Gas Laws
BOYLE
CHARLES
AVOGADRO
GAY-LUSSAC
Consider This
What happens to the volume
of a gas when you increase
the pressure? (e.g. Press a
syringe that is stoppered)
Push!
Consider This
What happens to the volume
of a gas when you increase
the pressure? (e.g. Press a
syringe that is stoppered)
Push!
Consider This
What happens to the Volume
of a Gas When you Increase
the Pressure? (e.g. Press a
syringe that is stoppered)
There is lots of space
between gas particles.
Therefore, gases are
compressible!
Why?
Let’s investigate the relationship between
pressure and volume if the quantity of gas and
temperature are 100
heldkPa
constant.
Pressure Volume
(kPa)
(L)
100 kPa
100
50
Volume = 50 L
Pressure Volume
(kPa)
(L)
100
200
50
25
200 kPa
Volume = 25 L
Pressure Volume
(kPa)
(L)
100
200
400
50
25
12.5
400 kPa
Volume = 12.5 L
What is the
mathematical
relationship
between P and V?
P x V = constant
Pressure Volume
(kPa)
(L)
5000
5000
5000
5000
100
200
400
800
50
25
12.5
6.25
800 kPa
Volume = 6.25 L
Boyle’s Law
In the 17th Century Robert Boyle
described this property as, “the spring
of air”.
Boyle showed that when temperature
and amount of gas were constant then:
P  1/V
OR:
PV = k
Boyle’s Law
For a fixed quantity of gas at a constant
temperature, the volume and pressure are
inversely proportional.
Boyle showed that when temperature
and amount of gas were constant then:
P  1/V
OR:
PV = k
Who Cares?
Scuba Divers!
At sea level air pressure
= 100 kPa
At 10 m deep in water pressure = 200 kPa
At 20 m deep
= 300 kPa
At 30 m deep
= 400 kPa
SCUBA provides air at the same pressure
A Scuba Diver goes to a depth of 90 m
and takes a breath of 3 L volume from
her tank.
Suddenly! A Dolphin lunges
at the diver and takes the SCUBA!
The Diver holds her breath and quickly
returns to the surface.
What will the volume of air in the diver’s
lungs be at the surface (100 kPa)?
What will happen to the diver?
A Scuba Diver goes to a depth of 90 m
and takes a breath of 3 L volume from
her tank.
Assuming that T is constant we can use
Boyle’s Law:
PV = k
At 90 m: k = P1V1 (1000 kPa)(3 L) = 100 kPa(V2)
At surface: k = P2V2
V2 = 300 L
Yikes Exploding lungs
Therefore!
P1V1 = P2V2
Consider This!
Two balloons are filled with equal
volumes of air.
What happens to the Volume
of each if one is heated and
the other is frozen?
HEATED BALLOON
50oC, V=1.18 L
FROZEN BALLOON
HEATED BALLOON
60oC, V=1.22 L
40oC, V=1.14 L
FROZEN BALLOON
HEATED BALLOON
70oC, V=1.25 L
30oC, V=1.11 L
FROZEN BALLOON
HEATED BALLOON
80oC, V=1.29 L
20oC, V=1.07 L
FROZEN BALLOON
HEATED BALLOON
90oC, V=1.32 L
10oC, V=1.04 L
FROZEN BALLOON
HEATED BALLOON
100oC, V=1.36 L
0oC, V=1.00 L
FROZEN BALLOON
To study this relationship let’s look at this data
in a table.
