1. Gases
Chapter 5
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2. Physical Characteristics of Gases
Gases assume the volume and shape of their containers.
Gases are the most compressible state of matter.
Gases will mix evenly and completely when confined to the same container.
Gases have much lower densities than liquids and solids.
NO2 gas

3. Kinetic Molecular Theory of Gases
A gas is composed of molecules that are separated from each other by distances far greater than their own dimensions. The molecules can be considered to be points; that is, they possess mass but have negligible volume.
Gas molecules are in constant motion in random directions, and they frequently collide with one another. Collisions among molecules are perfectly elastic.
Gas molecules exert neither attractive nor repulsive forces on one another.
The average kinetic energy of the molecules is proportional to the temperature of the gas in kelvins. Any two gases at the same temperature will have the same average kinetic energy
KE = ½ mu2

4. Elements that exist as gases at 250C and 1 atmosphere

6. VARIABLES USED TO DEFINE PROPERTIES OF GASES
PRESSURE: (P)
MOLES: (n)
TEMPERATURE: (T)
VOLUME: (V)
EQUATIONS

7. PRESSURE TERMS USED TO MEASURE PRESSURE Atmosphere (atm) Mm of Hg Torr
Pascal (Pa)

8. (force = mass x acceleration)
Area
Pressure =
(force = mass x acceleration)
Units of Pressure
1 pascal (Pa) = 1 N/m2
1 atm = 760 mmHg = 760 torr
1 atm = 101,325 Pa

9. Manometers Used to Measure Gas Pressures
closed-tube
open-tube

10. Conversion of Pressure Units
Problem 5.14
The atmospheric pressure at the summit of Mt. McKinley is 606 mm Hg on a certain day. What is the pressure in atm and in kPa?

11. BOYLE’S LAW FOCUSES ON RELATIONSHIP BETWEEN PRESSURE AND VOLUME
PRESSURE AND VOLUME ARE INVERSELY PROPORTIONAL

12. P x V = constant P1 x V1 = P2 x V2 P1 = 726 mmHg P2 = ? V1 = 946 mL
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?
P x V = constant
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

13. Problem 5.19
A gas occupying a volume of 725 mL at a pressure of atm is allowed to expand at constant temperature until its pressure reaches atm. What is the final volume?

14. 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?

15. Boyles law Temperature does have an effect on pressure and volume.
If temperature increases, volume increases and pressure decreases.

16. Variation of Gas Volume with Temperature at Constant Pressure
Charles’ & Gay-Lussac’s Law
V a T
Temperature must be
in Kelvin
V = constant x T
V1/T1 = V2 /T2
T (K) = t (0C)

17. A sample of carbon monoxide gas occupies 3. 20 L at 125 0C
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 = ?
T1 = 125 (0C) (K) = K
V2 x T1 V1
1.54 L x K
3.20 L =
T2 =
= 192 K

18. VARIABLES USED TO DEFINE PROPERTIES OF GASES
PRESSURE
MOLES
TEMPERATURE
VOLUME
EQUATIONS

19. VARIABLES USED TO DEFINE PROPERTIES OF GASES
PRESSURE: (P)
MOLES: (n)
TEMPERATURE: (T)
VOLUME: (V)
EQUATIONS

20. RELATIONSHIP BETWEEN NUMBER OF MOLES AND OTHER VARIABLES
AVOGARDO’S LAW
FOCUSES ON RELATIONSHIP
MOLES
PRESSURE
VOLUME

21. AVOGARDO’S LAW ASSUMING TEMPERATURE AND PRESSURE ARE CONSTANT
VOLUME DIRECTLY PROPORTIONAL TO NUMBER OF MOLES

22. Avogadro’s Law V a number of moles (n) V = constant x n
Constant temperature
Constant pressure
V a number of moles (n)
V = constant x n
V1 / n1 = V2 / n2

23. Relationship between Volume and Temperature
Charles’s and Gay-Lussac’s Law
Temperature in Kelvin
V1/T1 = 2

24. Relationship Between Pressure and Temperature
Another form of Charles’s and Gay Lussac’s Law
P1/T1 = P2/T2

25. Problem 5.22 (b)
A sample of air occupies 3.8 liters when the pressure is 1.2 atm. (b) What pressure is required in order to compress it to L.

