Transcript The Gaseous State of Matter Chapter 12 Hein and Arena Version 1.1 Chapter Outline 12.2 The Kinetic MolecularTheory 12.9 Combined Gas Laws 12.7 Gay Lussac’s Law 12.15 Gas Stoichiometry 12.8

The Gaseous State of Matter
Chapter 12
Hein and Arena
Version 1.1
1
Chapter Outline
12.2 The Kinetic
MolecularTheory
12.9
Combined Gas Laws
12.7 Gay Lussac’s Law
12.15 Gas Stoichiometry
12.8 Standard Temperature and
Pressure
12.16 Real Gases
12.10 Dalton’s Law of Partial
12.3 Measurement of Pressure of
Pressures
Gases
12.11 Avogadro’s Law
12.4 Dependence of Pressure on
Number of Molecules and 12.12 Mole-Mass-Volume
Relationships of Gases
Temperature
12.13 Density of Gases
12.5 Boyle’s Law
12.14 Ideal Gas Equation
12.6 Charles’ Law
2
The Kinetic-Molecular Theory
• KMT is based on the motions of gas
particles.
• A gas that behaves exactly as outlined
by KMT is known as an ideal gas.
• While no ideal gases are found in
nature, real gases can approximate
ideal gas behavior under certain
conditions of temperature and pressure.
3
Principle Assumptions of the KMT
1. Gases consist of tiny subatomic
particles.
2. The distance between particles is
large compared with the size of the
particles themselves.
3. Gas particles have no attraction for
one another.
4
Principle Assumptions of the KMT
4. Gas particles move in straight lines
in all directions, colliding frequently
with one another and with the walls
of the container.
5. No energy is lost by the collision of a
gas particle with another gas particle
or with the walls of the container.
All collisions are perfectly elastic.
5
Principle Assumptions of the KMT
6. The average kinetic energy for
particles is the same for all gases at
the same temperature, and its value is
directly proportional to the Kelvin
temperature.
6
Kinetic Energy
1 2
KE = mv
2
7
Kinetic Energy
• All gases have the same average
kinetic energy at the same temperature.
• As a result lighter molecules move
faster than heavier molecules.
mH2 = 2
vH2 4
=
vO2 1
mO2= 32
8
Diffusion
The ability of two or more gases to mix
spontaneously until they form a uniform
mixture.
Stopcock closed
No diffusion occurs
Stopcock open
Diffusion occurs
9
Effusion
A process by which gas molecules pass
through a very small orifice from a
container at higher pressure to one at
lower pressure.
10
Graham’s Law of Effusion
The rates of effusion of two gases at the
same temperature and pressure are
inversely proportional to the square roots
of their densities, or molar masses.
rate of effusion of gas A
=
rate of effusion of gas B
dB
=
dA
molar mass B
molar mass A
11
What is the ratio of the rate of effusion of CO to CO2?
effusion rate CO
=
effusion rate CO 2
molar mass CO 2
=
molar mass CO
44.0 g
 1.25
28.0 g
12
Measurement of
Pressure of Gases
13
Pressure equals force per unit area.
Force
Pressure =
Area
14
The pressure resulting
from the collisions of
gas molecules with
the walls of the
balloon keeps the
balloon inflated.
15
The pressure exerted by a gas depends on the
• Number of gas molecules present.
• Temperature of the gas.
• Volume in which the gas is confined.
16
Mercury Barometer
A tube of mercury
is inverted and
placed in a dish of
mercury.
The barometer is
used to measure
atmospheric
pressure.
17
18
Average Composition of Dry Air
Gas Volume Percent
N2
O2
78.08%
20.95%
Ar
0.93%
CO2
0.033%
Ne
0.0018%
Gas Volume Percent
He
CH4
0.0005%
0.0002%
Kr
0.0001%
Xe, H2, and N2O Trace
19
Dependence of Pressure on Number of
Molecules and Temperature
• Pressure is produced by gas molecules
colliding with the walls of a container.
• At a specific temperature and volume,
the number of collisions depends on
the number of gas molecules present.
• For an ideal gas the number of
collisions is directly proportional to the
number of gas molecules present.
20
V = 22.4 L
T = OoC
The pressure exerted by a gas is
