Chapter 5 Gases - Suffolk County Community College

Report
Chemistry: A Molecular Approach, 2nd Ed.
Nivaldo Tro
Chapter 5
Gases
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Copyright © 2011 Pearson Education, Inc.
The Structure of a Gas
• Gases are composed of particles
•
•
that are flying around very fast in
their container(s)
The particles in straight lines until
they encounter either the container
wall or another particle, then they
bounce off
If you were able to take a snapshot
of the particles in a gas, you would
find that there is a lot of empty
space in there
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Gases Pushing
• Gas molecules are constantly in
•
motion
As they move and strike a surface,
they push on that surface
 push = force
• If we could measure the total
amount of force exerted by gas
molecules hitting the entire surface
at any one instant, we would know
the pressure the gas is exerting
 pressure = force per unit area
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The Effect of Gas Pressure
• The pressure exerted by a gas can cause
some amazing and startling effects
• Whenever there is a pressure difference, a
gas will flow from an area of high pressure
to an area of low pressure
the bigger the difference in pressure, the
stronger the flow of the gas
• If there is something in the gas’s path, the
gas will try to push it along as the gas flows
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Air Pressure
• The atmosphere exerts a
pressure on everything it
contacts
the atmosphere goes
up about 370 miles, but
80% is in the first 10 miles
from the earth’s surface
• This is the same pressure
that a column of water
would exert if it were
about 10.3 m high
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Atmospheric Pressure Effects
• Differences in air pressure result
•
in weather and wind patterns
The higher in the atmosphere
you climb, the lower the
atmospheric pressure
is around you
at the surface the
atmospheric pressure
is 14.7 psi, but at 10,000 ft
it is only 10.0 psi
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Pressure Imbalance in the Ear
If there is a difference
in pressure across
the eardrum
membrane,
the membrane will be
pushed out – what we
commonly call a
“popped eardrum”
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The Pressure of a Gas
• Gas pressure is a result of the
constant movement of the gas
molecules and their collisions
with the surfaces around them
• The pressure of a gas
depends on several factors
number of gas particles in a
given volume
volume of the container
average speed of the gas
particles
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Measuring Air Pressure
• We measure air
•
•
pressure with a
barometer
Column of mercury
supported by air
pressure
Force of the air on the
surface of the mercury
counter balances the
force of gravity on the
column of mercury
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gravity
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Practice – What happens to the height of the column
of mercury in a mercury barometer as you climb to
the top of a mountain?
1.
The height of the column increases because atmospheric
pressure decreases with increasing altitude
2.
The height of the column decreases because atmospheric
pressure decreases with increasing altitude
3.
The height of the column decreases because atmospheric
pressure increases with increasing altitude
4.
The height of the column increases because atmospheric
pressure increases with increasing altitude
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Common Units of Pressure
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Example 5.1: A high-performance bicycle tire has a
pressure of 132 psi. What is the pressure in mmHg?
Given:
132 psi
Find:
mmHg
Conceptual
Plan:
psi
atm
mmHg
Relationships: 1 atm = 14.7 psi, 1 atm = 760 mmHg
Solution:
Check:
because mmHg are smaller than psi, the
answer makes sense
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Practice—Convert 45.5 psi into kPa
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Practice—Convert 45.5 psi into kPa
Given:
Find:
Conceptual
Plan:
645.5 psi
kPa
psi
atm
kPa
Relationships: 1 atm = 14.7 psi, 1 atm = 101.325 kPa
Solution:
Check:
because kPa are smaller than psi, the
answer makes sense
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Manometers
• The pressure of a gas trapped in a container can
•
•
•
be measured with an instrument called a
manometer
Manometers are U-shaped tubes, partially filled
with a liquid, connected to the gas sample on one
side and open to the air on the other
A competition is established between the
pressures of the atmosphere and the gas
The difference in the liquid levels is a measure of
the difference in pressure between the gas and
the atmosphere
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Manometer
for this sample, the gas
has a larger pressure
than the atmosphere, so
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Boyle’s Law
Robert Boyle (1627–1691)
• Pressure of a gas is inversely proportional to its
volume
 constant T and amount of gas
 graph P vs V is curve
 graph P vs 1/V is straight line
• As P increases, V decreases by the same factor
• P x V = constant
• P1 x V1 = P2 x V2
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Boyle’s Experiment
• Added Hg to a J-tube
•
with air trapped inside
Used length of air
column as a measure of
volume
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Boyle’s Experiment, P x V
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Boyle’s Law: A Molecular View
• Pressure is caused by the molecules striking the
•
•
sides of the container
When you decrease the volume of the container
with the same number of molecules in the
container, more molecules will hit the wall at the
same instant
This results in increasing the pressure
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Boyle’s Law and Diving
• Because water is more
dense than air, for each
10 m you dive below the
surface, the pressure on
your lungs increases
1 atm
 at 20 m the total pressure
is 3 atm
• If your tank contained air
at 1 atm of pressure, you
would not be able to
inhale it into your lungs
 you can only generate
enough force to overcome
about 1.06 atm
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Scuba tanks have a
regulator so that the
air from the tank is
delivered at the same
pressure as the water
surrounding you.
