1. The
Gas
Laws

2. Gases - 4 min

3. with the volume of gases
The Gas Laws
Gas laws are concerned
with the volume of gases

4. Gas volume can be expressed in any of the metric units

5. Volume
1 liter L = 3 dm
1 cubic decimeter =
1000 milliliters ml = 3
1000 cubic centimeters cm

6. Gas volume is determined by temperature & pressure
The Gas Laws
Gas volume is determined by
temperature & pressure

7. Changes in temperature cause changes in particle motion

8. Changes in particle motion cause changes in volume
Temperature
Changes in particle motion
cause changes in volume

9. Standard temperature is
zero degrees Celsius o C

10. Temperature
All gas calculations
must be done using
Kelvin temperatures

11. Temperature o C
273 K + =

12. Standard atmospheric pressure is one atmosphere at sea level

13. Pressure
1 atm
760 mm Hg = =
101 kPa
101,300 Pascles = =
1013 mb
29.92 in Hg =

14. STP
Standard Temperature
and Pressure

15. Boyle's Law
Charles' Law
Combined Gas Law
Dalton's Law

16. At a constant temperature, the pressure exerted by a
Boyle's Law
At a constant
temperature, the
pressure exerted by a
gas varies inversely
with its volume.

17. Whatever pressure does, volume does the opposite.
Boyle's Law
Used when only
pressure changes.
Whatever pressure does,
volume does the opposite.

18. Boyle's Law

19. Boyle's Law
V2 = V1 P1 P2

20. Boyle's Law
Practice
Problem

21. Boyle's Law
A gas occupies 15.5 dm3 at a pressure of 30 mm Hg. What is its volume when the pressure is increased to 50 mm Hg. Assume temp does not change.

22. Write the proper equation.
Boyle's Law
Write the proper equation.
V2 = V1 P1 P2

23. Substitute the given numbers.
Boyle's Law
Substitute the given numbers.
V2 = dm3 30 mm Hg
50 mm Hg

24. Since pressure increases, what will happen to volume?
Boyle's Law
V2 = dm3 30 mm Hg
50 mm Hg
Since pressure increases,
what will happen to volume?

25. DECREASE V2 will be less than V1.
Boyle's Law
V2 = dm3 30 mm Hg
50 mm Hg
DECREASE
V2 will be less than V1.

26. Boyle's Law
Do the math.
V2 = dm3 30 mm Hg
50 mm Hg
V2 = 9.3 dm3

27. All Boyle's Law problems should look like this:
V2 = V1 P1 P2
= dm3 30 mm Hg
50 mm Hg
= 9.3 dm3
stop

28. varies directly with the Kelvin temperature.
Charles' Law
At a constant pressure,
the volume of a gas
varies directly with the
Kelvin temperature.

29. Whatever temperature does, volume does the same.
Charles' Law
Used only when
temperature changes.
Whatever temperature does,
volume does the same.

30. V2 = V1 T2 T1 K e l v i n t e m p e r a t u r e K = o C + 2 7 3
Charles' Law
K e l v i n t e m p e r a t u r e
K = o C
V2 = V1 T2 T1

31. Charles' Law
Practice
Problem

32. 500 ml of air at 20 oC is heated to 60 oC with no
Charles' Law
500 ml of air at 20 oC is
heated to 60 oC with no
pressure change. What
is the new gas volume?

33. Write the proper equation.
Charles' Law
Write the proper equation.
V2 = V1 T2 T1

34. Substitute the given numbers.
Charles' Law
Substitute the given numbers.
V2 = 500 ml 60 oC
20 oC

35. Temps must be in Kelvins.
Charles' Law
Temps must be in Kelvins.
V2 = 500 ml 333 K
293 K
Add 273 to Celsius temps.

36. Since temperature increases, what will happen to volume?
Charles' Law
V2 = 500 ml 333 K
293 K
Since temperature increases,
what will happen to volume?

