Gases
Chapter 13
13.1 – The Gas Laws
Kinetic Theory = assumes that gas particles:
do not repel or attract each other
are much smaller than the distances between them
(particles have no volume)
are in constant, random motion (straight lines)
have completely elastic collisions (no loss of KE)
Have the same average KE at a given temp.
Nature of gases = determined by pressure,
temperature, volume, and number of particles
13.1 – The Gas Laws
Our variables =
Pressure (P)
Temperature (T) – Must be in Kelvin!
Volume (V)
Amount of particles/number of moles (n)
Gas constant - R
13.1 – The Gas Laws
 Boyle’s Law =
Studied relationship
between pressure and
volume of a gas
At a given temp., volume
and pressure are inversely
related
 P1V1 = P2V2
P1 and V1 are initial
conditions
P2 and V2 are new
conditions
13.1 – The Gas Laws (Boyle’s Law)
13.1 – The Gas Laws
 Steps for solving gas law problems =
1. Identify all variables.
2. Analyze the problem. “Which equation should
I use?”
3. Rearrange the equation to solve for the
unknown variable.
4. Plug in the numbers from step 1 into the
equation from step 3  solve!
13.1 – The Gas Laws (Boyle’s Law)
 A sample of helium gas in a balloon is
compressed from 4.0 L to 2.5 L at a
constant temperature. If the pressure of
the gas in the 4.0-L volume is 210 kPa,
what will the pressure be at 2.5 L?
13.1 – The Gas Laws (Boyle’s Law)
1. Identify all variables.
T=constant
V1=4.0 L
V2=2.5 L
P1=210 kPa
P2=?
2. Which equation should I use?
We know P and V. Boyle’s Law: P1V1=P2V2
3. Rearrange the equation.
To solve for P2, divide both sides by V2:
P1V1
 P2
V2
4. Plug in numbers from #1 into equation from #3:
(210kPa)(4.0 L)
 ? kPa
(2.5L)
13.1 – The Gas Laws (Boyle’s Law)
The volume of a gas at 99.0 kPa is 300.0
mL. If the pressure is increased to 188
kPa, what will be the new volume?
Air trapped in a cylinder fitted with a piston
occupies 147.5 mL at 1.08 atm pressure.
What is the new volume of air when the
pressure is increased to 1.43 atm by
applying force to the piston?
13.1 – The Gas Laws (Charles’s Law)
 Charles’s Law =
Studied relationship between
volume and temperature of a
gas
At a given pressure, volume
and temperature are directly
related
 . V1
V2

T1 T2
V1 and T1= initial cond.
V2 and T2= new cond.
13.1 – The Gas Laws (Charles’s Law)
13.1 – The Gas Laws (Charles’s Law)
A gas sample at 40.0°C occupies a volume
of 2.32 L. If the temperature is raised to
75.0°C, what will the volume be, assuming
the pressure remains constant?
A gas at 89°C occupies a volume of 0.67 L.
At what Celsius temperature will the volume
increase to 1.12 L?
13.1 – The Gas Laws (Gay-Lussac’s Law)
 Gay-Lussac’s Law =
 Studied the relationship
between temperature and
pressure of a gas
 At a given volume,
temperature and pressure
are directly related
 .P
P2

T1 T2
1
 P1 and T1 = initial cond.
 P2 and T2 = new cond.
13.1 – The Gas Laws (Gay-Lussac’s Law)
13.1 – The Gas Laws (Gay-Lussac’s Law)
The pressure of a gas in a tank is 3.20 atm
at 22.0°C. If the temperature rises to
60.0°C, what will be the gas pressure in
the tank?
A gas in a sealed container has a pressure
of 125 kPa at a temperature of 30.0°C. If
the pressure in the container is increased
to 201 kPa, what is the new temperature?
13.2 – The Combined Gas Law
Combines all four equations together into
one that relates temperature, pressure,
and volume:
P1V1 P2V2

T1
T2
13.2 – The Combined Gas Law
 A gas at 110 kPa and 30.0°C fills a flexible
container with an initial volume of 2.00 L. If the
temperature is raised to 80.0°C and the pressure
increased to 440 kPa, what is the new volume?
 At 0.00°C and 1.00 atm pressure, a sample of gas
occupies 30.0 mL. If the temperature is increased
to 30.0°C and the entire gas sample is transferred
to a 20.0-mL container, what will be the gas
pressure inside the container?
13.2 – Avogadro’s Principle
Avogadro’s principle = equal volumes of
gases at the same temperature and
pressure contain equal numbers of
particles
STP=Standard Temperature and Pressure, 0°C
(273 K) and 1 atm
**One mole of any gas will occupy 22.4 L at
STP
Now we can convert from liters to moles!
1 mole
0.25 L O 2 
 ? moles O 2
22.4 L
13.2 – Avogadro’s Principle
Calculate the volume that 0.881 mol of a
gas at standard temperature and pressure
(STP) will occupy.
How many moles of nitrogen gas will be
contained in a 2.00-L flask at STP?
13.2 – The Ideal Gas Law
All other gas laws apply to “a fixed mass”
or “a given amount”
Changing the number of gas particles
affects other variables
Increasing the number of particles will…
Increase P (if T and V are constant)
Increase V (if T and P are constant)
We need a new equation that includes
amount of gas present
13.2 – The Ideal Gas Law
PV = nRT
 P = pressure
 V = volume
 n = number of moles
of gas present
 R = ideal gas
constant (depends on
units of P)
 T = temperature
UNITS
of P
UNITS
of R
VALUE
of R
atm
L·atm
mol·K
0.0821
kPa
L·kPa
mol·K
8.314
mm Hg L·mm Hg
mol·K
62.4
13.2 – The Ideal Gas Law
Calculate the number of moles of gas
contained in a 3.0-L vessel at 3.00 x 102 K
with a pressure of 1.50 atm.
Determine the kelvin temperature required
for 0.0470 mol of gas to fill a balloon to
1.20 L under 0.988 atm pressure.
13.2 – The Ideal Gas Law
So what’s an ideal gas anyway?
Its particles don’t take up space and have no
intermolecular attractive forces
Follows the gas laws under all conditions of T
and P
**In the real world, NO gas is truly ideal! 
When do real gases not behave as “ideal”
gases?
At high P and low T  we can compress them
into liquids
Ex. Propane and liquid nitrogen
13.3 – Gas Stoichiometry
Volume – Volume
Solve same as mole-mole problems
How many liters of propane gas (C3H8) will
undergo complete combustion with 34.0L of
oxygen gas?
C3H8 + O2 → H2O + CO2
13.3 – Gas Stoichiometry
Volume – Mass
If 5.00L of nitrogen reacts completely with hydrogen at a pressure of 3.00atm
and a temperature of 298K, how much ammonia, in grams, is produced?
N2 + 3H2 → 2NH3
 Calculate as a Volume to Volume Problem
5.00L N2 x (2NH3/1N2) = 10.00L NH3
 Utilize PV=nRT to solve for the number of moles (n)
(3.00atm)(10.00L) = n (0.0821L∙amt/mol∙K)(298K)
 Convert moles to mass
1.23mol NH3 ÷ 17.04g/mol NH3 = 21.0g NH3
n=1.23mol NH3
13.3 Gas Stoichiometry
 When 3.00L of propane gas is completely combusted to
form water vapor and carbon dioxide at 350°C and
0.990atm, what mass of water vapor results?
C3H8 + O2 → H2O + CO2