Kinetic Molecular Theory Teacher: H. Michael Hayes
Slide 2
Kinetic Theory of Gases Seeks to explain the similarities
observed in the behaviour of gases based on the movement of
particles that compose them.
Slide 3
Particle model of matter All matter is composed of particles
with more or less space between them depending on their phase.
Particles attract or repel each other and that force depends on the
distance between the particles. Particles are always moving.
Slide 4
Particle model of matter Particle behaviour in solids Weak
vibration. Particle behaviour in liquids Vibration, rotation and
weak translation. Particle behaviour in gases Vibration, rotation
and translation.
Slide 5
Particle model of matter Summary Table
Slide 6
Kinetic Energy of Gas Particles and Temperature E k varies
directly with mass and velocity. What variable has the greater
effect on E k and why?
Slide 7
Kinetic Energy of Gas Particles and Temperature In a sample of
any gas, not all particles have the same kinetic energy. Therefore
not all particles have the same velocity. This is why it is
appropriate to consider mean velocity to determine the mean kinetic
energy of a gas.
Slide 8
Kinetic Energy of Gas Particles and Temperature
Slide 9
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Slide 11
Hypotheses of the kinetic theory of gases H YPOTHESIS 1 T HE
PARTICLES OF A GAS ARE INFINITELY SMALL AND THE SIZE OF A PARTICLE
IS NEGLIGIBLE COMPARED TO THE VOLUME OF THE CONTAINER THAT HOLDS
THE GAS. H YPOTHESIS 2 T HE PARTICLES OF A GAS ARE IN CONSTANT
MOTION AND MOVE IN A STRAIGHT LINE IN ALL DIRECTIONS. H YPOTHESIS 3
T HE PARTICLES OF A GAS DO NOT EXERT ANY FORCE OF ATTRACTION OR
REPULSION ON EACH OTHER. H YPOTHESIS 4 T HE MEAN KINETIC ENERGY OF
A GAS IS DIRECTLY PROPORTIONAL TO THE ABSOLUTE TEMPERATURE.
Slide 12
Kinetic theory of gases Questions Numbers 1 - 11 Page 62 of
Quantum Be able to answer and understand all! Computer simulation
of particle motion and mean kinetic energy.
http://youtu.be/A0May2m6cIc
Slide 13
Behavior of gases. Compressibility Supported by Hypothesis 1.
Due to the large inter particle distance gases can be stored in
small spaces under high pressure. Expansion Supported by Hypothesis
2. Gases dilate indefinitely by filling any accessible space.
Greater dilation leads to greater inter particle distance.
Expansion varies directly with atmospheric pressure.
Slide 14
Behavior of gases. Diffusion The random dispersion of aimlessly
colliding particles.
Slide 15
Behavior of gases. Diffusion Not all gases diffuse at the same
rate. The rate depends on the velocity of particles. Therefore: At
the SAME TEMPERATURE particles with a small mass move more quickly
than particles with a large mass.
Slide 16
Behavior of gases. Effusion When gas flows across a barrier
through a small opening, it is termed EFFUSION. Example is a
deflating Helium balloon In this picture two balloons were
inflated. One with Nitrogen and the other with Helium. Due to the
small radius of He it effuses more readily through the balloon
(membrane).
Slide 17
Behavior of gases. Diffusion & Effusion of Two Different
Gases at the Same Temperature and Pressure. Hypothesis 4 tells us:
At the same temperature and pressure, two different gases will have
the same Kinetic Energy If we use the following expression for the
kinetic energy of the two gases: Given the statement above
(hypothesis 4)
Slide 18
Behavior of gases. Diffusion & Effusion of Two Different
Gases at the Same Temperature and Pressure. If we use the following
expression for the kinetic energy of the two gases: We can express
the relationship like this: Simplify to this:(cancel out ) Simplify
to this:(divide both sides m 1 v 2 2 ) Simplify to this:GRAHAMS
LAW
Slide 19
Behavior of gases. Diffusion & Effusion of Two Different
Gases at the Same Temperature and Pressure. If we use the following
expression for the kinetic energy of the two gases: We can express
the relationship like this: Simplify to this:(cancel out ) Simplify
to this:(divide both sides m 1 v 2 2 ) Simplify to this:GRAHAMS
LAW
Slide 20
Behavior of gases. Diffusion & Effusion of Two Different
Gases at the Same Temperature and Pressure. If we use the following
expression for the kinetic energy of the two gases: We can express
the relationship like this: Simplify to this:(cancel out ) Simplify
to this:(divide both sides m 1 v 2 2 ) Simplify to this:GRAHAMS
LAW
Slide 21
Behavior of gases. Diffusion & Effusion of Two Different
Gases at the Same Temperature and Pressure. If we use the following
expression for the kinetic energy of the two gases: We can express
the relationship like this: Simplify to this:(cancel out ) Simplify
to this:(divide both sides m 1 v 2 2 ) Simplify to this:GRAHAMS
LAW
Slide 22
Behavior of gases. Diffusion & Effusion of Two Different
Gases at the Same Temperature and Pressure. Grahams Law Simplify to
this:(cancel out ) Simplify to this:(divide both sides m 1 v 2 2 )
Simplify to this:GRAHAMS LAW Under identical conditions of
temperature and pressure, the relative rates of diffusion and
effusion of two gases are inversely proportional to the square
roots of their molar masses.
