Gas Laws: Introduction
At the conclusion of our time together,
you should be able to:
1. List 5 properties of gases
2. Identify the various parts of the kinetic
molecular theory
3. Define pressure
4. Convert pressure into 3 different units
5. Define temperature
6. Convert a temperature to Kelvin
Importance of Gases
• Airbags fill with N2 gas in an
accident.
• Gas is generated by the
decomposition of sodium
azide, NaN3.
• 2 NaN3 --->
2 Na + 3 N2
THREE STATES OF MATTER
General Properties of Gases
• There is a lot of “free” space in a
gas.
• Gases can be expanded infinitely.
• Gases fill containers uniformly and
completely.
• Gases diffuse and mix rapidly.
To Review
Gases expand to fill their containers
Gases are fluid – they flow
Gases have low density
1/1000 the density of the equivalent
liquid or solid
Gases are compressible
Gases effuse and diffuse
Properties of Gases
Gas properties can be modeled using
math. This model depends on —
• V = volume of the gas (L)
• T = temperature (K)
– ALL temperatures in the entire
unit MUST be in Kelvin!!! No
Exceptions!
• n = amount (moles)
• P = pressure
(atmospheres)
Ideal Gases
Ideal gases are imaginary gases that perfectly
fit all of the assumptions of the kinetic
molecular theory.
 Gases consist of tiny particles that are
far apart relative to their size.
 Collisions between gas particles and
between particles and the walls of the
container are elastic collisions
 No kinetic energy is lost in elastic
collisions
Ideal Gases
(continued)
 Gas particles are in constant, rapid motion.
They therefore possess kinetic energy, the
energy of motion
 There are no forces of attraction between
gas particles
 The average kinetic energy of gas particles
depends on temperature, not on the identity
of the particle.
Pressure
Is caused by the collisions of molecules with the
walls of a container
Is equal to force/unit area
SI units = Newton/meter2 = 1 Pascal (Pa)
1 atmosphere = 101,325 Pa
1 atmosphere = 1 atm = 760 mm Hg = 760 torr
1 atm = 29.92 in Hg = 14.7 psi = 0.987 bar =
10 m column of water.
Measuring Pressure
The first device for
measuring atmospheric
pressure was developed
by Evangelista Torricelli
during the 17th century.
The device was called a
“barometer”
Baro = weight
Meter = measure
An Early Barometer
The normal
pressure due to
the atmosphere at
sea level can
support a column
of mercury that is
760 mm high.
Pressure
Column height measures
Pressure of atmosphere
• 1 standard atmosphere (atm)
*
= 760 mm Hg (or torr) *
= 29.92 inches Hg *
= 14.7 pounds/in2 (psi)
= 101.325 kPa
(SI unit is PASCAL)
= about 34 feet of water!
And now, we pause for this commercial
message from STP
OK, so it’s really not THIS
kind of STP…
STP in chemistry stands for
Standard Temperature and
Pressure
Standard Pressure
= 1 atm (or an
equivalent)
Standard Temperature
= 0 deg C (273 K)
STP allows us to compare
amounts of gases between
different pressures and
temperatures
Let’s Review:
Standard Temperature and Pressure
“STP”
P = 1 atmosphere, 760 torr
T = 0C, 273 Kelvins
The molar volume of an ideal gas is
22.42 liters at STP
Pressure Conversions
A. What is 475 mm Hg expressed in atm?
475 mm Hg x
1 atm
760 mm Hg
= 0.625 atm
B. The pressure of a tire is measured as 29.4 psi.
What is this pressure in mm Hg?
29.4 psi x
760 mm Hg
14.7 psi
= 1.52 x 103 mm Hg
Pressure Conversions – Your Turn
A. What is 2.00 atm expressed in torr?
2.00 atm x
760 torr
1 atm
= 1520 torr
B. The pressure of a tire is measured as 32.0 psi.
What is this pressure in kPa?
32.0 psi x
101.325 kPa
14.7 psi
= 221 kPa
Converting Celsius to Kelvin
Gas law problems involving temperature
require that the temperature be in
KELVINS!
Kelvins = C + 273
°C = Kelvins - 273
Gas Laws: Boyle’s and
Charles’ Law
At the conclusion of our time together,
you should be able to:
1. Describe Boyle’s Law with a formula.
2. Use Boyle’s Law to determine either a
pressure or volume
3. Describe Charles’ Law with a formula.
4. Use Charles’ Law to determine either a
temperature or volume
Boyle’s Law
P α 1/V
This means Pressure and
Volume are INVERSELY
PROPORTIONAL if moles and
temperature are constant
(do not change). For
example, P goes up as V goes
down.
PV

