Chapter 12
The Gaseous State of Matter
The air in a hot air balloon expands
When it is heated. Some of the air
escapes from the top of the balloon,
lowering the air density inside the
balloon, making the balloon buoyant.
Introduction to General, Organic, and Biochemistry 10e
John Wiley & Sons, Inc
Morris Hein, Scott Pattison, and Susan Arena
Chapter Outline
12.1 General Properties
12.8 Combined Gas Laws
12.2 The Kinetic-Molecular
Theory
12.9 Dalton’s Law of Partial
Pressures
12.3 Measurement of Pressure 12.10 Avogadro’s Law
12.4 Dependence of Pressure
on Number of Molecules
and Temperature
12.11 Mole-Mass-Volume
Relationships of Gases
12.5 Boyle’s Law
12.13 Ideal Gas Law
12.6 Charles’ Law
12.14 Gas Stoichiometry
12.7 Gay-Lussac’s Law
12.15 Real Gases
12.12 Density of Gases
Copyright 2012 John Wiley & Sons, Inc
Objectives for Today



Kinetic Molecular Theory of Gases
Gas Measurements
Boyle’s, Charles’ and Gay-Lussac’s Laws
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GASES AND KINETIC MOLECULAR
THEORY
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General Properties
• Gases
• Have an indefinite volume
Expand to fill a container
• Have an indefinite shape
Take the shape of a container
• Have low densities
dair  1.2 g / L at 25C
d H2O  1.0 g / mL
• Have high kinetic energies
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Kinetic Molecular Theory (KMT)
Assumptions of the KMT and ideal gases include:
1. Gases consist of tiny particles
2. The distance between particles is large
compared with the size of the particles.
3. Gas particles have no attraction for each other
4. Gas particles move in straight lines in all
directions, colliding frequently with each other
and with the walls of the container.
Copyright 2012 John Wiley & Sons, Inc
Kinetic Molecular Theory
Assumptions of the KMT (continued):
5. Collisions are perfectly elastic (no energy is lost
in the collision).
6. The average kinetic energy for particles is the
same for all gases at the same temperature.
1 2
KE = mv where m is mass and v is velocity
2
7. The average kinetic energy is directly
proportional to the Kelvin temperature.
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Diffusion
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Effusion
• Gas molecules pass through a very small
opening from a container at higher pressure of
one at lower pressure.
• Graham’s law of effusion:
rate of effusion of gas A
density B
molar mass B
=
=
rate of effusion of gas B
density A
molar mass A
Copyright 2012 John Wiley & Sons, Inc
Your Turn!
rate of effusion of gas A
density B
molar mass B
=
=
rate of effusion of gas B
density A
molar mass A
Which gas will diffuse most rapidly?
a. He
b. Ne
c. Ar
d. Kr
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Measurement of Pressure
Pressure =
Force
Area
Pressure depends on the
•Number of gas molecules
•Temperature of the gas
•Volume the gas occupies
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Atmospheric Pressure
• Atmospheric pressure is due to the mass of the
atmospheric gases pressing down on the earth’s
surface.
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Barometer
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Pressure Conversions
• Convert 675 mm Hg to atm. Note: 760 mm Hg = 1 atm
1 atm
675 mm Hg 
= 0.888 atm
760 mm Hg
Convert 675 mm Hg to torr. Note: 760 mm Hg = 760
torr.
760 torr
675 mm Hg 
= 675 torr
760 mm Hg
Copyright 2012 John Wiley & Sons, Inc
Your Turn!
A pressure of 3.00 atm is equal to
a. 819 torr
b. 3000 torr
c. 2280 torr
d. 253 torr
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Dependence of Pressure on
Number of Molecules
P is proportional to n
(number of molecules)
at Tc (constant T) and
Vc (constant V).
The increased pressure
is due to more
frequent collisions with
walls of the container
as well increased force
of each collision.
Copyright 2012 John Wiley & Sons, Inc
Dependence of Pressure on
Temperature
P is proportional to T at nc
(constant number of
moles) and Vc.
The increased pressure is
due to
• more frequent collisions
•higher energy collisions
Copyright 2012 John Wiley & Sons, Inc
Your Turn!
If you change the temperature of a sample of
gas from 80°C to 25°C at constant volume, the
pressure of the gas
a. will increase.
b. will decrease.
c. will not change
Copyright 2012 John Wiley & Sons, Inc
Boyle’s Law
1
At Tc and nc : V α
P
or
PV
1 1  PV
2 2
What happens to V if you double P?
•V decreases by half!
What happens to P if you double V?
•P decreases by half!
Copyright 2012 John Wiley & Sons, Inc
Boyle’s Law
PV
1 1  PV
2 2
• A sample of argon gas occupies 500.0 mL at
920. torr. Calculate the pressure of the gas if
the volume is increased to 937 mL at constant
temperature.
Knowns V1 = 500 mL P1 = 920. torr V2 = 937 mL
Set-Up
PV
P2  1 1
V2
Calculate
920. torr  500. mL
P2 =
= 491 torr
937 mL
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Boyle’s Law
• Another approach to the same problem:
• Since volume increased from 500. mL to 937
ml, the pressure of 920. torr must decrease.
• Multiply the pressure by a volume ratio that
decreases the pressure:
 500. mL
P2 = 920. torr 
 937 mL