Temperature(o C)
0
10
20
30
40
50
60
70
80
90
100
Volume (L)
1.00
1.04
1.07
1.11
1.14
1.18
1.22
1.25
1.29
1.32
1.36
Graph this data
using
temperature as
the
independent
variable
Relationship Between Temperature
and Volume for a Fixed Quantity of
Gas at a Constant Pressure
Volume (L)
1.60
1.40
1.20
1.00
0.80
0.60
0.40
0.20
0.00
-280
-230
-180
-130
-80
-30
Temperature oC
20
70
Volume (L)
Relationship Between Temperature
and Volume for a Fixed Quantity of
Gas at a Constant Pressure
If this line is extended
backwards the volume of
0 L of gas is found to be
-273 oC
-280
-230
-180
-130
-80
1.60
1.40
1.20
1.00
0.80
0.60
0.40
0.20
0.00
-30
Temperature oC
20
70
-273.15 oC is known as absolute zero.
When using gas laws, temperature must be
expressed using a temperature scale where
0 is -273.15oC.
This is called the Kelvin scale of absolute
temperature.
0 K = -273.15 oC
When changing oC to K simply add 273.15
What is 12.3oC in K?
12.3 + 273.15
= 285.45
= 285.5 K
Now let’s look at this table with temperatures
in Kelvin (K).
Temperature(K)
273
283
293
303
313
323
333
343
353
363
373
Volume (L)
1.00
1.04
2.07
2.10
2.14
2.18
2.21
2.25
2.28
2.32
2.36
Can you spot a
mathematical
relationship
between T and
V.
V/T in Kelvin is a constant.
Temperature(K)
273
283
293
303
313
323
333
343
353
363
373
Volume (L)
1.00
1.04
2.07
2.10
2.14
2.18
2.21
2.25
2.28
2.32
2.36
0.0037
0.0037
0.0037
0.0037
0.0037
0.0037
0.0037
0.0037
0.0037
0.0037
0.0037
Charles’ Law
In the 17th Century Jacques Charles
examined the relationship between
Temperature and Volume
Charles showed that when Pressure
and amount of gas were
constant then:
VT
OR:
V/T = k
Charles’ Law
For a fixed quantity of gas at a constant
pressure, absolute temperature and volume
are directly proportional.
Charles showed that when Pressure
and amount of gas were
constant then:
VT
OR:
V/T = k
If
then
Combined Gas Law
V1P1 = V2P2 and V1 =
T1
V1P1
V2P2
=
or
T1
T2
What does P2 = ?
V2
T2
V1P1 T2 =V2P2 T1
V2T1
V2T1
RECAP
Boyle’s Law
At Constant T and n:
VP = k
Charles’ Law
At Constant P and n:
V/T = k
Combined Gas Law
For a fixed quantity of gas:
VP/T = k
If 12.5 L of a gas at a pressure of 125 kPa is
placed in an elastic container at 15oC what
volume would it occupy if the pressure is
increased to 145 kPa?
V1P1
V2P2
Given:
=
T1
T2
V1 = 12.5 L
P1 = 125 kPa Since T = T cancel them to get
1
2
o
T1 = 15 C
V1P1 = V2P2
V2 = ?
P2 = 145 kPa (12.5 L)(125 kPa) = V2(145 kPa)
V2 = (12.5 L)(125 kPa)/145 kPa
T2 = 15oC
V2 = 10.8 L
Does this answer make sense?
125 kPa
12.5 L @ 125 kPa
Does this answer make sense?
165 kPa
Does this answer make sense?
165 kPa
Does this answer make sense?
Yes, as the pressure increases at constant
temperature the volume decreases.
165 kPa
V reduced
to
10.8 L
If 15.6 L of a gas at a pressure of 165 kPa is
placed in an elastic container at 15oC what
volume would it occupy if the temperature
is increased to 98oC?
V1P1
V2P2
Given:
=
T1
T2
V1 = 15.6 L
P1 = 165 kPa Since P = P cancel them to get
1
2
o
288CK
T1 = 15
V1/T1 = V2 / T2
V2 = ?
(15.6
L)/(288
K)
=
V
/(371
K)
o
2
T2 = 98
371CK
P2 = 165 kPa V2 = (15.6 L)(371 K) / 288K
V2 = 20.1 L
If 5.3 L of a gas at a pressure of 75 kPa is
placed in an elastic container at 24oC what
volume would it occupy if the temperature is
increased to 62oC and pressure to 155 kPa?