directly proportional to the number
of molecules present.
21
Dependence of Pressure on
Temperature
• The pressure of a gas in a fixed volume
increases with increasing temperature.
• When the pressure of a gas increases,
its kinetic energy increases.
• The increased kinetic energy of the gas
results in more frequent and energetic
collisions of the molecules with the
walls of the container.
22
The pressure of a
gas in a fixed
volume increases
with increasing
temperature.
Lower T
Lower P
Increased pressure is
due to more frequent
and more energetic
collisions of the gas
molecules with the
walls of the container
at the higher
temperature.
Higher T
Higher P
24
Boyle’s Law
At constant temperature (T), the volume
(V) of a fixed mass of gas is inversely
proportional to the Pressure (P).
1
V 
P
P1V1 = P2 V2
25
Graph of pressure versus volume. This shows the
26
inverse PV relationship of an ideal gas.
The effect of pressure on the volume of a gas.27
An 8.00 L sample of N2 is at a pressure of 500
torr. What must be the pressure to change the
volume to 3.00 L? (T is constant).
Method A. Conversion Factors
Step 1. Determine whether volume is
being increased or decreased.
Initial volume = 8.00 L
Final volume = 3.00 L
volume decreases  pressure increases
28
An 8.00 L sample of N2 is at a pressure of 500
torr. What must be the pressure to change the
volume to 3.00 L? (T is constant).
Step 2. Multiply the original pressure by
a ratio of volumes that will result in an
increase in pressure.
new pressure = original pressure x ratio of volumes
8.00 L
3
P = 500 torr x
= 1333 L = 1.33 x 10 L
3.00 L
29
An 8.00 L sample of N2 is at a pressure of 500
torr. What must be the pressure to change the
volume to 3.00 L? (T is constant).
Method B. Algebraic Equation
Step 1. Organize the given information:
P1 = 500 torr
V1 = 8.00 L
P2 = ?
V2 = 3.00 L
30
An 8.00 L sample of N2 is at a pressure of 500
torr. What must be the pressure to change the
volume to 3.00 L? (T is constant).
Step 2. Write and solve the equation for
the unknown.
P1V1 = P2 V2
P1V1
P2 =
V2
31
An 8.00 L sample of N2 is at a pressure of 500
torr. What must be the pressure to change the
volume to 3.00 L? (T is constant).
Step 3. Put the given information into
the equation and calculate.
P1V1
(500
torr)(8.00
L)
1
P2 =
=
= 1.33 x 10 torr
V2
3.00 L
32
Charles’ Law
Absolute Zero on the Kelvin Scale
• If a given volume of any gas at 0oC is
cooled by 1oC the volume of the gas
decreases by 1 .
273
• If a given volume of any gas at 0oC is
cooled by 20oC the volume of the gas
decreases by 20 .
273
33
Absolute Zero on the Kelvin Scale
• If a given volume of any gas at 0oC is
cooled by 273oC the volume of the gas
decreases by 273.
273
• -273oC (more precisely –273.15oC) is
the zero point on the Kelvin scale. It is
the temperature at which an ideal gas
would have 0 volume.
34
35
Volume-temperature relationship of methane (CH4).
Charles’ Law
At constant pressure the volume of a
fixed mass of gas is directly proportional
to the absolute temperature.
V  T
V1 V2
=
T1
T2
36
Effect of temperature on the volume of a gas. Pressure
is constant at 1 atm. When temperature increases at
37
constant pressure, the volume of the gas increases.
A 255 mL sample of nitrogen at 75oC is confined at
a pressure of 3.0 atmospheres. If the pressure
remains constant, what will be the volume of the
nitrogen if its temperature is raised to 250oC?
Method A. Conversion Factors
Step 1. Change oC to K:
oC + 273 = K
75oC + 273 = 348 K
250oC + 273 = 523 K
38
A 255 mL sample of nitrogen at 75oC is confined at
a pressure of 3.0 atmospheres. If the pressure
remains constant, what will be the volume of the
nitrogen if its temperature is raised to 250oC?
Step 2: Multiply the original volume by
a ratio of Kelvin temperatures that will
result in an increase in volume:
523K 

