1. Chapter 13 – Gases 13.1 The Gas Laws
13.2 The Ideal Gas Law (Part I – Equation Only)
13.3 Gas Stoichiometry

2. Section 13.2 The Ideal Gas Law
The ideal gas law relates the number of particles to pressure, temperature, and volume.
Relate the amount of gas present to its pressure, temperature, and volume using the ideal gas law.

3. Section 13.2 The Ideal Gas Law
Key Concept
The ideal gas law relates the amount of a gas present to its pressure, temperature, and volume
PV = nRT

4. Gas Laws – Mechanistic and Empirical
“Named” laws described in section were empirically determined from experiments – stated relationships “fit the data”
Example: Boyle experimentally determined that P & V are inversely related but there was no underlying understanding of why this occurs – this is an empirical model

5. Gas Laws – Mechanistic and Empirical
Ideal gas law is mechanistic – it can be derived by using the basic theory (kinetic- molecular) of how gases behave; the underlying mechanism that leads to the equation is known and understood
Ideal gas law was formulated after the empirical (named) laws were discovered
It predicted the same behavior as the empirical laws but provided a way to understand why they were true

6. Kinetic-Molecular Theory (KMT) of Gases (see section 12.1)
Mostly empty space
No attractions/repulsions between particles
Constant straight-line motion until collisions with other particles or walls
Collisions perfectly elastic: KE constant
T constant, KE constant
All gases have same KEavg at a given T

7. R = Gas Constant – value depends upon units used for P
Ideal Gas Law
Relationship between pressure P, volume V, Kelvin temperature T, and number of moles n
PV = nRT
T must be expressed in units of kelvin
Equation derived using KMT assumptions
R = Gas Constant – value depends upon units used for P

8. R (Gas Constant) – Table 13.2
Value
Units
0.0821
L atm
mol K
8.314
L kPa
62.4
L mm Hg

9. Ideal Gas Law – Dependence on n
PV = nRT
Fixed V, T then
P  n
Fixed P, T
V  n

10. Ideal Gas Law: Example Problem 13.6
Calculate number of moles of NH3 for V = 3.0 L, T = 3.00x102 K, P = 1.50 atm
PV = nRT  n = PV / RT
n = (1.50 atm  3.0 L) / (8.21x10-2 L atm/mol K  3.00x102 K)
n = 0.18 mol
Note that value of R was chosen that has atm as the pressure unit

11. Practice (Ideal Gas Law)
Problems 26 – 30, page 455
Problems 68, 69, 74, 75, 77 – 78, pages 468 – 9
Problems 15, 16, 18, page 985

12. Ideal Gas Law and the Combined Gas Law
PV = nRT
If the # of moles of gas (n) is fixed, then
PV/T = nR = constant
If P,V,T conditions change (from 1 to 2)
P1V1/T1 = P2V2/T2
(have derived the Combined Gas Law)

13. Combined & Named Gas Laws (Subject of Section 13.1)
For given mass (fixed # moles) of gas
P1V1/T1 = P2V2/T2
3 laws can be derived from this:
Fix T: P1V1 = P2V2 (Boyle’s)
Fix P: V1/T1 = V2/T2 (Charles’)
Fix V: P1/T1 = P2/T2 (Gay-Lussac’s)

14. Chapter 13 – Gases 13.1 The Gas Laws 13.2 The Ideal Gas Law (Part I)
13.3 Gas Stoichiometry

15. Section The Gas Laws
For a fixed amount of gas, a change in one variable—pressure, temperature, or volume—affects the other two.
State the relationships among pressure, temperature, and volume of a constant amount of gas.
Apply the gas laws to problems involving the pressure, temperature, and volume of a constant amount of gas.

16. Section 13.1 The Gas Laws Key Concepts
Boyle’s law states that the volume of a fixed amount of gas is inversely proportional to its pressure at constant temperature.
P1V1 = P2V2
Charles’s law states that the volume of a fixed amount of gas is directly proportional to its kelvin temperature at constant pressure.

17. Section 13.1 The Gas Laws Key Concepts
Gay-Lussac’s law states that the pressure of a fixed amount of gas is directly proportional to its kelvin temperature at constant volume.
The combined gas law relates pressure, temperature, and volume in a single statement.