PV
1 1
2 2
Robert Boyle
(1627-1691).
Son of Earl of
Cork, Ireland.
Boyle’s Law Summary
Pressure is inversely proportional to volume
when temperature is held constant.
PV

PV
1 1
2 2
Boyle’s Law Example
A bicycle pump is a good
example of Boyle’s law.
As the volume of the air
trapped in the pump is
reduced, its pressure
goes up, and air is forced
into the tire.
Example
P1
V1
(0.947 atm) (0.15 L)
P2
V2
(0.987 atm) V2
0.14 L
or 140 mL
Boyle’s Law Practice #1
P1
V1
(1 atm) (0.5 L)
P2
V2
(0.5 atm)
1L
V2
Charles’ Law
If n and P are constant,
then V α T
V and T are directly
proportional.
V1 V2

T1 T2
• If one temperature goes
up, the volume goes up!
Jacques Charles
(1746-1823). Isolated
boron and studied
gases. Balloonist.
Charles’s Original Balloon
Modern LongDistance Balloon
Charles’ Law Summary
The volume of a gas is directly
proportional to temperature, and
extrapolates to zero at zero Kelvin.
(P = constant)
V1 V2

T1 T2
Charles’ Law
Example
V1
V2
T1
T2
0.075 L
V2
298 K
323 K
0.081 L
Charles’ Law Practice #1
V1
V2
T1
T2
2.75 L
2.46 L
293.0 K
T2
262 K or
-10.9 oC
Gas Laws: Gay-Lussac’s
and Combined
At the conclusion of our time together,
you should be able to:
1. Describe Gay-Lussac’s Law with a formula
2. Use Gay-Lussac’s Law to determine either
a temperature or volume
3. Combine all three laws into the Combined
Gas Law
4. Use the Combined Gas Law to determine
either temperature, volume or pressure
Gay-Lussac’s Law
If n and V are constant,
then P α T
P and T are directly
proportional.
P1 P2

T1 T2
• If one temperature goes
up, the pressure goes up!
Joseph Louis GayLussac (1778-1850)
Gay Lussac’s Law Summary
The pressure and temperature of a gas are
directly related, provided that the volume
remains constant.
P1 P2

T1 T2
Example
P1
P2
T1
T2
3.00 atm
298 K
P2
325 K
3.3 atm
Gay-Lussac’s Law Practice #1
P1
P2
T1
T2
1.8 atm
1.9 atm
293 K
T2
310 K or
36 oC
Now let’s put all 3 laws
together into one big
law…….
Combined Gas Law
• The good news is that you don’t have to
remember all three gas laws! Since they
are all related to each other, we can
combine them into a single equation.
BE SURE YOU KNOW THIS EQUATION!
PV
PV
1 1
2 2

T1
T2
No, it’s not related to R2D2
The Combined Gas Law
The combined gas law expresses the
relationship between pressure, volume and
temperature of a fixed amount of gas.
PV
PV
1 1
2 2