This allows you to
take in air even when
the outside pressure
is large.
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Boyle’s Law and Diving
• If a diver holds her breath and rises to the
•
•
surface quickly, the outside pressure drops to
1 atm
According to Boyle’s law, what should happen
to the volume of air in the lungs?
Because the pressure is decreasing by a
factor of 3, the volume will expand by a factor
of 3, causing damage to internal organs.
Always Exhale When Rising!!
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Example 5.2: A cylinder with a movable piston has a volume
of 7.25 L at 4.52 atm. What is the volume at 1.21 atm?
Given: V1 =7.25 L, P1 = 4.52 atm, P2 = 1.21 atm
Find: V2, L
Conceptual
Plan:
V1, P1, P2
Relationships: P1
V2
∙ V1 = P2 ∙ V2
Solution:
Check: because P and V are inversely proportional, when the pressure
decreases ~4x, the volume should increase ~4x, and it does
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Practice – A balloon is put in a bell jar and the
pressure is reduced from 782 torr to 0.500 atm. If
the volume of the balloon is now 2.78 x 103 mL,
what was it originally?
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A balloon is put in a bell jar and the pressure is reduced
from 782 torr to 0.500 atm. If the volume of the balloon is
now 2.78x 103 mL, what was it originally?
Given: V2 =2780 mL, P1 = 762 torr, P2 = 0.500 atm
Find: V1, mL
Conceptual
Plan:
V2, P1, P2
Relationships: P1
V1
∙ V1 = P2 ∙ V2 , 1 atm = 760 torr (exactly)
Solution:
Check: because P and V are inversely proportional, when the pressure
decreases ~2x, the volume should increase ~2x, and it does
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Charles’s Law
Jacques Charles (1746–1823)
• Volume is directly proportional to temperature
constant P and amount of gas
graph of V vs. T is straight line
• As T increases, V also increases
• Kelvin T = Celsius T + 273
• V = constant x T
if T measured in Kelvin
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If you plot volume vs.
temperature for any gas at
constant pressure, the points
will all fall on a straight line
If the lines are
extrapolated back
to a volume of “0,”
they all show the
same temperature,
−273.15 °C, called
absolute zero
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Charles’s Law – A Molecular View
• The
The pressure
pressureofofgas
gas
••
inside
inside and
andoutside
outsidethe
the
balloon
balloon are
arethe
thesame
same
At
the
At low
hightemperatures,
temperatures,
gas
molecules
are not
the gas
molecules
are
moving
fast, so
so they
moving as
faster,
they
don’t hit the sides of the
hit the sides of the
balloon as hard –
balloon
harder –
therefore the volume is
causing
the
volume
to
small
become larger
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Example 5.3: A gas has a volume of 2.57 L at 0.00 °C.
What was the temperature at 2.80 L?
Given: V1 =2.57 L, V2 = 2.80 L, t2 = 0.00 °C
Find: t1, K and °C
Conceptual
Plan:
V1, V2, T2
T1
Relationships:
Solution:
Check: because T and V are directly proportional, when the volume
decreases, the temperature should decrease, and it does
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Practice – The temperature inside a balloon is
raised from 25.0 °C to 250.0 °C. If the volume of
cold air was 10.0 L, what is the volume of hot air?
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The temperature inside a balloon is raised from 25.0 °C to
250.0 °C. If the volume of cold air was 10.0 L, what is the
volume of hot air?