18. Boyle’s Law
Volume of given mass (fixed # of moles) of gas held at constant T varies inversely with pressure
P1V1 = P2V2
PV = constant
V2 = V1 (P1/P2)

19. 10 L
1 atm
5 L
2 atm
2.5 L
4 atm
Constant = PV = 10 atm L

21. Boyle’s Law – Example Problem 13.1
Diver blows bubble of volume = 0.75 L at 10 m depth & pressure = 2.25 atm. Volume of bubble at surface when pressure = 1.03 atm?
P1V1 = P2V2
V2 = V1 (P1/P2)
V2 = 0.75 L (2.25 atm / 1.03 atm)
V2 = 1.6 L

22. Practice (Boyle’s Law)
Problems 1- 3 page 443
Problems 59 – 60, page 468
Problems 1 – 2, page 984

23. Charles’s Law
Volume of given mass (fixed # of moles) of gas held at constant P is directly proportional to its kelvin temperature
V1/T1 = V2/T2
V2 = V1(T2/T1)
TK = TC

24. V change with T at fixed P

25. Volume vs Celsius Temperature

26. Volume vs Kelvin Temperature

27. Volume vs T (Kelvin) – Different Gases
Gases follow Charles’s Law but liquefy at different T– dashed lines are extrapolations of gas phase behavior
Ideal gases occupy no volume at absolute zero

28. Charles’s Law: Example Problem 13.2
Balloon has volume of 2.32 L at 40.0 C. Volume at 75.0 C if P remains constant?
V1/T1 = V2/T2
T1 = 40.0 C = K
T2 = 75.0 C = K
V2 = V1 (T2/T1)
V2 = 2.32 L (348.2 K/ K)
V2 = 2.58 L

29. Practice (Charles’s Law)
Problems 4 – 7, page 446
Problems 55, 58, 59 page 448
Problems 3 – 4, page 984

30. Joseph Louis Gay-Lussac (1778 – 1850)
French chemist & physicist. Known mostly for 2 laws related to gases, & for his work on alcohol-water mixtures, which led to the degrees Gay-Lussac used to measure alcoholic beverages in many countries.
Prof. of physics at Sorbonne (1808 to 1832), a post which he only resigned for the chair of chemistry at the Jardin
des Plantes. In 1802, first formulated law stating that if mass and pressure of gas are held constant, then gas volume increases linearly as temperature rises.

31. P change with T at fixed V
Gay-Lussac’s Law
P change with T at fixed V

32. Gay-Lussac’s Law
Pressure of given mass (fixed # of moles) of gas in a constant volume is directly proportional to its kelvin temperature
P1/T1 = P2/T2
P2 = P1(T2/T1)
TK = TC

33. At 0 K, ideal gas exerts zero pressure
Gay-Lussac’s Law
At 0 K, ideal gas exerts zero pressure

34. Gay-Lussac’s Law: Example Problem 13.3
Pressure of O2 in canister is 5.00 atm at 25.0 C. New pressure at T of C?
P1/T1 = P2/T2
T1 = 25.0 C = K
T2 = C = K
P2 = P1(T2/T1)
P2 = 5.00 atm (263.2 K / K)
P2 = 4.41 atm

35. Practice (Gay-Lussac’s Law)
Problems 8 – 10, page 448
Problem 18, page 451
Problem 59 page 468
Problems 5 – 7, page 984

36. If P,V,T conditions change (from 1 to 2)
Combined Gas Law
If P,V,T conditions change (from 1 to 2)
P1V1/T1 = P2V2/T2

37. Combined Gas Law: Practice Problem 13.4
Gas at 110 kPa and 30.0 C fills flexible container with initial V of 2.00 L. If T and P raised to 80.0 C and 440 kPa, new V?
T1 = 30.0 C = K
T2 = 80.0 C = K
P1V1/T1 = P2V2/T2
V2 = V1 (P2 /P1) (T1/T2)
V2 = 2.00 L (440 kPa/ 110 kPa) (303.2 K/ K)
V2 = 0.58 L

38. Practice (Combined Gas Law)
Problems 11 – 13, page 450
Problem 60 page 468
Problems 8 – 9, page 984

39. Gas Laws Summary – Table 13.1
All require fixed n (moles of gas)

40. Chapter 13 – Gases 13.1 The Gas Laws (includes Combined)
13.2 The Ideal Gas Law (Part II - Avogadro’s Principle, Gas Density, Real Gases)
13.3 Gas Stoichiometry

41. Section 13.2 The Ideal Gas Law
The ideal gas law relates the number of particles to pressure, temperature, and volume.
Relate number of particles and volume using Avogadro’s principle.
Compare the properties of real and ideal gases.

42. Section 13.2 The Ideal Gas Law
Key Concepts
Avogadro’s principle states that equal volumes of gases at the same pressure and temperature contain equal numbers of particles.
The ideal gas law can be used to find molar mass if the mass of the gas is known, or the density of the gas if its molar mass is known.
At very high pressures and very low temperatures, real gases behave differently than ideal gases.

