TEACHER NOTES Chemistry
U N I T 1 3 | Page 1
UNIT 10 NOTES: GAS LAWS
STUDENT OBJECTIVES: Your fascinating teachers would like you amazing learners to be able to…
1. Distinguish between real and ideal gas behavior, and identify the criteria in the kinetic molecular theory that
conflict with the properties of real gases.
2. Describe the relationships between volume, pressure, number of moles, and temperature for an ideal gas.
3. Perform calculations with the following laws and formulas:
a. Boyle’s Law
b. Charles’ Law
c. Gay-Lussac’s Law
d. Combined Gas Law
e. Dalton’s Law
f. Ideal Gas Law
4. List the independent and dependent variables for each of the gas laws.
5. Identify graphs and/or graph data for Boyle’s, Charles’, and Gay-Lussac’s laws.
6. Explain how a gas is collected over water and how the gas pressure is corrected for the vapor pressure of water.
7. Determine the density of a gas using a form of the Ideal Gas Law.
8. Determine the Molar Mass of a gas using a form of the Ideal Gas Law.
9. Integrate gas law calculations with stoichiometric calculations.
I. KINETIC – MOLECULAR THEORY OF GASES
The kinetic molecular theory is based on the idea that particles of matter
are in constant motion. This theory helps us understand why gases
behave the way they do and give us insight into the behavior of solids and
liquids. There are five basic postulates (assumptions) of this theory.
1. Gases consist of molecules whose separation is much larger than the
size of the molecules themselves. Most of the volume of a gas is
empty space.
2. Particles in a gas constantly move in straight line paths and random
directions.
3. Particles in a gas collide frequently with the sides of the container
and less frequently with each other. All collisions are elastic (energy
may be transferred between particles, but no energy is gained or lost as a result of the collisions).
4. Particles in a gas do not attract or repel one another. They do not sense any intermolecular forces.
5. The average kinetic energy of all of the gas particles in a sample is proportional to the temperature of that
gas sample. (Meaning that as kinetic energy increases, temperature increases.) All gases at the same
temperature have the same amount of kinetic energy.
These postulates describe ideal gases. Real gases act like ideal gases at normal temperatures and pressures. However
at low temperatures or high pressures, the behavior of real gases is significantly different than ideal gases. Real gas
particles do interact to some degree. This becomes a “problem” at low temperatures as the gas particles lose kinetic
energy and start to attract each other and condense to form a liquid. At high pressures, the particles get much closer
together and the gas is no longer mostly empty space. Again, the particles can interact and the gas can condense.
TEACHER NOTES Chemistry
U N I T 1 3 | Page 2
II. DEFINITIONS,VARIABLES, AND THEIR UNITS
Gas
-phase of matter with no definite shape or volume
Kinetic Energy -energy of motion
Pressure
-the force exerted by a gas per unit area on a surface
Partial Pressure-pressure of one gas in a mixture of gases
Temperature -(Kelvin) directly related to the amount of kinetic energy of a substance
STP
-Standard Temperature and Pressure = 1.0 atm and 0.0˚C.
The state of a gas can be described by four variables:
QUANTITY (Symbol):
Pressure (P)
Volume (V)
Units (symbol):
Atmospheres (atm)
kilopascals (kPa)
millimeters of Mercury (mmHg)
Liters (L)
milliliters (mL)
cubic centimeters (cm3)
Remember, 1 mL = 1 cm3
Temperature (T)
Amount (n)
Kelvin (K)
Moles (mol)
Conversions:
*The temperature MUST be in KELVIN UNITS for any calculation in this unit!
Remember: K = oC + 273
*It doesn’t matter what unit Pressure and Volume are in, as long as you use the SAME UNIT
within a single calculation.
Remember: 1 atm = 760 mm Hg = 101.3 kPa
TEACHER NOTES Chemistry
U N I T 1 3 | Page 3
III. THE GAS LAWS
LAW
INDEP/DEP
VARIABLES
CONTROL
VARIABLES
(held
constant)
BOYLE’S
V, P
T, n
↑V, ↓ P
inverse, indirect
V1 P1  V2 P2
CHARLES’S
T, V
P, n
↑T ↑V
direct
V1 V2