Given: V1 =10.0 L, t1 = 25.0 °C L, t2 = 250.0 °C
Find: V2, L
Conceptual
Plan:
V1, T1, T2
V2
Relationships:
Solution:
Check:
when the temperature increases, the volume should
increase, and it does
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Avogadro’s Law
Amedeo Avogadro (1776–1856)
• Volume directly proportional to
the number of gas molecules
V = constant x n
constant P and T
more gas molecules = larger
volume
• Count number of gas
•
molecules by moles
Equal volumes of gases
contain equal numbers of
molecules
the gas doesn’t matter
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Example 5.4:A 0.225 mol sample of He has a volume of 4.65
L. How many moles must be added to give 6.48 L?
Given: V1 = 4.65 L, V2 = 6.48 L, n1 = 0.225 mol
Find: n2, and added moles
Conceptual
Plan:
V1, V2, n1
n2
Relationships:
Solution:
Check: because n and V are directly proportional, when the volume
increases, the moles should increase, and they do
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Practice — If 1.00 mole of a gas occupies 22.4 L
at STP, what volume would 0.750 moles occupy?
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Practice — If 1.00 mole of a gas occupies 22.4 L at STP,
what volume would 0.750 moles occupy?
Given: V1 =22.4 L, n1 = 1.00 mol, n2 = 0.750 mol
Find: V2
Conceptual
Plan:
V1, n1, n2
V2
Relationships:
Solution:
Check:
because n and V are directly proportional, when the moles
decrease, the volume should decrease, and it does
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Ideal Gas Law





By combining the gas laws we can write a general
equation
R is called the gas constant
The value of R depends on the units of P and V
we will use 0.08206
and convert P to atm
and V to L
The other gas laws are found in the ideal gas law if two
variables are kept constant
Allows us to find one of the variables if we know the
other three
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Example 5.6: How many moles of gas are in a basketball
with total pressure 24.3 psi, volume of 3.24 L at 25°C?
Given: V = 3.24 L, P = 24.3 psi, t = 25 °C
Find: n, mol
Conceptual
P, V, T, R
n
Plan:
Relationships:
Solution:
Check:
1 mole at STP occupies 22.4 L, because there is a much
smaller volume than 22.4 L, we expect less than 1 mole of gas
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Standard Conditions
• Because the volume of a gas varies with
pressure and temperature, chemists have
agreed on a set of conditions to report our
measurements so that comparison is easy –
we call these standard conditions
STP
• Standard pressure = 1 atm
• Standard temperature = 273 K
0 °C
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Practice – A gas occupies 10.0 L at 44.1 psi and
27 °C. What volume will it occupy at standard
conditions?
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A gas occupies 10.0 L at 44.1 psi and 27 °C. What
volume will it occupy at standard conditions?
Given: V1 = 10.0L, P1 = 44.1 psi, t1 = 27 °C, P2 = 1.00 atm, t2 = 0 °C
Find: V2, L
Conceptual
P1, V1, T1, R
Plan:
n
P2, n, T2, R
V2
Relationships:
Solution:
Check: 1 mole at STP occupies 22.4 L, because there is more than 1
mole, we expect more than 22.4 L of gas
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Practice — Calculate the volume occupied by
637 g of SO2 (MM 64.07) at 6.08 x 104 mmHg
and –23 °C
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Practice—Calculate the volume occupied by 637 g of
SO2 (MM 64.07) at 6.08 x 104 mmHg and –23 °C.
Given: mSO2 = 637 g, P = 6.08 x 104 mmHg, t = −23 °C,
Find: V, L
Conceptual
g
Plan:
n
P, n, T, R
V
Relationships:
Solution:
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Molar Volume
• Solving the ideal gas equation for the volume
of 1 mol of gas at STP gives 22.4 L
6.022 x 1023 molecules of gas
notice: the gas is immaterial
• We call the volume of 1 mole of gas at STP
the molar volume
it is important to recognize that one mole measures
of different gases have different masses, even
though they have the same volume
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Molar Volume
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Practice — How many liters of O2 @ STP can be made
from the decomposition of 100.0 g of PbO2?
2 PbO2(s) → 2 PbO(s) + O2(g)
(PbO2 = 239.2, O2 = 32.00)
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Practice — How many liters of O2 @ STP can be made from
the decomposition of 100.0 g of PbO2?
