Gases and Gas Laws
• Properties of gases
• STP
– Are considered fluids“any substance that
flows”
– Have low density
– Are highly compressible
– Fill the container they
occupy
Kinetic Molecular Theory
Postulates of the Kinetic Molecular Theory of Gases
1.
Gases consist of tiny particles (atoms or molecules)
2.
These particles are so small, compared with the distances between
them, that the volume (size) of the individual particles can be assumed
to be negligible (zero).
3.
The particles are in constant random motion, colliding with the walls of
the container. These collisions with the walls cause the pressure exerted
by the gas.
4.
The particles are assumed not to attract or to repel each other.
5.
The average kinetic energy of the gas particles is directly proportional
to the Kelvin temperature of the gas
Kinetic Molecular Theory (KMT)
 explains why gases behave as they do
 deals w/“ideal” gas particles…
1.
…are so small that they are assumed to have zero volume
2. …are in constant, straight-line motion
3. …experience elastic collisions in which no energy is lost
4. …have no attractive or repulsive forces toward each other
5. …have an average kinetic energy (KE) that is proportional
to the absolute temp. of gas (i.e., Kelvin temp.)
AS TEMP.
, KE
Kinetic Molecular Theory- KMT
• Describes relationships between pressure, volume,
temperature, velocity, frequency and force of
collisions
• 1- gases contain particles in constant, random,
straight line motion
• 2- collisions are elastic- E is completely transferred
(not converted to heat or light etc) when particles
collide with each other and the walls of the container
Collisions of Gas Particles
KMT con’t.
• 3- particles are very far apart- volume occupied by
the particles themselves is negligible
– The distance between them is more important than their
size
• 4- particles don’t attract each other
– Assume no interaction
Real vs. Ideal Gases
• Ideal gases obey the assumptions of KMT at all time
– It’s a model
– It doesn’t exist
– A gas is most ideal at low pressure and high temperature
• Real gases follow the KMT most of the time, but NOT
at high pressure and low temperature
– KMT breaks down when molecules are close together
– A real gas is most ideal at low pressure and high
temperature
The Gas Laws
• Describe HOW gases behave
• Can be predicted by the KMT
• Amount of change can be calculated by
mathematical equations
– Boyle’s Law
– Charles’s Law
– Gay-Lussac’s Law
– Avogadro’s Law
– Combined Gas Law
To describe a gas- 4 things to measure:
•
•
•
•
Pressure (P)
Temperature (T)
Volume (V)
Number of particles (n)
Gas Relationships:
• Pressure and number of particles
– Direct relationship
– More molecules = more collisions
– Double molecules  double pressure
2 atm
1 atm

Relationships con’t.
• Pressure and volume- Boyle’s Law
– Indirect relationship
– Double the pressure  ½ the volume
1 atm
2 atm
***- at constant
temperature

4 Liters
2 Liters
Relationships con’t.
• Pressure and temperature- Gay-Lussac’s Law
– Direct relationship
– As temperature increases, KE increases, velocity of
particles hitting walls increases
- If volume isn’t
constant, as
temperature
increases volume
will increase
300K
*** at constant volume
600K
Relationships con’t.
• Temperature and volume- Charles’s Law
– Direct relationship (if pressure is constant)
Relationships con’t.
• Temperature and velocity
– Direct relationship
– As temperature increases particles move faster
Boyle’s Law (Pressure and Volume)
• You can observe the relationship
between pressure and the volume
of a gas by constructing a
Cartesian diver from a soda
bottle and a medicine dropper.
• When you squeeze on the soda
bottle, the dropper dives.
• This is because as the pressure
increases, the volume of the air
bubble in the medicine dropper
decreases.
Boyle’s Law
• As the pressure on a gas
increases at a constant
temperature, volume
decreases
• If P doubles, V is halved
• The product of P and V is a constant (k)
• PV = k
• P1V1 = P2V2
Examples:
• A balloon is filled with
• A balloon is filled with 73
25 L of air at 1.0 atm
L of air at 1.3 atm
pressure. If the pressure
pressure. What pressure
is changed to 1.5 atm
is needed to change to
what is the new
volume to 43 L?
volume?
Charles’s Law (Volume and Temperature)
• A balloon is placed over
the lip of a flask. Then
the flask is placed on a
hotplate and heated.
• The balloon blows up.
• As a gas is heated at
constant pressure, its
volume increases.
Charles’s Law
• As the temperature of a
gas increases at a
constant pressure , the
volume increases
• In fact if the Kelvin
temperature doubles,
volume doubles
• The ratio of V and T is a constant K
• V/T = k
• V1= V2
T1 T 2
Examples:
• What is the
temperature of a gas
that is expanded from
2.5 L at 25ºC to 4.1L at
constant pressure.
• What is the final
volume of a gas that
starts at 8.3 L and 17ºC
and is heated to 96ºC?
Gay-Lussac’s Law (Pressure and Temperature)
• As the temperature of a
gas increases at a
constant volume,
pressure increases
• The ratio of P to T is a
constant
• P/T = k
• P1/T1 = P2/T2
Examples:
• What is the pressure
inside a 0.250 L can of
deodorant that starts at
25ºC and 1.2 atm if the
temperature is raised to
100ºC?
• At what temperature
will the can above have
a pressure of 2.2 atm?
Avogadro’s Law
• The volume of a gas is directly proportional to
the number of moles
• When the T, P and V of 2 gases are the same
they contain the same number of molecules
• 22.4L of any gas at STP = 1 mole of the gas
• V= volume n= # of particles k=constant
• V/n = k
• V1/n1 = V2/n2
The Combined Gas Law
• If: PV = k and P/T = k and V/T = k
• And: the initial pressure, volume, and
temperature of a gas are called P1, V1, and T1
• And: after conditions change, the new
pressure, volume, and temperature of the gas
are called P2, V2, and T2.
• We determine that:
P1V1 ____
P2V2
____
=
T1
T2
Example:
A gas at 27°C and 100. kPa occupies 250. mL. How much space
will the gas occupy if the temperature is reduced to 0.0°C and
the pressure is increased to 150. kPa?
• STEP 1: Identify the variables
o P1 = 100. kPa
o V1 = 250 mL
o T1 = 27°C + 273 = 300. K
o P2 = 150. kPa
o V2 = V2
o T2 = 0°C + 273 = 273 K
• STEP 2: Plug the variables into the equation
P2 V2
P1 V1 = ____
____
T2
T1
(100.
kPa)(250. mL) = (150.
kPa)(V2 )
_______________
___________
(300. K)
(273 K)
• STEP 3: Solve for the unknown
(273
K)(100. kPa)(250. mL) = V2 = 152 mL
_____________________
(150. kPa)(300. K)
Gas Stoichiometry
• Mass – Volume Gas Problems
– g  moles  moles  L
• Volume – Volume Gas Problems
– L  moles  moles  L
• Don’t forget the mole crutch!!