1. The Gas Laws
Chemistry
Dr. May

2. Gaseous Matter Indefinite __________ and ___ fixed shape
Particles move ____________ of each other
Particles have gained enough energy to overcome the attractive forces that held them together as _________ ______ __________

3. Avogadro’s Number
One mole of a gas contains Avogadro’s number of molecules
Avogadro’s number is _______________
602,000,000,000,000,000,000,000

4. Diatomic Gas Elements Gas Hydrogen (H2) Nitrogen (N2) Oxygen (O2)
Fluorine (F2)
Chlorine (Cl2)
Molar Mass
2 grams/mole
28 grams/mole
32 grams/mole
38 grams/mole
70 grams/mole

5. Inert Gas Elements Gas Helium Neon Argon Krypton Xenon Radon
Molar Mass
4 grams/mole
20 grams/mole
40 grams/mole
84 grams/mole
131 grams/mole
222 grams/mole

6. Other Important Gases Gas Carbon Dioxide Carbon Monoxide
Sulfur Dioxide
Methane
Ethane
Freon 14
Formula Molar Mass
CO2 44 g/mole
CO 28 g/mole
SO2 64 g/mole
CH4 16 g/mole
CH3CH3 30 g/mole
CF4 88 g/mole

7. One Mole of Oxygen Gas (O2)
Has a mass of _____ grams
Occupies _______ liters at STP
273 Kelvins (0oC)
One atmosphere ( kPa)(760 mm)
Contains _____________ molecules
(Avogadro’s Number)

8. Mole of Carbon Dioxide (CO2)
Has a mass of ___ grams
Occupies _______ liters at STP
Contains _________________ molecules

9. One Mole of Nitrogen Gas (N2)
Has a mass of ___ grams
Occupies _______ liters at STP
Contains _____________ molecules

10. Mole of Hydrogen Gas (H2)
Mass
Volume at STP
Molecules
______ grams
_______ liters
6.02 x 1023

11. Standard Conditions (STP)
Molar Volume
Standard
Temperature
Standard Pressure
_______ liters/mole
_____ oC
_____ Kelvins
____ atmosphere
________ kilopascals
_____ mm Hg

12. Gas Law Unit Conversions
liters  milliliters
milliliters  liters
o C  Kelvins
Kelvins  o C
mm  atm
atm  mm
atm  kPa
kPa  atm
Multiply by 1000
Divide by 1000
Add 273
Subtract 273
Divide by 760
Multiply by 760
Multiply by
Divide by

13. Charles’ Law
At constant pressure, the volume of a gas is directly proportional to its temperature in Kelvins
___________________________________
As the temperature goes _____, the volume goes_______

14. As the pressure goes _______ the volume goes ________
Boyle’s Law
At constant temperature, the volume of a gas is inversely proportional to the pressure.
__________________________________
As the pressure goes _______ the volume goes ________

15. Combined Gas Law ________________________________________
Standard Pressure (P) = kPa, 1 atm, or 760 mm Hg
Standard Temperature (T) is 273 K
Volume (V) is in liters, ml or cm3

16. Charles’ Law Problem
A balloon with a volume of 2 liters and a temperature of 25oC is heated to 38oC. What is the new volume?
1. Convert oC to Kelvins
= 298 K
= 311 K
2. Insert into formula

17. Charles’ Law Solution T1 T2 V1 = 2 liters V2 = Unknown
V1 = V2
T T2
V1 = 2 liters V2 = Unknown
T1 = 298 K T2 = 311 K
2 = V2

18. Charles’ Law Solution 2 = V2 298 311 298 V2 = (2) 311 V2 = 622 298
298 V2 = (2) 311
V2 = 622
298
V2 =

19. Charles’ Law Problem Answer
A balloon with a volume of 2 liters and a temperature of 25oC is heated to 38oC. What is the new volume?

20. Boyle’s Law Problem A balloon has a volume of 2.0 liters at
743 mm. The pressure is increased to 2.5 atmospheres (atm). What is the new volume?
1. Convert pressure to the same units
743  760 = .98 atm
2. Insert into formula

21. Boyle’s Law Solution P1V1 = P2V2 P1 = 0.98 atm P2 = 2.5 atm
V1 = 2.0 liters V2 = unknown
0.98 (2.0) = 2.5 V2

22. Boyle’s Law Solution P1V1 = P2V2 0.98 (2.0) = 2.5 V2 V2 = 0.98 (2.0)

23. Boyle’s Law Problem Answer
A balloon has a volume of 2.0 liters at
743 mm. The pressure is increased to 2.5 atmospheres (atm). What is the new volume?

24. Combined Gas Law Problem
A balloon has a volume of 2.0 liters at a pressure of 98 kPa and a temperature of 25 oC. What is the volume under standard conditions?
1. Convert 25 oC to Kelvins = 298 K
2. Standard pressure is kPa
3. Standard temperature is 273 K
4. Insert into formula

25. Combined Gas Law Solution
P1V1 = P2V2
T T2
P1 = 98 kPa P2 = kPa
V1 = 2.0 liters V2 = unknown
T1 = 298 K T2 = 273 K

26. Combined Gas Law Solution
P1V1 = P2V2
T T2
98 (2.0) = V2
(298) (101.32) V2 = (273) (98) (2.0)

27. Combined Gas Law Solution
P1V1 = P2V2
T T2
(298) (101.32) V2 = (273) (98) (2.0)
V2 = (98) (2.0)
(298) (101.32)

28. Combined Gas Law Solution
P1V1 = P2V2
T1 T2
V2 = (98) (2.0)
(298) (101.32)
V2 =

29. Combined Gas Law Problem Answer
A balloon has a volume of 2.0 liters at a pressure of 98 kPa and a temperature of 25 oC. What is the volume under standard conditions?

30. Combined Gas Law – V2 T1 T2 P1V1T2 = P2V2T1 P2T1 P1V1 = P2V2
P1V1T2 = V2
P2T1

31. The End
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