T1 T2
GAY-LUSSAC’S
P, T
V, n
↑T ↑P
direct
P1 P2

T1 T2
AVOGADRO’S
V, n
P, T
↑n ↑V
direct
V1 V2

n1 n2
COMBINED
V, P, T, n
NA
NA
V1 P1 V2 P2

n1T1 n2T2
MATH
RELATIONSHIP
FORMULA
*The combined gas law is often written without the moles, since moles are often held
constant. However, on the STAAR Chemistry Reference page, moles are included in the
equation.
*Notice that Boyle’s, Charles’, Gay-Lussac’s, and Avogadro’s Laws are “simplified” versions of
the combined gas law! In these laws, certain variables are held constant.
V1
= initial (beginning) volume
V2
= final volume
P1
= initial (beginning) pressure
P2
= final pressure
T1
= initial (beginning) temperature
T2
= final temperature
n1
= initial (beginning) amount in moles
n2
= final amount in moles
TEACHER NOTES Chemistry
U N I T 1 3 | Page 4
Boyle’s Law
It states that volume and pressure vary inversely when temperature and amount of
gas is constant. This means that when the pressure increases, the volume decreases
and vice versa. It’s helpful to think about the fact that pressure is caused by the
number of collisions of particles with the wall of the container. In a small volume, the
particles will hit the walls of the container more frequently, which leads to a higher
pressure.
P1 V1  P2 V2
Any units of pressure are acceptable, but P1 and P2 must be in the same units. Any
units of volume are acceptable, but again, you must use the same units.
Real life application: When you breathe, your diaphragm moves downward,
increasing the volume of the lungs. This causes the pressure inside the lungs to be
less than the outside pressure so air rushes in.
Charles’s Law
It states that volume and temperature vary directly when pressure and
amount of gas is constant. This means that when the temperature
increases, the volume increases and vice versa.
However, this only happens proportionally if temperature is recorded in
KELVIN. So, anytime you use this law, temperature must be in Kelvin!!!
I’m Jacques Charles, and
even though my friend
Gay-Lussac published
the official law, he was
so nice to name it after
me, since I originally had
the idea in 1787!
V1 V2

T1 T2
Real life application: Bread dough rises because yeast produces carbon
dioxide. When placed in the oven, the heat causes the gas to expand, and
the bread rises even further.
I’m Robert Boyle,
and I came up
with my law way
back in 1622.
TEACHER NOTES Chemistry
U N I T 1 3 | Page 5
Gay-Lussac’s Law
It states that pressure and temperature vary directly when volume and
amount of gas is constant. This means that when the temperature
increases, the pressure increases and vice versa. To keep volume
constant, a rigid container must be used.
However, this only happens proportionally if temperature is recorded in
KELVIN. So, anytime you use this law, temperature must be in Kelvin!!!
P1 P2