V = (255mL) 
 = 383 mL
 348K 
39
A 255 mL sample of nitrogen at 75oC is confined at
a pressure of 3.0 atmospheres. If the pressure
remains constant, what will be the volume of the
nitrogen if its temperature is raised to 250oC?
Method B. Algebraic Equation
Step 1. Organize the information (remember
to make units the same):
V1 = 255 mL
T1 = 75oC = 348 K
V2 = ?
T2 = 250oC = 523 K
40
A 255 mL sample of nitrogen at 75oC is confined at
a pressure of 3.0 atmospheres. If the pressure
remains constant, what will be the volume of the
nitrogen if its temperature is raised to 250oC?
Step 2. Write and solve the equation for
the unknown:
V1 V2
=
T1
T2
V1T2
V2 =
T1
41
A 255 mL sample of nitrogen at 75oC is confined at
a pressure of 3.0 atmospheres. If the pressure
remains constant, what will be the volume of the
nitrogen if its temperature is raised to 250oC?
Step 3. Put the given information into the
equation and calculate:
V1 = 255 mL
T1 = 75oC = 348 K
V2 = ?
T2 = 250oC = 523 K
V1T2
(255mL)(523K)
V2 =
= 383 mL
=
T1
348K
42
Gay-Lussac’s Law
The pressure of a fixed mass of gas, at
constant volume, is directly proportional
to the Kelvin temperature.
P = kT
P1
P2
=
T1 T2
43
At a temperature of 40oC an oxygen container is at a pressure
of 2.15 atmospheres. If the temperature of the container is
raised to 100oC what will be the pressure of the oxygen?
Method A. Conversion Factors
Step 1. Change oC to K:
oC + 273 = K
40oC + 273 = 313 K
100oC + 273 = 373 K
Determine whether temperature is being
increased or decreased.
temperature increases  pressure increases
44
At a temperature of 40oC an oxygen container is at a pressure
of 2.15 atmospheres. If the temperature of the container is
raised to 100oC what will be the pressure of the oxygen?
Step 2: Multiply the original pressure by
a ratio of Kelvin temperatures that will
result in an increase in pressure:
373K 

P = (21.5 atm) 
 = 25.6 atm
 313K 
45
At a temperature of 40oC an oxygen container is at a pressure
of 2.15 atmospheres. If the temperature of the container is
raised to 100oC what will be the pressure of the oxygen?
A temperature ratio
greater than 1 will
increase the pressure
373K 

P = (21.5 atm) 
 = 25.6 atm
 313K 
46
At a temperature of 40oC an oxygen container is at a pressure
of 2.15 atmospheres. If the temperature of the container is
raised to 100oC what will be the pressure of the oxygen?
Method B. Algebraic Equation
Step 1. Organize the information
(remember to make units the
same):
P1 = 21.5 atm
T1 = 40oC = 313 K
P2 = ?
T2 = 100oC = 373 K
47
At a temperature of 40oC an oxygen container is at a pressure
of 2.15 atmospheres. If the temperature of the container is
raised to 100oC what will be the pressure of the oxygen?
Step 2. Write and solve the equation for
the unknown:
P1
P2
=
T1 T2
P1T2
P2 =
T1
48
At a temperature of 40oC an oxygen container is at a pressure
of 2.15 atmospheres. If the temperature of the container is
raised to 100oC what will be the pressure of the oxygen?
Step 3. Put the given information into the
equation and calculate:
P1 = 21.5 atm
T1 = 40oC = 313 K
P2 = ?
T2 = 100oC = 373 K
P1T2 (21.5 atm)(373 K)
P2 =
=
= 25.6 atm
T1
313 K
49
Standard Temperature and
Pressure
Selected common reference points of temperature and
pressure.
Standard Conditions
Standard Temperature and Pressure
STP
273.15 K or 0.00oC
1 atm or 760 torr or 760 mm Hg
50
Combined Gas Laws
P1V1
P2 V2
=
T1
T2
• A combination of Boyle’s and Charles’
Law.
• Used when pressure and temperature
change at the same time.
• Solve the equation for any one of the 6
variables
52
calculate final
volume
final volume = initial volume
ratio of
pressures
ratio of
temperatures
53
increases or
decreases volume
final volume = initial volume
ratio of
pressures
ratio of
temperatures
54
increases or
decreases volume
final volume = initial volume
ratio of
pressures
ratio of
temperatures
55
A sample of hydrogen occupies 465 ml at STP. If the
pressure is increased to 950 torr and the temperature is
decreased to –15oC, what would be the new volume?
Step 1. Organize the given information,
putting temperature in Kelvins:
oC
+ 273 = K
0oC + 273 = 273 K
-15oC + 273 = 258 K
56
A sample of hydrogen occupies 465 ml at STP. If the
pressure is increased to 950 torr and the temperature is
decreased to –15oC, what would be the new volume?
Step 1. Organize the given information,
putting temperature in Kelvins:
P1 = 760 torr
P2 = 950 torr
V1 = 465 mL
V2 = ?
T1 = 273 K
T2 = 258 K
57
A sample of hydrogen occupies 465 ml at STP. If the
pressure is increased to 950 torr and the temperature is
decreased to –15oC, what would be the new volume?
Method A Conversion Factors
Step 2. Set up ratios of T and P
258 K
(decrease in T decreases V)
T ratio =
273 K
760 torr
(increase in P decreases V)
P ratio =
950 torr
58
A sample of hydrogen occupies 465 ml at STP. If the
pressure is increased to 950 torr and the temperature is
decreased to –15oC, what would be the new volume?
Step 3. Multiply the original volumes by
the ratios:
P1 = 760 torr
P2 = 950 torr
V1 = 465 mL
V2 = ?
T1 = 273 K
T2 = 258 K
760 torr   258K 

