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.
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
Gas Stoichiometry


In the previous unit, we used relationships
between amounts (moles) and masses
(grams) of reactants and products to solve
stoichiometry problems.
When the reactants and/or products are
gases, we can also use the relationships
between amounts (moles, n) and volume
(liters, V) to solve such problems.
Problem

Calculate the volume of O2 (in liters)
required for the complete
combustion of 7.64 L of acetylene
(C2H2) measured at the same
temperature and pressure.
2 C2H2(g) + 5 O2(g) → 4 CO2(g) + 2 H2O(l)
Problem
According to Avogadro’s law, at the
same T and P, the number of moles
of gases are directly related to their
volumes.
2 C2H2(g) + 5 O2(g) → 4 CO2(g) + 2 H2O(l)
Since 2 moles C2H2 = 5 moles O2,
therefore 2 L C2H2 = 5 L O2

Problem

Assuming no change in temperature
and pressure, calculate the volume
of O2 (in liters) required for the
complete combustion of 14.9 L of
butane C4H10.
2 C4H10(g) + 13 O2(g) → 8 CO2(g) + 10 H2O(l)
Problem

Calculate the volume of N2(g) created
from 60.0 g NaN3(s) at STP.
2 NaN3(s) → 2 Na(s) + 3 N2(g)

Strategy: How is this problem
different from the previous
combustion examples?
Problem

Calculate the volume of N2(g) created
from 60.0 g NaN3(s) at STP.
2 NaN3(s) → 2 Na(s) + 3 N2(g)
g
NaN3
mol
NaN3
mol
N2
L
N2
Pressure
 Pressure
is the
force applied by
molecules per
unit of area.
 Air pressure is
measured using
a barometer.
Standard Atmospheric Pressure
 Air
pressure at
higher altitudes,
such as on a
mountaintop, is
slightly lower
than air pressure
at sea level.
Standard Pressure (Units)
 One
atmosphere
is equal to
760 mm Hg,
760 torr, or
101.3 kPa
(kilopascals).
Pressure
 Perform
the following pressure
conversions.
a) 144 kPa = _____ atm
(1.42)
b) 795 mm Hg = _____ atm
(1.05)
Pressure
 Perform
the following pressure
conversions.
c) 669 torr = ______ kPa
(89.2)
d) 1.05 atm = ______ mm Hg
(798)
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.
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!
Temperature
 Temperature
is a measure of the
average kinetic energy of the
particles in a sample of matter.
 Temperature must be in Kelvin (K)
for the IDG Law.
 K = °C + 273
Example
 How
many moles of air are there in a
2.0 L bottle at 19 ºC and 0.983 atm?
 Choose
the value of R based on the
pressure unit.
 Since
atm are use, R = 0.0821.
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 0.983 atm?
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 0.983 atm?
292 K
0.983
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
V L) = n (0.0821
atm·L/mol·K) T
(300 K)
P (4.3
R
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
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
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 4.7 atm. 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
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.
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.
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
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.
729.35 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