1. The Gas Laws

2. 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.

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

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
Always use absolute temperature (Kelvin) when working with gases. ºF ºC K
-459 32
212
-273
100
273
373
K = ºC + 273

6. Kelvin Practice
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
Absolute zero is – 273C or 0 K
273
373
– 173
243 27
130
298
– 273

7. 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.
Which shoes create the most pressure?

8. Pressure 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 = Pa.

9. Pressure Barometer measures atmospheric pressure Aneroid Barometer
Mercury Barometer
Aneroid Barometer

10. Standard Temperature & Pressure
STP
STP
Standard Temperature & Pressure
273 K
kPa

11. Standard Laboratory Conditions
STP
SLC
Standard Laboratory Conditions
25°C or 298 K
kPa

12. -BOYLES -CHARLE -GAY-LUSSAC
The Gas Laws P V T
-BOYLES -CHARLE -GAY-LUSSAC

13. Boyle’s Law
The pressure and volume of a gas are inversely related at constant mass & temp.
PV = k P V

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. P1 x V1 = P2 x V2 P1 = 726 mmHg P2 = ? V1 = 946 mL V2 = 154 mL P1 x V1
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
P2 = ?
V1 = 946 mL
V2 = 154 mL
P1 x V1 V2
726 mmHg x 946 mL
154 mL =
P2 =
= 4460 mmHg

17. Charles’ Law
The volume and absolute temperature (K) of a gas are directly related at constant mass & pressure V T

18. Charles’ Law

19. Practice
A sample of gas occupies 3.5 L at 300 K. What volume will it occupy at 200 K?
If a 1 L balloon is heated from 22°C to 100°C, what will its new volume be?

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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
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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21. V1/T1 = V2/T2 V1 = 3.20 L V2 = 1.54 L T1 = 398.15 K T2 = ? V2 x T1 V1
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
V1 = 3.20 L
V2 = 1.54 L
T1 = K
T2 = ?
V2 x T1 V1
1.54 L x K
3.20 L =
T2 =
= 192 K

22. Gay-Lussac’s Law
The pressure and absolute temperature (K) of a gas are directly related at constant mass & volume P T

23. Gay-Lussac’s Law

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

25. E. Gas Law Problems
A gas occupies 473 cm3 at 36°C. Find its volume at 94°C.
CHARLES’ LAW
GIVEN:
V1 = 473 cm3
T1 = 36°C = 309K
V2 = ?
T2 = 94°C = 367K T V
WORK:
P1V1T2 = P2V2T1
(473 cm3)(367 K)=V2(309 K)
V2 = 562 cm3

26. E. Gas Law Problems
A gas occupies 100. mL at kPa. Find its volume at 200. kPa.
BOYLE’S LAW
GIVEN:
V1 = 100. mL
P1 = 150. kPa
V2 = ?
P2 = 200. kPa P V
WORK:
P1V1T2 = P2V2T1
(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. E. Gas Law Problems COMBINED GAS LAW P T V V1 = 7.84 cm3
A gas occupies 7.84 cm3 at 71.8 kPa & 25°C. Find its volume at STP.
COMBINED GAS LAW
GIVEN:
V1 = 7.84 cm3
P1 = 71.8 kPa
T1 = 25°C = 298 K
V2 = ?
P2 = kPa
T2 = 273 K
P T V
WORK:
P1V1T2 = P2V2T1
(71.8 kPa)(7.84 cm3)(273 K)
=( kPa) V2 (298 K)
V2 = 5.09 cm3

29. 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:
P1 = 765 torr
T1 = 23°C = 296K
P2 = 560. torr
T2 = ? P T
WORK:
P1V1T2 = P2V2T1
(765 torr)T2 = (560. torr)(309K)
T2 = 226 K = -47°C

30. Avogadro’s Principle
Equal volumes of all gases contain equal numbers of moles at constant temp & pressure. V n

31. The Ideal Gas Equation
11.5
The gas laws can be combined into a general equation that describes the physical behavior of all gases.
Boyle’s law
Charles’s law
Avogadro’s law
PV = nRT
rearrangement
R is the proportionality constant, called the gas constant.

32. PV=nRT UNIVERSAL GAS CONSTANT R = 8.3145 J/mol·K R=0.0821 Latm/molK
B. Ideal Gas Law
PV=nRT
UNIVERSAL GAS CONSTANT
R = J/mol·K R= Latm/molK

33. R = 0. 0821 liter·atm/mol·K R = 8. 3145 J/mol·K R = 8
R = liter·atm/mol·K R = J/mol·K R = m3·atm/mol·K R = L·Torr/mol·K or L·mmHg/mol·K

34. B. Ideal Gas Law IDEAL GAS LAW P = ? atm n = 0.412 mol
Calculate the pressure in atmospheres of mol of He at 16°C & occupying L.
IDEAL GAS LAW
GIVEN:
P = ? atm
n = mol
T = 16°C = 289 K
V = 3.25 L
R = L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

35. B. Ideal Gas Law IDEAL GAS LAW V = ? n = 85 g T = 25°C = 298 K
Find the volume of 85 g of O2 at 25°C and kPa.
IDEAL GAS LAW
GIVEN:
V = ?
n = 85 g
T = 25°C = 298 K
P = kPa
R = dm3kPa/molK
WORK:
85 g 1 mol = 2.7 mol
32.00 g
= 2.7 mol
PV = nRT
(104.5)V=(2.7) (8.315) (298)
kPa mol dm3kPa/molK K
V = 64 dm3