T1
T2
Boyle’s law, Gay-Lussac’s law, and Charles’
law are all derived from this by holding a
variable constant.
Combined Gas Law
If you should only need one of the other gas laws,
you can cover up the item that is constant and
you will get that gas law!
P1 V1=
T1
P2 V2
T2
Boyle’s Law
Charles’ Law
Gay-Lussac’s
Law
Combined Gas Law Problem
A sample of helium gas has a volume of 0.180 L, a
pressure of 0.800 atm and a temperature of 29°C. What
is the new temperature(°C) of the gas at a volume of
90.0 mL and a pressure of 3.20 atm?
Set up Data Table
P1 = 0.800 atm
V1 = 180 mL
T1 = 302 K
P2 = 3.20 atm
V2 = 90 mL
T2 = ??
Calculation
• P1 = 0.800 atm
V1 = 180 mL
T1 = 302 K
• P2 = 3.20 atm
V2 = 90 mL
T2 = ??
P1 V1
T1
=
P2 V2
T2
P1 V1 T2 = P2 V2 T1
T2 = 3.20 atm x 90.0 mL x 302 K
0.800 atm x 180.0 mL
T2 = 604 K - 273 = 331 °C
= 604 K
Gas Laws: Avogadro’s and
Ideal
At the conclusion of our time together,
you should be able to:
1. Describe Avogadro’s Law with a formula.
2. Use Avogadro’s Law to determine either
moles or volume
3. Describe the Ideal Gas Law with a formula.
4. Use the Ideal Gas Law to determine either
moles, pressure, temperature or volume
5. Explain the Kinetic Molecular Theory
Avogadro’s Law
Equal volumes of gases at the same T and P have
the same number of molecules.
V = an
V and n are directly related.
twice as many
molecules
Avogadro’s Law Summary
For a gas at constant temperature and
pressure, the volume is directly
proportional to the number of moles of
gas (at low pressures).
V = an
a = proportionality constant
V = volume of the gas
n = number of moles of gas
Standard Molar Volume
Equal volumes of all
gases at the same
temperature and
pressure contain the
same number of
molecules.
- Amedeo
Avogadro
Avogadro’s Law Practice #1
V1
V2
n1
n2
4.00 L
7.12 L
0.21 mol
n2
0.37 mol total
0.16 mol added
IDEAL GAS LAW
PV=nRT
Brings together gas
properties.
Can be derived from
experiment and theory.
BE SURE YOU KNOW THIS
EQUATION!
Ideal Gas Law
PV = nRT
P = pressure in atm
V = volume in liters
n = moles
R = proportionality constant
= 0.08206 L atm/ mol·K
T = temperature in Kelvins
Holds closely at P < 1 atm
Review of
Kinetic Molecular Theory
Particles of matter are ALWAYS in motion
Volume of individual particles is  zero.
Collisions of particles with container walls
cause pressure exerted by gas.
Particles exert no forces on each other.
Average kinetic energy  Kelvin
temperature of a gas.
Deviations from Ideal Gas Law
• Real molecules have volume.
The ideal gas consumes the entire amount
of available volume. It does not
account for the volume of the
molecules themselves.
• There are intermolecular forces.
An ideal gas assumes there are no
attractions between molecules.
Attractions slow down the molecules
and reduce the amount of collisions.
– Otherwise a gas could not condense
to become a liquid.
R is a constant, called the Ideal Gas Constant
Instead of learning a different value for R for all the
possible unit combinations, we can just memorize
one value and convert the units to match R.
R = 0.08206
L • atm
mol • K
Using PV = nRT
How much N2 is required to fill a small room with a
volume of 960 cubic feet (27,000 L) to 745 mm Hg at
25 oC?
Solution
1. Get all data into proper units
V = 27,000 L
T = 25 oC + 273 = 298 K
P = 745 mm Hg (1 atm/760 mm Hg)
= 0.98 atm
