1. Chapter 7 Gases

2. Kinetic Theory of Gases
Particles of a gas
Move rapidly in straight lines and are in constant motion.
Have kinetic energy that increases with an increase in temperature.
Are very far apart.
Have essentially no attractive (or repulsive) forces.
Have very small volumes compared to the volume of the container they occupy.

3. Properties of Gases
Gases are described in terms of four properties: pressure (P), volume (V), temperature (T), and amount (n).

4. Barometer
A barometer measures the pressure exerted by the gases in the atmosphere.
The atmospheric pressure is measured as the height in mm of the mercury column.

5. Learning Check
A. The downward pressure of the Hg in a barometer is _____ than/as the weight of the atmosphere.
1) greater 2) less 3) the same
B. A water barometer is 13.6 times taller than a Hg barometer (DHg = 13.6 g/mL) because
1) H2O is less dense
2) H2O is heavier
3) air is more dense than H2O

6. Solution
A.The downward pressure of the Hg in a barometer is 3) the same as the weight of the atmosphere.
B. A water barometer is 13.6 times taller than a Hg barometer (DHg = 13.6 g/mL) because
1) H2O is less dense

7. Pressure
A gas exerts pressure, which is defined as a force acting on a specific area. Pressure (P) = Force Area
One atmosphere (1 atm) is 760 mm Hg.
1 mm Hg = 1 torr 1.00 atm = 760 mm Hg = 760 torr

8. Units of Pressure
In science, pressure is stated in atmospheres (atm), millimeters of mercury (mm Hg), and Pascals (Pa).

9. Learning Check
A. What is 475 mm Hg expressed in atm? 1) 475 atm 2) atm 3) x 105 atm
B. The pressure in a tire is 2.00 atm. What is this pressure in mm Hg? 1) mm Hg 2) mm Hg 3) 22,300 mm Hg

10. Solution A. What is 475 mm Hg expressed in atm? 2) 0.638 atm
485 mm Hg x atm = atm
760 mm Hg
B. The pressure of a tire is measured as 2.00 atm. What is this pressure in mm Hg?
2) mm Hg
2.00 atm x mm Hg = mm Hg
1 atm

11. Boyle’s Law
The pressure of a gas is inversely related to its volume when T and n are constant.
If volume decreases, the pressure increases.

12. PV Constant in Boyle’s Law
The product P x V remains constant as long as T and n do not change. P1V1 = atm x L = atm L P2V2 = atm x L = atm L P3V3 = atm x L = atm L
Boyle’s Law can be stated as P1V1 = P2V2 (T, n constant)

13. Solving for a Gas Law Factor
The equation for Boyle’s Law can be rearranged to solve for any factor.
To solve for V2, divide both sides by P2.
P1V1 = P2V Boyle’s Law
P2 P2
P1V1 = V2 P2
Solving for P2
P2 = P1 V1 V2

14. PV in Breathing Mechanics
When the lungs expand, the pressure in the lungs decreases.
Inhalation occurs as air flows towards the lower pressure in the lungs.

15. PV in Breathing Mechanics
When the lung volume decreases, pressure within the lungs increases.
Exhalation occurs as air flows from the higher pressure in the lungs to the outside.

16. Calculation with Boyle’s Law
Freon-12, CCl2F2, is used in refrigeration systems. What is the new volume (L) of a 8 L sample of Freon gas initially at 50 mm Hg after its pressure is changed to 200 mm Hg at constant T?
1. Set up a data table
Conditions 1 Conditions 2
P1 = 50 mm Hg P2 = 200 mm Hg
V1 = 8 L V2 = ?

17. Calculation with Boyle’s Law (continued)
2. When pressure increases, volume decreases.
Solve Boyle’s Law for V2:
P1V1 = P2V2
V = V1P1 P2
V2 = 8 L x mm Hg = L
200 mm Hg
pressure ratio decreases volume

18. Learning Check
The helium in a cylinder has a volume of 120 mL and a pressure of 840 mm Hg. A change in the volume results in a lower pressure inside the cylinder. Does cylinder A or B represent the final volume?
At a new pressure of 420 mm Hg, what is the new volume?
1) 60 mL 2) 120 mL 3) 240 mL

19. Solution
The helium in a cylinder has a volume of 120 mL and a pressure of 840 mm Hg. A change in the volume results in a lower pressure inside the cylinder. Does cylinder A or B represent the final volume?
B) If P decreases, V increases.
At a new pressure of 420 mm Hg, what is the new volume of the cylinder?
3) 240 mL

20. Learning Check
A sample of helium gas has a volume of 6.4 L at a pressure of atm. What is the new volume when the pressure is increased to 1.40 atm (T constant)?
A) 3.2 L B) 6.4 L C) 12.8 L

21. Solution A) 3.2 L Solve for V2: P1V1 = P2V2 V2 = V1P1 P2
V2 = L x atm = L
1.40 atm
Volume decreases when there is an increase in the pressure (Temperature is constant).