Slide 23
Grahams Law Problem. A 3.00 L sample of helium was placed in
container fitted with a porous membrane. Half of the helium effused
through the membrane in 25 hours. A 3.00 L sample of oxygen was
placed in an identical container. How many hours will it take for
half of the oxygen to effuse though the membrane?: Given: Applying
Grahams Law: How long for half of O 2 to effuse?
Slide 24
Grahams Law Problem. A 3.00 L sample of helium was placed in
container fitted with a porous membrane. Half of the helium effused
through the membrane in 25 hours. A 3.00 L sample of oxygen was
placed in an identical container. How many hours will it take for
half of the oxygen to effuse though the membrane?: Given: Applying
Grahams Law: How long for half of O 2 to effuse?
Slide 25
Grahams Law Problem. A 3.00 L sample of helium was placed in
container fitted with a porous membrane. Half of the helium effused
through the membrane in 25 hours. A 3.00 L sample of oxygen was
placed in an identical container. How many hours will it take for
half of the oxygen to effuse though the membrane?: Given: Applying
Grahams Law: How long for half of O 2 to effuse? See more problems
in Quantum text page 67
Slide 26
Pressure of Gases The pressure of a gas corresponds to the
force they exert on a surface. Supported by Hypothesis 2 and 4
(particles are in constant motion & the greater their kinetic
energy the faster they move.
Slide 27
Pressure of Gases From Hypothesis 4 the mean kinetic energy of
the particles of 2 gases is the same Therefore: a light gas will
collide more frequently with less force and a heavy gas colliding
less frequently with more force given the same pressure.
Slide 28
Atmospheric Pressure Atmospheric pressure is measured using a
barometer invented by Evangelista Torricelli in 1643.
Slide 29
Atmospheric Pressure - measurement Covert the lowest ever
pressure recorded on earth in kPa to mm of Hg.
Slide 30
Atmospheric Pressure - measurement Covert the lowest ever
pressure recorded on earth in kPa to mm of Hg.
Slide 31
Measuring Gas Pressure Barometers are exclusively used to
measure atmospheric pressure. Gas pressure is measured using a
manometer or pressure gauge. The pictures below illustrate DIAL
Manometers.
Slide 32
Measuring Gas Pressure Manometers U-Tube 2 types 1.Closed-end
2.Open-end ClosedOpen
Slide 33
Measuring Gas Pressure in a close-end manometer Real pressure
of a gas using a closed-end manometer. P gas = h P gas = Pressure
of the gas in the container expressed in mm of Hg. h = Height of
the column of Hg expressed in mm of Hg.
