The document discusses the key gas laws and concepts: 1) Gases are highly compressible, occupy their container fully, exert uniform pressure, diffuse easily, and have low densities according to the characteristics of gases. 2) The kinetic molecular theory states that gases consist of molecules in random motion where volume and intermolecular forces are negligible compared to temperature and collisions. 3) The gas laws of Boyle, Charles, Gay-Lussac, Avogadro, and the combined ideal gas law relate the pressure, volume, temperature, and moles of a gas at a given time. 4) Problems can be solved using the appropriate gas law equations to find pressure, volume, temperature or

1. The Gas Laws

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

12. -BOYLES
-CHARLE
-GAY-LUSSAC
The Gas Laws

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 



14. A. Boyle’s Law

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


18. Charles’ Law

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
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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

23. Gay-Lussac’s Law

24. Combined Gas Law
P1V1
T1
=
P2V2
T2
P1V1T2 = P2V2T1

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