The
Gas
Laws
The Gas Laws
 The
gas laws describe HOW gases
behave.
 They can be predicted by theory.
 The amount of change can be
calculated with mathematical
equations.
Standard Atmospheric Pressure
 One
atmosphere
is equal to
760 mm Hg,
760 torr, or
101.3 kPa
(kilopascals).
Standard Atmospheric Pressure
 Perform
the following pressure
conversions.
a) 144 kPa = _____ atm
(1.42)
b) 795 mm Hg = _____ atm
(1.05)
Standard Atmospheric Pressure
 Perform
the following pressure
conversions.
c) 669 torr = ______ kPa
(89.2)
d) 1.05 atm = ______ mm Hg
(798)
Standard Atmospheric Pressure
 Air
pressure at
higher altitudes,
such as on a
mountaintop, is
slightly lower
than air pressure
at sea level.
Standard Atmospheric Pressure
 Air
pressure is
measured using
a barometer.
Pressure and the
Number of Molecules
 More
molecules mean more collisions
between the gas molecules
themselves and more collisions
between the gas molecules and the
walls of the container.
 Number of molecules is DIRECTLY
proportional to pressure.
Pressure and the
Number of Molecules
 Doubling
the
number of gas
particles in a
basketball
doubles the
pressure.
Pressure and the
Number of Molecules
 Gases
naturally move from areas of
high pressure to low pressure
because there is empty space to
move in.
 If
you double the number of
molecules,
1 atm
 If
you double the number of
molecules, you double the
pressure.
2 atm
4 atm
 As
you remove
molecules from a
container,
2 atm
 As
you remove
molecules from a
container, the pressure
decreases.
1 atm
 As
you remove
molecules from a
container, the pressure
decreases until the
pressure inside equals
the pressure outside.
Changing the Size (Volume)
of the Container
 In
a smaller container, molecules
have less room to move.
 The molecules hit the sides of the
container more often, striking a
smaller area with the same force.
Changing the Size (Volume)
of the Container
 As
volume decreases, pressure
increases.
 Volume and pressure are INVERSELY
proportional.
1 atm
 As
the
pressure on
a gas
increases,
4 Liters
 As
the
pressure on
a gas
increases,
the volume
decreases.
2 atm
2 Liters
Temperature and Pressure
 Raising
the temperature of a gas
increases the pressure if the
volume is held constant.
 At higher temperatures, the
particles in a gas have greater
kinetic energy.
Temperature and Pressure
 They
move faster and collide with
the walls of the container more
often and with greater force, so the
pressure rises.
300 K
 If
you start with 1 liter of gas at 1 atm
pressure and 300 K and heat it to
600 K, one of 2 things happens.
600 K
300 K
 Either
the volume will
increase to 2 liters at
1 atm,
300 K
• or the pressure will
increase to 2 atm
while the volume
remains constant.
600 K
Ideal Gases
 In
this unit we will assume the gases
behave ideally.
 Ideal gases do not really exist, but
this makes the math easier and is a
close approximation.
Kinetic Molecular Theory
of Gases
 Gas
particles are much smaller than
the spaces between them. The
particles have negligible volume.
 There are no attractive or repulsive
forces between gas molecules.
Kinetic Molecular Theory of Gases
• Gas particles are in constant, random
motion. Until they bump into
something (another particle or the
side of a container), particles move in
a straight line.
Kinetic Molecular Theory
of Gases
• No kinetic energy is lost when gas
particles collide with each other or
with the walls of their container.
• All gases have the same kinetic
energy at a given temperature.
Temperature
 Temperature
is a measure of the
average kinetic energy of the
particles in a sample of matter.
Ideal Gases
 There
are no gases for which this is
true.
 Real gases behave more ideally at
high temperature and low pressure.
Ideal Gases
 At
low temperature, the gas molecules
move more slowly, so attractive forces are
no longer negligible.