Slide 34
Measuring Gas Pressure in a close-end manometer What is the
real pressure of the gas in this manometer? F IND THE DIFFERENCE IN
HEIGHTS. P gas = h h = 88 cm Hg 32 cm Hg h= 56 cm Hg = 560 mm Hg P
gas = 560 mm of HG
Slide 35
Measuring Gas Pressure in a open-end manometer If P gas > P
atm Then P gas = P atm + h If P gas
Measuring Gas Pressure in a open-end manometer What is the gas
pressure? F IND THE DIFFERENCE IN HEIGHTS. h= 75 cm of Hg 45 cm Hg
h=30 cm = 300 mm Since P gas > P atm then P gas = P atm + 300 mm
of Hg P gas = 760 mm + 300 mm P gas = 1060 mm of Hg Ex #1
Slide 37
Measuring Gas Pressure in a open-end manometer What is the gas
pressure? F IND THE DIFFERENCE IN HEIGHTS. h= 62 cm of Hg 25 cm Hg
h=37 cm = 370 mm Since P gas < P atm then P gas = P atm - 370 mm
of Hg P gas = 769.8 mm - 370 mm P gas = 399.8 mm of Hg Ex #2 In
this example you first need to convert 102.6 kPa to mm of Hg
Slide 38
Measuring Gas Pressure in a open-end manometer What is the
pressure of Ne? F IND THE DIFFERENCE IN HEIGHTS. h = 47 cm Hg 39 cm
Hg h = 8 cm = 80 mm Hg Since P Ne < P atm then P Ne = P atm - 80
mm of Hg P Ne = 650 mm - 80 mm P Ne = 570 mm of Hg Ex #3
Slide 39
Measuring Gas Pressure in a open-end manometer What is the
pressure of Ne? F IND THE DIFFERENCE IN HEIGHTS. h = 47 cm Hg 39 cm
Hg h = 8 cm = 80 mm Hg Since P Ne < P atm then P Ne = P atm - 80
mm of Hg P Ne = 650 mm - 80 mm P Ne = 570 mm of Hg Ex #3
Slide 40
Measuring Gas Pressure in a open-end manometer What is the
pressure of Ne? F IND THE DIFFERENCE IN HEIGHTS. h = 47 cm Hg 39 cm
Hg h = 8 cm = 80 mm Hg Since P Ne < P atm then P Ne = P atm - 80
mm of Hg P Ne = 650 mm - 80 mm P Ne = 570 mm of Hg Ex #3
Slide 41
Measuring Gas Pressure in a open-end manometer What is the
pressure of Ne? F IND THE DIFFERENCE IN HEIGHTS. h = 47 cm Hg 39 cm
Hg h = 8 cm = 80 mm Hg Since P Ne < P atm then P Ne = P atm - 80
mm of Hg P Ne = 650 mm - 80 mm P Ne = 570 mm of Hg Ex #3
Slide 42
Simple Gas Laws The simple gas laws help to solve problems that
establish a relationships between TWO of the FOUR variables that
describe gases. Pressure (P) Volume (V) Absolute Temperature (T)
Quantity of gas number of moles (n) .while the remaining two
variables remain constant These laws are the results of several
independent scientific investigations that began in the 17 th
century. At the time no International System of Units existed and
laboratory conditions varied greatly. (Temperature and pressure
varied from lab to lab). This brought about standards so data could
be compared. These standard conditions are now known as STP
StandardTemperaturePressure Standard Temperature and Pressure (STP)
0 o C / 273 o K101.3 kPa Standard Ambient Temperature and Pressure
(SATP) 25 o C / 298 o K101.3 kPa
Slide 43
Simple Gas Laws By combining the simple gas laws, a more
general gas law evolved: Simple Gas Laws The Ideal Gas Law The
General Gas Law This will be our focus over the coming weeks.
Slide 44
Relationship between pressure and volume
Slide 45
When the pressure on a gas increases the volume of a gas
decreases proportionately In the expression to the left we have
introduced a proportionality constant. The constant permits us to
create a simple equation. The equation tells us that Pressure times
Volume is constant.
Slide 46
Relationship between pressure and volume If we calculate the
slope of this line we get the value for the constant. For any pair
of co- ordinates on the line the following is true. If we now plot
pressure versus the inverse of volume (1/v) we get the graph to the
right.
Slide 47
Relationship between pressure and volume
Slide 48
Relationship between pressure and volume sample problem.
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Slide 53
Relationship between volume and absolute temperature. Jacques
Charles observed that the volume of a gas increased by 1/273 of its
original volume when heated by 1 o C. He concluded that when a gas
is heated from 0 o C to 273 o C its volume doubles.
Slide 54
Gas Volume versus Absolute Temperature It is Charles
extrapolated data (dotted lines) that leads Lord Kelvin to the
discovery of the Kelvin scale for temperature and the concept of
absolute zero.
Slide 55
Charles Law
Slide 56
Slide 57
Relationship between pressure and temperature.