And we always know R, 0.08206 L atm / mol K
Using PV = nRT
How much N2 is required to fill a small room with a
volume of 960 cubic feet
(27,000 L) to P = 745 mm Hg at 25 oC?
Solution
2. Now plug in those values and solve for the
unknown.
PV = nRT
RT
RT
(0.98 atm)(2.7 x 104 L)
n =
(0.0821 L• atm/K • mol)(298 K)
n = 1.1 x 103 mol (or about 30 kg of gas)
Ideal Gas Law Problems #1
(5.6 atm) (12 L)
(4 mol) (0.08206 atm*L/mol*K)(T)
200 K
Gas Laws: Dalton’s Law
At the conclusion of our time together,
you should be able to:
1. Explain Dalton’s Law
2. Use Dalton’s Law to solve a problem
Dalton’s Law
John Dalton
1766-1844
Dalton’s Law of Partial Pressures
For a mixture of gases in a container,
PTotal = P1 + P2 + P3 + . . .
This is particularly useful in calculating the
pressure of gases collected over water.
Dalton’s Law of Partial Pressures
2 H2O2 (l) ---> 2 H2O (g) + O2 (g)
0.32 atm
0.16 atm
What is the total pressure in the flask?
Ptotal in gas mixture = PA + PB + ...
Therefore,
Ptotal = PH2O + PO2 = 0.48 atm
Dalton’s Law:
total P is sum of PARTIAL pressures.
Gases in the Air
The % of gases in air
Partial pressure (STP)
78.08% N2
593.4 mm Hg
20.95% O2
159.2 mm Hg
0.94% Ar
7.1 mm Hg
0.03% CO2
0.2 mm Hg
PAIR = PN + PO + PAr + PCO = 760 mm Hg
2
2
2
Total Pressure = 760 mm Hg
Collecting a gas “over water”
• Gases, since they mix with other gases readily, must be collected in
an environment where mixing can not occur. The easiest way to do
this is under water because water displaces the air. So when a gas is
collected “over water”, that means the container is filled with water
and the gas is bubbled through the water into the container. Thus,
the pressure inside the container is from the gas AND the water
vapor. This is where Dalton’s Law of Partial Pressures becomes
useful.
Table of Vapor Pressures for Water
Solve This!
A student collects
some hydrogen
gas over water at
20 degrees C and
768 torr. What is
the pressure of the
H2 gas?
768 torr – 17.5 torr = 750.5 torr
Dalton’s Law of Partial Pressures
Also, for a mixture of gases in a container,
because P, V and n are directly
proportional if the other gas law variables
are kept constant:
nTotal = n1 + n2 + n3 + . . .
VTotal = V1 + V2 + V3 + . . .
This is useful in solving problems with differing
numbers of moles or volumes of the gases that
are mixed together.
Gas Laws: Gas
Stoichiometry
At the conclusion of our time together, you
should be able to:
1. Use the Ideal Gas Law to solve a gas
stoichiometry problem.
Gases and Stoichiometry
2 H2O2 (l) ---> 2 H2O (g) + O2 (g)
Decompose 1.1 g of H2O2 in a flask with a volume of
2.50 L. What is the volume of O2 at STP?
Bombardier beetle
uses decomposition of
hydrogen peroxide to
defend itself.
Gases and Stoichiometry
2 H2O2 (l) ---> 2 H2O (g) + O2 (g)
Decompose 1.1 g of H2O2 in a flask with a volume of
2.50 L. What is the volume of O2 at STP?
Solution
1.1 g H2O2 1 mol H2O2 1 mol O2 22.4 L O2
34 g H2O2 2 mol H2O2 1 mol O2
= 0.36 L O2 at STP
Gas Stoichiometry: Practice!
How many grams of He are present in 8.0 L of gas at STP?
8.0 L He
x
1 mol He
x
22.4 L He
= 1.4 g He
4.00 g He
1 mol He
Gas Stoichiometry Trick
If reactants and products are at the same
conditions of temperature and pressure, then
mole ratios of gases are also volume ratios.
3 H2(g)
3 moles H2
+
N2(g)
+ 1 mole N2