V2 = (465 ml)
 = 352 mL

 950 torr   273K 
59
A sample of hydrogen occupies 465 ml at STP. If the
pressure is increased to 950 torr and the temperature is
decreased to –15oC, what would be the new volume?
Method B Algebraic Equation
Step 2. Write and solve the equation for
the unknown V2.
T2 PV
PV
T
PV
1 1
2 2 T2
2
1 1





P2 T1
T2 P2 P2 T1
V1 PT
1 2
V2 
P2T1
V2
60
A sample of hydrogen occupies 465 ml at STP. If the
pressure is increased to 950 torr and the temperature is
decreased to –15oC, what would be the new volume?
Step 2 Put the given information into the
equation and calculate.
V1 PT
1 2
V2 
P2T1
(465 ml)  760 torr  (258 K)
V2 =
= 352 mL
(950 torr)(273 K)
61
Dalton’s Law of
Partial Pressures
Each gas in a mixture exerts a pressure
that is independent of the other gases
present.
The total pressure of a mixture of gases is the
sum of the partial pressures exerted by each of
the gases in the mixture.
Ptotal = Pa + Pb + Pc + Pd + ….
62
A container contains He at a pressure of 0.50 atm, Ne
at a pressure of 0.60 atm, and Ar at a pressure of 1.30
atm. What is the total pressure in the container?
Ptotal = PHe + PNe+ PAr
Ptotal = 0.5 atm + 0.6 atm + 1.30 atm = 2.40 atm
63
Collecting a Gas Sample Over Water
• The pressure in the collection container
is equal to the atmospheric pressure.
• The pressure of the gas collected plus
the pressure of water vapor at the
collection temperature is equal to the
atmospheric pressure.
Ptotal = Patm = Pgas + PH O
2
64
Oxygen collected over water.
65
A sample of O2 was collected in a bottle over water at a
temperature of 25oC when the atmospheric pressure was 760
torr. The vapor pressure of water at 25oC is 23.8 torr.
Ptotal = 760 torr = PO +PH O
2
2
PO = 760 torr - PH O
2
2
PO = 760 torr - 23.8 torr = 736 torr
2
66
Gay Lussac’s Law of Combining Volumes
When measured at the same temperature
and pressure, the ratio of the volumes of
reacting gases are small whole numbers.
N2
1 volume
+
+
→
3 volumes →
3 H2
V
VNH
H 2 3 32
==
VNH 113
22
2 NH3
2 volumes
67
Gay Lussac’s Law of Combining Volumes
When measured at the same temperature
and pressure, the ratio of the volumes of
reacting gases are small whole numbers.
68
Avogadro’s Law
Equal volumes of different gases at the
same temperature and pressure contain
the same number of molecules.
69
AVOGADRO'S LAW
Explained Gay Lussac's
Law of Combining Volumes
Provided a method
for
the determination of
molar masses of gases
Served as a foundation
for the devolopment of the
Kinetic-Molecular Theory
comparing densities of gases
of known molar mass
70
hydrogen + chlorine  hydrogen chloride
1 volume 1 volume
2 volumes
1 molecule 1 molecule
2 molecules
1 mol
1 mol
2 mol
There
are 2 molecules
of hydrogen
Each
molecule
of hydrogen
chloride
chloride.
contains
at least 1 atom of hydrogen
71
and 1 atom of chlorine.
H2 + Cl2 → 2 HCl
hydrogen + chlorine → hydrogen chloride
1 volume 1 volume
2 volumes
1 molecule 1 molecule
2 molecules
1 mol
1 mol
Each molecule of hydrogen and each
molecule of chlorine must contain at
least 2 atoms.
2 mol
72
Mole-Mass-Volume
Relationships
• Volume of one mole of any gas at STP
= 22.4 L.
• 22.4 L at STP is known as the molar
volume of any gas.
73
• 22.4 L at STP is known as the molar
volume of any gas.
74
75
The density of neon at STP is 0.900 g/L. What is the
molar mass of neon?
g
 0.900 g   22.4 L 