T1 T2
I’m Joseph-Louis
Gay-Lussac, and
I named a law
after myself in
1802.
I also picked the
names of the
buret and pipet!
Real life application: The air pressure inside a tire increases on a hot summer day.
Some things to think about…
Example 11-1.
Charles did a lot of work with hot-air balloons. A hot air balloon works by
heating up the air in the balloon, and then the balloon rises. Explain why this
works using gas laws.
Heat up the air, volume increases, molecules spread out, density becomes less, floats higher
Example 11-2.
If you sit on a closed bag of potato chips, it will pop. Why? Whose gas law
does this demonstrate?
Sitting on the bag will decrease volume, pressure increases, too much for the bag so it pops! - Boyle
Example 11-3.
Even though an airplane cabin is “pressurized”, it’s still at a lower pressure than at sea level. If you
take a bag of potato chips on a plane, it will expand. Why? Whose gas law does this demonstrate?
As pressure pushing on outside of bag decreases, molecules inside can expand more! - Boyle
Example 11-4.
Tires on a vehicle need a certain level of pressure in order to run optimally. Why are people
encouraged to check the pressure of their tires in both the summer and the winter? Whose gas
law does this demonstrate?
During the summer, pressure inside tires will increase as temperature increases. In the winter, pressure
inside tires will decrease as temperature decreases. – Gay-Lussac
Example 11-5.
A gas doubles its temperature from 100.0C to 200.0C does this mean that the volume has also
doubled if the pressure is constant? Explain.
NO! Going from 100°C to 200°C is like going from 273 K to 373 K – which is NOT doubling in Kelvin. The
relationships only hold true in Kelvin.
TEACHER NOTES Chemistry
U N I T 1 3 | Page 6
Combined Gas Law
When you work your math, it is best to start with the Combined Gas Law Equation. It includes pressure, volume and
temperature relationships and is really a combination of all three-gas laws – even adding in Avogadro’s Law with “n”. All
you do is write down the combined equation and then cross off the variables that remain constant (if any). If a variable
is not mentioned in a problem, you can assume that variable has remained constant.
P1 V1 P2 V2

n1 T1 n2 T2
DON’T FORGET: The term STP stands for Standard
Temperature and Pressure. Mathematically, this means
1.0 atmosphere pressure (or 760 mm Hg or 101.3 kPa)
and 0.0oC (or 273 K).
WORKED EXAMPLE: A gas is known to have the volume of 7.81 liters when the pressure is 754 mmHg. What would be
the volume in liters when the pressure is changed to 1.23 atm? Temperature and amount is constant.
List Givens
V1 = 7.81 L
P1 = 754 mmHg
P2 = 1.23 atm
V2 = ? in liters
This is where you list out the variables given to you in the problem. Watch out for “STP” – it also tells
you variable information!
Check your Units
754 mmHg 
1 atm
 0.992 atm = P1
760 mmHg
You must make sure that volumes and pressures are in the SAME UNIT! If they are not, it might be
necessary to convert!
Write your
equation
P1 V1 P2 V2

n1 T1 n2 T2
P1 V1 P2 V2

Since “T” and “n” are constant 
n1 T1 n2 T2

P1 V1  P2 V2
Pick which equation you are going to use. Remember, if you like using the combined formula instead
of Boyle’s/Charles’s/Gay-Lussac’s individually, check for constant variables, as they cross out!
Plug & Chug!
Law Name
0.9927.81  1.23V2
Boyle’s Law
V2 = 6.30 L
TEACHER NOTES Chemistry
Example 11-6.
U N I T 1 3 | Page 7
A 36.2 mole sample of carbon dioxide occupies a volume of 5.20 liters. If 15.4 more liters are
added, what is the new volume if the pressure and temperature remains constant? (7.51 L)
List Givens
V1 = 5.20 L
n1 = 36.2 mol
n2 = 36.2 + 15.4 = 51.6 mol
V2 = ? in liters
Check your Units
OKAY !!!!!
Write your
equation
Plug & Chug!
V1
P1 V1 P2 V2
V2


n1 T1 n2 T2 Since P and T are constant  n1 n2
5.20 L
36.2mol
Law Name
Example 11-7.

V2
51.6mol
V2 = 7.41 L
Avogadro’s Law (direct relationship)
A gas is found to have a pressure of 700. mmHg at a temperature of 54.00C. If the volume of the
gas under these conditions is 3.21 L, what was the original volume in liters when the temperature
was 20.00C and pressure was 700. mmHg? (2.88 L)
List Givens
P2 = 700 mmHg V2 = 3.21 L
T2 = 327 K
P1 = 700 mmHg
T1 = 293 K V1 = ? in liters
Check your Units
OKAY !!!!! Hey…. The Pressure is constant
Write your
equation
Plug & Chug!
Law Name
V1 V2
P1 V1 P2 V2


