Gas Laws
Properties of Gases
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Have mass
Have low densities
Easily compressed
Completely fill their container
Can move through each other
rapidly

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Diffusion: the movement of one
substance through another
Exert pressure
Gas Pressure

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Caused by collisions of the gas
molecules with the sides of their
container
Pressure depends on temperature

Directly proportional: will both go
up or go down simultaneously
Kinetic-Molecular Model

Gas particles are:
Constant, random, straight line
motion
 Space between gas particles is
large
 Negligible volume
 No forces of attraction
 Elastic collisions

Gas Variables

Temp (determines average kinetic
energy of the particles) must be
expressed in Kelvin


Volume (usually in liters)

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Kelvin = °C + 273
1 cm3 = 1 mL
Pressure (force exerted by
particles per unit area)
Gas Pressure
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P = F/A, where P = pressure, F =
force of the gas particles, and A =
area
Is dependent on number of
collisions of gas particles in a given
time
Units include Pa (pascals – SI
units), atm (atmospheres), torr,
mm Hg, bars, and psi
Gas Pressure

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1 atm is defined as the force the air
in the atmosphere exerts on the
Earth at sea level
Conversions:
1 atm = 760 mm Hg
 1 atm = 760 torr
 1 atm = 101,325 Pa
 1 atm = 101.325 kPa

Air Pressure
STP and Avogadro’s Principle

STP
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Stands for “standard temperature and
pressure”
Equals 1.00 atm at 0.00 °C
Avogadro’s Principle


States that equivalent volumes of gases
under the same conditions have an equal
number of particles
1 mole of any gas at STP has a volume
of 22.4 L
Boyle’s Law

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The pressure and volume of a
sample of gas at constant temp are
inversely proportional to each other
P1V1 = P2V2
Check this out
Solving Gas Problems Using Boyle's Law
A sample of helium gas in a balloon has a
pressure of 240 kPa in a 1.4-L container. What
will be the new pressure if the gas sample is
transferred to a 3.0-L container?
Step 1. Analyze the problem.
Known Variables
P1 = 240 kPa
V1 = 1.4 L
Unknown Variable P2 = ? kPa
V2 = 3.0 L
Step 2. Solve for the unknown.
Divide both sides of the equation for Boyle's law by V2 to solve for P2.
P2 = P1V1
V2
Substitute the known values into the rearranged equation.
P2 = 240 kPa ( 1.4 L )
(3.0 L)
Multiply and divide numbers and units to solve for P2.
P2 = 240 kPa ( 1.4 L) = 110 kPa
(3.0 L)
Step 3. Evaluate the answer.
Graph depicting change in
pressure with change in
volume of a gas

(a) A plot of V versus P for a gas sample is a hyperbola, but a plot
of V versus 1/P is a straight line (b).
Sample Problem
A gas occupies a volume of 2.45 L at
a pressure of 1.03 atm and a
temperature of 293 K. What
volume will the gas occupy if the
pressure changes to 0.980 atm
and the temperature remains
unchanged?
Sample Problem
The gas in a balloon has a volume of
4.0 L at 100. kPa. The balloon is
released into the atmosphere, and
the pressure decreases to 50. kPa.
What is the new volume of the
balloon?
Charles’s Law
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At constant pressure, the volume
of a fixed amount of gas is directly
proportional to its absolute
temperature
V1 = V2
T1
T2
V1T2 = V2T1
Check this out
Charles’s Law
Charles’s Law
Absolute Zero
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Predicted by Charles’s Law
-273.15 °C = 0 K
The hypothetical temperature at
which all motion stops
It cannot be attained by gases
because the gas will change to a
liquid before it reaches this
temperature
Sample Problem
A gas at 27 °C occupies 6.50 dm3.
What will its volume be at -23 °C?
Sample Problem
A gas occupies 0.105 dm3 at
-173 °C. At what Celsius
temperature will its volume be
0.140 dm3? Assume that pressure
is constant.
Gay-Lussac’s Law
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The pressure of a gas is directly
proportional to the temperature if
the volume remains constant
P1 = P2
T1
T2
Check this out
Gay-Lussac’s Law
Sample Problem
What is the pressure of a fixed
volume of a gas at 30.0 °C if it has
a pressure of 1.1 atm at 15.0 °C?
Sample Problem
What is the temperature of a fixed
volume of a gas at 0.96 atm if it
has a pressure of 1.0 atm at
37 °C?
Combined Gas Laws
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Combines Charles’s, Boyle’s and
Gay-Lussac’s Laws
P1V1 = P2V2
T1
T2
How can you solve gas problems using the
combined gas law?
A gas at 100.0 kPa and 25.0°C fills a flexible
container with an initial volume of 2.00 L. If the
temperature is raised to 60.0° and the pressure
increased to 320.0 kPa, what is the new volume?
Step 1. Analyze the problem.
Known Variables
P1 = 100.0 k Pa
P2 = 320.0 kPa
T2 = 60.0°C
T1 = 25.0°C
V1 = 2.00 L
Unknown Variable
V2 = ? L
Step 2. Solve for the unknown.
Add 273 to the Celsius temperature for T1 and T2 to
obtain the kelvin temperature.
T1= 273 + 25.0°C = 298 K; T2 = 273 + 60.0°C = 333 K
Multiply both sides of the equation for the combined
law by T2 and divide by P2 to solve for V2.
P1V1 = P2 V2
T1
T2
V2 = V1(P1)(T2)
(P2)(T1)
Substitute the known values into the rearranged
equation; multiply and divide numbers and units to
solve for V2.
V2 = 2.00 L ( 100.0 kPa)( 333 K) = 0.698 L
(320.0 kPa) (298 K)
Step 3. Evaluate the answer.
Sample Problem
A sample of oxygen gas has a
volume of 7.84 cm3 at a pressure
of 71.8 kPa and a temperature of
25.0 °C. What will be the volume
of the gas if the pressure is
changed to 101 kPa and the
temperature is changed to 0.0 °C?
Sample Problem