V1P1
V2P2
Given:
=
OR V1P1T2 =V2P2 T1
T2
V1 = 5.3 L T1
P1 = 75 kPa
(5.3
L)(75
kPa)(335
K)
o
T1 = 297
24 CK
V2 =
(155 kPa)(297K)
V2 = ?
T2 = 335
62oCK
V2= 2.9 L
P2 = 155 kPa
Ideal Gas Law
An Ideal Gas is a hypothetical gas that obeys all
the gas laws perfectly under all conditions.
PV = nRT
Where n is the number of moles of gas, P is
pressure in kPa, R = 8.313 kPaL/molK and T is
temperature in K.
Find the mass of helium gas which would be
introduced into a 0.95 L container to produce a
pressure of 125 kPa at 25oC.
Find the volume occupied by 25 g of chlorine
gas at SATP.
4.2 g of propane gas is introduced into a 325 mL
container at 45oC. What is the pressure of the
container.
Propane is C3H8.
What is the density of NH3 gas at STP if 1.0 mol
of this gas occupies 22.4 L.
At what temperature does methane gas have a
density of 1.2 mg/L if its pressure is 65 kPa.
Methane is CH4.
This gas
exerts a
pressure of 100 kPa
inside this container.
If another gas
is
injected into the
same container and
it exerts a pressure
of 70 kPa what is
the total pressure in
the container?
170 kPa
This gas
exerts a
pressure of 100 kPa
inside this container.
If another gas
is
injected into the
same container and
it exerts a pressure
of 70 kPa what is
the total pressure in
the container?
170 kPa
This gas
exerts a
pressure of 100 kPa
inside this container.
If another gas
is
injected into the
same container and
it exerts a pressure
of 70 kPa what is
the total pressure in
the container?
170 kPa
The total pressure of a gas mixture is the
sum of the partial pressures of each of
the gases in the mixture.
Example
If a container of air has a pressure of
100 kPa and the % of N2 in the container
is 78%, % of O2 is 21%, what are the
partial pressures of each of these gases
inside this container.
PN2= 78 kPa, PO2= 21 kPa
What would the total
pressure be if the gas
in container 1 was
injected into container
2.
6.0 L, 400 K,
155 kPa
2
1
5.0 L
300 K
125 kPa
The total pressure is the sum of
the pressures of gas 1 and gas
2.Since gas 1 changed volume
1
and temperature its pressure
changed.
5.0 L
6.0 L, 400 K,
300 K
155 kPa
125 kPa
2
V1=5.0 L, V2=6.0 L, T1=300 K, T2 = 400 K
P1=125 kPa, P2=?
V1P1T2=V2P2T1
1
P2 = 5.0 L x 125 kPa x 400 K = 139 kPa
6.0 L x 300 K
5.0 L
6.0 L, 400 K,
300 K
155 kPa
125 kPa
2
Total pressure is
155 kPa + 139 kPa = 294 kPa
= 2.9 x 102 kPa
6.0 L, 400 K,
155 kPa 139 kPa
2
1
5.0 L
300 K
125 kPa
Find the total pressure in container 1 if the
gas in container 2 is injected into container 1.
1
6.0 L, 400 K,
155 kPa
2
5.0 L
300 K
125 kPa
V1=6.0 L, V2=5.0 L, T1=400 K, T2 = 300 K
P1=155 kPa, P2=?
V1P1T2=V2P2T1
1
P2 = 6.0 L x 155 kPa x 300 K = 139.5 kPa
5.0 L x 400 K
5.0 L
6.0 L, 400 K,
300 K
155 kPa
125 kPa
2
Total pressure is
125 kPa + 139.5 kPa = 264.5 kPa
= 2.6 x 102 kPa
6.0 L, 400 K,
155 kPa
2
1
5.0 L
300 K
125 kPa
139.5 kPa