22. Learning Check
A sample of oxygen gas has a volume of 12.0 L at 600. mm Hg. What is the new pressure when the volume changes to 36.0 L? (T and n constant.)
1) mm Hg
2) mm Hg
3) 1200 mm Hg

23. Solution
1) mm Hg Data table Conditions 1 Conditions 2 P1 = mm Hg P2 = ??? V1 = L V2 = L
P2 = P1 V1 V2
600. mm Hg x L = mm Hg L

24. Charles’ Law
The Kelvin temperature of a gas is directly related to the volume (P and n are constant).
When the temperature of a gas increases, its volume increases.

25. Charles’ Law V and T For two conditions, Charles’ Law is written
V1 = V (P and n constant)
T T2
Rearranging Charles’ Law to solve for V2
V2 = V1T2 T1

26. Learning Check
Solve Charles’ Law expression for T2.
V1 = V2
T T2

27. Solution V1 = V2 T1 T2 Cross multiply to give V1T2 = V2T1
Isolate T2 by dividing through by V1
V1T = V2T1
V1 V1
T2 = V2T1 V1

28. Calculations Using Charles’ Law
A balloon has a volume of 785 mL at 21°C. If
The temperature drops to 0°C, what is the new
volume of the balloon (P constant)?
1. Set up data table:
Conditions 1 Conditions 2
V1 = 785 mL V2 = ?
T1 = 21°C = 294 K T2 = 0°C = 273 K
Be sure that you always use the Kelvin (K)
temperature in gas calculations.

29. Calculations Using Charles’ Law (continued)
2. Solve Charles’ law for V2
V1 = V2
T T2
V2 = V1 T2 T1
V2 = 785 mL x K = 729 mL
294 K

30. Learning Check
A sample of oxygen gas has a volume of 420 mL at a temperature of 18°C. At what temperature (in °C) will the volume of the oxygen be 640 mL (P and n constant)?
1) 443°C
2) 170°C
3) – 82°C

31. Solution 170°C T2 = T1V2 V1 T2 = 291 K x 640 mL = 443 K 420 mL
= 443 K – 273 K = 170°C

32. Gay-Lussac’s Law: P and T
The pressure exerted by a gas is directly related to the Kelvin temperature of the gas at constant V and n.
P1 = P2
T1 T2

33. Calculation with Gay-Lussac’s Law
A gas has a pressure at 2.0 atm at 18°C. What
is the new pressure when the temperature is
62°C? (V and n constant)
Set up a data table.
Conditions 1 Conditions 2
P1 = 2.0 atm P2 =
T1 = 18°C T2 = 62°C + 273
= 291 K = K ?

34. Calculation with Gay-Lussac’s Law (continued)
2. Solve Gay-Lussac’s Law for P2
P1 = P2
T T2
P2 = P1 T2 T1
P2 = 2.0 atm x 335 K = atm
291 K

35. Learning Check Use the gas laws to complete with
Increases 2) Decreases
A. Pressure _______ when V decreases.
B. When T decreases, V _______.
C. Pressure _______ when V changes
from 12.0 L to 24.0 L.
D. Volume _______when T changes from
15.0 °C to 45.0°C.

36. Solution Use the gas laws to complete with 1) Increases 2) Decreases
A. Pressure 1) Increases, when V decreases.
B. When T decreases, V 2) Decreases.
C. Pressure 2) Decreases when V changes
from 12.0 L to 24.0 L
D. Volume 1) Increases when T changes from
15.0 °C to 45.0°C

37. Next Time
We complete Chapter 7

38. Combined Gas Law
The combined gas law uses Boyle’s Law, Charles’ Law, and Gay-Lussac’s Law (n is constant).
P1 V1 = P2 V2
T1 T2

39. Combined Gas Law Calculation
A sample of helium gas has a volume of L, a pressure of atm and a temperature of 29°C. At what temperature (°C) will the helium have a volume of 90.0 mL and a pressure of 3.20 atm (n constant)?
1. Set up Data Table
Conditions 1 Conditions 2
P1 = atm P2 = atm
V1 = L (180 mL) V2 = 90.0 mL
T1 = 29°C = 302 K T2 = ??