 As the pressure on a gas increases, the
molecules are forced closer together and
attractive forces are no longer negligible.
 Therefore, real gases behave more ideally
at high temperature and low pressure.
Avogadro’s Law
 Avogadro’s
law states that equal
volumes of different gases (at the
same temperature and pressure)
contain equal numbers of atoms or
molecules.
Avogadro’s Law
2 Liters
of
Helium
2 Liters
of
Oxygen
 has
the same
number of
particles as ..
Avogadro’s Law
 The
molar volume for a gas is the
volume that one mole occupies at
0.00ºC and 1.00 atm.
 1 mole = 22.4 L at STP (standard
temperature and pressure).
 As a result, the volume of gaseous
reactants and products can be
expressed as small whole numbers in
reactions.
Problem

How many moles are in 45.0 L of a
gas at STP?
2.01 moles
Problem

How many liters are in 0.636 moles of
a gas at STP?
14.2 L
Avogadro’s Law
=Kxn
V / n = K
V
 Easier
to use:
(K is some constant)
V1
n1
=
V2
n2
Example
 Consider
two samples of nitrogen gas.
Sample 1 contains 1.5 mol and has a
volume of 36.7 L. Sample 2 has a volume
of 16.5 L at the same temperature and
pressure. Calculate the number of moles
of nitrogen in sample 2.
Example
 Sample
1 contains 1.5 mol and has a
volume of 36.7 L. Sample 2 has a volume
of 16.5 L. Calculate the number of moles
of nitrogen in sample 2.
36.7
V1 L
n1mol
1.5
=
16.5
V2L
n2
n2 = 0.67 mol
Problem
 If
0.214 mol of argon gas occupies a
volume of 652 mL at a particular
temperature and pressure, what
volume would 0.375 mol of argon
occupy under the same conditions?
V2 = 1140 mL
Problem
 If
46.2 g of oxygen gas occupies a
volume of 100. L at a particular
temperature and pressure, what
volume would 5.00 g of oxygen gas
occupy under the same conditions?
V2 = 10.8 L
Boyle’s Law
Boyle’s law states that the
pressure and volume
of a gas at constant
temperature are
inversely proportional.
 Inversely proportional
means as one goes
up the other goes
down.
 At
Boyle’s Law
Boyle’s Law
P
xV=K
P 1
(K is some constant)
V1 = P 2 V2
Boyle’s Law
P-V graph for Boyle’s law
results in a hyperbola because
pressure and volume are
inversely proportional.
 The
P
V
Example
A
balloon is filled with 25 L of air at
1.0 atm pressure. If the pressure is
changed to 1.5 atm, what is the new
volume?
Example
 First,
make sure the pressure and
volume units in the question match.
A
balloon is filled with 25 L of air at
1.0
atm pressure. If the pressure is changed
to 1.5 atm, what is the new volume?
THEY DO!
Example
A
balloon is filled with 25 L of air at
1.0 atm pressure. If the pressure is
changed to 1.5 atm, what is the new
volume?
P1 (25
V1L)
1.0 atm
= 1.5
P2 atm
V2 V2
V2 = 17 L
Problem
A
balloon is filled with 73 L of air at
1.3 atm pressure. What pressure is
needed to change the volume to
43 L?
P2 = 2.2 atm
Problem
gas is collected in a 242 cm3
container. The pressure of the gas in
the container is measured and
determined to be 87.6 kPa. What is
the volume of this gas at standard
pressure?
A
V2 = 209 cm3
Problem
A
gas is collected in a 24.2 L
container. The pressure of the gas in
the container is determined to be
756 mm Hg. What is the pressure of
this gas if the volume increases to
30.0 L?
P2 = 610. mm Hg
Charles’ Law
 The
volume of a gas is directly
proportional to the Kelvin
temperature if the pressure is
held constant.
