GAS LAWS
Kinetic Molecular Theory
• Particles in an ideal gas…
– have no volume.
– have elastic collisions.
– are in constant, random, straight-line motion.
– don’t attract or repel each other.
– have an average KE directly related to Kelvin
temperature.
Real Gases
• Particles in a REAL gas…
– have their own volume
– attract and repel each other
• Gas behavior is most ideal…
– at low pressures
– at high temperatures
***Most real gases act like ideal gases
except under high pressure and low
temperature.
Characteristics of Gases
• Gases expand to fill any container.
– Take the shape and volume of their
container.
• Gases are fluids (like liquids).
– Little to no attraction between the particles
• Gases have very low densities.
= lots of empty space between the particles
Characteristics of Gases
• Gases can be compressed.
– lots of empty space between the particles
– Indefinite density
• Gases undergo diffusion.
– random motion
– scatter in all directions
Pressure
force
pressure 
area
Which shoes create the most pressure?
Pressure- how much a gas is pushing
on a container.
• Atmospheric pressure- atmospheric gases
push on everything on Earth
• UNITS AT SEA LEVEL
1 atm =101.3 kPa (kilopascal)= 760 mmHg =760 torr
Pressure
• Barometer
– measures atmospheric pressure
Aneroid Barometer
Mercury Barometer
Pressure
• Manometer
– measures contained gas pressure
C. Johannesson
U-tube Manometer
Bourdon-tube gauge
Temperature= how fast the molecules
are moving
• Always use absolute temperature (Kelvin)
when working with gases.
ºF
-459
ºC
-273
K
0
32
212
0
100
273
373
K = ºC + 273
C. Johannesson
STP
Standard Temperature & Pressure
0°C
1 atm
-OR-
273 K
-OR-
101.3 kPa
760 mm Hg
Volume = how much space a gas occupies
Units
– L, mL, cm3
• 1000 mL = 1 L
• 1 mL = 1 cm3
BASIC GAS LAWS
CharlesLaw.exe
Charles’ Law
• T  V (temperature is directly proportional to
volume)
• T ↑ V↑ & T↓ V↓
• V1 = V2
T1
T2 T is always in K
– K = °C + 273
– P and n = constant
V
T
Charles’ Law
V1
V2
=
T1
T2
(Pressure is held constant)
Timberlake, Chemistry 7th Edition, page 259
Timberlake, Chemistry 7th Edition, page 254
Charles’ Law
The egg out of the bottle
Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem
Charles’ Law Problem
• Mrs. Rodriguez inflates a balloon for a party. She is in an airconditioned room at 27.0oC, and the balloon has a volume of
4.0 L. Because she is a curious and intrepid chemistry teacher,
she heats the balloon to a temperature of 57.0oC. What is the
new volume of the balloon if the pressure remains constant?
Given
Unkown
Equation
T1 = 27.0oC +273= 300 K
V1 = 4.0 L
T2 = 57.0oC +273= 330 K
V2 = ? L
P1V1 = P2V2
T1 V1T2
 Substitute and Solve

4.0 L = V2
=
300 K 330K
4.4 L
Charles’ Law Learning Check
• A 25 L balloon is released into the air on a warm afternoon
(42º C). The next morning the balloon is recovered on the
ground. It is a very cold morning and the balloon has
shrunk to 22 L. What is the temperature in º C?
Given
Unkown
Equation
V1 = 25 L
T1 = 42 oC +273= 315 K
V2 = 22 L
T2 = ? ºC
P1V1 = P2V2
T1 V1T2
 Substitute and Solve

25 L = 22 L
315 K T2
=
277.2 K -273 = 4.2 ºC
Boyle'sLaw.exe
Boyle’s Law
• P↓ V ↑ & P↑ V ↓
• P  1/V (pressure is inversely proportional to volume)
• P1V1 = P2V2
– T and n = constant
P
V
Boyle’s Law
P1V1 = P2V2
(Temperature is held constant)
Boyle’s Law
 Marshmallows in a vacuum
Timberlake, Chemistry 7th Edition, page 254
Boyle’s Law
Mechanics of
Breathing
Timberlake, Chemistry 7th Edition, page 254
Boyle’s Law Problem
A balloon is filled with 30.L of helium gas at 1.00 atm.
What is the volume when the balloon rises to an
altitude where the pressure is only 0.25 atm?
Given
V1 = 30 L
P1 = 1 atm
P2 = .25atm
Unkown
V2 = ? L
• Substitute and Solve
 V2 0.25 atm = 30 L x 1.0 atm = 120 L
Equation
P1V1 = P2V2
T1
T2
Boyle’s Law Learning Check
A gas occupies 100. mL at 150. kPa. Find its volume
at 200. kPa.
Given
V1 = 100. mL
= 0.100 L
P1 = 150. kPa
P2 = 200. kPa
Unkown
V2 = ? L
Equation
P1V1 = P2V2
T1
T2
• Substitute and Solve
 V2 x 200. kPa = 0.100 L x 150. kPa= 75.0 mL