26. Problem 5.36
The temperature of 2.5 L of a gas initially at STP is raised to 250 C at at constant volume. Calculate the final pressure of the gas in atm.
ST = 0 degrees C
SP = 1 atm

27. PV = nRT PV (1 atm)(22.414L) R = = nT (1 mol)(273.15 K)
The conditions 0 0C and 1 atm are called standard temperature and pressure (STP).
Experiments show that at STP, 1 mole of an ideal gas occupies L.
PV = nRT
R = PV nT =
(1 atm)(22.414L)
(1 mol)( K)
R = L • atm / (mol • K)

28. Summary of Gas Laws
Boyle’s Law

29. Charles Law

30. Avogadro’s Law

31. Ideal Gas Equation 1 Boyle’s law: P a (at constant n and T) V
Charles’ law: V a T (at constant n and P)
Avogadro’s law: V a n (at constant P and T)
V a nT P
V = constant x = R nT P
R is the gas constant
PV = nRT

32. PV = nRT PV (1 atm)(22.414L) R = = nT (1 mol)(273.15 K)
The conditions 0 0C and 1 atm are called standard temperature and pressure (STP).
Experiments show that at STP, 1 mole of an ideal gas occupies L.
PV = nRT
R = PV nT =
(1 atm)(22.414L)
(1 mol)( K)
R = L • atm / (mol • K)

33. PV = nRT nRT V = P 1.37 mol x 0.0821 x 273.15 K V = 1 atm V = 30.7 L
What is the volume (in liters) occupied by 49.8 g of HCl at STP?
T = 0 0C = K
P = 1 atm
PV = nRT
n = 49.8 g x
1 mol HCl
36.45 g HCl
= 1.37 mol
V =
nRT P
V =
1 atm
1.37 mol x x K
L•atm
mol•K
V = 30.7 L

34. PV = nRT n, V and R are constant nR V = P T = constant P1 T1 P2 T2 =
Argon is an inert gas used in lightbulbs to retard the vaporization of the filament. A certain lightbulb containing argon at 1.20 atm and 18 0C is heated to 85 0C at constant volume. What is the final pressure of argon in the lightbulb (in atm)?
PV = nRT
n, V and R are constant nR V = P T
= constant
P1 = 1.20 atm
T1 = 291 K
P2 = ?
T2 = 358 K P1 T1 P2 T2 =
P2 = P1 x T2 T1
= 1.20 atm x
358 K
291 K
= 1.48 atm

35. d is the density of the gas in g/L
Density (d) Calculations
m is the mass of the gas in g
d = m V = PM RT
M is the molar mass of the gas
Molar Mass (M ) of a Gaseous Substance
dRT P
M =
d is the density of the gas in g/L

36. dRT P M = d = m V = = 2.21 1 atm x 0.0821 x 300.15 K M = M =
A 2.10-L vessel contains 4.65 g of a gas at 1.00 atm and C. What is the molar mass of the gas?
dRT P
M =
d = m V
4.65 g
2.10 L =
= 2.21 g L
2.21 g L
1 atm
x x K
L•atm
mol•K
M =
M =
54.5 g/mol

37. Gas Stoichiometry
What is the volume of CO2 produced at 37 0C and 1.00 atm when 5.60 g of glucose are used up in the reaction:
C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l)

38. Dalton’s Law of Partial Pressures Gas Mixtures
Most of the volume occupied by a gas is empty space.
Ideally no attractive forces between gas molecules.
Mixtures of gases are common.

39. Dalton’s Law of Partial Pressures Gas Mixtures
Pi is pressure of gas in a mixture.
PT = PiA + PiB + PiC …….

40. Ideal Gas Law and Partial Pressures
PV =nRT
For a gas in a gas mixture, ideal gas law applies to each gas as well as to the mixture.

41. Consider a case in which two gases, A and B, are in a container of volume V.
PT = PA + PB
PA = nART/V
PB = nBRT/V
Partial Pressure exerted by Gas A and by Gas B dependent on concentration.

42. Mole Fraction
Mole Fraction of Gas (Xi) = Moles of gas/total moles in a mixture
XA = nA /nA + n B
XB = nB /nA + n B

43. mole fraction (Xi ) = ni/nT
Pi = Xi PT
When
mole fraction (Xi ) = ni/nT

44. Problem 5.22 (b)
A sample of air occupies 3.8 liters when the pressure is 1.2 atm. (b) What pressure is required in order to compress it to L.

45. Problem 5.70
A sample of ammonia (NH3) gas is completely decomposed to nitrogen and hydrogen gas over heated iron wool. If the total pressure is 866 mm Hg, calculate the partial pressure of N2 and H2.

46. A sample of natural gas contains 8. 24 moles of CH4, 0
A sample of natural gas contains 8.24 moles of CH4, moles of C2H6, and moles of C3H8. If the total pressure is 1.37 atm, what is the partial pressure of propane (C3H8)?

47. Vapor Pressure
Pressure exerted by a gas over a liquid in a closed container.

48. Collecting a Gas over Water
2KClO3 (s) KCl (s) + 3O2 (g)
PT = PO + PH O 2

49. Vapor of Water and Temperature

50. Kinetic Molecular Theory of Gases
A gas is composed of molecules that are separated from each other by distances far greater than their own dimensions. The molecules can be considered to be points; that is, they possess mass but have negligible volume.
Gas molecules are in constant motion in random directions, and they frequently collide with one another. Collisions among molecules are perfectly elastic.
Gas molecules exert neither attractive nor repulsive forces on one another.
The average kinetic energy of the molecules is proportional to the temperature of the gas in kelvins. Any two gases at the same temperature will have the same average kinetic energy
KE = ½ mu2