37. INCREASE V2 will be greater than V1.
Charles' Law
V2 = 500 ml 333 K
293 K
INCREASE
V2 will be greater than V1.

38. Charles' Law
Do the math.
V2 = 500 ml 333 K
293 K
V2 = 600 ml

39. All Charles' Law problems should look like this:
V2 = V1 T2 T1
= 500 ml 333 K
293 K
= 600 ml
stop

40. In the real world,
pressure and
temperature BOTH
will change.

41. What do
we do
then??

42. Combined Gas Law
Used when both temperature and
pressure change

43. Combined Gas Law
V2 = V1 P1 T 2
P2 T1

44. Combined Gas Law
Practice
Problem

45. would be the gas volume at 227 oC and 650 mm Hg?
Combined Gas Law
A gas has a volume of 810 ml
at 44 oC and 325 mm Hg. What
would be the gas volume at
227 oC and 650 mm Hg?

46. Combined Gas Law
Write the proper equation.
V2 = V1 P1 T 2
P2 T1

47. Substitute the given numbers.
Combined Gas Law
Substitute the given numbers.
V2 = 810 ml 325 mm Hg 500 K
650 mm Hg K

48. Do the math. = 640 ml V2 = 810 ml 325 mm Hg 500 K 650 mm Hg 317 K
Combined Gas Law
Do the math.
V2 = 810 ml 325 mm Hg 500 K
650 mm Hg K
= 640 ml

49. Calculate the volume of
a gas at STP if 5.05 dm3
of the gas are collected
at 27.5 oC and 95.0 kPa.

50. V1 P1 T2
V2 =
P2 T1
5.05 dm kPa K =
101 kPa K
= 4.32 dm3

51. The total pressure in a container is the sum of the partial
Dalton's Law
The total pressure
in a container is the
sum of the partial
pressures of all the
gases in the container

52. calculations with gases collected over water.
Dalton's Law
This law is used most
often when doing
calculations with gases
collected over water.

53. Dalton's Law
Practice
Problem #1

54. A container holds three gases: O2, CO2, and N2. The partial
Dalton's Law
A container holds three gases:
O2, CO2, and N2. The partial
pressures are 2 atm, 3 atm,
and 4 atm respectively. What
is the total pressure inside the
container?

55. Dalton's Law
O atm
CO2 - 3 atm
N atm
Total Pressure = 9 atm

56. Collecting a Gas
Over Water

57. As gas bubbles
through
the
water ...

58. water vapor
mixes
with the
gas being
collected.

59. Gas pressure in
container
is equal
to air
pressure ...

60. when water
levels
are the
SAME.

61. Dalton's Law
Practice
Problem #2

62. If 60 liters of nitrogen gas is collected over water when the
Dalton's Law
If 60 liters of nitrogen gas is
collected over water when the
temperature is 40 oC and
atmospheric pressure is 760
mm Hg, what is the partial
pressure of the nitrogen?

63. From research, the vapor pressure of H2O at 40 oC is 7.4 kPa.
Dalton's Law
From research, the vapor
pressure of H2O at 40 oC
is 7.4 kPa.
101.0 kPa Total
kPa H2O
93.6 kPa N2

64. Gas Diffusion

65. At equal temperature and pressure, equal volumes
Avogadro's Principle
At equal temperature and
pressure, equal volumes
of gases contain the same
number of molecules.

66. At STP, 22.4 dm3 of any gas contains one mole of molecules,
Molar Volume
At STP, dm3 of
any gas contains one
mole of molecules,
6.02 X 1023

67. Ideal Gas
Theoretical gas
molecules that
have mass,
but no volume

68. Ideal Gas Equation
P V = n R T

69. Ideal Gas Equation
P V = n R T
P - standard pressure

70. Ideal Gas Equation
P V = n R T
V - molar volume

71. Ideal Gas Equation
P V = n R T
n - number of moles

72. Ideal Gas Equation
P V = n R T
R dm3 . kPa
mole . K

73. Ideal Gas Equation
P V = n R T
T - standard temp (K)

74. How many moles of gas
are found in a 500 dm3
container if the conditions
inside the container are
25 oC and 200 kPa?

75. PV PV = nRT n = RT = 40.4 moles 200 kPa 500 dm3 mole K
298 K dm3 kPa
= 40.4 moles

76. Substituting for n: moles = mass (m) molecular mass (M)
Ideal Gas Equation
Modification
Substituting for n:
moles = mass (m)
molecular mass (M)

77. Ideal Gas Equation
Modification
P V = m R T M

78. The relative rates at which two gases diffuse under
Graham's Law
The relative rates at which
two gases diffuse under
identical conditions vary
inversely as the square roots
of their molecular masses.

79. Fifteenth Lab
Fuel in a Butane Lighter

80. END
Gas
Laws