2NH3(g)
2 moles NH3
67.2 liters H2 + 22.4 liter N2  44.8 liters NH3
Gas Stoichiometry Trick Example
How many liters of ammonia can be produced
when 12 liters of hydrogen react with an
excess of nitrogen in a closed container at
constant temperature?
3 H2(g)
+
12 L H2
N2(g)
2 L NH3
3 L H2
2NH3(g)

=
8.0
L NH3
What if the problem is NOT at STP?
• 1. You will need to use PV = nRT
Gas Stoichiometry Example on HO
How many liters of oxygen gas, at 1.00 atm and
25 oC, can be collected from the complete
decomposition of 10.5 grams of potassium
chlorate?
2 KClO3(s)  2 KCl(s) + 3 O2(g)
10.5 g KClO3
1 mol KClO3
122.55 g KClO3
3 mol O2
2 mol KClO3
0.13 mol O2
Gas Stoichiometry Example on HO
How many liters of oxygen gas, at 1.00 atm and
25 oC, can be collected from the complete
decomposition of 10.5 grams of potassium
chlorate?
2 KClO3(s)  2 KCl(s) + 3 O2(g)
(1.0 atm)(V)
(0.13 mol) (0.08206 atm*L/mol*K) (298 K)
3.2 L O2
Gas Laws: Dalton, Density and Gas
Stoichiometry
At the conclusion of our time together,
you should be able to:
1. Explain Dalton’s Law and use it to solve a
problem.
2. Use the Ideal Gas Law to solve a gas
density problem.
3. Use the Ideal Gas Law to solve a gas
stoichiometry problem.
GAS DENSITY
Lower
density
22.4 L of ANY gas
AT STP = 1 mole
High
density
Gas Density
mass
molar mass
Density 

volume molar volume
… so at STP…
molar mass
Density 
22.4 L
Density and the Ideal Gas Law
Combining the formula for density with the
Ideal Gas law, substituting and rearranging
algebraically:
M = Molar Mass
MP
D
RT
P = Pressure
R = Gas Constant
T = Temperature in
Kelvins
Gas Stoichiometry #4
How many liters of oxygen gas, at 37.0C and
0.930 atmospheres, can be collected from the
complete decomposition of 50.0 grams of
potassium chlorate?
2 KClO3(s)  2 KCl(s) + 3 O2(g)
50.0 g KClO3
1 mol KClO3
122.55 g KClO3
nRT
V
P
3 mol O2
2 mol KClO3
= “n” mol O2
= 0.612 mol O2
L  atm
(0.612mol)(0.0821
)(310 K)
mol  K
= 16.7 L

0.930 atm
Try this one!
How many L of O2 are needed to react 28.0 g NH3 at
24°C and 0.950 atm?
4 NH3(g) + 5 O2(g)
4 NO(g) + 6 H2O(g)
Gas Laws: Finding “R”
At the conclusion of our time together, you
should be able to:
1. Use the Ideal Gas Law to find the value of
“R”.
See the problem on p. 50
Partial Pressure of CO2
Carbon Dioxide gas
over water at 17.0
degrees C and
97.932 kPa. What
is the pressure of
the CO2 gas?
1.90
kPa
97.932 kPa – 1.90 kPa = 96.032 kPa = 0.948 atm
Determine the Moles of CO2 Used:
Mass of canister before-mass of
canister after
17.099 g – 16.524 g= 0.575 g
0.575 g CO2
1 mol CO2
44.01 g CO2
0.0131 mol CO2
(0.948 atm) (0.347 L)
(0.0131 mol) (R) (290 K)
0.0868 atm*L/mol*K
0.08206 atm*L/mol*K
0.0868
atm*L/
mol*K
- 0.00474 atm*L/mol*K
- 0.0048
atm*L/
mol*K
(standard)
(experimental results)
(experimental error)
0.08206 atm*L/mol*K x 100
- 5.85 % error