Gas Laws
Characteristics of Gases
• highly compressible.
• occupy the full volume of their containers.
• exert a uniform pressure on all inner surfaces of a
container
• diffuse (mix) easily and quickly
• have very low densities.
Kinetic Molecular Theory
– Gases consist of a large number of molecules in
constant random motion.
– Volume of individual molecules negligible
compared to volume of container.
– Intermolecular forces (forces between gas
molecules) negligible.
– Collision of gas particles are elastic so no
kinetic energy is lost
– As temperature increases the gas particles
move faster, hence increased kinetic energy.
Four Physical Quantities for
Gases
Phys. Qty. Symbol
SI unit
Other common units
pressure
P
Pascal
(Pa)
atm, mm Hg, torr,
psi
volume
V
m3
dm3, L, mL, cm3
temp.
T
K
°C, °F
moles
n
mol
Temperature
Always use absolute temperature
(Kelvin) when working with gases.
ºF
-459
ºC
-273
K
0
C  F  32
5
9
32
212
0
100
273
373
K = ºC + 273
Practise
Absolute zero is – 273C or 0 K
Calculate the missing temperatures
0C
= _______ K
100C = _______ K
100 K = _______ C
– 30C = _______ K
300 K = _______ C
403 K = _______ C
25C = _______ K
0K
= _______ C
Kelvin Practice
What is the approximate temperature for
absolute zero in degrees Celsius and kelvin?
Absolute zero is – 273C or 0 K
Calculate the missing temperatures
273 K
373 K
0C = _______
100C = _______
– 173 C
100 K = _______
243 K
– 30C = _______
300 K = _______
C
27
403 K = _______
130 C
298 K
25C = _______
0K
– 273 C
= _______
Pressure
Pressure (P ) is defined as the force exerted per unit area.
The atmospheric pressure is measured using a barometer.
force
pressure 
area
Which shoes create the most pressure?
Pressure
 KEY
UNITS AT SEA LEVEL
101.325 kPa (kilopascal)
1 atm
760 mm Hg
760 torr
N
kPa  2
m
14.7 psi
•1 atm = 760 mmHg = 760 torr = 101325 Pa.
Pressure
 Barometer
• measures atmospheric pressure
Aneroid Barometer
Mercury Barometer
STP
STP
Standard Temperature & Pressure
273 K
101.325 kPa
STP
SLC
Standard Laboratory Conditions
25°C or 298 K
101.325 kPa
The Gas Laws
-BOYLES
-CHARLE
-GAY-LUSSAC
Boyle’s Law
The pressure and volume of
a gas are inversely related at
constant mass & temp.
PV = k
P
V
P1  V1  P2  V2
A. Boyle’s Law
Practise
A sample of chlorine gas occupies a volume of
946 mL at a pressure of 726 mmHg. What is
the pressure of the gas (in mmHg) if the
volume is reduced at constant temperature to
154 mL?
A sample of chlorine gas occupies a volume of 946 mL
at a pressure of 726 mmHg. What is the pressure of
the gas (in mmHg) if the volume is reduced at constant
temperature to 154 mL?
P1 x V1 = P2 x V2
P2 =
P1 = 726 mmHg
P2 = ?
V1 = 946 mL
V2 = 154 mL
P1 x V1
V2
726 mmHg x 946 mL
=
= 4460 mmHg
154 mL
Charles’ Law
The volume and absolute
temperature (K) of a gas are
directly related at constant
mass & pressure
V
k
T
V
T
V1 V2

T1 T2
Charles’ Law
Practise
2.
A sample of gas occupies 3.5 L at
300 K. What volume will it occupy
at 200 K?
3. If a 1 L balloon is heated from 22°C to
100°C, what will its new volume be?
2.
A sample of gas occupies 3.5 L at 300
K. What volume will it occupy at 200 K?
V1 = 3.5 L, T1 = 300K, V2 = ?, T2 = 200K
3.5 L / 300 K = V2 / 200 K
V2 = (3.5 L/300 K) x (200 K) = 2.3 L
3. If a 1 L balloon is heated from 22°C to 100°C,
what will its new volume be?
