Chapter 13: Gases
Getting started with gas calculations:
Before we can start talking about how gases behave in numerical terms, we need
to define some of the quantitative properties that are characteristic of gases:
 Pressure (P): The force of gas molecules as they hit the sides of the
container in which they are placed.
o Common units of pressure:
 atmospheres (atm): The average air pressure at sea level.
 kilopascals (kPa): The SI unit for pressure; 101.325 kPa = 1
atm.
 mm Hg (Torr): 760 Torr = 1 atm.
 Volume (V): The amount of space in which a gas is enclosed.
o The only commonly used unit of volume is liters (L).
 Temperature (T): A measurement of the amount of energy that molecules
have. The higher the energy, the higher the temperature.
o Common units of pressure:
 Kelvin (K): The only units that can be used when doing
numerical problems with gases.
 Degrees Celsius (0C): Must be converted to Kelvin before
doing problems (by adding 273).
Other terms frequently used:
 STP: Stands for “standard temperature and pressure”, namely 273 K (00
C) and 1.00 atm.
 “Room temperature”: 298 K (250 C)
A whirlwind tour through the early gas laws:
Boyle’s Law: P1V1 = P2V2
 For any gas, the product of the pressure and the volume before a change is
equal to the product of the pressure and the volume after a change.
 In plain English, what this means is that if you put pressure on a gas, it gets
smaller. If you decrease pressure on a gas, it gets larger.
 Demonstration: Have them sit on balloons to demonstrate that decreasing
the volume increases the pressure inside the balloon so much that it pops!
o Another demonstration (if the equipment is available): Use a vacuum
pump to decrease the pressure around a balloon – the balloon will
get bigger (until it pops, probably).
 Sample problems:
o If I have 10 L of gas at a pressure of 1 atm and double the pressure,
what will the new volume of the gas be?
5L
o If 250 L of a gas is in a sealed container at a pressure of 1.5 atm and
I decrease the volume of the container to 100 L, what will the gas
pressure inside the container be?
3.75 atm.
Charles’s Law:
V1 V2

T1 T2
 If you increase the temperature of a gas, the volume also increases. (Note:
The temperature must be in Kelvin, NOT degrees centigrade)
 Why? The KMT tells us that the amount of energy that a gas has is
determined by the temperature of the gas. The more energy a gas has, the
faster the gas molecules move away from each other, causing more space
between the molecules and a larger overall volume.
 Examples:
o If you heat a 1.25 L balloon from a temperature of 250 C to 400 C,
what will the new volume of the balloon be?
1.31 L
o What temperature will be required to raise the volume of a 1.0 L
balloon to 1.25 L if the initial temperature is 250 C?
373 K
Gay-Lussac’s Law:
P1 P2

T1 T2
 When you increase the temperature of an enclosed gas, the pressure of
the gas goes up.
 This is why it’s a bad idea to put a spray can into a campfire – eventually
the pressure rises so much that the sides of the can split and the can
explodes.
 Example: If you have a spray can at a pressure of 20 atm at room
temperature and put it into a campfire at a temperature of 12000 C, what
will the pressure in the canister be right before it explodes?
98.9 atm
The combined gas law:
 If we put the last three gas laws together, we can devise another law that
encompasses all three of them (making it unnecessary to memorize the
three):
P1V1 P2V2