2 PbO2(s) → 2 PbO(s) + O2(g)
Given:
Find:
100.0 g PbO2, 2 PbO2 → 2 PbO + O2
L O2
Conceptual g PbO2
Plan:
mol PbO2
mol O2
L O2
Relationships: 1 mol O = 22.4 L, 1 mol PbO = 239.2g, 1 mol O ≡ 2 mol PbO
2
2
2
2
Solution:
Check:
because less than 1 mole PbO2, and ½ moles of O2 as
PbO2, the number makes sense
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Density at Standard Conditions
•
•
•
•
Density is the ratio of mass to volume
Density of a gas is generally given in g/L
The mass of 1 mole = molar mass
The volume of 1 mole at STP = 22.4 L
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Practice – Calculate the density of N2(g) at STP
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Practice – Calculate the density of N2(g) at STP
Given:
Find:
Conceptual
Plan:
N2,
dN2, g/L
MM
d
Relationships:
Solution:
Check:
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Gas Density
• Density is directly proportional to molar mass
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Example 5.7: Calculate the density of N2
at 125°C and 755 mmHg
Given:
Find:
Conceptual
Plan:
P = 755 mmHg, t = 125 °C,
dN2, g/L
P, MM, T, R
d
Relationships:
Solution:
Check:
because the density of N2 is 1.25 g/L at STP, we expect the
density to be lower when the temperature is raised, and it is
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Practice — Calculate the density of a gas at 775 torr and
27 °C if 0.250 moles weighs 9.988 g
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Practice — Calculate the density of a gas at 775 torr
and 27 °C if 0.250 moles weighs 9.988 g
Given:
Find:
Conceptual
Plan:
m
= 9.988g, n=0.250
n = 0.250mol,
mol, P=775
P = 1.0197
atm, t=27
T = 300.K
m=9.988g,
mmHg,
°C,
density, g/L
P, n, T, R
V
V, m
d
Relationships:
Solution:
Check:
the unit is correct and the value is reasonable
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Molar Mass of a Gas
• One of the methods chemists use to determine
the molar mass of an unknown substance is to
heat a weighed sample until it becomes a gas,
measure the temperature, pressure, and
volume, and use the ideal gas law
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Example 5.8: Calculate the molar mass of a gas with mass
0.311 g that has a volume of 0.225 L at 55°C and 886 mmHg
Given: m=0.311g, V=0.225 L, P=1.1658
P=886 mmHg,
atm, t=55°C,
T=328 K,
Find: molar mass, g/mol
Conceptual
P, V, T, R
n
Plan:
n, m
MM
Relationships:
Solution:
T (K ) = 5 5 °C + 2 7 3 .1 5
T = 328 K
Check:
the unit is correct, the value is reasonable
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Practice — What is the molar mass of a gas if 12.0 g
occupies 197 L at 3.80 x 102 torr and 127 °C?
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Practice — What is the molar mass of a gas if 12.0 g
occupies 197 L at 380 torr and 127 °C?
Given: m
m=12.0
= 12.0g,
g, V=
V =197
197L,L,P=380
P = 0.50
torr,atm,
t=127°C,
T =400 K,
Find: molar mass, g/mol
Conceptual
Plan:
P, V, T, R
n
n, m
MM
Relationships:
Solution:
Check:
the unit is correct and the value is reasonable
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Mixtures of Gases
• When gases are mixed together, their molecules
behave independent of each other
 all the gases in the mixture have the same volume
 all completely fill the container  each gas’s volume = the volume
of the container
 all gases in the mixture are at the same temperature
 therefore they have the same average kinetic energy
• Therefore, in certain applications, the mixture can
be thought of as one gas
 even though air is a mixture, we can measure the pressure,
volume, and temperature of air as if it were a pure substance
 we can calculate the total moles of molecules in an air sample,
knowing P, V, and T, even though they are different molecules
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Composition of Dry Air
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Partial Pressure
• The pressure of a single gas in a mixture of gases
•
is called its partial pressure
We can calculate the partial pressure of a gas if
 we know what fraction of the mixture it composes and
the total pressure
 or, we know the number of moles of the gas in a
container of known volume and temperature
• The sum of the partial pressures of all the gases in
the mixture equals the total pressure
 Dalton’s Law of Partial Pressures
 because the gases behave independently
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The partial pressure of each gas in a
mixture can be calculated using the ideal
gas law
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Example 5.9: Determine the mass of Ar in the
mixture
= 0.275
mmHg,
atm, V
PNe
= =112
1.00 L,
mmHg,
T=298PK
Given: PAr
He=341
tot = 662 mmHg,
V = 1.00 L, T=298 K
massArAr,,gg
Find: mass
Conceptual P , P , P
PAr
Plan: tot He Ne
PAr = Ptot – (PHe + PNe)
PAr, V, T
nAr
mAr
Relationships:
Solution:
Check:
the units are correct, the value is reasonable
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Practice – Find the partial pressure of neon in a
mixture with total pressure 3.9 atm, volume 8.7 L,
temperature 598 K, and 0.17 moles Xe.