Slide 58
Gay-Lussacs Law provides us with a means to predict/compare
pressures and temperatures for a given sample of gas at a CONSTANT
volume
Slide 59
Relationship between pressure and temperature. Connection with
the Kinetic Molecular Theory of gases. Increase in temperature
translates to more rapid movement of particles Hypothesis 4
Increased temperature leads to more rapid rate of translation,
leading to more collisions with the wall of the container causing a
direct increase in pressure.
Slide 60
Relationship between volume and quantity of gas in moles.
Gay-Lussacs fundamental law of chemistry: When gases react, the
volumes of the reactants and the products are ALWAYS present in the
form of simple whole number ratios if they are measured at constant
temperatures and pressures.
Slide 61
Relationship between volume and quantity of gas in moles.
Gay-Lussacs fundamental law of chemistry: When gases react, the
volumes of the reactants and the products are ALWAYS present in the
form of simple whole number ratios if they are measured at constant
temperatures and pressures.
Slide 62
Relationship between volume and quantity of gas in moles.
Avogadros Hypothesis: Given the same conditions of temperature and
pressure, equal volumes of different gases contain the same number
of particles. Avogadros Law: Given the same temperature and
pressure conditions, the volume of a gas is directly proportional
to its quantity expressed in moles.
Slide 63
Relationship between volume and quantity of gas in moles. -
Sample problem A helium balloon occupies a volume of 15L and
contains 0.50 moles of helium at SATP. What will the new volume of
the balloon be if 0.20 mol of helium is added to the balloon be if
0.20 mol of helium is added to the balloon under the same
conditions? V 1 = 15L n 1 = 0.50 mol V 2 = ? n 2 = 0.50 mol + 0.20
mol What we know.
Slide 64
Relationship between volume and quantity of gas in moles. -
Sample problem A helium balloon occupies a volume of 15L and
contains 0.50 moles of helium at SATP. What will the new volume of
the balloon be if 0.20 mol of helium is added to the balloon be if
0.20 mol of helium is added to the balloon under the same
conditions? V 1 = 15L n 1 = 0.50 mol V 2 = ? n 2 = 0.50 mol + 0.20
mol What we know.
Slide 65
Relationship between volume and quantity of gas in moles. -
Sample problem A helium balloon occupies a volume of 15L and
contains 0.50 moles of helium at SATP. What will the new volume of
the balloon be if 0.20 mol of helium is added to the balloon be if
0.20 mol of helium is added to the balloon under the same
conditions? V 1 = 15L n 1 = 0.50 mol V 2 = ? n 2 = 0.50 mol + 0.20
mol What we know.
Slide 66
Relationship between volume and quantity of gas in moles. -
Sample problem A helium balloon occupies a volume of 15L and
contains 0.50 moles of helium at SATP. What will the new volume of
the balloon be if 0.20 mol of helium is added to the balloon be if
0.20 mol of helium is added to the balloon under the same
conditions? V 1 = 15L n 1 = 0.50 mol V 2 = ? n 2 = 0.50 mol + 0.20
mol What we know.
Slide 67
Relationship between volume and quantity of gas in moles.
Connection with the Kinetic Molecular Theory of gases. as the
number of particles of a gas increases, they collide more
frequently with each other and the walls of the container which
increases the pressure. At a constant temperature the additional
pressure increases the volume of the container until the internal
pressure of the gas equals the external pressure on the container.
- Hypothesis 2
Slide 68
Molar Volume of Gases Definition: The volume occupied by ONE
mole of any gas, under fixed conditions of temperature and
pressure. The molar volume (V m ) is expressed in L/mol. Avogadros
Law states that under the same conditions of temperature and
pressure, equal volumes of any gas contain the same number of
particles. Therefore it follows that 1 mole (6.02 x 10 23 )
particles of a gas under the same conditions will occupy the same
volume. The size of the gas particles and the molar mass of the gas
particles have NO INFLUENCE on the volume occupied by the gas.