T1
T2
 How to use this law: Whenever you have a problem in which you change
the pressure, volume, and/or temperature, just plug the values into it.
o If one of the variables isn’t mentioned, we can assume that it’s kept
constant and we can just cross it out of the equation.
o Demonstrate how the three gas laws can be easily derived from this
law.
 Examples:
o If I have 25 mL of a gas at a pressure of 2.1 atm and a temperature of
300 K, what will the pressure become if I raise the temperature to 400
K and decrease the volume to 10 mL?
7 atm
o If I have a container with an internal pressure of 1.5 atm and
temperature of 250 C, what will the pressure be if I heat the container
to 1500 C?
2.13 atm
Ideal Gases:
Now that we know how gases behave when we manipulate P, V, and T, it’s time
to start thinking about how to deal with things like moles and grams. After all, if
we’re going to do chemical reactions with gases, we’ll need to know how to
calculate these!
Avogadro’s law: One mole of every gas has the same volume.
 This law assumes that all gases behave perfectly and identically according
to the rules of the kinetic molecular theory. Though not precisely true, it
gives us very good answers under most conditions.
 Ideal gas: A gas that behaves according to the kinetic molecular
theory.
o No intermolecular forces, infinitely small, etc.
o There is no ideal gas in the real world, but some gases come closer than
others:
 The gas molecules are small.
 The gas molecules have very weak intermolecular forces.
 The gas molecules are very hot, so they move quickly around and
don’t interact with each other much.
 The gas is at low pressure, so the molecules have a lot of space
between them.
Assuming that all gases are ideal, we can use an equation to relate the number
of moles to the pressure (P), volume (V), and temperature (T), giving us the…
Ideal gas law: PV = nRT
P = pressure (in atm of kPa)
V = volume (L)
n = number of moles
T = temperature (Kelvin)
R = ideal gas constant (depends on the unit of pressure used)
 8.314 L kPa/mol K
 0.08206 L atm/mol K
 Examples:
 If I have 10 liters of a gas at a pressure of 1.5 atm and a temperature of 250
C, how many moles of gas do I have?
0.61 mol.
 If I have 3.5 moles of a gas at a pressure of 895 kPa and they take up a
volume of 50 L, what’s the temperature?
1538 K
Dalton’s Law of Partial Pressures: General Chemistry
So far, we’ve assumed that all the gases we’ve been working with are pure.
However, this isn’t true most of the time. Most gases we use are actually
mixtures of several different elements or compounds.
To deal with this, we’ve come up with the idea of partial pressure:
Partial pressure: The pressure that each of the components in a mixture of
gases would exert if the other gases in the mixture were removed.
To work with mixtures, we usually treat each gas as if it were alone, rather than
in a mixture. This is a reasonable assumption because gases don’t interact with
each other much anyway. As a result, the total pressure of the mixture of gases
would be the same as the sums of all the partial pressures.
Ptot = P1 + P2 + P3 + …
Dalton’s Law of Partial Pressures
Example: A container contains three gases. If the partial pressure of carbon
dioxide is 0.50 atm, the partial pressure of nitrogen is 0.30 atm, and the partial
pressure of methane is 0.40 atm, what’s the total pressure inside the container?
 Ptot = 0.50 atm + 0.30 atm + 0.40 atm = 1.20 atm
 Example: A container has a total internal pressure of 1.50 atm. This
container contains two gases, nitrogen (which has a partial pressure of
0.80 atm) and oxygen (with unknown partial pressure). Given this
information, what’s the partial pressure of oxygen in the container?
Ptot = Pnitrogen + Poxygen
1.50 atm = 0.80 atm + X
X = 0.70 atm
Dalton’s Law of Partial Pressures: Honors Chemistry
So far, we’ve assumed that all the gases we’ve been working with are pure.
However, this isn’t true most of the time. Most gases we use are actually
mixtures of several different elements or compounds.
To deal with this, we’ve come up with the idea of partial pressure:
Partial pressure: The pressure that each of the components in a mixture of
gases would exert if the other gases in the mixture were removed.
To work with mixtures, we usually treat each gas as if it were alone, rather than
in a mixture. This is a reasonable assumption because gases don’t interact with
each other much anyway. As a result, the total pressure of the mixture of gases
would be the same as the sums of all the partial pressures.
Ptot = P1 + P2 + P3 + …
Dalton’s Law of Partial Pressures
Example: If I have a container that contains both oxygen and nitrogen, what’s
the total pressure inside the container of the partial pressure of oxygen is 0.50
atm and the partial pressure of nitrogen is 0.75 atm?
 Answer: 0.50 atm + 0.75 atm = 1.25 atm
Example: What’s the pressure in atmospheres if I put 2 moles of N2 and 3 moles
of H2 in a 20 L container at 310 K?
 Answer:
o Ptot = Pnitrogen + Phydrogen
o To solve for each of the partial pressures, we need to use the
equation PV=nRT to find the partial pressures of each gas, then add
them up.
 Pnitrogen(20 L) = (2 mol)(0.08206 Latm/molK)(310 K) = 2.54 atm
 Phydrogen(20 L) = (3 mol)(0.08206 Latm/molK)(310 K) = 3.82 atm
o Ptot = 2.54 atm + 3.82 atm = 6.36 atm
Example: A 3 L flask containing hydrogen gas at a pressure of 1.5 atm is
connected to a 5 L flask containing nitrogen gas at a pressure of 2.5 atm. When
the tube between the two flasks is opened, what will the total pressure inside the
system be?
 Answer:
o Use the combined gas law to solve the partial pressure of each gas
when you open the connector between the flasks, then add the partial
pressures together.
 Hydrogen: P1V1 = P2V2; P2 = 0.563 atm.
 Nitrogen: P1V1 = P2V2; P2 = 1.56 atm.
o Overall, Ptot = 0.563 atm + 1.56 atm = 2.12 atm
Gas Stoichiometry – General Chemistry
 Explain to the general chemistry classes that at STP, the conversion factor
between liters and moles is 22.4 L = 1 mole.
What this means is that, whatever method of stoichiometry we use, we need only
replace the molar masses of the gases with the number 22.4 – for compounds
that are solids, we still use the molar mass.
Example: Using the equation 2 H2 + O2  2 H2O, how many liters of water can
be made from 25 liters of oxygen at STP?
Answer: 50 L
Example (same equation): How many liters of water can be made from 240
grams of hydrogen?
Answer: 2688 L
Gas Stoichiometry – Honors Chemistry
Explain to the honors chemistry classes that you need to convert between liters
 moles in both cases using PV = nRT.
 At STP, one mole has a volume of 22.4 liters.
If the compounds you’re working with on both sides of the equation are
measured in liters, you can use the mole ratio conversion directly between them.
Example: Using the equation 2 H2 + O2  2 H2O, how many liters of water can
be made from 25 liters of oxygen?
Answer: 50 L
Example (same equation): If we do the reaction above using electrochemistry,
we can make liquid water from H2 and O2 gas. If I use 20 L of oxygen gas at a
pressure of 1.0 atm and a temperature of -350 C, how many grams of water will I
make?
Answer: 36.9 grams