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Find the partial pressure of neon in a mixture with total pressure
3.9 atm, volume 8.7 L, temperature 598 K, and 0.17 moles Xe
Given:
Find:
Ptot = 3.9 atm, V = 8.7 L, T = 598 K, Xe = 0.17 mol
PNe, atm
Conceptual nXe, V, T, R
Plan:
PXe
Ptot, PXe
PNe
Relationships:
Solution:
Check:
the unit is correct, the value is reasonable
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Mole Fraction
The fraction of the total pressure that
a single gas contributes is equal to
the fraction of the total number of
moles that a single gas contributes
The ratio of the moles of a single
component to the total number of
moles in the mixture is called the
mole fraction, c
for gases, = volume % / 100%
The partial pressure of a gas is equal
to the mole fraction of that gas times
the total pressure
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Ex 5.10: Find the mole fractions and partial pressures in
a 12.5 L tank with 24.2 g He and 4.32 g O2 at 298 K
Given: nHe
mHe
==
6.05
24.2
mol,
g, m
nO2
4.32 g,mol,
V =V12.5
= 12.5
L, TL,= T298
= 298
K K
O2 = 0.135
cHe
=0.97817,
, cO2, PHe,catm,
PO2, atm,PPHe
, atm,
, atmPO2, atm, Ptotal, atm
O2=0.021827,
total
Find: cHe
Conceptual mgas
Plan:
ngas
cgas, Ptotal
cgas
ntot, V, T, R
Ptot
Pgas
Relationships:
Solution:
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Practice – Find the mole fraction of neon in a mixture
with total pressure 3.9 atm, volume 8.7 L,
temperature 598 K, and 0.17 moles Xe
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Find the mole fraction of neon in a mixture with total pressure
3.9 atm, volume 8.7 L, temperature 598 K, and 0.17 moles Xe
Given:
Find:
Ptot = 3.9 atm, V = 8.7 L, T = 598 K, Xe = 0.17 mol
PNe, atm
Conceptual n , V, T, R
Xe
Plan:
PXe
Ptot, PXe
PNe
Ptot, PNe
cNe
Relationships:
Solution:
Check:
the unit is correct, the value is reasonable
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Collecting Gases
• Gases are often collected by having them
•
•
displace water from a container
The problem is that because water evaporates,
there is also water vapor in the collected gas
The partial pressure of the water vapor, called
the vapor pressure, depends only on the
temperature
 so you can use a table to find out the partial pressure
of the water vapor in the gas you collect
 if you collect a gas sample with a total pressure of
758.2 mmHg* at 25 °C, the partial pressure of the
water vapor will be 23.78 mmHg – so the partial
pressure of the dry gas will be 734.4 mmHg
Table 5.4*
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Collecting Gas by Water Displacement
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Vapor Pressure of Water
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Ex 5.11: 1.02 L of O2 collected over water at 293 K with
a total pressure of 755.2 mmHg. Find mass O2.
Given: V=1.02 L, P
P=755.2
mmHg,
atm,T=293
T=293K K
O2=0.97059
Find: mass O2, g
Conceptual Ptot, PH2O
Plan:
PO2
PO2,V,T
nO2
gO2
Relationships:
Solution:
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Practice – 0.12 moles of H2 is collected over water in
a 10.0 L container at 323 K. Find the total pressure.
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0.12 moles of H2 is collected over water in a 10.0 L
container at 323 K. Find the total pressure.
Given: V=10.0 L, nH2 = 0.12 mol, T = 323 K
Find: Ptotal, mmHg
Conceptual
Plan:
nH2,V,T
PH2
PH2, PH2O
Ptotal
Relationships:
Solution:
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Reactions Involving Gases
• The principles of reaction stoichiometry from
•
Chapter 4 can be combined with the gas laws for
reactions involving gases
In reactions of gases, the amount of a gas is often
given as a volume
 instead of moles
 as we’ve seen, you must state pressure and temperature
• The ideal gas law allows us to convert from the
•
volume of the gas to moles; then we can use the
coefficients in the equation as a mole ratio
When gases are at STP, use 1 mol = 22.4 L
P, V, T of Gas A
mole A
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mole B
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P, V, T of Gas B
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Ex 5.12: What volume of H2 is needed to make 35.7 g of
CH3OH at 738 mmHg and 355 K?
CO(g) + 2 H2(g) → CH3OH(g)
Given: n
mH2
= 2.2284
= 37.5g,
mol,P=738
P=0.97105
mmHg,
atm,
T=355
T=355
K K
CH3OH
Find: VH2, L
Conceptual g CH3OH
Plan:
mol CH3OH
mol H2
P, n, T, R
V
Relationships:
Solution:
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Ex 5.13: How many grams of H2O form when 1.24 L H2
reacts completely with O2 at STP?