Slide 69
Molar Volume of Gases Connection to the Kinetic Molecular
Theory. The size of the particles of gases is negligible in
relation to the volume of the particles occupy. Hypothesis 1
Slide 70
Molar Volume of Gases Molar Volume of any gas at STP = 22.4
L/mol. Molar Volume of any gas at SATP = 24.5 L/mol. Sample
Problem: How many moles are there in a container holding 69.2 L of
methane (CH4) gas at STP? What do we know? V = 69.2L n = ? V m =
22.4 L/mol. (at STP)
Slide 71
Molar Volume of Gases Molar Volume of any gas at STP = 22.4
L/mol. Molar Volume of any gas at SATP = 24.5 L/mol. Sample
Problem: How many moles are there in a container holding 69.2 L of
methane (CH4) gas at STP? What do we know? V = 69.2L n = ? V m =
22.4 L/mol. (at STP)
Slide 72
Relationship between pressure and the quantity of gas in moles.
Under the same conditions of TEMPERATURE and VOLUME, the pressure
of a gas is directly proportional to the number of moles. If the
number of moles is doubled the pressure doubles.
Slide 73
Relationship between pressure and the quantity of gas in moles.
Under the same conditions of TEMPERATURE and VOLUME, the pressure
of a gas is directly proportional to the number of moles. If the
number of moles is doubled the pressure doubles.
Slide 74
Relationship between pressure and the quantity of gas in moles.
Under the same conditions of TEMPERATURE and VOLUME, the pressure
of a gas is directly proportional to the number of moles. If the
number of moles is doubled the pressure doubles. Connection with
Kinetic Theory of Gases Hypothesis 2 explains the relationship
between gas pressure and the number of moles. As the number of
particles (moles) of gas increase, they collide with each other and
the walls of the container more frequently. Therefore the number of
collisions per unit of surface area increases which increases the
pressure.
Slide 75
Relationship between pressure and the quantity of gas in moles.
Under the same conditions of TEMPERATURE and VOLUME, the pressure
of a gas is directly proportional to the number of moles. Sample
Problem
Slide 76
Relationship between pressure and the quantity of gas in moles.
Under the same conditions of TEMPERATURE and VOLUME, the pressure
of a gas is directly proportional to the number of moles. Sample
Problem
Slide 77
Ideal Gas Law It describes the interrelationship between the
Four (4) variables that characterize a gas at a given moment in
time. The variables are: (P) pressure, (T) absolute temperature,
(V) volume, (n) quantity of gas in moles. An ideal gas does not
exist in reality. It is an hypothetical gas that adheres to all the
simple gas laws and conforms to Kinetic Molecular Theory. Ideal
gases do not condense at low temperatures or high pressures. In
contrast real gases do not resemble ideal gases at extreme
temperatures and pressures.
Slide 78
Ideal Gases vs Real Gases
Slide 79
Ideal Gases Law Formula The Ideal Gas law is a combination of
the 4 simple gas laws. R is the proportionality constant. It
replaces the proportionality symbol (). We determine the value of R
by finding its value for 1 mol of gas at STP.
Slide 80
Ideal Gases Law Formula R is the proportionality constant. It
replaces the proportionality symbol (). We determine the value of R
by finding its value for 1 mol of gas at STP. To determine the
value for R:
Slide 81
Ideal Gases Law Problem When inflated to their maximum
capacity, human lungs contain approximately 4.09 L of air at 37
degrees C. How many moles of air do lungs contain if the air
pressure is 100kPa? Known Data P= 100kPa n= ? R= 8.31 (kPaL)/(molK)
T= 37 o C V= 4.09L 1)Convert Temperature to K T = 37 o C + 273 =
310K 2)Calculate the number of moles
Slide 82
Ideal Gases Law Problem When inflated to their maximum
capacity, human lungs contain approximately 4.09 L of air at 37
degrees C. How many moles of air do lungs contain if the air
pressure is 100kPa? 1)Convert Temperature to K T = 37 o C + 273 =
310K 2)Calculate the number of moles Known Data P= 100kPa n= ? R=
8.31 (kPaL)/(molK) T= 37 o C V= 4.09L
Slide 83
Ideal Gases Law Problem When inflated to their maximum
capacity, human lungs contain approximately 4.09 L of air at 37
degrees C. How many moles of air do lungs contain if the air
pressure is 100kPa? 1)Convert Temperature to K T = 37 o C + 273 =
310K 2)Calculate the number of moles Known Data P= 100kPa n= ? R=
8.31 (kPaL)/(molK) T= 37 o C V= 4.09L
Slide 84
Molar Mass of a Gas The molar mass of a gas (M) can be
determined by dividing the mass of the gas (m) by its number of
moles: We can use the Ideal Gas Law to determine the molar mass of
a gas using the simple transformation on the left. Formula for
Molar Mass derived from the Ideal Gas Law
Slide 85
Molar Mass of a Gas Sample Problem What is the molar mass of a
sample of an unknown gas if, at a temperature of 0 o C and under a
pressure of 102 kPa, a volume of 2.30 L of the gas weighs 4.23
grams? Known Data M=? T= 0 o C P=102 kPa V=2.30L m=4.23 g R=8.31
(kPaL)/(molK) 1)Convert Temperature to K T = 0 o C + 273 = 273K
2)Calculate the Molar Mass
Slide 86
General Gas Law The General Gas Law establishes a relationship
between the 4 variables that describe P, V, T and n. It is used to
predict the final conditions of a gas once its initial conditions
have been modified. From this final equation for the General Gas
Law we can deduce all formulas for the simple gas laws.