 = 491 torr

Copyright 2012 John Wiley & Sons, Inc
Your Turn!
A 6.00 L sample of a gas at a pressure of 8.00
atm is compressed to 4.00 L at a constant
temperature. What is the pressure of the gas?
a. 4.00 atm
b. 12.0 atm
c. 24.0 atm
d. 48.0 atm
Copyright 2012 John Wiley & Sons, Inc
Your Turn!
• A 400. mL sample of a gas is at a pressure of
760. torr. If the temperature remains
constant, what will be its volume at 190. torr?
•
A. 100. mL
•
B. 400. mL
•
C. 25.0 mL
•
D. 1.60x102 mL
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Charles’ Law
• The volume of an ideal gas at
absolute zero (-273°C) is zero.
• Real gases condense at their
boiling point so it is not
possible to have a gas with zero
volume.
• The gas laws are based on
Kelvin temperature.
• All gas law problems must be
worked in Kelvin!
At Pc and nc :
V α T or
Copyright 2012 John Wiley & Sons, Inc
V1
V2

T1
T2
Charles’ Law
V1
V2

T1
T2
• A 2.0 L He balloon at 25°C is taken outside on
a cold winter day at -15°C. What is the
volume of the balloon if the pressure remains
constant?
Knowns V1 = 2.0 L T1 = 25°C= 298 K T2 = -15°C = 258 K
Set-Up
Calculate
V1T2
V1
V2
rearranged gives V2 

T1
T1
T2
(2.0 L)(258 K)
V2 =
= 1.7 L
298 K
Copyright 2012 John Wiley & Sons, Inc
Charles’ Law
• Another approach to the same problem:
• Since T decreased from 25°C to -15°C, the
volume of the 2.0L balloon must decrease.
• Multiply the volume by a Kelvin temperature
ratio that decreases the volume:
 258K
P2 = 2.0L 
 298K

 = 1.7L

Copyright 2012 John Wiley & Sons, Inc
Your Turn
The volume of a gas always increases when
a. Temperature increases and pressure
decreases
b. Temperature increases and pressure
increases
c. Temperature decreases and pressure
increases
d. Temperature decreases and pressure
decreases
Copyright 2012 John Wiley & Sons, Inc
Your Turn!
A sample of CO2 has a volume of 200. mL at 20.0
° C. What will be its volume at 40.0 °C,
assuming that the pressure remains constant?
a. 18.8 mL
b. 100. mL
c. 213 mL
d. 400. mL
Copyright 2012 John Wiley & Sons, Inc
Your Turn!
A sample of gas has a volume of 3.00 L at 10.0
°C. What will be its temperature in °C if the
gas expands to 6.00 L at constant pressure?
a. 20.0°C
b. 293°C
c. 566°C
d. 142°C
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Gay-Lussac’s Law
At Vc and nc : P α T
or
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P1
P2

T1
T2
At a temperature of 40oC an oxygen container is at a pressure
of 21.5 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 31
At a temperature of 40oC an oxygen container is at a pressure
of 21.5 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 
32
At a temperature of 40oC an oxygen container is at a pressure
of 21.5 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 
33
At a temperature of 40oC an oxygen container is at a pressure
of 21.5 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
34
At a temperature of 40oC an oxygen container is at a pressure
of 21.5 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
35
At a temperature of 40oC an oxygen container is at a pressure
of 21.5 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
P2 = ?
T1 = 40oC = 313 K
T2 = 100oC = 373
K
P1T2 (21.5 atm)(373 K)
P2 =
=
= 25.6 atm
T1
313 K
36
Objectives for Today



Kinetic Molecular Theory of Gases
Gas Measurements
Boyle’s, Charles’ and Gay-Lussac’s Laws
Copyright 2012 John Wiley & Sons, Inc
Objectives for Today
Combined Gas Law
Dalton’s Law of Partial Pressure
Avogadro’s Law
Density of Gases
Copyright 2012 John Wiley & Sons, Inc
Combined Gas Laws
Used for calculating the results
of changes in gas conditions.
•Boyle’s Law where Tc
PV
1 1
T1