0.0750 L
AVOGADRO’S LAW
• Vn Vn
• V n (direct)
• V1 = V2
n1
n2
V
– T & P Constant
n
Avogadro'sLaw.exe
Avogadro’s Law Problem
A 3.0 liter sample of gas contains 7.0 moles. How
much gas will there be, in order for the sample to be
2.3 liters? P & T do not change
Given
V1 = 3.0 L
n1 = 7.0 mol
V2 = 2.3 L
Unkown
n2 = ? mol
• Substitute and Solve
3.0 L = 2.3 L =
7.0 mol n2 mol
5.4 mol
Equation
P1V1 = P2V2
n1T1
n2T2
Gay-Lussac’s Law
Gay-Lussac'sLaw.exe
• P1 = P2
T1 T2
– V & n constant
• Direct relationship
• PT PT 
P
T
Gay-Lussac Law
• Collapsing Barrel
Gay-Lussac Law
• Tank car implosion
COMBINED IDEAL GAS LAW
• P1V1 = P2V2
n1T1
n2T2
• If P, V, n, or T are constant then they cancel
out of the equation.
• n usually constant (unless you add or
remove gas), so
• P1V1 = P2V2
T1
T2
Combined Gas Law Problem
Ms. Evans travels to work in a hot air balloon from the Rocky
Mountains. At her launch site, the temperature is 5.00 °C, the
atmospheric pressure is 0.801 atm, and the volume of the air in
the balloon is 120.0 L. When she lands in Plano, the temperature
is 28.0 °C and the atmospheric pressure is 101.3 kPa. What is
the new volume of the air in the balloon?
Given
Unkown
T1 = 5.0oC +273= 278 K
P1 = 0.801 atm
V1 = 120.0 L
T2 = 28.0oC +273= 301 K
P2 = 101.3 kPa = 1 atm
V2 = ? L
Equation
V1 x P1 = V 2 x P2
T1
T2
 Substitute and Solve
 V2 x 1 atm = 120.0 L x 0.801 atm = 104 L

301K
278 K
Combined Gas Law Learning Check
Nitrogen gas is in a 7.51 L container at 5.C and 0.58 atm.
What is the new volume of the gas at STP?
Given
Unkown
T1 = 5.0oC +273= 278 K
P1 = 0.58 atm
V1 = 7.51 L
T2 = 273 K
P2 = 1 atm
V2 = ? L
Equation
V1 x P1 = V 2 x P2
T1
T2
 Substitute and Solve
 V2 x 1.0 atm = 7.51L x 0.58 atm = 4.3 L