 = 20.2
mol
 1 L   1 mol 
76
Density of Gases
mass
density =
volume
77
Density of Gases
grams
m
d=
v
liters
78
Density of Gases
m
d=
v
depends
on T and P
79
The molar mass of SO2 is 64.07 g/mol. Determine the
density of SO2 at STP.
1 mole of any gas
occupies 22.4 L at
STP
64.07 g   1 mol 
g

d= 

 = 2.86
L
 mol   22.4 L 
80
Ideal Gas Equation
nRT
nT
VV=a
PV
= nRT
P
P
81
atmospheres
nRT
nT
VV=a
PV
= nRT
P
P
82
liters
nRT
nT
VV=a
PV
= nRT
P
P
83
moles
nRT
nT
VV=a
PV
= nRT
P
P
84
Kelvin
nRT
nT
VV=a
PV
= nRT
P
P
85
Ideal L-atm
Gas
0.0821
mol-K
Constant
nRT
nT
VV=a
PV
= nRT
P
P
86
A balloon filled with 5.00 moles of helium gas is at a
temperature of 25oC. The atmospheric pressure is 750.
torr. What is the balloon’s volume?
Step 1. Organize the given information.
Convert temperature to kelvins.
K = oC + 273
K = 25oC + 273 = 298K
Convert pressure to atmospheres.
1 atm
P = 750. torr x
= 0.987 atm
760 torr
87
A balloon filled with 5.00 moles of helium gas is at a
temperature of 25oC. The atmospheric pressure is 750.
torr. What is the balloon’s volume?
Step 2. Write and solve the ideal gas
equation for the unknown.
nRT
PV = nRT
V=
P
Step 3. Substitute the given data into the
equation and calculate.
(5.00 mol) (0.0821 L×atm/mol×K)(298 K)
= 124 L
V=
88
(0.987 atm)
Determination of Molecular Weights
Using the Ideal Gas Equation
g
molar mass =
mol
M = molar mass
g
mol =
molar mass
g
n = mol =
M
g
RT
PV = nRT PV =
M
gRT
M=
PV
89
Calculate the molar mass of an unknown gas, if 0.020
g occupies 250 mL at a temperature of 305 K and a
pressure of 0.045 atm.
V = 250 mL = 0.250 L
g = 0.020 g
T = 305 K
P = 0.045 atm
gRT
M=
PV
(0.020 g)(0.082 L × atm/mol × K)(305 K)
g
M=
= 44
90 mol
(0.045 atm) (0.250 L)
Gas Stoichiometry
• All calculations are done at STP.
• Gases are assumed to behave as ideal
gases.
• A gas not at STP is converted to STP.
91
Definition
Stoichiometry: The area of chemistry that
deals with the quantitative relationships
among reactants and products in a chemical
reaction.
92
Gas Stoichiometry
Primary conversions involved in stoichiometry.
93
Mole-Volume Calculations
Mass-Volume Calculations
94
What volume of oxygen (at STP) can be formed
from 0.500 mol of potassium chlorate?
• Step 1 Write the balanced equation
2 KClO3  2 KCl + 3 O2
• Step 2 The starting amount is 0.500
mol KClO3. The conversion is
moles KClO3  moles O2  liters O2
95
What volume of oxygen (at STP) can be formed
from 0.500 mol of potassium chlorate?
2 KClO3  2KCl + 3 O2
• Step 3. Calculate the moles of O2, using the
mole-ratio method.
 3 mol O2 
(0.500 mol KClO3 ) 
= 0.750 mol O 2

