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The Gas Laws
• 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.
Characteristics of Gases
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
ºF
ºC
K
-459 32 212
-273 0 100
0 273 373
 
32
F
C 9
5



 K = ºC + 273
Always use absolute temperature
(Kelvin) when working with gases.
What is the approximate temperature for
absolute zero in degrees Celsius and kelvin?
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 0 K = _______ C
Kelvin Practice
273 373
–173 243
27 130
298 –273
Absolute zero is –273C or 0 K
Pressure
area
force
pressure 
Which shoes create the most pressure?
Pressure (P ) is defined as the force exerted per unit area.
The atmospheric pressure is measured using a barometer.
Pressure
2
m
N
kPa 
KEY UNITS AT SEA LEVEL
101.325 kPa (kilopascal)
1 atm
760 mm Hg
760 torr
14.7 psi
•1 atm = 760 mmHg = 760 torr = 101325 Pa.
Pressure
Barometer
• measures atmospheric pressure
Mercury Barometer
Aneroid Barometer
STP
Standard Temperature & Pressure
273 K
101.325 kPa
STP
STP
Standard Laboratory Conditions
25°C or 298 K
101.325 kPa
SLC
-BOYLES
-CHARLE
-GAY-LUSSAC
The Gas Laws
Boyle’s Law
The pressure and volume of
a gas are inversely related at
constant mass & temp.
P
V
PV = k
2
2
1
1 V
P
V
P 


A. Boyle’s Law
Practice
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
P1 = 726 mmHg
V1 = 946 mL
P2 = ?
V2 = 154 mL
P2 =
P1 x V1
V2
726 mmHg x 946 mL
154 mL
= = 4460 mmHg
k
T
V

V
T
Charles’ Law
The volume and absolute
temperature (K) of a gas are
directly related at constant
mass & pressure
2
2
1
1
T
V
T
V

Charles’ Law
Practice
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?
3. If a 1 L balloon is heated from 22°C to 100°C,
what will its new volume be?
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
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
For more lessons, visit
www.chalkbored.com
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 = 3.20 L
T1 = 398.15 K
V2 = 1.54 L
T2 = ?
T2 =
V2 x T1
V1
1.54 L x 398.15 K
3.20 L
= = 192 K
V1/T1 = V2/T2
k
T
P

P
T
Gay-Lussac’s Law
The pressure and absolute
temperature (K) of a gas are
directly related at constant
mass & volume
Gay-Lussac’s Law
Combined Gas Law
P1V1
T1
=
P2V2
T2
P1V1T2 = P2V2T1
GIVEN:
V1 = 473 cm3
T1 = 36°C = 309K
V2 = ?
T2 = 94°C = 367K
WORK:
P1V1T2 = P2V2T1
E. Gas Law Problems
A gas occupies 473 cm3 at 36°C.
Find its volume at 94°C.
CHARLES’ LAW
T V
(473 cm3)(367 K)=V2(309 K)
V2 = 562 cm3
GIVEN:
V1 = 100. mL
P1 = 150. kPa
V2 = ?
P2 = 200. kPa
WORK:
P1V1T2 = P2V2T1
E. Gas Law Problems
A gas occupies 100. mL at 150.
kPa. Find its volume at 200. kPa.
BOYLE’S LAW
P V
(150.kPa)(100.mL)=(200.kPa)V2
V2 = 75.0 mL
Practice
A gas occupies 7.84 cm3 at 71.8 kPa &
25°C. Find its volume at STP.
GIVEN:
V1 = 7.84 cm3
P1 = 71.8 kPa
T1 = 25°C = 298 K
V2 = ?
P2 = 101.325 kPa
T2 = 273 K
WORK:
P1V1T2 = P2V2T1
(71.8 kPa)(7.84 cm3)(273 K)
=(101.325 kPa) V2 (298 K)
V2 = 5.09 cm3
E. Gas Law Problems
A gas occupies 7.84 cm3 at 71.8 kPa &
25°C. Find its volume at STP.
P T V
COMBINED GAS LAW
GIVEN:
P1 = 765 torr
T1 = 23°C = 296K
P2 = 560. torr
T2 = ?
WORK:
P1V1T2 = P2V2T1
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
P T
(765 torr)T2 = (560. torr)(309K)
T2 = 226 K = -47°C
k
n
V

V
n
Avogadro’s Principle
 Equal volumes of all gases
contain equal numbers of
moles at constant temp &
pressure.
2
2
1
1
n
V
n
V

The Ideal Gas Equation
The gas laws can be combined into a general equation that
describes the physical behavior of all gases.
11.5
nT
V
P

V
P

1
Boyle’s law
V n

Avogadro’s law
V T

Charles’s law
nT
V R
P
 PV = nRT
rearrangement
R is the proportionality constant, called the gas constant.
B. Ideal Gas Law
UNIVERSAL GAS
CONSTANT
R = 8.3145 J/mol·K
R=0.0821 Latm/molK
PV=nRT
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
GIVEN:
P = ? atm
n = 0.412 mol
T = 16°C = 289 K
V = 3.25 L
R = 0.0821Latm/molK
WORK:
PV = nRT
P(3.25)=(0.412)(0.0821)(289)
L mol Latm/molK K
P = 3.01 atm
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:
V = ?
n = 85 g
T = 25°C = 298 K
P = 104.5 kPa
R = 8.315 dm3kPa/molK
B. Ideal Gas Law
Find the volume of 85 g of O2 at 25°C
and 104.5 kPa.
= 2.7 mol
WORK:
85 g 1 mol = 2.7 mol
32.00 g
PV = nRT
(104.5)V=(2.7) (8.315) (298)
kPa mol dm3kPa/molK K
V = 64 dm3
IDEAL GAS LAW