273 K
278 K
Ideal Gas Law (“Pivnert”)
PV=nRT
R = The Ideal Gas Constant
R = 0.0821 (L*atm)
(mol*K)
R = 8.31 (L*kPa)
(mol*K)
•
•
•
R = 62.4 (L*mm Hg)
(mol*K)
V has to be in Liters, n in Moles, T in Kelvin,
P can be in atm, kPa or mmHg
* Choose which R to used based on the units of your pressure.
P
V
(atm) (L)
(kPa) (L)
mm Hg (L)
= n
= (moles)
= (moles)
= (moles)
R
T
(L*atm/mol*K)
(K)
(L*kPa/mol*K)
(K)
(L*mmHg/mol*K) (K)
Ideal Gas Law Problem
A rigid steel cylinder with a volume of 20.0 L is filled with nitrogen
gas to a final pressure of 200.0 atm at 27.0 oC. How many moles
of gas does the cylinder hold?
Given
Unkown
V = 20.0 L
moles of
P = 200.0 atm
nitrogen?
T =27.0oC +273= 300 K
Equation
PV=nRT
R= .0821 atm L/K Mole
 Substitute and Solve
n 0821 atm L/K Mole x 300 K = 200.0 atm x 20.0L= 162 moles
Ideal Gas Law Learning Check
• A balloon contains 2.00 mol of nitrogen at a pressure of 0.980
atm and a temperature of 37C. What is the volume of the
balloon?
Given
Unkown
n = 2.00 mol
V in L?
P = 0.980 atm
T =37.0oC +273= 310 K
Equation
PV=nRT
R= .0821 atm L/K Mole
 Substitute and Solve
0.980 atm x V= 2.00 mol x .0821 atm L/K Mole x 310 K = 51.9 L
Dalton’s Law of Partial Pressure
• The total pressure of a mixture of gases is
equal to the sum of the partial pressures of the
component gases.
• Ptotal = Pgas 1 + Pgas 2 + Pgas 3 + …
A metal container holds a mixture of 2.00 atm of nitrogen, 1.50
atm of oxygen and 3.00 atm of helium. What is the total pressure
in the canister?
6.5 atm
Welcome to Mole Island
1 mol =
6.02 x 1023
particles
Welcome to Mole Island
1 mol = molar mass
Welcome to Mole Island
1 mole = 22.4 L
@ STP
Gas Stoichiometry
Moles  Liters of a Gas:
– 2C4H10 (g) + 13O2(g)
2 mol
2L
͢ 8CO2(g) + 10H2O(g)
+ 13 mol
͢
8 mol
+
10 mol
+ 13 L
͢
8L
+
10 L
Recall:
The coefficients in a chemical reaction represent molar amounts of
substances taking part in the reaction.
Avogadro’s principle states that one mole of any gas occupies 22.4 L
at STP.
Thus when gases are involved, the coefficients in a balanced chemical
equation represent not only molar amounts but also relatives volumes
Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem
Gas Stoichiometry Problem
In the following combustion reaction, what volume of
methane (CH4) is needed to produce 26 L of water vapor?
͢
xL
– CH4 (g) + 2O2(g)
1 mol
1L
26 L
͢
͢
͢
x L = 26 L
1L
2L
CO2(g) + 2H2O(g)
2 mol
2L
x= 13 L
Gas Stoichiometry
use ideal gas law
PV=nRT
– Looking for grams or moles of gas?
• Step 1: start with ideal gas law to find moles of gas
• Step 2: 1change to grams of gas
Grams/mol? 1) Use Ideal Gas Law
2) Do stoichiometry calculations
Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem
Example 1
How many grams of Al2O3 are formed from 15.0 L of O2 at
97.3 kPa & 21°C?
PV=nRT
4 Al(s) +
3 O2(g) 
2 Al2O3(s)
Given
Unkown
V O2 = 15.0 L O2
grams of Al2O3?
P O2 = 97.3 kPa= 0.9605 atm
T O2 =21oC +273= 294 K
Step 1: Calculate moles of O2
R= .0821 atm L/K Mole
Given liters: Start with
Ideal Gas Law and
calculate moles of O2.
n = PV =
0.9605 atm x 15.0 L
= 0.5969 mol O2
RT 0.0821 atm L/K Mole 294 K
Step 2: Calculate mass of Al2O3
Use stoich to convert moles of
O2 to grams Al2O3.
0.5969 mol O2 = X mol Al2O3 = 0.3979 mol Al2O3
3moleO2
2 mole Al2O3
0.3979 mol Al2O3 x 101.96 g Al2O3=
41 g Al2O3
1 mol Al O
Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem
Gas Stoichiometry
use ideal gas law
• Looking for volume of gas?
PV=nRT
•Step 1: start with stoichiometry conversion to find moles of gas
•Step 2: use ideal gas law to find the volume
Liters ?
1) Do stoichiometry calculations
2) Use Ideal Gas Law
Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem
Example 2
What volume of CO2 forms from 5.25 g of CaCO3 at 101.3 kPa &
25ºC?
CaCO3
Given
m = 5.25 g CaCO3
P = 101.3 kPa = 1 atm
T =25.0oC +273= 298 K

CaO
+
CO2
Unkown
volume of CO2?
PV=nRT
R= .0821 atm L/K Mole
Looking for liters: Start with
stoich and calculate moles of CO2.
Step 1: Calculate moles of CO2
5.25 g CaCO3 x 1 mole CaCO3 = 0.0525 mol CaCO3
100 g CaCO3
1 mole CO2 = 1mole CaCO3 ; 0.0525 mol CO2 Plug this into the Ideal
Gas Law to find volume.
Step 2: Calculate volume of CO2
V = nRT = 0.0525 mol CO2 x .0821 atm L/K Mole x 298 K = 1.28 L
P
1 atm