O2(g) + 2 H2(g) → 2 H2O(g)
Given: VH2 = 1.24 L, P = 1.00 atm, T = 273 K
Find: massH2O, g
Concept Plan:
L H2
mol H2
mol H2O
g H2 O
Relationships: H2O = 18.02 g/mol, 1 mol = 22.4 L @ STP
2 mol H2O : 2 mol H2
Solution:
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Practice – What volume of O2 at 0.750 atm and 313 K
is generated by the thermolysis of 10.0 g of HgO?
2 HgO(s)  2 Hg(l) + O2(g)
(MMHgO = 216.59 g/mol)
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What volume of O2 at 0.750 atm and 313 K is generated
by the thermolysis of 10.0 g of HgO?
2 HgO(s)  2 Hg(l) + O2(g)
Given: n
mO2
= =0.023085
10.0g, P=0.750
mol, P=0.750
atm, T=313
atm, T=313
K
K
HgO
Find: VO2, L
Conceptual
Plan:
g HgO
mol HgO
mol O2
P, n, T, R
V
Relationships:
Solution:
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Properties of Gases
• Expand to completely fill their container
• Take the shape of their container
• Low density
much less than solid or liquid state
• Compressible
• Mixtures of gases are always homogeneous
• Fluid
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Kinetic Molecular Theory
• The particles of the gas (either
atoms or molecules) are
constantly moving
• The attraction between particles
is negligible
• When the moving gas particles
hit another gas particle or the
container, they do not stick; but
they bounce off and continue
moving in another direction
like billiard balls
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Kinetic Molecular Theory
• There is a lot of empty space between the gas
particles
compared to the size of the particles
• The average kinetic energy of the gas particles
is directly proportional to the Kelvin
temperature
as you raise the temperature
of the gas, the average
speed of the particles
increases
but don’t be fooled
into thinking all the
gas particles are
moving at the same speed!!
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Gas Properties Explained –
Pressure
• Because the gas particles are constantly
•
moving, they strike the sides of the container
with a force
The result of many particles in a gas sample
exerting forces on the surfaces around them is
a constant pressure
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Gas Properties Explained –
Indefinite Shape and Indefinite Volume
Because the gas
molecules have
enough kinetic
energy to overcome
attractions, they
keep moving around
and spreading out
until they fill the
container.
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As a result, gases
take the shape and
the volume of the
container they
are in.
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Gas Properties Explained Compressibility
Because there is a lot of unoccupied space in the structure
of a gas, the gas molecules can be squeezed closer together
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Gas Properties Explained –
Low Density
Because there is a lot of
unoccupied space in the
structure of a gas, gases
do not have a lot of mass
in a given volume; the
result is they have low
density
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Density & Pressure
• Pressure is the result of the constant movement
of the gas molecules and their collisions with the
surfaces around them
• When more molecules
are added, more
molecules hit the
container at any one
instant, resulting in
higher pressure
also higher density
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Gas Laws Explained –
Boyle’s Law
• Boyle’s Law says that the volume of a gas is
•
•
inversely proportional to the pressure
Decreasing the volume forces the molecules into
a smaller space
More molecules will collide with the container at
any one instant, increasing the pressure
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Gas Laws Explained –
Charles’s Law
• Charles’s Law says that the volume
•
of a gas is directly proportional to
the absolute temperature
Increasing the temperature
increases their average speed,
causing them to hit the wall
harder and more frequently
 on average
• To keep the pressure constant, the
volume must then increase
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Gas Laws Explained –
Avogadro’s Law
• Avogadro’s Law says that the volume of a gas
•
•
is directly proportional to the number of gas
molecules
Increasing the number of gas molecules
causes more of them to hit the wall at the same
time
To keep the pressure constant, the volume
must then increase
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Gas Laws Explained –
Dalton’s Law of Partial Pressures
• Dalton’s Law says that the total pressure of a
•
•
•
mixture of gases is the sum of the partial pressures
Kinetic-molecular theory says that the gas
molecules are negligibly small and don’t interact
Therefore the molecules behave independently of
each other, each gas contributing its own collisions
to the container with the same average kinetic
energy
Because the average kinetic energy is the same,
the total pressure of the collisions is the same
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Dalton’s Law & Pressure
• Because the gas
molecules are not
sticking together,
each gas molecule
contributes its own
force to the total
force on the side
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Kinetic Energy and
Molecular Velocities
• Average kinetic energy of the gas molecules
