1. The Gas Laws
Chemistry
Dr. May

2. Gaseous Matter Indefinite volume and no fixed shape
Particles move independently of each other
Particles have gained enough energy to overcome the attractive forces that held them together as solids and liquids

3. Avogadro’s Number
One mole of a gas contains Avogadro’s number of molecules
Avogadro’s number is
6.02 x or
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 32 grams
Occupies 22.4 liters at STP
273 Kelvins (0oC)
One atmosphere ( kPa)(760 mm)
Contains 6.02 x 1023 molecules
(Avogadro’s Number)

8. Mole of Carbon Dioxide (CO2)
Has a mass of 44 grams
Occupies 22.4 liters at STP
Contains 6.02 x 1023 molecules

9. One Mole of Nitrogen Gas (N2)
Has a mass of 28 grams
Occupies 22.4 liters at STP
Contains 6.02 x 1023 molecules

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

11. Standard Conditions (STP)
Molar Volume
Standard
Temperature
Standard Pressure
22.4 liters/mole
0 oC
273 Kelvins
1 atmosphere
kilopascals
760 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
V1 = V2
T T2
As the temperature goes up , the volume goes up 

14. As the pressure goes up , the volume goes down 
Boyle’s Law
At constant temperature, the volume of a gas is inversely proportional to the pressure.
P1V1 = P2V2
As the pressure goes up , the volume goes down 

15. Combined Gas Law
P1V1 = P2V2
T1 T2
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 = 2.09 liters

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?
V2 = 2.09 liters

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)
V2 = 0.78 liters

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?
V2 = 0.78 liters

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 = = liters
30193

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?
V2 = 1.77 liters

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

31. The End
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32. The Ideal Gas Law
Chemistry
Dr. May

33. Kinetic Molecular Theory
Molecules of an ideal gas
Are dimensionless points
Are in constant, straight-line motion
Have kinetic energy proportional to their absolute temperature
Have elastic collisions
Exert no attractive or repulsive forces on each other

34. Ideal Gas Law
PV = nRT
P = pressure in kilopascals (kPa) or atmospheres (atm)
V = volume in liters
n = moles
T = temperature in Kelvins
R = universal gas constant

35. Ideal Gas Law: PV = nRT Pressure (P) Volume (V) Moles (n)
Temperature (T)
The universal gas constant (R)
Atm or kPa
Always liters
Moles
Kelvins
( P in atm) or
8.3 (P in kPa)

36. Universal Gas Constant
R = if P = atmospheres
R = 8.3 if P = kilopascals
R = PV nT

37. Deriving R for P in Atmospheres
R = PV nT
Assume n = 1 mole of gas
Standard P = 1 atmosphere
Standard V = molar volume = 22.4 liters
Standard T = 273 Kelvins

38. R Value When P Is In Atmospheres
R = PV nT
R = (1) (22.4)
(1) 273
R = atm Liters
mole Kelvins

39. Deriving R For P In Kilopascals
R = PV nT
Assume n = 1 mole of gas
Standard P = kilopascals
Standard V = molar volume = 22.4 liters
Standard T = 273 Kelvins

40. R Value When P Is In Kilopascals
R = PV nT
R = (101.32) (22.4)
(1) 273
R = 8.3 kPa Liters
mole Kelvins

41. Ideal Gas Law - Pressure
PV = nRT
P = nRT V
Solves for pressure when moles, temperature, and volume are known

42. Solves for volume when moles, temperature, and pressure are known
Ideal Gas Law - Volume
PV = nRT
V = nRT P
Solves for volume when moles, temperature, and pressure are known

43. Ideal Gas Law - Temperature
PV = nRT
T = PV nR
Solves for temperature when moles, pressure, and volume are known

44. Solves for moles when pressure, temperature, and volume are known
Ideal Gas Law - Moles
PV = nRT
n = PV RT
Solves for moles when pressure, temperature, and volume are known

45. Ideal Gas Law Problem
What is the mass of nitrogen in a 2.3 liter container at 1.2 atmospheres, and 25 oC ?
V = 2.3 liters
P = 1.2 atmospheres
T = 25 oC = 298 Kelvins
R = since P is in atms.
Find moles (n), then grams

46. Ideal Gas Law Solution (moles)
PV = nRT
1.2 (2.3) = n (0.0821) (298)
n = ( 2.3) = moles
(0.0821) (298)

47. Ideal Gas Law Solution (Grams)
Grams = moles x molecular weight (MW)
Moles = 0.11
Molecular Weight of N2 = 28 g/mole
Grams = 0.11 x 28 = 3.1 grams

48. Ideal Gas Law Answer
What is the mass of nitrogen in a 2.3 liter container at 1.2 atmospheres, and 25 oC ?
The answer is 0.11 moles and
3.1 grams

49. The End
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