n1 T1 n2 T2 Since P and n are constant  T1 T2
3.21L
V1

293K
327 K
Charles’ Law (direct relationship)
V1 = 2.88 L
TEACHER NOTES Chemistry
Example 11-8.
U N I T 1 3 | Page 8
What happens to the volume of a gas if the pressure is increased 2.41 times, assuming temperature
remains constant? (0.414 L)
List Givens
V1 = 1 L P2 = 1 atm
Check your Units
Write your
equation
Plug & Chug!
P2 = 2.41 atm V2 = ? in L
Good job making up numbers…….
P1 V1 P2 V2

n1 T1 n2 T2 Since T and n are constant  P1V1  P2V2
(1 L)( 1 atm) = (V2 )(2.41 atm)
V2 = 0.414 L
Law Name
Example 11-9.
Boyle’s Law (inverse relationship)
What would be the volume in liters of an 8.90 liter sample of gas at 100.oC and 113 kPa if
conditions were changed to STP? (7.26 L)
List Givens
V1 = 8.90 L T1 = 373 K
P1 = 113 kPa
STP: 1 atm and 0.0 0C
(sneaky numbers)
Check your Units
P2 = 101.3 kPa T2 = 273 K V2 ? in Liters
BUT…use 101.3 kPa since other pressure was in kPa
Write your
equation
Plug & Chug!
P1 V1 P2 V2

n1 T1 n2 T2
V1P1 V2 P2

since n is constant  T
T2
1
(8.90 L) (113kPa)
373K
V2 = 7.26 L
Law Name
Combined Gas Law

(101.3kPa)V2
273K
.
TEACHER NOTES Chemistry
U N I T 1 3 | Page 9
Example 11-10. If the pressure on a gas in a rigid container changes from 800. mmHg to 2.15 atm, and at the lower
pressure, the temperature is found to be 40.0°C, what was the original temperature in Kelvin? (T2
= 641 K)
List Givens
P1 = 800 mmHg
T1 = 313 K
P2 = 11.0 atm T2 = ? in K
Check your Units
800mmHg 
1atm
 1.05atm = P1
760mmHg
Write your
equation
P1
P1 V1 P2 V2
P2