Slide 87
General Gas Law sample problem At SATP, 0.150 mol of water
vapour occupies a volume of 55.0 mL. What is the new temperature in
degrees Celsius if 0.030 mol of water vapour is removed while
increasing the pressure to 115.0 kPa and decreasing the volume to
40.0 ml? Known Data P 1 = 101.3 kpa V 1 = 55.0 mL n 1 = 0.150 mol T
1 = 25 C P 2 = 115.0 kPa V 2 = 40.0mL n 2 = ? T 2 = ? 1 Convert
Temperature to K 2 - Calculate the final number of moles 3- Use
General Gas Law to solve problem
Slide 88
General Gas Law sample problem At SATP, 0.150 mol of water
vapour occupies a volume of 55.0 mL. What is the new temperature in
degrees Celsius if 0.030 mol of water vapour is removed while
increasing the pressure to 115.0 kPa and decreasing the volume to
40.0 ml? Known Data P 1 = 101.3 kpa V 1 = 55.0 mL n 1 = 0.150 mol T
1 = 25 C P 2 = 115.0 kPa V 2 = 40.0mL n 2 = ? T 2 = ?
Slide 89
Stoichiometry of Gases..is a calculation method based on the
ratios between the quantities of gas involved in a chemical
reaction. This method is used to predict the quantity of a reactant
or product involved in a chemical reaction in which at least one of
the components is a gas. Sample problem: Ethanol vapour (C 2 H 5
OH) burns in air according to the following equation: 2 C 2 H 5 OH
(g) + 6O 2 - 4 CO 2(g) + 6 H 2 O (l) a)If 2.5 L of ethanol burn at
STP, what volume of oxygen (O2 ) is required? b)What volume of
carbon dioxide (CO2) will be produced?
Slide 90
Stoichiometry of Gases Sample problem: Ethanol vapour (C 2 H 5
OH) burns in air according to the following equation: 2 C 2 H 5 OH
(g) + 6O 2 - 4 CO 2(g) + 6 H 2 O (l) a)If 2.5 L of ethanol burn at
STP, what volume of oxygen (O2 ) is required? b)What volume of
carbon dioxide (CO2) will be produced?
Slide 91
Stoichiometry of Gases Sample problem: Ethanol vapour (C 2 H 5
OH) burns in air according to the following equation: 2 C 2 H 5 OH
(g) + 6O 2 - 4 CO 2(g) + 6 H 2 O (l) a)If 2.5 L of ethanol burn at
STP, what volume of oxygen (O2 ) is required? b)What volume of
carbon dioxide (CO2) will be produced?
Slide 92
Daltons Law Law of Partial Pressures At a given temperature,
the total pressure of a mixture of gases equals the sum of the
pressures of each of the gases.
Slide 93
Daltons Law Law of Partial Pressures At a given temperature,
the total pressure of a mixture of gases equals the sum of the
pressures of each of the gases.
Slide 94
Daltons Law Law of Partial Pressures At a given temperature,
the total pressure of a mixture of gases equals the sum of the
pressures of each of the gases.
Slide 95
Daltons Law Law of Partial Pressures At a given temperature,
the total pressure of a mixture of gases equals the sum of the
pressures of each of the gases.
Slide 96
Daltons Law Law of Partial Pressures At a given temperature,
the total pressure of a mixture of gases equals the sum of the
pressures of each of the gases.
Slide 97
Significant figures - Rules
Slide 98
Physical Properties of Gases Covers up to page 113 of Quantum
Text