answer key in the back for these
– Do these on your own as review
Exercise Answers
14) The pressure will double, since there is twice
as much gas occupying the same space. (I
answered this using logic and Avagadro’s
hypothesis rather than math. It stands to reason
that twice as much gas in the same space will
increase the pressure.)
15) The pressure will be four times as high, since
the volume is one quarter what it was before:
P1V1 = P2V2 so… P1V1 = 4P1 x ¼V1 (again,
although you can do it with math, logic works
better)
16) The pressure will be one third as great as it
was before, since there is three times the
volume: P1V1 = P2V2, so = 1/3 P1 x 3V1
17) The gas cannot expand, so it exerts force on its
container. As the temperature increases, the
gas particles move faster, hitting the container
sides more frequently and with more force.
This causes greater pressure. You can also
explain this using Gay-Lussac’s law;
P1/T1 = P2/T2
18) Make sure you use the KELVIN temperatures.
The formula is P1/T1 = P2/T2 or 300 kpa/300K =
xkPa/100K, so the pressure will be 100 kPa
19) An ideal gas obeys the gas laws at all
temperatures and pressures (no real gas is
perfectly ideal. More ideal properties will be
discussed in the next section).
20) PCO2 = 3.33 kPa, since all the partial
pressures will add up to the total
pressure (3.33+23.3+6.67=33.3)
21) Use Boyle’s law: P1V1=P2V2, therefore
91.2kpa4.0L=20.3kpaxL so therefore
x=91.2x4÷20.3 the new volume is 17.9 L
22) Use Boyle’s law: P1V1=P2V2 ,so
x=100kPa6L÷25.3kPa. The new volume
will be 23.7L
• 23) Use Charles’Law: V1/T1=V2/T2, convert the
temperature from °CK, so -50°C223K and
100 °C373K so… 5L/223K = x/373K so…
x=5373÷223.
The new volume will be about 8.36 L
• 24) Use Gay-Lussack’s law: P1/T1=P2/T2, don’t
forget to change 27°C300K. So…
200kPa/300K=223kPa/x.
The new temperature will be 61.5°C
(converted from 334.5K)
ANSWERS
25) The combined gas laws:
(this answer is straight from the lesson)
P1V1 P2V2

n1T1 n2T2
107kPa 5L
P  7L
26) Convert the temperatures to
 2
223K
373K
kelvin, set up equation, leaving
out n1 and n2 (moles don’t
change), cross multiply:
Answer:
The new pressure is 127.8Multiply
kPa these together
Then divide by these
27) Data given:
need to find:
32g/mol
m=12g(O2)
M(O2)
P=52.7kPa
V=x L
R=8.31LkPa/Kmol n in mol 0.375mol
T= 25°C
T in kelvin 298K
Find the number of moles of O2: n=m/M
M(O2)=32g/mol so: 12g ÷ 32g/mol = 0.375mol.
Convert CK, 25°C+273=298K
formula:
PV=nRT
so: 52.7kPaxL=0.375mol8.31Lk•Pa/Kmol298K
so: x = (0.375 mol  8.31L•kPa  298 K) • __1_
K mol
Answer: The volume will be about 17.6 L
52.7 kpa
#28-30, answers (with brief explanation)
(see me at lunch if you need more explanation)
28) Litres at STP
a) 56 L
b) 6.72 L
c) 7.84 L
(remember: each mole of gas @STP=22.4L)
29) Answer: The pressure will be 1714 kPa
(use the formula PV=nRT)
30) Answer: The volume will be 16.8 L
(use the formula PV=nRT)
Lesson 2.8
Dalton’s Law of partial pressures
John Dalton
Besides being the founder
of modern atomic theory,
John Dalton experimented
Born
on gases. He was the
first to reasonably
estimate the composition Died
of the atmosphere at 21% Notable students
oxygen, 79% Nitrogen
Known for
Influences
6 September 1766
Eaglesfield,
Cumberland, England
27 July 1844
Manchester, England
James Prescott Joule
Atomic Theory, Law of
Multiple Proportions,
Dalton's Law of Partial
Pressures, Daltonism
John Gough
Partial Pressure
‚
Many gases are mixtures,
‚
eg. Air is 78% nitrogen, 21% Oxygen, 1% other gases
‚ Each gas in a mixture contributes a partial
pressure towards the total gas pressure.
‚ The total pressure exerted by a mixture of
gases is equal to the sum of the partial pressures
of the individual gases in the mixture.