 2 mol KClO3 
• Step 4. Convert moles of O2 to liters of O2
22.4 L 

(0.750 mol O 2 ) 
 = 16.8 L O 2
 1 mol 
96
What volume of oxygen (at STP) can be formed
from 0.500 mol of potassium chlorate?
The problem can also be solved in
one continuous calculation.
2 KClO3  2KCl + 3 O2
 3 mol O2   22.4 L 
(0.500 mol KClO3 ) 

 = 16.8 L O 2

 2 mol KClO3   1 mol 
97
What volume of hydrogen, collected at 30.0oC and
700. torr, will be formed by reacting 50.0 g of
aluminum with hydrochloric acid?
2 Al(s) + 6 HCl(aq)  2AlCl3(aq) + 3 H2(g)
Step 1 Calculate moles of H2.
grams Al  moles Al  moles H2
 1 mol Al   3 mol H 2 
50.0 g Al 

 = 2.78 mol H 2

 26.98 g Al   2 mol Al 
98
What volume of hydrogen, collected at 30.0oC and
700. torr, will be formed by reacting 50.0 g of
aluminum with hydrochloric acid?
2 Al(s) + 6 HCl(aq)  2AlCl3(aq) + 3 H2(g)
Step 2 Calculate liters of H2.
• Convert oC to K: 30.oC + 273 = 303 K
• Convert torr to atm:
 1 atm 
700 torr 
 = 0.921 atm
 760 torr 
99
What volume of hydrogen, collected at 30.0oC and
700. torr, will be formed by reacting 50.0 g of
aluminum with hydrochloric acid?
• Solve the ideal gas equation for V
PV = nRT
nRT
V=
P
(2.78 mol H 2 )(0.0821 L-atm)(303 K)
= 75.1 L H 2
V=
(0.921 atm)(mol-K)
100
Volume-Volume
Calculations
101
For reacting gases at constant temperature and
pressure: Volume-volume relationships are the same as
mole-mole relationships.
H2(g)
+
Cl2(g)

2HCl(g)
1 mol H2
1 mol Cl2
2 mol HCl
22.4 L
STP
1 volume
22.4 L
STP
1 volume
2 x 22.4 L
STP
2 volumes
Y volume
Y volume
2Y volumes
102
What volume of nitrogen will react with 600. mL of
hydrogen to form ammonia? What volume of
ammonia will be formed?
N2(g) + 3H2(g)  2NH3(g)
 1 vol N 2 
600. ml H 2 
= 200. mL N 2

 3 vol H 2 
 2 vol NH 3 
= 400. mL NH 3
600. ml H 2 

 3 vol H 2 
103
Real Gases
Ideal Gas
• An ideal gas obeys the gas laws.
– The volume the molecules of an ideal gas
occupy is negligible compared to the
volume of the gas. This is true at all
temperatures and pressures.
– The intermolecular attractions between the
molecules of an ideal gas are negligible at
104
all temperatures and pressures.
Real Gases
• Deviations from the gas laws occur at
high pressures and low temperatures.
– At high pressures the volumes of the real
gas molecules are not negligible
compared to the volume of the gas
– At low temperatures the kinetic energy of
the gas molecules cannot completely
overcome the intermolecular attractive
forces between the molecules.
105
Key Concepts
12.2 The Kinetic
MolecularTheory
12.9
Combined Gas Laws
12.7 Gay Lussac’s Law
12.15 Gas Stoichiometry
12.8 Standard Temperature and
Pressure
12.16 Real Gases
12.10 Dalton’s Law of Partial
12.3 Measurement of Pressure of
Pressures
Gases
12.11 Avogadro’s Law
12.4 Dependence of Pressure on
Number of Molecules and 12.12 Mole-Mass-Volume
Relationships of Gases
Temperature
12.13 Density of Gases
12.5 Boyle’s Law
12.14 Ideal Gas Equation
12.6 Charles’ Law
106