P2V2
T2
PV
1 1  PV
2 2
•Charles’ Law where Pc
V1
V2

T1
T2
•Gay Lussacs’ Law where Vc
P1
P2

T1
T2
P1 and P2 , V1 and V2 can be any units as long as they are
the same. T1 and T2 must be in Kelvin.
Copyright 2012 John Wiley & Sons, Inc
Combined Gas Law
PV
1 1
T1

PV
2 2
T2
If a sample of air occupies 500. mL at STP, what is the
volume at 85°C and 560 torr?
STP: Standard Temperature 273K or 0°C
Standard Pressure 1 atm or 760 torr
Knowns
Set-Up
Calculate
V1 = 500. mL T1 =273K
T2 = 85°C = 358K
P1= 760 torr
P2= 560 torr
PV
1 1T2
V2 
T1 P2
(760 torr)(500. mL)(358K)
V2 
= 890. ml
(273K)(560 torr)
Copyright 2012 John Wiley & Sons, Inc
Combined Gas Law
PV
1 1
T1

PV
2 2
T2
• A sample of oxygen gas occupies 500.0 mL at
722 torr and –25°C. Calculate the
temperature in °C if the gas has a volume of
2.53 L at 491 mmHg.
Knowns
Set-Up
Calculate
V1 = 500. mL T1 = -25°C = 248K
P1= 722 torr
V2 = 2.53 L = 2530 mL
P2= 560 torr
T PV
T2  1 2 2
PV
1 1
T2
491 torr  2530 ml  248K 

=
=853K 
 722 torr  500.0 ml 
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580C
Your Turn!
A sample of gas has a volume of 8.00 L at 20.0 ° C
and 700. torr. What will be its volume at STP?
a. 1.20 L
b. 9.32 L
c. 53.2 L
d. 6.87 L
Copyright 2012 John Wiley & Sons, Inc
Dalton’s Law of Partial Pressures
• 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 + ….
• Atmospheric pressure is the result of the
combined pressure of the nitrogen and oxygen
and other trace gases in air.
PAir  PN2  PO2  PAr  PCO2  PH 2O  .
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Collecting Gas Over Water
• Gases collected over water contain both the
gas and water vapor.
• The vapor pressure of water is
•
constant at a given temperature
• Pressure in the bottle is
•
equalized so that the Pinside = Patm
Patm  Pgas  PH 2 O
Copyright 2012 John Wiley & Sons, Inc
Your Turn!
A sample of oxygen is collected over water at 22
° C and 762 torr. What is the partial pressure
of the dry oxygen? The vapor pressure of
water at 22°C is 19.8 torr.
a. 742 torr
b. 782 torr
c. 784 torr
d. 750. torr
Copyright 2012 John Wiley & Sons, Inc
Avogadro’s Law
• Equal volumes of different gases at the same T
and P contain the same number of molecules.
The ratio is
the same:
1 volume
1 molecule
1 mol
1 volume
1 molecule
1 mol
Copyright 2012 John Wiley & Sons, Inc
2 volumes
2 molecules
2 mol
Mole-Mass-Volume Relationships
• Molar Volume: One mole of any gas occupies
22.4 L at STP.
• Determine the molar mass of a gas, if 3.94 g of
the gas occupied a volume of 3.52 L at STP.
Knowns m = 3.94 g
Set-Up
Calculate
V = 3.52 L T = 273 K P = 1 atm
22.4 L
1 mol = 22.4 L so the conversion factor is
1mol
3.94 g  22.4 L 


1
mol

3.52 L 
= 25.1 g/mol
Your Turn!
What is the molar mass of a gas if 240. mL of the
gas at STP has a mass of 0.320 grams?
a. 8.57 g
b. 22.4 g
c. 16.8 g
d. 29.9 g
Density of Gases
mass
g
d=
=
volume
L
• Calculate the density of nitrogen gas at STP.
 1 mol 
dSTP = molar mass 