40. Combined Gas Law Calculation (continued)
2. Solve for T2 P1 V1 = P2 V2
T1 T2
T2 = T1 P2V2
P1V1
T2 = 302 K x atm x mL = K
0.800 atm mL
T2 = 604 K – = °C

41. Learning Check
A gas has a volume of 675 mL at 35°C and atm pressure. What is the volume(mL) of the gas at –95°C and a pressure of 802 mm Hg (n constant)?

42. Solution Data Table T1 = 308 K T2 = -95°C + 273 = 178K
V1 = 675 mL V2 = ???
P1 = 646 mm Hg P2 = 802 mm Hg
Solve for T2
V2 = V1 P1 T2
P2T1
V2 = mL x 646 mm Hg x 178K = mL mm Hg x 308K

43. Avogadro's Law: Volume and Moles
The volume of a gas is directly related to the number of moles of gas when T and P are constant. V1 = V2 n n2

44. Learning Check
If 0.75 mole of helium gas occupies a volume of 1.5 L, what volume will 1.2 moles of helium occupy at the same temperature and pressure?
1) L
2) 1.8 L
3) 2.4 L

45. Solution 3) 2.4 L Conditions 1 Conditions 2 V1 = 1.5 L V2 = ???
n1 = mole He n2 = 1.2 moles He
V2 = V1n n1 V2 = 1.5 L x moles He = L
0.75 mole He

46. STP
The volumes of gases can be compared when they have the same conditions of temperature and pressure (STP).
Standard temperature (T)
0°C or 273 K
Standard pressure (P)
1 atm (760 mm Hg)

47. Molar Volume At STP, 1 mole of a gas occupies a volume of 22.4 L.
The volume of one mole of a gas is called the molar volume.

48. Molar Volume as a Conversion Factor
The molar volume at STP can be used to form conversion factors L and mole 1 mole L

49. Learning Check A. What is the volume at STP of 4.00 g of CH4?
1) L 2) L 3) 44.8 L
B. How many grams of He are present in 8.00 L of gas at STP?
1) g 2) g 3) 1.43 g

50. Solution 1) 5.60 L 4.00 g CH4 x 1 mole CH4 x 22.4 L (STP) = 5.60 L
16.0 g CH mole CH4
B. 3) 1.43 g
8.00 L x 1 mole He x g He = g He
22.4 L mole He

51. Ideal Gas Law
The relationship between the four properties (P, V, n, and T) of gases can be written equal to a constant R.
PV = R nT
Rearranging this expression gives the expression called the ideal gas law.
PV = nRT

52. Universal Gas Constant, R
The universal gas constant, R, can be calculated using the molar volume of a gas at STP.
At STP (273 K and 1.00 atm), 1 mole of a gas occupies 22.4 L.
P V
R = PV = (1.00 atm)(22.4 L) nT (1 mole) (273K)
n T
= L atm
mole K
Note there are four units associated with R.

53. Learning Check
Another value for the universal gas constant is obtained using mm Hg for the STP pressure. What is the value of R when a pressure of 760 mm Hg is placed in the R value expression?

54. Solution What is the value of R when the STP value for P is 760 mmHg?
R = PV = (760 mm Hg) (22.4 L)
nT (1 mole) (273K)
= L mm Hg mole K

55. Learning Check
Dinitrogen oxide (N2O), laughing gas, is used by dentists as an anesthetic. If a 20.0 L tank of laughing gas contains 2.8 moles N2O at 23°C, what is the pressure (mm Hg) in the tank?