51.  M2 M1 NH4Cl NH3 17 g/mol HCl 36 g/mol
Gas diffusion is the gradual mixing of molecules of one gas with molecules of another by virtue of their kinetic properties. r1 r2 M2 M1  =
molecular path
NH4Cl
NH3
17 g/mol
HCl
36 g/mol

52. Gas effusion is the is the process by which gas under pressure escapes from one compartment of a container to another by passing through a small opening. = r1 r2 t2 t1 M2 M1 
Nickel forms a gaseous compound of the formula Ni(CO)x What is the value of x given that under the same conditions methane (CH4) effuses 3.3 times faster than the compound?
M2 = r1 r2 ( ) 2
x M1
r1 = 3.3 x r2
= (3.3)2 x 16 = 174.2
M1 = 16 g/mol
x • 28 = 174.2
x = 4.1 ~ 4

53. Kinetic Molecular Theory of Gases
A gas is composed of molecules that are separated from each other by distances far greater than their own dimensions. The molecules can be considered to be points; that is, they possess mass but have negligible volume.
Gas molecules are in constant motion in random directions, and they frequently collide with one another. Collisions among molecules are perfectly elastic.
Gas molecules exert neither attractive nor repulsive forces on one another.
The average kinetic energy of the molecules is proportional to the temperature of the gas in kelvins. Any two gases at the same temperature will have the same average kinetic energy
KE = ½ mu2

54. Deviations from Ideal Behavior
1 mole of ideal gas
Repulsive Forces
PV = nRT
n = PV RT
= 1.0
Attractive Forces

55. Effect of intermolecular forces on the pressure exerted by a gas.

56. ( ) } } Van der Waals equation nonideal gas an2 P + (V – nb) = nRT V2
corrected
pressure }
corrected
volume

57. Problem 5.22 (b)
A sample of air occupies 3.8 liters when the pressure is 1.2 atm. (b) What pressure is required in order to compress it to L.

58. Problem 5.70
A sample of ammonia (NH3) gas is completely decomposed to nitrogen and hydrogen gas over heated iron wool. If the total pressure is 866 mm Hg, calculate the partial pressure of N2 and H2.

59. A sample of natural gas contains 8. 24 moles of CH4, 0
A sample of natural gas contains 8.24 moles of CH4, moles of C2H6, and moles of C3H8. If the total pressure is 1.37 atm, what is the partial pressure of propane (C3H8)?

60. Problem 5.38
A gas evolved during the fermentation of glucose has a volume of 0.78 L at 20.1 C and 1.00 atm. What was the volume of this gas at the fermentation temperature of 36.5 C and 1.00 atm?

61. Problem 5.54
In alcohol fermentation, yeast converts glucose to ethanol and carbon dioxide:
C6H12O6(s)  2C2H5OH(l) + 2CO2(g)
If 5.97 g of glucose are reacted and 1.44 L of CO2 gas are collected at 293 K and atm, what is the percent yield of the reaction.

62. Problem 5.44
At 741 torr and 44 C, 7.10 g of a gas occupt a volume of 5.40 L. What is the molar mass of the gas?

63. d is the density of the gas in g/L
Density (d) Calculations
m is the mass of the gas in g
d = m V = PM RT
M is the molar mass of the gas
Molar Mass (M ) of a Gaseous Substance
dRT P
M =
d is the density of the gas in g/L

64. Problem 5.70
A sample of ammonia (NH3) gas is completely decomposed to nitrogen and hydrogen gas over heated iron wool. If the total pressure is 866 mm Hg, calculate the partial pressure of N2 and H2.

65. Problem 5.70
A sample of ammonia (NH3) gas is completely decomposed to nitrogen and hydrogen gas over heated iron wool. If the total pressure is 866 mm Hg, calculate the partial pressure of N2 and H2.

66. Problem 5.70
Pi = Xi (PT)
Use coefficients in equation as moles to calculate Xi
Don’t have any ammonia left. Doesn’t enter into mole fraction calculation.

67. Problem 5.90
At 27 C, 10.0 moles of a gas in a 1.50 L container exert a pressure of 130 atm. Is this an ideal gas.
Hint. Compare actual pressure to ideal gas pressure.

68. ( ) } } Van der Waals equation nonideal gas an2 P + (V – nb) = nRT V2
corrected
pressure }
corrected
volume

69. Problem 5.100
A healthy adult exhales about 5.0 x 102 ml of a gaseous mixture with each breath. Calculate the number of molecules present in this volume at 37 C and 1.1 atm.

70. 5.100
Need moles of the gas. Then convert moles to molecules.

71. A sample of natural gas contains 8. 24 moles of CH4, 0
A sample of natural gas contains 8.24 moles of CH4, moles of C2H6, and moles of C3H8. If the total pressure is 1.37 atm, what is the partial pressure of propane (C3H8)?

72. Deviations from Ideal Behavior
1 mole of ideal gas
Repulsive Forces
PV = nRT
n = PV RT
= 1.0
Attractive Forces