depends on the average mass and velocity
 KE = ½mv2
• Gases in the same container have the same
•
temperature, therefore they have the same
average kinetic energy
If they have different masses, the only way for
them to have the same kinetic energy is to have
different average velocities
 lighter particles will have a faster average velocity than
more massive particles
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Molecular Speed vs. Molar Mass
• To have the same average kinetic energy,
heavier molecules must have a slower average
speed
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Temperature_ and Molecular Velocities
• KEavg = ½NAmu2
NA is Avogadro’s number
• KEavg = 1.5RT
R is the gas constant in energy units, 8.314 J/mol∙K
1 J = 1 kg∙m2/s2
• Equating and solving we get
NA∙mass = molar mass in kg/mol
• As temperature increases, the average velocity increases
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Molecular Velocities
• All the gas molecules in a sample can travel at
•
•
•
different speeds
However, the distribution of speeds follows a
statistical pattern called a Boltzman distribution
Ee talk about the “average velocity” of the
molecules, but there are different ways to take
this kind of average
The method of choice for our average velocity is
called the root-mean-square method, where the
rms average velocity, urms, is the square root of
the average of the sum of the squares of all the
molecule velocities
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Boltzman Distribution
Distribution Function
Fraction of Molecules
O2 @ 300 K
Molecular Speed
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Temperature vs. Molecular Speed
• As the absolute
temperature
increases, the
average velocity
increases
the distribution function
“spreads out,” resulting
in more molecules with
faster speeds
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Ex 5.14: Calculate the rms velocity of O2 at 25 °C
Given: O2, t = 25 °C
Find: urms
Conceptual
Plan:
MM, T
urms
Relationships:
Solution:
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Practice – Calculate the rms velocity of CH4
(MM 16.04) at 25 °C
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Practice – Calculate the rms velocity of CH4 at 25 °C
Given: CH4, t = 25 °C
Find: urms
Conceptual
Plan:
MM, T
urms
Relationships:
Solution:
The mass of O2 (32.00) is
2x the mass of CH4
(16.04).
The rms of CH4 (681 m/s)
is √2x the rms of O2
(482 m/s)
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Mean Free Path
• Molecules in a gas travel in
•
•
straight lines until they
collide with another
molecule or the container
The average distance a
molecule travels between
collisions is called the mean
free path
Mean free path decreases
as the pressure increases
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Diffusion and Effusion
• The process of a collection of molecules
•
•
•
spreading out from high concentration to low
concentration is called diffusion
The process by which a collection of molecules
escapes through a small hole into a vacuum is
called effusion
The rates of diffusion and effusion of a gas are
both related to its rms average velocity
For gases at the same temperature, this means
that the rate of gas movement is inversely
proportional to the square root of its molar mass
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Effusion
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Graham’s Law of Effusion
Thomas Graham (1805–1869)
• For two different gases at the same
temperature, the ratio of their rates of effusion
is given by the following equation:
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Ex 5.15: Calculate the molar mass of a gas that
effuses at a rate 0.462 times N2
Given:
Find: MM, g/mol
Conceptual
Plan:
rateA/rateB, MMN2
MMunknown
Relationships:
Solution:
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Practice – Calculate the ratio of rate of
effusion for oxygen to hydrogen
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Practice – Calculate the ratio of rate of effusion
for oxygen to hydrogen
Given: O2, 32.00 g/mol; H2 2.016 g/mol
Find:
Conceptual
Plan:
MMO2, MMH2
rateA/rateB
Relationships:
Solution:
This means that, on average, the O2 molecules
are traveling at ¼ the speed of H2 molecules
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Ideal vs. Real Gases
• Real gases often do not behave like ideal gases
•
at high pressure or low temperature
Ideal gas laws assume
1. no attractions between gas molecules
2. gas molecules do not take up space
 based on the kinetic-molecular theory
• At low temperatures and high pressures these
assumptions are not valid
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Real Gas Behavior
• Because real
molecules take up
space, the molar
volume of a real gas
is larger than
predicted by the ideal
gas law at high
pressures
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The Effect of Molecular Volume
Johannes van der Waals (1837–1923)
• At high pressure, the amount of space
occupied by the molecules is a significant
amount of the total volume
• The molecular volume makes the real volume
larger than the ideal gas law would predict
• van der Waals modified the ideal gas
equation to account for the molecular volume
b is called a van der Waals constant and is