‚
101.3 kPa (Pair) = 79.1 kPa (N2)+ 21.2 kPa (O2) + 1.0 kPa(Other)
Next slide:
Kinetic Theory Connection
• Hypothesis 3 of the kinetic theory states
that gas particles do not attract or repel
each other.
• Dalton established that each type of gas in
a mixture behaved independently of the
other gases.
• The pressure of each gas contributes
towards the total pressure of the mixture.
Dalton’s Law
The Law of Partial Pressures of Gases
PT  P1  P2  ...
Where:
PT is the total pressure of mixed gases
P1 is the pressure of the 1st gas
P2 is the pressure of the 2nd gas
etc...
Variant of Dalton’s Law
(used for finding partial pressure of a gas in a mixture)
nA
PA 
 PT
nT
Where: PA=Pressure of gas A
nA = moles of gas A
nT= total moles of all gases
PT= Total Pressure of all gases
Uses of Dalton’s Law
In the 1960s NASA used the law of partial
pressures to reduce the launch weight of their
spacecraft. Instead of using air at 101 kPa, they
used pure oxygen at 20kPa.
Breathing low-pressure pure oxygen gave the
astronauts just as much “partial pressure” of
oxygen as in normal air.
Lower pressure spacecraft reduced the chances of
explosive decompression, and it also meant
their spacecraft didn’t have to be as strong or
heavy as those of the Russians (who used
normal air).. This is one of the main reasons the
Americans beat the Russians to the moon.
Carelessness with pure oxygen, however,
lead to the first major tragedy of the
American space program…
At 20 kPa, pure oxygen is very safe to
handle, but at 101 kPa pure oxygen
makes everything around it extremely
flammable, and capable of burning five
times faster than normal.
On January 27, 1967, during a pre-launch
training exercise, the spacecraft Apollo-1
caught fire. The fire spread instantly, and
the crew died before they could open the
hatch.
Crew of Apollo 1
Gus Grissom, Ed White, Roger Chaffee
Exercises
:
• Page 113 in new textbook, # 1 to 8
Extra practice (if you haven’t already started):
• Study guide: pp 2.12 to 2.17 # 1 to 22
– There is an answer key in the back for these
– Do these on your own as review
Summary:
• Dalton’s Law: The total pressure of a
gas mixture is the sum of the partial
pressures of each gas.
PT = P1 + P2 + …
• Graham’s Law: light molecules diffuse
faster than heavy ones
M2
Rate1

Rate2
M1
• Avogadro’s hypothesis
– A mole of gas occupies 22.4L at STP and
contains 6.02x1023 particles
Summary of Kinetic Theory
• Hypotheses (re. Behaviour of gas molecules):
1. Gases are made of molecules moving randomly
2. Gas molecules are tiny with lots of space between.
3. They have elastic collisions (no lost energy).
4. Molecules don’t attract or repel each other (much)
• Results:
• The kinetic energy of molecules is related to their
temperature (hot molecules have more kinetic energy
because they move faster)
– Kinetic theory is based on averages of many molecules
(graphed on the Maxwell distribution “bell” curve)
– Pressure is caused by the collision of molecules with the
sides of their containers.
– Hotter gases and compressed gases have more collisions,
therefore greater pressure.
Pressure
Gases are made of particles
Particles move randomly!
Energy of a particle:
KE = ½ mV 2
Pressure is the result of
particles colliding with
the container walls.
P = F /A
Assigned Activities
• References:
– Read Textbook pp.197-203
• Practice problems:
– Textbook: p199 #1-3
– Student study guide: pp. 2-19 to 2-20
(practice problems are for self-correction)
• Assignments (to be collected in your folder):
– Page 241: all questions from 25 to 34
– Handout #1: “combined gas law” #52-58
– Handout #2: “gases & gas laws” 5 questions
(on the back.)
•
•
•
•
•
•
•
Answers (sheet 1)
52: The volume of gas will be 36.5 L
53: The temperature will be 908K or 635C
54: The volume will be 250 mL or 0.25L
55: The pressure will be 251 kPa
56: The pressure will stay the same
57: The pressure will be 42.2 kPa
58: The volume will be 10.2 L
Answers (sheet 2)
•
•
•
•
1: The volume is about 32.5 L
2: The mass is about 1.53 x 10-7 g
3: The pressure is about 61909 kPa
4: The pressure will increase by 168 kPa
(tricky: most students say 268kPa, but that’s
what it ends at, NOT how much it changes!)
• 5: The total pressure is about 172kPa
• The end of module 2