K
= °C + 273
Charles’ Law
Charles’ Law
=KxT
V / T = K
V
 Easier
to use:
(K is some constant)
V1
T1
=
V2
T2
Charles’ Law
V-T graph for Charles’ law
results in a straight line because
pressure and volume are directly
proportional.
 The
V
T
Example
 What
is the temperature of a gas that
is expanded from 2.5 L at 25 ºC to
4.1 L at constant pressure?
Example
 First,
make sure the volume units in
the question match.
 What
is the temperature of a gas that is
expanded from 2.5 L at 25 ºC to 4.1 L at
constant pressure?
THEY DO!
Example
 Second,
make sure to convert
degrees Celsius to Kelvin.
 What
is the temperature of a gas that is
expanded from 2.5 L at 25 ºC to 4.1 L at
constant pressure?
K = °C
25 + 273
K = 298 K
Example
 What
is the temperature of a gas that is
expanded from 2.5 L at 25 ºC to 4.1 L at
constant pressure?
V1L
2.5
298
TK
1
=
V2L
4.1
T2
T2 = 489 K
Problem
 What
is the final volume of a gas that
starts at 8.3 L and 17 ºC and is
heated to 96 ºC?
V2 = 11 L
Problem
225 cm3 volume of gas is collected
at 57 ºC. What volume would this
sample of gas occupy at standard
temperature?
A
V2 = 186 cm3
Problem
225 cm3 volume of gas is collected
at 42 ºC. If the volume is decreased
to 115 cm3, what is the new
temperature?
A
T2 = 161 K
Gay-Lussac’s Law
 The
temperature and the pressure
of a gas are directly related at
constant volume.
Gay-Lussac’s Law
=KxT
P / T = K
P
 Easier
to use:
(K is some constant)
P1
T1
=
P2
T2
P
T
Example
 What
is the pressure inside a 0.250 L
can of deodorant that starts at 25 ºC
and 1.2 atm if the temperature is
raised to 100 ºC? Volume remains
constant.
Example
 First,
make sure the pressure units
in the question match.
 What
is the pressure inside a 0.250 L can
of deodorant that starts at 25 ºC and
1.2 atm if the temperature is raised to
100 ºC?
There is only one
pressure unit!
Example
 Second,
make sure to convert
degrees Celsius to Kelvin.
 What
is the pressure inside a 0.250 L can
of deodorant that starts at 25 ºC and
1.2 atm if the temperature is raised to
100 ºC?
K = °C
25 + 273
K = 298 K
Example
 What
is the pressure inside a 0.250 L can
of deodorant that starts at 25 ºC and
1.2 atm if the temperature is raised to
100 ºC?
K = 100
°C + 273
K = 373 K
Example
 What
is the pressure inside a 0.250 L can
of deodorant that starts at 25 ºC and
1.2 atm if the temperature is raised to
100 ºC?
1.2Patm
1
298
TK
1
=
P2
373
TK
2
P2 = 1.5 atm
Problem
A
can of deodorant starts at 43 ºC
and 1.2 atm. If the volume remains
constant, at what temperature will
the can have a pressure of 2.2 atm?
T2 = 579 K
Problem
A
can of shaving cream starts at 25
ºC and 1.30 atm. If the temperature
increases to 37 ºC and the volume
remains constant, what is the
pressure of the can?
P2 = 1.35 atm
Problem
A
12 ounce can of a soft drink starts
at STP. If the volume stays constant,
at what temperature will the can have
a pressure of 2.20 atm?
T2 = 601 K
The Combined Gas Law
 The
gas laws may be combined into
a single law, called the combined
gas law, which relates two sets of
conditions of pressure, volume, and
temperature by the following
equation.

P1 V1
T1
=
P2 V2
T2
Example
A
15 L cylinder of gas at 4.8 atm
pressure at 25 ºC is heated to 75 ºC
and compressed to 17 atm. What is
the new volume?
Example
 First,
make sure the volume units in
the question match.
