Section 4.1 pg. 150 – 160
Homework:
Boyle’s Law: pg. 152 # 6-9
Charles’ Law: pg. 156 # 14-17
Combined Gas Law: pg. 159 # 20-23
 
They are based on the temperature, pressure and
volume relationships that all gases have in
common
1.  Boyle’s Law P1V1 = P2V2
2.  Charles’ Law
V1 = V2
T1 = T2
3.  Combined Gas Law
P1V1 =P2V2
T1 = T2
Boyle’s Law
Anglo- Irish chemist Robert Boyle (1627-1691) was a
founding member of the Royal Society of London.
 
Considers the effects of pressure on the volume of a
gas only while temperature is held constant
 
This is an inverse relationship:
P = V
P =
V
 
Boyle’s Law states: “as the pressure on a gas increases,
the volume of the gas decreases proportionally if
temperature and mass are constant”
 
Boyle’s Equation: P1V1 = P2V2
 
“As the pressure on a gas increases, the volume of the gas decreases
proportionally if temperature and mass are constant”
Pressure
(kPa)
100
200
300
400
500
Volume
(L)
3.00
1.52
1.01
0.74
0.60
PV
(kPa•L)
300
304
303
296
300
  Cartesian
Diver Demo
◦  “The Cartesian diver is named after the French philosopher, Rene
Descartes (1596-1650), and is a very old experiment. The volume of
a gas decreases as the pressure on the gas increases. As you squeeze
the bottle, the pressure is transferred from your hand to the water
and from the water to the air trapped inside the diver. As the
volume of air in the diver gets smaller, more water enters the diver,
making it heavier and less buoyant, and the diver sinks to the
bottom. As the pressure is released, the air inside the diver expands
and increases the buoyancy so that the diver rises.”   Expanding
Marshmallow Demo
◦  “The marshmallow expands as the volume in the syringe increases
and the pressure decreases. It shrinks as the volume is reduced and
the pressure is increased”
1. 
A sample of gas at 1.0 atm is in a 1.0 L container. What
is the pressure when the volume is changed to 2.0L?
(Assume T and m are constant)
P1V1 = P2V2
(1.0 atm) (1.0 L) = (x) (2.0 L)
1.0 atm•L = 2.0 L • (x)
1.0 atm•L = 2.0 L • (x)
2.0L
2.0L
0.5 atm = P2
According to Boyle’s Law, the gas would now have a
pressure of 0.5 atm
2. 
A 2.0 L party balloon at 98 kPa is taken to the top of a
mountain where the pressure is 75 kPa. Assume that
the temperature and mass of the gas remain the same.
What is the new volume of the balloon.
P1V1 = P2V2
(98 kPa) (2.0 L) = (75 kPa) (x)
196 kPa•L = 75 kPa • (x)
196 kPa•L = 75 kPa • (x)
75 kPa
75 kPa
2.6 L = V2
According to Boyle’s Law, the balloon would have a
new volume of 2.6L
  Pg.
152 #6-9
Charles’ Law
Jacque Charles (1746- 1823)
He made the first flight of a hydrogen
balloon on August 27, 1783. This balloon was
destroyed by terrified peasants when it
landed outside of Paris.
 
Shows the relationship between temperature
(must be in Kelvin) and volume of gas if pressure
and mass are constant
 
This is a direct relationship:
T = V
T = V
 
Charles’ Law states: “as the temperature of a gas
increases, the volume increases proportionally,
provided that the pressure and mass remain
constant”
 
Charles’ Equation:
V1 = V2
T1 = T2
 
“As the temperature of a gas increases, the volume increases
proportionally, provided that the pressure and mass remain
constant”
When the graphs of several careful
volume-temperature experiments are
extrapolated, all the lines meet at
absolute zero, 0K or -273° C
  Ivory
Soap Demo
◦  “It demonstrates Charles' Law, which states the volume of a gas
increases with its temperature. The microwaves impart energy
into the soap, water, and air molecules, causing them to move
faster and further away from each other. The result is that the
soap puffs up. Other brands of soap don't contain as much
whipped air and simply melt in the microwave.”
1. 
A gas inside a cylinder with a movable piston is
heated to 315°C. The initial volume of gas in the
cylinder is 0.30 L at 25°C. What will be the final
volume when the temperature is 315°C?
T1 = (25 + 273) = 298K
T2 = (315 + 273) = 588K
* Remember temperature
has to be in Kelvin
According to Charles’ Law,
the final volume will be 0.59 L
V1 = V2
T1 = T2
V2 = V1 T2
T1
= (0.30 L)(588K)
298K
= 0.59 L
  pg.
156 #14 - 17
The
Combined
Gas Law
You can get Boyle’s Law
back by assuming
temperature is constant.
Combined Gas Law
You can get Charles’
Law back by assuming
pressure is constant
 
When Boyle’s and Charles’ laws are combined,
the resulting combined gas law produces a
relationship among the volume, temperature, and
pressure of any fixed mass of gas.
 
The combined gas law is a useful starting point
for all cases with gases, even if one of the
variables is a constant.
◦  A variable that is constant can easily be eliminated from
the combined gas law equation, reducing it back to
Boyle’s or Charles’ Law
1. 
A gas cylinder with a fixed volume contains a gas
at a pressure of 652 kPa and a temperature of 25°C.
If the cylinder is heated to 150°C, use the combined
gas law to calculate the new pressure.
Because volume is constant we can
cancel V1 and V2 because V1 = V2
T1 = (25 + 273) = 298 K
T2 = (150 + 273) = 423 K
The gas will have a new
pressure of 925 kPa
P1V1 = P2V2
T1 = T2
P2 = P1T2
T1
P2 = (625kPa) (423K)
298 K
P2 = 925 kPa
2. A balloon containing helium gas at 20 °C and a pressure of
100 kPa has a volume of 7.50 L. Calculate the volume of the
balloon after it rises 10 km into the upper atmosphere, where the
temperature is –36 °C and the outside air pressure is 28 kPa.
(Assume that no gas escapes and that the balloon is free to expand so that
the gas pressure within it remains equal to the air pressure outside.)
T1 = (20 + 273) = 293 K
T2 = (-36 + 273) = 237 K
According to the Combined
Gas law, the volume of the
balloon in the upper
atmosphere will be 22L
P1V1 = P2V2
T1 = T2
V2 = P1V1T2
T1P2
V2 = (100kPa)(7.50 L)(237 K)
(293 K)(28 kPa)
V2 = 21.6 L = 22 L
 
STP: 0 °C and 101.325 kPa (exact values)
 
SATP: 25 °C and 100 kPa (exact values)
 
101.325 kPa = 1 atm = 760 mm Hg (exact values)
 
absolute zero = 0 K or –273.15 °C
 
K = (°C) + 273 (for calculation)
 
Boyle’s Law P1V1 = P2V2
 
Charles’ Law
 
Combined Gas Law
V1 = V2
T1 = T2
P1V1 = P2V2
T1 = T2
  Pg.
159 # 20 - 23
  Boyle’s
Law: pg. 152 # 6-9
  Charles’ Law: pg. 156 # 14-17
  Combined Gas Law: pg. 159 # 20 – 23
  Gas Laws Worksheet