n1 T1 n2 T2 Since V and n are constant  T1 T2
Plug & Chug!
1.05atm 2.15atm

313K
T2
Law Name
Gay-Lussac’s Law (direct relationship)
T2 = 641 K
TEACHER NOTES Chemistry
U N I T 1 3 | Page 10
Dalton’s Law of Partial Pressure
This law is used when you have a mixture (solution) of 2 or more gases which do not react chemically. It states that the
TOTAL pressure of a mixture of gases is simply the sum of the individual pressures of each of the gases (known as partial
pressures).
However, this law only holds true if the pressures you are adding together were all recorded at the SAME TEMPERATURE
AND VOLUME CONDITIONS! You may have to do some converting if they are not in the same conditions!
Ptotal = PA + PB + PC + …for however many gases you have!
Example 11-11. There is a container which has oxygen, xenon and helium in it. Its total pressure is known to be
972 mmHg. If the pressure of the helium is 0.458 atm and the pressure of the oxygen is 74.1 kPa,
what is the pressure of the Xenon in mmHg? (68 mmHg)
List Givens
Ptotal = 972 mmHg
PHe = 0.458 atm
PO2 = 74.1 kPa
Check your Units
0.458atm 
74.1kPa 
760mmHg
 348mmHg = PHe
1atm
760mmHg
 556mmHg = PO2
101.33kPa
Write your
equation
Ptotal = PHe + PO2 + PXe
Plug & Chug!
972 mmHg = 348 mmHg + 556 mmHg + PXe
PXe = 68 mmHg
Law Name
Dalton’s law of partial pressures
PXe in mmHg
TEACHER NOTES Chemistry
U N I T 1 3 | Page 11
Dalton’s Law Over Water
Dalton’s Law can also be used for problems where a gas was collected “over water”
through water displacement. The gas that you collect is not a pure gas, but rather a
mixture of the gas you were trying to collect AND water vapor that is escaping the surface
of the water. This means that the pressure of the gas you collected was actually a
combination of the two.
If you see the words “over water” in a problem, you must subtract the water vapor
pressure from the total pressure of the gas collected over water. The water vapor values
in mmHg can be found on the next page of the notes. If the gas pressure in the problem is
in a unit other than mmHg, convert it to mmHg so you can subtract off the water vapor
value.
Here’s an example of
what collecting gas
“over water” looks like…
The gas collected
includes the desired gas
AND water vapor… so…
Ptotal = Pwater + Pgas
Example 11-12. 50.0 ml of hydrogen collected over water and has a pressure of 850. mmHg at 27.0oC. What is the
pressure of the dry gas at STP? (823.3 mmHg)
List Givens
Ptotal = 850. mmHg
Look up the wet pressure: Pwet = 26.7 mmHg
Check your Units
OK
Write your
equation
Ptotal = Pwet + Pdry
Plug & Chug!
Ptotal = Pwet + Pdry 850 mmHg = 26.7 mmHg + Pdry
PDry =823.3 mmHg
Law Name
Dalton over Water
TEACHER NOTES Chemistry
U N I T 1 3 | Page 12
WATER VAPOR PRESSURE VALUES IN mmHg
(torr = mmHg)
Vapor
Vapor
Vapor
Vapor
Temperature
Temperature
Temperature
Temperature
Pressure
Pressure
Pressure
Pressure
(°C)
(°C)
(°C)
(°C)
(torr)
(torr)
(torr)
(torr)
-10
-9
-8
-7
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
2.1
2.3
2.5
2.7
2.9
3.2
3.4
3.7
4.0
4.3
4.6
4.9
5.3
5.7
6.1
6.5
7.0
7.5
8.0
8.6
9.2
9.8
10.5
11.2
12.0
12.8
13.6
14.5
15.5
16.5
17.5
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
18.7
19.8
21.1
22.4
23.8
25.2
26.7
28.3
30.0
31.8
33.7
35.7
37.7
39.9
42.2
44.6
47.1
49.7
52.4
55.3
58.3
61.5
64.8
68.3
71.9
75.7
79.6
83.7
88.0
92.5
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
97.2
102.1
107.2
112.5
118.0
123.8
129.8
136.1
142.6
149.4
156.4
163.8
171.4
179.3
187.5
196.1
205.0
214.2
223.7
233.7
243.9
254.6
265.7
277.2
289.1
301.4
314.1
327.3
341.0
355.1
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
369.7
384.9
400.6
416.8
433.6
450.9
468.7
487.1
506.1
525.8
546.1
567.0
588.6
610.9
633.9
657.6
682.1
707.3
733.2
760.0
787.6
815.9
845.1
875.1
906.1
937.9
970.6
1004.4
1038.9
1074.6
TEACHER NOTES Chemistry
U N I T 1 3 | Page 13
Ideal Gas Law
The ideal gas law allows us to make relationships between pressure, volume, temperature, and amount under a single
set of stated conditions.
The way to identify that you have this type of problem (and not one of the previous types) is that CONDITIONS WILL NOT
BE CHANGING!!! You will not have both initial and final values for the variables.
P = pressure (it must match R)
V = volume (for this formula, it must be in LITERS)
n = number of moles of gas
R = universal gas constant
T = temperature (it must be in KELVIN)
PV = nRT
Out of these variables, “R” is the new one… the gas constant values are given to you on your reference table. You must
pick an R value which matches the pressure.
L mmHg
Latm
L kPa
R  0.0821 mol
or
62.4
or 8.314 mol
K
mol  K
K
There are other forms of PV=nRT as well, depending on what you’re trying to solve for. One of the formulas
makes the substitution of n=mass/MM, while the other formula is rearranged to include density (D) in g/L.
 MASS 
PV  
RT
 MM 
DRT
MM 
P
The “Meow-Meow” formula… cats
put “DiRT” over their “Pee”!!!
Example 11-13.
How many moles of oxygen must be put in a 0.500 L flask at 20.00C to have a pressure in the flask
of 1.77 atm? (0.0368 mol)
List Givens
V = 0.500 L T = 293 K P = 1.77 atm looking for n (number of moles)
Check your Units
Pick an R value !!!
Write your
equation
Plug & Chug!
0.0821 (liter-atm)/(mole-K)
PV = nRT
(1.77 atm)(.500 L) = n (.0821)(293 K)
n = .0368 moles of O2
Law Name
Ideal Gas Law
TEACHER NOTES Chemistry
Example 11-14.
U N I T 1 3 | Page 14
0
If 1.17 grams of helium are put into a 500.0 ml container at 10 C, what would be the pressure of
the container (in mmHg)? (10300 mmHg)
List Givens
V = 500 mL T = 283 K m = 1.17 g looking for P in mmHg
Check your Units
1x10 3 L
We need a volume in Liters to use PV=nRT
500mL 
 .500 L
1mL
1molHe
We need the number of moles to use PV=nRT
1.17 gHe 
 .2925molHe
4.00 gHe
Pick an R value !!!
Write your
equation
62.4 (liter-mmHg)/(mole-K)
PV = nRT
Plug & Chug!
(P)(.500 L) = (.2925moles ) (62.4)(283 K)
P = 10300 mmHg
Law Name
Ideal Gas Law
Example 11-15.
What is the molar mass of a diatomic gas if 1.60 grams of it in a 0.500 L container exert a
pressure of 2500 mm Hg at 350. K? What element does this gas represent? (28.0 g/mol; N2)
List Givens
m = 1.60 g V = 0.500 L P =2500 mm Hg T = 350 K
Check your Units
Pick an R value !!!
Write your
equation
62.4 (liter-mmHg)/(mole-K)
 MASS 
PV  
 RT
 MM 
 MASS 
 RT
 PV 
Rearranged  MM  
Plug & Chug!