The volume of a sample of gas is
200. mL at 275 K and 92.1 kPa.
What will the new pressure be if
the gas occupies 238 mL at
350. K?
Ideal Gas Law
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Describes the physical behavior of
an ideal gas in terms of the
pressure, volume, temperature,
and the # of moles of gas
There is no ideal gas, but the
formula is helpful in describing the
behavior of real gases.
Ideal Gas Law
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PV = nRT
R is called the gas constant, also
0.0821 atm·L/mol·K
(Value of R can change based on
its units)
N = # of moles
Solving Gas Problems
Using the Ideal Gas Law
Calculate the number of moles of gas
contained in a 2.0-L container at 200 K
with a pressure of 120 kPa.
Step 1. Analyze the problem.
Known Variables
V = 2.0 L
T = 200.0 K
P = 120 kPa R = 8.314 L•kPa/(mol•K)
Unknown Variable
n = ? mol
Step 2. Solve for the unknown.
Divide both sides of the ideal gas law equation by RT to
solve for n.
PV = nRT
n=PV
RT
Substitute the known values into the rearranged
equation.
n = (120 kPa)(2.0 L)
(8.314 L•kPa)(200.0 K)
(mol•K)
Multiply and divide numbers and units to solve for n.
n = (120 kPa)(2.0 L) = 0.14 mol
(8.314 L•kPa)(200.0 K)
(mol•K)
Step 3. Evaluate the answer.
Sample Problem
How many grams of argon would it
take to fill a light bulb with a
volume of 0.475 L at STP?
Sample Problem

What is the pressure exerted by
0.625 moles of a gas in a 45.4 L
container at -24.0 °C?
Dalton’s Law of Partial
Pressures
The sum of the partial pressures of
all the components in a gas
mixture is equal to the total
pressure of the gas mixture
 PT = pa + pb + pc + . . .
 Mole fractions can be used to
determine partial pressure
mole gas A
x PTotal = Pa
total # moles gas

Dalton’s Law
Sample Problem

What is the atmospheric pressure
if the partial pressures of nitrogen,
oxygen, and argon are 77.75 kPa,
19.94 kPa, and 1.99 kPa,
respectively?
Sample Problem

What partial pressure of oxygen is
a scuba diver breathing if the total
pressure is 6.3 atm, and 20. % of
the air is oxygen?
What is meant by “water
displacement?”

Sample Problem
A gas is collected over water at
30. °C. The total pressure inside
the apparatus is 160. kPa. What is
the partial pressure of the dry gas?
Graham’s Law

Under the same conditions
(constant temp and press), gases
diffuse at a rate inversely
proportional to the square roots of
their densities (or molecular
masses)
Graham’s Law
v = velocity of gas
m = mass
Graham’s Law
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Effusion - The movement of atoms or
molecules through a hole so tiny that
they do not stream through, but
instead pass through one particle at a
time
Similar to diffusion
Lighter gases effuse and diffuse
faster than heavier gases because the
particles of the lighter gas move
faster than the particles of the heavier
gas.
Graham’s Law
Sample Problem
What is the relative rate of diffusion
between hydrogen and oxygen
gases?
Sample Problem

A molecule of oxygen gas travels
about 250. m/s at 0 °C. How fast
would an atom of xenon travel at
the same temperature?
Sample Problem

The rate of diffusion of an unknown
gas is 4.00 times faster than
oxygen gas. Calculate the molar
mass of the unknown gas.