different for every gas because their molecules
are different sizes
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Real Gas Behavior
• Because real molecules attract each other, the
molar volume of a real gas is smaller than
predicted by the ideal gas law at low
temperatures
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The Effect of Intermolecular Attractions
• At low temperature, the attractions between
•
•
the molecules is significant
The intermolecular attractions makes the real
pressure less than the ideal gas law would
predict
van der Waals modified the ideal gas equation
to account for the intermolecular attractions
a is another van der Waals constant and is
different for every gas because their molecules
have different strengths of attraction
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van der Waals’
Equation
• Combining the equations to
account for molecular
volume and intermolecular
attractions we get the
following equation
used for real gases
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Real Gases
• A plot of PV/RT vs. P for 1 mole of a gas
•
•
shows the difference between real and ideal
gases
It reveals a curve that shows the PV/RT ratio
for a real gas is generally lower than ideal for
“low” pressures – meaning the most important
factor is the intermolecular attractions
It reveals a curve that shows the PV/RT ratio
for a real gas is generally higher than ideal for
“high” pressures – meaning the most important
factor is the molecular volume
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PV/RT Plots
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Structure of the Atmosphere
• The atmosphere shows several
•
layers, each with its own
characteristics
The troposphere is the layer
closest to the Earth’s surface
 circular mixing due to thermal
currents – weather
• The stratosphere is the next
layer up
 less air mixing
• The boundary between the
•
troposphere and stratosphere
is called the tropopause
The ozone layer is a layer of
high O3 concentration located
in the stratosphere
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Air Pollution
• Air pollution is materials added to the atmosphere
that would not be present in the air without, or are
increased by, man’s activities
 though many of the “pollutant” gases have natural
sources as well
• Pollution added to the troposphere has a direct
effect on human health and the materials we use
because we come in contact with it
 and the air mixing in the troposphere means that we all
get a smell of it!
• Pollution added to the stratosphere may have
indirect effects on human health caused by
depletion of ozone
 and the lack of mixing and weather in the stratosphere
means that pollutants last longer before “washing” out
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Pollutant Gases, SOx
• SO2 and SO3, oxides of sulfur, come from coal
combustion in power plants and metal refining
as well as volcanoes
• Lung and eye irritants
• Major contributors to acid rain
2 SO2 + O2 + 2 H2O  2 H2SO4
SO3 + H2O  H2SO4
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Pollutant Gases, NOx
• NO and NO2, oxides of nitrogen, come from burning of
fossil fuels in cars, trucks, and power plants
 as well as lightning storms
•
•
•
•
NO2 causes the brown haze seen in some cities
Lung and eye irritants
Strong oxidizers
Major contributors to acid rain
4 NO + 3 O2 + 2 H2O  4 HNO3
4 NO2 + O2 + 2 H2O  4 HNO3
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Pollutant Gases, CO
• CO comes from incomplete burning of fossil
•
•
fuels in cars, trucks, and power plants
Adheres to hemoglobin in your red blood cells,
depleting your ability to acquire O2
At high levels can cause sensory impairment,
stupor, unconsciousness, or death
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Pollutant Gases, O3
• Ozone pollution comes from other pollutant
gases reacting in the presence of sunlight
as well as lightning storms
known as photochemical smog and ground-level
ozone
• O3 is present in the brown haze seen in some
•
•
cities
Lung and eye irritant
Strong oxidizer
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Major Pollutant Levels
• Government
regulation has
resulted in a
decrease in the
emission levels for
most major
pollutants
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Stratospheric Ozone
• Ozone occurs naturally in the stratosphere
• Stratospheric ozone protects the surface of the
•
earth from over-exposure to UV light from the
Sun
O3(g) + UV light  O2(g) + O(g)
Normally the reverse reaction occurs quickly, but
the energy is not UV light
O2(g) + O(g)  O3(g)
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Ozone Depletion
• Chlorofluorocarbons became popular as aerosol
•
•
•
propellants and refrigerants in the 1960s
CFCs pass through the tropopause into the stratosphere
There, CFCs can be decomposed by UV light, releasing
Cl atoms
CF2Cl2 + UV light  CF2Cl + Cl
Cl atoms catalyze O3 decomposition and remove O atoms
so that O3 cannot be regenerated
 NO2 also catalyzes O3 destruction
Cl + O3  ClO + O2
O3 + UV light  O2 + O
ClO + O  O2 + Cl
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Ozone Holes
• Satellite data over
the past 3 decades
reveals a marked
drop in ozone
concentration over
certain regions
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