56. Solution
1. Adjust the units of the given properties to match the units of R.
V = 20.0 L, T = 296 K, n = 2.8 moles, P = ?
2. Rearrange the ideal gas law for P.
P = nRT V
P = (2.8 moles)(62.4 L mm Hg)(296 K)
(20.0 L) (mole K)
= 2.6 x 103 mm Hg

57. Learning Check
A cylinder contains 5.0 L of O2 at 20.0°C and 0.85 atm. How many grams of oxygen are in the cylinder?

58. Solution 1. Determine the given properties.
P = 0.85 atm, V = 5.0 L, T = 293 K, n (or g =?)
2. Rearrange the ideal gas law for n (moles).
n = PV RT
= (0.85 atm)(5.0 L)(mole K) = mole O2
(0.0821atm L)(293 K)
3. Convert moles to grams using molar mass.
= mole O2 x g O2 = 5.8 g O2
1 mole O2

59. Molar Mass of a Gas
What is the molar mass of a gas if g of the gas
occupy 215 mL at atm and 30.0°C?
1. Solve for the moles (n) of gas.
n = PV = (0.813 atm) (0.215 L) RT ( L atm/mole K)(303K)
= mole
2. Set up the molar mass relationship.
Molar mass = g = g = g/mole mole mole

60. Gases in Equations
The amounts of gases reacted or produced in a chemical reaction can be calculated using the ideal gas law and mole factors.
Problem:
What volume (L) of Cl2 gas at 1.2 atm and 27°C is needed to completely react with 1.5 g of aluminum?
2Al(s) Cl2 (g) AlCl3(s)

61. Gases in Equations (continued)
2Al(s) + 3Cl2 (g) AlCl3(s)
1.5 g ? L 1.2 atm, 300K
Calculate the moles of Cl2 needed.
1.5 g Al x 1 mole Al x 3 moles Cl2 = mole Cl2
27.0 g Al moles Al
2. Place the moles Cl2 in the ideal gas equation.
V = nRT = (0.083 mole Cl2)( Latm/moleK)(300K)
P atm
= 1.7 L Cl2

62. Learning Check
What volume (L) of O2 at 24°C and atm are needed to react with 28.0 g NH3?
4NH3(g) + 5O2(g) NO(g) + 6H2O(g)

63. Solution 1. Calculate the moles of O2 needed.
28.0 g NH3 x 1 mole NH3 x 5 mole O2
17.0 g NH mole NH3
= 2.06 mole O2
2. Place the moles O2 in the ideal gas equation.
V = nRT = (2.06 moles)( L atm/moleK)(297K) P atm
= L O2

64. Partial Pressure
In a mixture of gases, the partial pressure of each gas is the pressure that gas would exert if it were by itself in the container.

65. Dalton’s Law of Partial Pressures
The total pressure exerted by a gas mixture is the sum of the partial pressures of the gases in that mixture.
PT = P1 + P

66. Partial Pressures
The total pressure of a gas mixture depends on the total number of gas particles, not on the types of particles.

67. Total Pressure
For example, at STP, one mole of gas particles in a volume of 22.4 L will exert the same pressure as one mole of a mixture of gas particles in 22.4 L.
V = 22.4 L
1.0 mole N2
0.4 mole O2
0.6 mole He
1.0 mole
0.5 mole O2
0.3 mole He
0.2 mole Ar
1.0 mole
1.0 atm
1.0 atm
1.0 atm

68. Learning Check
A scuba tank contains O2 with a pressure of atm and He at 855 mm Hg. What is the total pressure in mm Hg in the tank?

69. Solution 1. Convert the pressure in atm to mm Hg
0.450 atm x 760 mm Hg = 342 mm Hg = PO 1 atm
2. Calculate the sum of the partial pressures.
Ptotal = PO + PHe 2
Ptotal = 342 mm Hg mm Hg
= mm Hg

70. Gases We Breathe

71. Health Note: Scuba Diving
When a scuba diver makes a deep dive, the increased pressure causes more N2 (g) to dissolve in the blood.
If a diver rises too fast, the dissolved N2 will form bubbles in the blood, a dangerous and painful condition called "the bends."
Helium, which does not dissolve in the blood, is mixed with O2 to prepare breathing mixtures for deep descents.

72. Learning Check
For a deep dive, some scuba divers are using a mixture of helium and oxygen gases with a pressure of atm. If the oxygen has a partial pressure of 1280 mm Hg, what is the partial pressure of the helium?
1) 520 mm Hg
2) mm Hg
3) mm Hg

73. Solution 4800 mm Hg PTotal = 8.00 atm x 760 mm Hg = 6080 mm Hg 1 atm
PTotal = PO PHe 2
PHe = PTotal - PO
PHe = mm Hg mm Hg
= mm Hg