V1 = 1 L, T1 = 22°C = 295 K
V2 = ?, T2 = 100 °C = 373 K
V1/T1 = V2/T2, 1 L / 295 K = V2 / 373 K
V2 = (1 L/295 K) x (373 K) = 1.26 L
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A sample of carbon monoxide gas occupies 3.20 L at
125 0C. At what temperature will the gas occupy a
volume of 1.54 L if the pressure remains constant?
V1/T1 = V2/T2
T2 =
V1 = 3.20 L
V2 = 1.54 L
T1 = 398.15 K
T2 = ?
V2 x T1
V1
=
1.54 L x 398.15 K
3.20 L
= 192 K
Gay-Lussac’s Law
The pressure and absolute
temperature (K) of a gas are
directly related at constant
mass & volume
P
k
T
P
T
Gay-Lussac’s Law
Combined Gas Law
P 1V 1
T1
=
P 2V 2
T2
P 1 V 1T 2 = P 2V 2 T 1
E. Gas Law Problems
gas occupies 473 cm3 at 36°C.
Find its volume at 94°C.
A
CHARLES’ LAW
GIVEN: T V
V1 = 473 cm3
T1 = 36°C = 309K
V2 = ?
T2 = 94°C = 367K
WORK:
P1V1T2 = P2V2T1
(473 cm3)(367 K)=V2(309 K)
V2 = 562 cm3
E. Gas Law Problems
A
gas occupies 100. mL at 150.
kPa. Find its volume at 200. kPa.
BOYLE’S LAW
GIVEN: P V
V1 = 100. mL
P1 = 150. kPa
V2 = ?
P2 = 200. kPa
WORK:
P1V1T2 = P2V2T1
(150.kPa)(100.mL)=(200.kPa)V2
V2 = 75.0 mL
Practise
occupies 7.84 cm3 at 71.8 kPa &
25°C. Find its volume at STP.
 A gas
E. Gas Law Problems
gas occupies 7.84 cm3 at 71.8 kPa &
25°C. Find its volume at STP.
A
COMBINED GAS LAW
GIVEN: P T V WORK:
V1 = 7.84 cm3
P1V1T2 = P2V2T1
P1 = 71.8 kPa
(71.8 kPa)(7.84 cm3)(273 K)
T1 = 25°C = 298 K
=(101.325 kPa) V2 (298 K)
V2 = ?
P2 = 101.325 kPa V2 = 5.09 cm3
T2 = 273 K
E. Gas Law Problems
A
gas’ pressure is 765 torr at 23°C.
At what temperature will the
pressure be 560. torr?
GAY-LUSSAC’S LAW
GIVEN: P T WORK:
P1 = 765 torr
P1V1T2 = P2V2T1
T1 = 23°C = 296K (765 torr)T2 = (560. torr)(309K)
P2 = 560. torr
T2 = 226 K = -47°C
T2 = ?
Avogadro’s Principle
 Equal
volumes of all gases
contain equal numbers of
moles at constant temp &
pressure.
V
k
n
V
n
V1
V2

n1
n2
11.5
The Ideal Gas Equation
The gas laws can be combined into a general equation that
describes the physical behavior of all gases.
1
V
P
V T
V n
Boyle’s law
Charles’s law
Avogadro’s law
nT
V
P
nT
V R
P
rearrangement
PV = nRT
R is the proportionality constant, called the gas constant.
B. Ideal Gas Law
PV=nRT
UNIVERSAL GAS
CONSTANT
R = 8.3145 J/mol·K
R=0.0821 Latm/molK
R
= 0.0821 liter·atm/mol·K
R = 8.3145 J/mol·K
R = 8.2057 m3·atm/mol·K
R = 62.3637 L·Torr/mol·K or
L·mmHg/mol·K
B. Ideal Gas Law
 Calculate
the pressure in atmospheres
of 0.412 mol of He at 16°C & occupying
3.25 L. IDEAL GAS LAW
GIVEN:
WORK:
P = ? atm
PV = nRT
n = 0.412 mol
P(3.25)=(0.412)(0.0821)(289)
L
mol Latm/molK K
T = 16°C = 289 K
V = 3.25 L
P = 3.01 atm
R = 0.0821Latm/molK
B. Ideal Gas Law
 Find
the volume of 85 g of O2 at 25°C
and 104.5 kPa.
IDEAL GAS LAW
GIVEN:
WORK:
V=?