43. Avogadro’s Principle
Equal volumes of gases at same T and P contain equal numbers of particles
Size of gas particles negligible compared to volume occupied, so it doesn’t matter which gas you are using

44. Molar Volume of Gas
Volume occupied by one mole of gas at 0.00 C and 1.00 atmosphere pressure
T, P conditions known as STP – Standard Temperature and Pressure
At STP, 1 mole of any gas occupies 22.4 L
Key to lots of problems involving gases at STP

45. Molar Volume of Gas: Example Problem 13.5
V of 2.00 kg of methane at STP?
2.00 kg CH4  1.00x103 g/kg  mol CH4/16.05 g CH4 = 125 mol CH4
125 mol CH4  22.4 L CH4 / mol CH4 (at STP)
V = 2.80x103 L

46. Practice (Avogadro’s Principle & Molar Volume)
Problems 20 – 25, page 453
Problems 67, 71, 73 pages 468-9
Problems , page 984

47. Ideal Gas Law: Calculate M or Density
PV = nRT
n = # moles = mass /molar mass = m / M
PV = mRT / M  MP = mRT / V 
M = mRT / PV
M = (m/V) RT / P (m/V) = density = D
M = D RT / P or D = MP / RT
For a given P, T , if know:
1. mass m & V, can compute molar mass M
2. density D, can compute molar mass M
3. molar mass M, can compute density D

48. Practice (Molar Mass, Density from Ideal Gas Law)
Problems 70, 72, 76 page 469
Problems 17 – 18, page 985

49. Ideal Gas Law - Deviations
No gas truly ideal, but most behave as ideal gases over wide range of T & P
Deviations happen at high P and low T where assumptions of KMT break down
Volume of gas no longer negligible
An ideal gas can’t be liquefied
Intermolecular forces more significant
Polar molecules especially affected

50. Ideal Gas Law - Deviations
Polar molecules have larger attractive forces between particles and generally do not behave as ideal gases at low T where slower-moving particles can be affected by these forces
Large gas particles occupy more space and deviate more from ideal gases at high density (high P, low T) where there is less empty space

51. Real Gases - PV/RT vs P for 1 Mole of Gas
Approach ideal gas at low P
2.0
1.0 PV RT
P (atm)

52. Chapter 13 – Gases 13.1 The Gas Laws 13.2 The Ideal Gas Law
13.3 Gas Stoichiometry

53. Section 13.3 Gas Stoichiometry
When gases react, the coefficients in the balanced chemical equation represent both molar amounts and relative volumes.
Determine volume ratios for gaseous reactants and products by using coefficients from chemical equations.
Apply gas laws to calculate amounts of gaseous reactants and products in a chemical reaction.

54. Section 13.3 Gas Stoichiometry
Key Concepts
The coefficients in a balanced chemical equation specify volume ratios for gaseous reactants and products.
The gas laws can be used along with balanced chemical equations to calculate the amount of a gaseous reactant or product in a reaction.

55. Balanced Equation & Gas Volumes
Gas laws can be applied to calculate stoichiometry of reactions in which gases are reactants or products
2H2(g) + O2(g) → 2H2O(g)
2 mol H2 reacts with 1 mol O2 to produce 2 mol water vapor
Coefficients in balanced equation represent volume ratios for gases if P and T are fixed during reaction

56. Balanced Equation & Gas Volumes
CH4(g) O2(g)  CO2(g) H2O(g)
There is a one-to-one relationship between number of moles and gas volume

57. Balanced Equation & Gas Volumes
2C4H10(g) + 13O2(g)  8CO2(g) + 10H2O(g)
Interpreted as moles:
2 mol butane + 13 mol oxygen 
8 mol carbon dioxide + 10 mol water
Avogadro’s principle  can also be volume
2 L butane + 13 L oxygen 
8 L carbon dioxide + 10 L water

58. Gas Stoichiometry Map

59. Practice – Gas Stoichiometry
How many mol of hydrogen gas are required to react with 1.50 mol oxygen gas in the following reaction?
2H2(g) + O2(g) → 2H2O(g) A B C D ?

60. V toV Stoichiometry: Practice Problem 13.7
Volume of oxygen gas needed for complete combustion of 4.00 L of propane (C3H8) assuming P and T remain constant?
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)
4.00 L C3H8  5 L O2 / L C3H8 = 20.0 L O2

61. Practice – Gas Stoich. (all V to V)
Problems 38 – 41, page 461
Problems 84 – 86, page 469
Problems 19 – 20, page 985

62. Combo Mass/Volume Problems
Need:
Balanced equation
At least one mass or V for product or reactant
Conditions under which volumes measured

63. Combo Mass/Volume – Prob 13.8
N2(g) + 3H2(g)  2NH3(g)
5.00 L N2 reacts completely
P = 3.00 atm T = 298 K
Mass of ammonia produced?
5.00 L N2  [2 L NH3/1 L N2] =
volume ratio from equation
10.0 L NH3

64. Combo Mass/Volume – Prob 13.8
N2(g) + 3H2(g)  2NH3(g)
10.0 L NH3 P = 3.00 atm T = 298 K
n = PV/RT = ( atm  10.0 L ) (298 K  L atm/mol K)
= 1.23 mol NH3
Molar mass NH3 = g/mol
1.23 mol NH3  g NH3/mol NH3
= 21.0 g NH3

65. Practice (Stoich. m & V combo)
Problems 42 – 45, page 463
Problems 88 – 90, page 469
Problems 21 – 22, page 985