The document discusses gas laws including Boyle's law, Charles' law, and the general gas equation. It provides sample data and calculations to illustrate these laws. Specifically, it shows how volume and pressure are inversely proportional based on Boyle's law. It also demonstrates how volume and temperature are directly proportional based on Charles' law. The general gas equation combines these relationships as PV=nRT. Sample questions at the end apply this equation to calculate pressure, volume, temperature or amount of a gas.

1. Gases
Gas equations

2. BOYLE’S LAW

3. P
1
V

4. have
Temperature
Volume
Pressure
Amount

5. Volume
Pressure
(dm3)
1
5
2
2,5
3
1,67
4
1,25
5
1
6
0,833
k
(units)
x
x
x
x
x
x
1
2
3
4
5
6
(proportionality
constant: PV)
5
=
=
=
=
=
=
5
5
5
5
5

6. Volume
k
Pressure
1 V1 = 5
P1 = 1
(proportionality
constant: PV)
5
2 2,5
2
5
3 1,67
3
5
4 1,25
4
5
5 1
5
5
6 0,833
6
5
(dm3)
(units)

7. V1 x P 1 = k

8. Volume
k
Pressure
1 V1 = 5
P1 = 1
(proportionality
constant: PV)
5
2 V2 = 2,5
P2 = 2
5
3 1,67
3
5
4 1,25
4
5
5 1
5
5
6 0,833
6
5
(dm3)
(units)

9. V1 x P 1 = k
V2 x P2 = k

10. V1 x P 1 = k
V2 x P2 = k
V1 x P1 = V2 x P2

11. V1 x P 1 = k
V2 x P2 = k
V1 x P1 = V2 x P2
P1V1 = P2V2

12. V1 x P 1 = k
V2 x P2 = k
V1 x P1 = V2 x P2
P1V1 = P2V2

13. P1V1 = P2V2

14. PRESSURE AND TEMPERATURE

15. P
T

16. have
Temperature
Volume
Pressure
Amount

17. Temperature
Pressure
(kPa)
( C)
(K)
1 20
293
1,00
0,0034
2 40
313
1,07
0,0034
3 60
333
1,14
0,0034
4 80
353
1,21
0,0034
5 100
373
1,28
0,0034
6 313
586
2,01
0,0034

18. Temperature
Pressure
(kPa)
( C)
(K)
1 20
T1=293
P1=1,00
0,0034
2 40
313
1,07
0,0034
3 60
333
1,14
0,0034
4 80
353
1,21
0,0034
5 100
373
1,28
0,0034
6 313
586
2,01
0,0034

20. Temperature
Pressure
(kPa)
( C)
(K)
1 20
T1=293
P1=1,00
0,0034
2 40
T2=313
P2=1,07
0,0034
3 60
333
1,14
0,0034
4 80
353
1,21
0,0034
5 100
373
1,28
0,0034
6 313
586
2,01
0,0034

24. CHARLES’ LAW

25. V
T

26. have
Temperature
Volume
Pressure
Amount

27. Temperature
Volume
(cm3)
( C)
(K)
1 1
274
34
0,12
2 14
287
36
0,13
3 53
326
37
0,11
4 76
349
41
0,12

31. GENERAL GAS EQUATION

32. have
Temperature
Volume
Pressure
Amount

33. have
Temperature
Volume
Pressure
Amount

36. QUESTION 1

37. Question
A gas bubble at the bottom of a lake has a
volume of 200cm3 at a temperature of 8 C.
This bubble rises and when it reaches the
surface where the temperature is 15 C and
the pressure is 100kPa, its volume
increases to 550cm3. Calculate the
pressure on the bubble when it was at the
bottom of the lake.

38. 8 C = 281 K
15 C = 288 K

42. have
Temperature
Volume
Pressure
Amount

43. have
Temperature
Volume
Pressure
Amount

44. Amount
(n)

45. Amount
(n)
measured in
moles
(mol)

46. movie

47. n
n
n
n

48. n
P
n
P
n
P
nP

49. P
n

51. QUESTION

52. Question
1 mole of gas at 101,3 kPa and 273
K has a volume of 22,4 dm3. What
is the volume of 5 moles of gas at
120 kPa and 280 K?

57. Standard Temperature: 273 K
Standard Pressure: 101,3 kPa

58. STP
Standard Temperature: 273 K
Standard Pressure: 101,3 kPa

59. STP
Standard Temperature: 273 K
Standard Pressure: 101,3 kPa
Volume of 1 mole of any gas at STP: 22,4 dm3

65. PV = nRT
GENERAL GAS EQUATION

66. Universal Gas Constant
R = 8,3 J • mol-1 • K-1

67. Universal Gas Equation
PV = nRT

71. 1 000 Pa = 1 kPa

72. 1 m3 = 1 000 dm3

73. 1 000 Pa = 1 kPa
1 m3 = 1 000 dm3

79. Universal Gas Constant

80. Universal Gas Constant

81. QUESTION

82. Question
What is the volume of 5 moles of
gas at 120 kPa and 280 K?

87. SI units
P : Pa
T:K
V : m3
n : mol

88. Question
What is the volume of 5 moles of
gas at 120 kPa and 280 K?

89. 120 kPa = 120 000 Pa

96. QUESTION

97. Question
A sample of nitrogen gas has a mass of
7 g and a volume of 5,6 dm3 at a
temperature of 27 °C. Calculate the
pressure of the gas.

98. PV = nRT

100. N2 (g)

101. N2
M = 2(14)g•mol-1
-1
M = 28 g•mol

104. T = 27 + 273 = 300 K