 22.4 L 
 28.02 g  1 mol 
dSTP = 

 = 1.25g/L
 1 mol  22.4 L 
• Note that densities are always cited for a
particular temperature, since gas densities
decrease as temperature increases.
Your Turn!
Which of the following gases is the most dense?
a. H2
b. N2
c. CO2
d. O2
Carbon dioxide fire extinguishers can be used to put
out fires because CO2 is more dense than air and can
be used to push oxygen away from the fuel source.
Objectives for Today
Combined Gas Law
Dalton’s Law of Partial Pressure
Avogadro’s Law
Density of Gases
Copyright 2012 John Wiley & Sons, Inc
Topics for Today
 The Ideal Gas Law
 Gas Stoichiometry
 Real Gases & Air Pollutants
IDEAL GAS LAW
L  atm
PV  nRT where R = 0.0821
mol  K
P
V
n
T
is in units of Atmospheres
is in units of Liters
is in units of moles
is in units of Kelvin
L  atm
PV  nRT where R = 0.0821
mol  K
Example 1: Calculate the volume of 1 mole of
any gas at STP.
Knowns
Set-Up
Calculate
n = 1 mole T = 273K P = 1 atm
nRT
V 
P
L  atm
(1 mol)(0.0821
)(273 K)
mol  K
V =
= 22.4 L
(1 atm)
Molar volume!
L  atm
PV  nRT where R = 0.0821
mol  K
Example 2: How many moles of Ar are
contained in 1.3L at 24°C and 745 mm Hg?
Knowns
V = 1.3 L T = 24°C = 297 K
P = 745 mm Hg = 0.980 atm
Set-Up
PV
n 
RT
Calculate
n=
(0.980 atm)(1.3 L)
=0.052 mol
L  atm
(0.0821
)(297 K)
mol  K
Example 3: Calculate the molar mass (M) of an
unknown gas, if 4.12 g occupy a volume of 943mL
at 23°C and 751 torr.
Knowns
Set-Up
m =4.12 g
T = 23°C = 296 K
g
n=
M
so
V = 943 mL = 0.943 L
P = 751 torr = 0.988 atm
g
PV = RT
M
gRT
M=
PV
L  atm
(4.12 g)(0.0821
)(296 K)
mol  K
Calculate M =
=107 g/mol
(0.988 atm)(0.943 L)
Your Turn!
What is the molar mass of a gas if 40.0 L of the
gas has a mass of 36.0 g at 740. torr and 30.0 °
C?
a. 33.1 g
b. 23.0 g
c. 56.0 g
d. 333 g
STOICHIOMETRY & GASES
•Convert between moles and volume using the Molar
Volume if the conditions are at STP : 1 mol = 22.4 L.
•Use the Ideal Gas Law if the conditions are not at STP.
Example 4: Calculate the number of moles of
phosphorus needed to react with 4.0L of
hydrogen gas at 273 K and 1 atm.
P4(s) + 6H2(g)  4PH3(g)
Knowns
V = 4.0 L
Solution Map
Calculate
T = 273 K P = 1 atm
L H2  mol H2  mol P4
 1 mol H2   1 mol P4 
4.0 L H2 
 
 = 0.030 mol P4
 22.4L   6 mol H2 
Example 5: What volume of oxygen at 760 torr and
25°C are needed to react completely with 3.2 g C2H6?
2 C2H6(g) + 7 O2(g)  4 CO2(g) + 6 H2O(l)
Knowns
m = 3.2 g C2H6
Solution Map
Calculate
T = 25°C = 298K P = 1 atm
m C2H6  mol C2H6  mol O2  volume O2



3.2g C2 H 6 1 mol C2 H6   7 mol O2  = 0.37mol O2
 30.08g C2 H6   2 mol C2 H 6 
L  atm
(0.37 mol)(0.0821
)(298 K)
mol  K
V =
= 9.1 L
(1 atm)
Your Turn!
How many moles of oxygen gas are used up during
the reaction with 18.0 L of CH4 gas measured at
STP?
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(l)
a.
b.
c.
d.
1.61 moles
2.49 moles
18.0 moles
36.0 moles
Example 6: Calculate the volume of nitrogen
needed to react with 9.0L of hydrogen gas at
450K and 5.00 atm.
N2(g) + 3H2(g)  2NH3(g)
Knowns
Solution Map
Calculate
V = 9.0 L
T = 450K P = 5.00 atm
Assume T and P for both gases are the same.
Use volume ratio instead of mole ratio!
L H2  L N2
9.0 L H2
 1 L N2 


 3 L H2 
= 3.0 L N2
Your Turn!
What volume of sulfur dioxide gas will react when
12.0 L of oxygen is consumed at constant
temperature and pressure?
2 SO2 + O2  2 SO3
a.
b.
c.
d.
6.00 L
12.0 L
24.0 L
60.0 L
REAL GASES & AIR
POLLUTANTS
• Most real gases behave like ideal gases under
ordinary temperature and pressure conditions.
• Conditions where real gases don’t behave
ideally:
• At high P because the distance between particles is
too small and the molecules are too crowded
together.
• At low T because gas molecules begin to attract
each other.
• High P and low T are used to condense gases.
•
•
•
•
•
Stratospheric Ozone
Tropospheric Ozone
Oxides of Nitrogen
Acid Rain
Greenhouse Gases
Topics for Today
 The Ideal Gas Law
 Gas Stoichiometry
 Real Gases & Air Pollutants