1.60 g
(62.4)(350 K )
MM  
 (2500mmHg )(.500 L) 
MM = 28.0 g/mol (It is nitrogen N2)
Law Name
Ideal Gas Law using Molar Mass
TEACHER NOTES Chemistry
Example 11-16.
U N I T 1 3 | Page 15
What would be the density of oxygen gas at 70.00C and 3.0 atm? (3.41 g/L)
List Givens
T = 343 K
P = 3.0 atm MM = 32.0 g/mol looking for density in g/L
Check your Units
Pick an R value!!!
Write your
equation
Plug & Chug!
.0821 (liter-atm)/(mole-K)
( D)( R )(T )
( P )( MM )
Rearranged  D 
( P)
RT
(3.0atm)(32.0 g / mol )
D
(.0821)(343K )
MM 
D= 3.41 g/L
Law Name
Ideal Gas Law using Density
Non-STP Gas Stoichiometry
Last unit, we learned that we can use the equality of 22.4 L = 1 mole of a gas collected at STP in order to convert
between volume and moles. However, what if our conditions are not at STP? In that instance, we can use PV=nRT to
convert between volume and moles!
Example 11-17.
How many liters of hydrogen gas can be collected in the lab (atmospheric pressure is 750. mmHg
and temperature is 22.00C) when 40.0 grams of zinc reacts with excess hydrochloric acid? (Zinc
reacts with hydrochloric acid to form hydrogen gas and zinc chloride.) (15.0 L)
V =???? L Cl2 P = 750. mmHg
Write and balance the equation:
40.0.gZn  1moleZn

 65.39 gZn

T = 295 K
m = 40.0 g Zn
We need to find moles of H2 with stoichiometry
Zn + 2 HCl  H2 + ZnCl2
  1moleH 2

  1 mole Zn
 

  0.61171 mol H 2


Pick an R value!!!! 62.4 (liter-mmHg)/(mole-K)
PV= nRT
(750. mmHg)(V) = (0.61171 mol) (62.4)(295 K)
V = 15.0 L H2