A
15 L cylinder of gas at 4.8 atm pressure
at 25 ºC is heated to 75 ºC and
compressed to 17 atm. What is the new
volume?
There is only one
volume unit!
Example
 Second,
make sure the pressure
units in the question match.
A
15 L cylinder of gas at 4.8 atm pressure
at 25 ºC is heated to 75 ºC and
compressed to 17 atm. What is the new
volume?
They do!
Example
 Third,
make sure to convert degrees
Celsius to Kelvin.
A
15 L cylinder of gas at 4.8 atm pressure
at 25 ºC is heated to 75 ºC and
compressed to 17 atm. What is the new
volume?
K = °C
25 + 273
K = 298 K
Example
A
15 L cylinder of gas at 4.8 atm pressure
at 25 ºC is heated to 75 ºC and
compressed to 17 atm. What is the new
volume?
K = °C
75 + 273
K = 348 K
Example
A
15 L cylinder of gas at 4.8 atm pressure
at 25 ºC is heated to 75 ºC and
compressed to 17 atm. What is the new
volume?
V1L)
4.8P
atm
1 (15
T1K
298
=
17P
atm
2
V2
T2K
348
V2 = 4.9 L
Problem
 If
6.2 L of gas at 723 mm Hg at 21 ºC
is compressed to 2.2 L at
4117 mm Hg, what is the temperature
of the gas?
T2 = 594 K
Problem
• A sample of nitrogen monoxide has a
volume of 72.6 mL at a temperature
of 16 °C and a pressure of 104.1 kPa.
What volume will the sample occupy
at 24 °C and 99.3 kPa?
V2 = 78.2 mL
Problem
• A hot air balloon rises to an altitude of
7000 m. At that height the atmospheric
pressure drops to 300. mm Hg and the
temperature cools to - 33 °C. Suppose on
the hot air balloon there was a small
balloon filled to 1.00 L at sea level and a
temperature of 27 °C. What would its
volume ultimately be when it reached the
height of 7000 m?
V2 = 2.03 L
Daltons’ Law of Partial Pressures
 Dalton’s
law of partial pressures
states that the total pressure of a
mixture of gases is equal to the sum
of the pressures of all the gases in
the mixture, as shown below.
PTotal = P1 + P2 + P3 +
 The partial pressure is the
contribution by that gas.
…
Example
 On
the next slide, determine the
pressure in the fourth container if all
of the gas molecules from the 1st
three containers are placed in the 4th
container.
2 atm
1 atm
3 atm
??
6 atm
Problem
 What
is the total pressure in a
balloon filled with air if the pressure
of the oxygen is 170 mm Hg and the
pressure of nitrogen is 620 mm Hg?
790 mm Hg
Example
 In
a second balloon the total
pressure is 1.30 atm. What is the
pressure of oxygen (in mm Hg) if the
pressure of nitrogen is 720. mm Hg?
Example
 The
two gas units do not match. We
must convert the 1.30 atm into
mm Hg.
1.30 atm 760 mm Hg
1 atm
= 988 mm Hg
Example
PTotal
= P1 + P2 + P3 + …
988 mm Hg = 720 mm Hg + Poxygen
268 mm Hg = Poxygen
Problem
A
container has a total pressure of
846 torr and contains carbon
dioxide gas and nitrogen gas. What
is the pressure of carbon dioxide (in
kPa) if the pressure of nitrogen is
50. kPa?
63 kPa
Problem
 When
a container is filled with
3 moles of H2, 2 moles of O2 and
4 moles of N2, the pressure in the
container is 8.7 atm. The partial
pressure of H2 is _____.
2.9 atm
Daltons’ Law of Partial Pressures
 It
is common to synthesize gases
and collect them by displacing a
volume of water.
Problem
 Hydrogen
was collected over water at
21°C on a day when the atmospheric
pressure is 748 torr. The volume of the
gas sample collected was 300. mL. The
vapor pressure of water at 21°C is
18.65 torr. Determine the partial pressure
of the dry gas.