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The Gas Laws Explained

  • 2. • 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. Characteristics of Gases
  • 3. 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.
  • 4. 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
  • 5. Temperature ºF ºC K -459 32 212 -273 0 100 0 273 373   32 F C 9 5     K = ºC + 273 Always use absolute temperature (Kelvin) when working with gases.
  • 6. What is the approximate temperature for absolute zero in degrees Celsius and kelvin? 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 0 K = _______ C Kelvin Practice 273 373 –173 243 27 130 298 –273 Absolute zero is –273C or 0 K
  • 7. Pressure area force pressure  Which shoes create the most pressure? Pressure (P ) is defined as the force exerted per unit area. The atmospheric pressure is measured using a barometer.
  • 8. Pressure 2 m N kPa  KEY UNITS AT SEA LEVEL 101.325 kPa (kilopascal) 1 atm 760 mm Hg 760 torr 14.7 psi •1 atm = 760 mmHg = 760 torr = 101325 Pa.
  • 9. Pressure Barometer • measures atmospheric pressure Mercury Barometer Aneroid Barometer
  • 10. STP Standard Temperature & Pressure 273 K 101.325 kPa STP
  • 11. STP Standard Laboratory Conditions 25°C or 298 K 101.325 kPa SLC
  • 13. Boyle’s Law The pressure and volume of a gas are inversely related at constant mass & temp. P V PV = k 2 2 1 1 V P V P   
  • 15. Practice 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?
  • 16. 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 P1 = 726 mmHg V1 = 946 mL P2 = ? V2 = 154 mL P2 = P1 x V1 V2 726 mmHg x 946 mL 154 mL = = 4460 mmHg
  • 17. k T V  V T Charles’ Law The volume and absolute temperature (K) of a gas are directly related at constant mass & pressure 2 2 1 1 T V T V 
  • 19. Practice 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?
  • 20. 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? 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 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 For more lessons, visit www.chalkbored.com
  • 21. 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 = 3.20 L T1 = 398.15 K V2 = 1.54 L T2 = ? T2 = V2 x T1 V1 1.54 L x 398.15 K 3.20 L = = 192 K V1/T1 = V2/T2
  • 22. k T P  P T Gay-Lussac’s Law The pressure and absolute temperature (K) of a gas are directly related at constant mass & volume
  • 25. GIVEN: V1 = 473 cm3 T1 = 36°C = 309K V2 = ? T2 = 94°C = 367K WORK: P1V1T2 = P2V2T1 E. Gas Law Problems A gas occupies 473 cm3 at 36°C. Find its volume at 94°C. CHARLES’ LAW T V (473 cm3)(367 K)=V2(309 K) V2 = 562 cm3
  • 26. GIVEN: V1 = 100. mL P1 = 150. kPa V2 = ? P2 = 200. kPa WORK: P1V1T2 = P2V2T1 E. Gas Law Problems A gas occupies 100. mL at 150. kPa. Find its volume at 200. kPa. BOYLE’S LAW P V (150.kPa)(100.mL)=(200.kPa)V2 V2 = 75.0 mL
  • 27. Practice A gas occupies 7.84 cm3 at 71.8 kPa & 25°C. Find its volume at STP.
  • 28. GIVEN: V1 = 7.84 cm3 P1 = 71.8 kPa T1 = 25°C = 298 K V2 = ? P2 = 101.325 kPa T2 = 273 K WORK: P1V1T2 = P2V2T1 (71.8 kPa)(7.84 cm3)(273 K) =(101.325 kPa) V2 (298 K) V2 = 5.09 cm3 E. Gas Law Problems A gas occupies 7.84 cm3 at 71.8 kPa & 25°C. Find its volume at STP. P T V COMBINED GAS LAW
  • 29. GIVEN: P1 = 765 torr T1 = 23°C = 296K P2 = 560. torr T2 = ? WORK: P1V1T2 = P2V2T1 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 P T (765 torr)T2 = (560. torr)(309K) T2 = 226 K = -47°C
  • 30. k n V  V n Avogadro’s Principle  Equal volumes of all gases contain equal numbers of moles at constant temp & pressure. 2 2 1 1 n V n V 
  • 31. The Ideal Gas Equation The gas laws can be combined into a general equation that describes the physical behavior of all gases. 11.5 nT V P  V P  1 Boyle’s law V n  Avogadro’s law V T  Charles’s law nT V R P  PV = nRT rearrangement R is the proportionality constant, called the gas constant.
  • 32. B. Ideal Gas Law UNIVERSAL GAS CONSTANT R = 8.3145 J/mol·K R=0.0821 Latm/molK PV=nRT
  • 33. 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
  • 34. GIVEN: P = ? atm n = 0.412 mol T = 16°C = 289 K V = 3.25 L R = 0.0821Latm/molK WORK: PV = nRT P(3.25)=(0.412)(0.0821)(289) L mol Latm/molK K P = 3.01 atm 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
  • 35. GIVEN: V = ? n = 85 g T = 25°C = 298 K P = 104.5 kPa R = 8.315 dm3kPa/molK B. Ideal Gas Law Find the volume of 85 g of O2 at 25°C and 104.5 kPa. = 2.7 mol WORK: 85 g 1 mol = 2.7 mol 32.00 g PV = nRT (104.5)V=(2.7) (8.315) (298) kPa mol dm3kPa/molK K V = 64 dm3 IDEAL GAS LAW

Editor's Notes

  1. Gases only occupy about 0.1 % of the volume of their containers.