85 g 1 mol = 2.7 mol
n = 85 g = 2.7 mol
32.00 g
T = 25°C = 298 K PV = nRT
P = 104.5 kPa
(104.5)V=(2.7) (8.315) (298)
kPa
mol
dm3kPa/molK K
R = 8.315 dm3kPa/molK
V = 64 dm3
Summary
Dalton found that the total pressure of mixed
gases is equal to the sum of their individual
pressures (provided the gases do not react).
50 kPa
100 kPa
150 kPa
Note: all
of these
+
=
volumes
are the
1 L oxygen 1 L nitrogen 1 L mixed gas same
Gas Mixtures
Each component of a gas mixture exerts a pressure independent of
the other components. The total pressure is the sum of the partial
pressures.
Practise
• Calculating partial pressures
Vapour Pressure Defined
• Vapour pressure is the pressure exerted by a
vapour. E.g. the H2O(g) in a sealed container.
Eventually the air above the water
is filled with vapour pushing down.
As temperature , more molecules
fill the air, and vapour pressure .
Collecting gases over water
• Many times gases are collected over H2O
• Often we want to know the volume of dry
gas at STP (useful for stoichiometry).
For this we must make 3 corrections:
1. The level of water inside and outside the
tube must be level (so pressure inside is
equal to the pressure outside).
2. The water vapour pressure must be
subtracted from the total pressure (to get the
pressure of the dry gas).
3. Finally, values are converted to STP using
the combined gas law.
Sample calculation
A gas was collected over 21°C H2O. After equalizing water levels, the volume was 325 mL. Give
the volume of dry gas at STP (Patm=102.9 kPa).
Step 1: Determine vapour pressure (pg. 464)
At 21°C vapour pressure is 2.49 kPa
Step 2: Calculate the pressure of dry gas
Pgas = Patm - PH2O = 102.9 - 2.49 = 100.41 kPa
Step 3: List all of the data
T1 = 294 K, V1 = 325 mL, P1 = 100.41 kPa
Step 4: Convert to STP
= 299 mL
(P1)(V1)(T2) (100.4 kPa)(325 mL)(273 K)
V2=
=
(P2)(T1)
(101.325 kPa)(294 K)
Assignment
1. 37.8 mL of O2 is collected by the downward
displacement of water at 24°C and an
atmospheric pressure of 102.4 kPa. What is
the volume of dry oxygen measured at STP?
2. Try questions 8 – 10 on page 465.
3. 236 mL of H2 is collected over water at 22°C
and at an atmospheric pressure of 99.8 kPa.
What is the volume of dry H2 at STP?
4. If H2 is collected over water at 22°C and an
atmospheric pressure of 100.8 kPa, what is
the partial pressure of the H2 when the water
level inside the gas bottle is equal to the
water level outside the bottle?
1)
Vapor pressure at 24C = 2.98 kPa
Pgas = Patm - Pvapor
= 102.4 kPa - 2.98 kPa
= 99.42 kPa = P1
V1 = 37.8 mL, P1 = 99.42 kPa, T1 = 297 K
V2 = ?, P2 = 101.3 kPa, T2 = 273 K
P1V1
=
T1
(99.42 kPa)(37.8 mL)
=
(297 K)
P2V2
T2
(101.3 kPa)(V2)
(273 K)
(99.42 kPa)(37.8 mL)(273 K)
(V2) =
= 34.1 mL
(297 K)(101.3 kPa)
3)
Vapor pressure at 22C = 2.64 kPa
Pgas = Patm - Pvapor
= 99.8 kPa - 2.64 kPa
= 97.16 kPa = P1
V1 = 236 mL, P1 = 97.16 kPa, T1 = 295 K
V2 = ?, P2 = 101.3 kPa, T2 = 273 K
P1V1
P2V2
=
T1
T2
(97.16 kPa)(236 mL)
(101.3 kPa)(V2)
=
(295 K)
(273 K)
(97.16 kPa)(236 mL)(273 K)
(V2) =
= 209 mL
(295 K)(101.3 kPa)
Answers
4 - Total pressure = PH2 + PH2O
100.8 kPa = PH2 + 2.64 kPa
100.8 kPa - 2.64 kPa = PH2 = 98.16 kPa
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Ideal and real gas
• Ideal gas:
• Ideal gas molecule have
- zero volume
-zero attraction between molecules
• Real gases behave ideally at
- high temperature
- low pressure