739.25 torr
Problem
A
sample of oxygen gas is
saturated with water vapor at 27ºC.
The total pressure of the mixture is
772 mm Hg and the vapor pressure
of water is 26.7 mm Hg at 27ºC.
What is the partial pressure of the
oxygen gas?
745.3 mm Hg
Remember Ideal Gases Don’t Exist
 Molecules
do take up space.
 There are attractive forces;
otherwise, there would be no liquids.
The Ideal Gas Law
P V = n R T
 Pressure times volume equals the
number of moles (n) times the ideal
gas constant (R) times the
temperature in Kelvin.
The Ideal Gas Law
R
= 0.0821 (L atm)/(mol K)
 R = 8.314 (L kPa)/(mol K)
 R = 62.4 (L mm Hg)/(mol K)
 The one you choose depends on the
unit for pressure!
Example
 How
many moles of air are there in a
2.0 L bottle at 19 ºC and 747 mm Hg?
 Choose
the value of R based on the
pressure unit.
 Since
mm Hg are use, R = 62.4.
Example
 Second,
make sure to convert
degrees Celsius to Kelvin.
 How
many moles of air are there in a
2.0 L bottle at 19 ºC and 747 mm Hg?
K = °C
19 + 273
K = 292 K
Example
 How
many moles of air are there in a
2.0 L bottle at 19 ºC and 747 mm Hg?
292 K
747
P (2.0)
V
= n 62.4
R (292)
T
n = 0.082 mol
Example
 What
is the pressure in atm exerted
by 1.8 g of H2 gas in a 4.3 L balloon at
27 ºC?
 Choose
the value of R based on the
pressure unit.
 Since
atm is requested, R = 0.0821.
Example
 Second,
make sure to convert
degrees Celsius to Kelvin.
 What
is the pressure in atm exerted by
1.8 g of H2 gas in a 4.3 L balloon at
27 ºC?
K = °C
27 + 273
K = 300. K
Example
 What
is the pressure in atm exerted by
1.8 g of H2 gas in a 4.3 L balloon at 27 ºC?
300. K
(300.)
P (4.3)
V = 0.90
n (0.0821)
R
T
P = 5.2 atm
Example
 Next,
convert grams to moles.
 What is the pressure in atm exerted by
1.8 g of H2 gas in a 4.3 L balloon at 300. K?
1.8 g H2
1
__ mol H2
2.0 g H2
__
= 0.90 mol H2
Example
 What
is the pressure in atm exerted by
0.90 mol 1.8 g of H2 gas in a 4.3 L balloon at
27 ºC?
300. K
(300.)
P (4.3)
V = 0.90
n (0.0821)
R
T
P = 5.2 atm
Problem
 Sulfur
hexafluoride (SF6) is a
colorless, odorless and very
unreactive gas. Calculate the
pressure (in atm) exerted by
1.82 moles of the gas in a steel vessel
of volume 5.43 L at 69.5 ºC.
P = 9.42 atm
Problem
 Calculate
the volume (in liters)
occupied by 7.40 g of CO2 at STP.
V = 3.77 L
Example
 Next,
you will have to change grams
to moles.
 What
is the pressure in atm exerted by
1.8 g of H2 gas in a 4.3 L balloon at
27 ºC?
1.8 g
1 mol
2.0 g
= 0.90 mol
Problem
A
sample of nitrogen gas kept in a
container of volume 2.30 L and at a
temperature of 32 ºC exerts a
pressure of 476 kPa. Calculate the
number of moles of gas present.
n = 0.432 mol
Problem
A
1.30 L sample of a gas has a mass
of 1.82 g at STP. What is the molar
mass of the gas?
31.4 g/mol
Problem
 Calculate
the mass of nitrogen gas
that can occupy 1.00 L at STP.
28.0 g