The document discusses the gas laws relating pressure, volume, and temperature, including Boyle's, Charles', and Gay-Lussac's laws. It explains the relationships between these variables through equations and provides examples of calculations involving gas behavior under different conditions. Additionally, it introduces the ideal gas law and the combined gas law, illustrating how these concepts apply to real-world scenarios.

1. The Gas
Laws

2. PRESSURE, VOLUME, TEMPERATURE are measurable
properties of gases that are related to each other. If
one of these variables changed, there is a
corresponding change in other variables depending
on its relationship. Gas law equations can be derived
whenever ne of these variables is altered.

3. GAS PRESSURE – described as a force acting on a
specific area.
P = Force over(divide) Area
Has units of
atm Atmosphere
Mm Hg Millimeter mercury
Torr Torr
lb/in^2 Pound per square inch
kPa Kilopascals
1 atm = 760 mm hg = 760 torr = 101.325 kPa

4. Volume – three dimensional space occupied by a gas.
Ml Milliliter
L Liter
m^3 Cubic meter
Cm^3 Cubic centimeter
1 L = 1000 ml, 1 ml = 1 cm^3

5. Temperature – measure of the warmth or coldness of a body
Measure of the average kinetic energy of the particles in an object
Kinetic energy is the movement of the particle. High KE, faster
movement
F Fahrenheit
C Celsius
K kelvin

6. Quantity – measured in moles (mol)
1 mol = 6.022 x 10^23 units of a
substance

7. MAIN
GAS LAWS
 Boyle’s Law
 Charles’ Law
 Gay-Lussac’s Law
These laws are
products of various
experiments that are
done many centuries ago.
The Ideal Gas Law can be
used to describe the
relationship between
variables used by the gas
laws.

8. BOYLE’S
LAW:
Volume-
Pressure
Relationship

9. ROBERT BOYLE
• An Anglo-Irish chemist.
• Investigated the relationship between pressure
and volume of a gas using a J-shaped tube
apparatus which was closed to one end.
• He then proposed that, the volume of a given
mass of gas held at constant temperature is
inversely proportional to its pressure.

10. Graph that shows the
relationship of Pressure and
Volume
Schematic Illustration of
Boyle’s Law

11. Boyle has shown us that as the volume decreases,
the pressure increases, which pertains to an inverse
relationship. This relationship is more commonly
called Boyle’s Law. This relationship can be
expressed in a mathematical way: V ∝
1
𝑝
where ∝
means “proportional to”. To change symbol ∝ to an
equal sign, a proportionality constant k, is
introduced. Hence, V = 𝑘1 (
1
𝑝
) or more simply:
PV = 𝒌𝟏

12. For a given sample of gas under two different
conditions at a constant temperature, the product
of pressure and volume is constant , thus, it is
written as follows:
𝑃1 𝑉1 = 𝑃2𝑉2
where 𝑃1 and 𝑉1 are the initial pressure and
volume, while, 𝑃2 and 𝑉2 are the final pressure and
volume of the same amount of gas at the same
temperature.

13. 𝑃1 𝑉1 = 𝑃2𝑉2
P1 = Initial Pressure
V1 = initial Volume
P2 = Final Pressure
V2 = Final volume

14. 𝑃1 𝑉1 = 𝑃2𝑉2
A gas occupies 12.3 liters at a pressure of
40.0 mm hg. What is the volume when the
pressure is increased to 120.0 mmhg?
1.
GIVEN:
V1 = 12.3 L
P1 = 40.0 mm hg
P2 = 120.0 mm hg
V2 = ???

15. 𝑃1 𝑉1 = 𝑃2𝑉2
1. GIVEN:
V1 = 12.3 L
P1 = 40.0 mm hg
P2 = 120.0 mm hg
V2 = ???
P1V1 = P2V2
(40.O mmhg) (12.3L) = (120.0mmhg) V2
492 mmhgL= (120.0mmhg) V2
492 𝑚𝑚ℎ𝑔𝐿
120.0𝑚𝑚ℎ𝑔
=
120.0 𝑚𝑚ℎ𝑔
120.0𝑚𝑚ℎ𝑔
V2
4.1 L = V2

16. 𝑃1 𝑉1 = 𝑃2𝑉2
If a gas at 25.0 °C occupies 3.60 liters at a
pressure of 1.00 atm, what will be its volume
at a pressure of 2.50 atm?
GIVEN:
2.
V1 = 3.60 L
P1 = 1 atm
P2 = 2.50 atm
V2 = ???

17. 𝑃1 𝑉1 = 𝑃2𝑉2
GIVEN:
2.
V1 = 3.60 L
P1 = 1 atm
P2 = 2.50 atm
V2 = ???
P1V1 = P2V2
(1 atm) (3.6 L) = (2.5 atm) V2
3.6 atm(L) = (2.5 atm) V2
3.6 𝑎𝑡𝑚 (𝐿)
2.5 𝑎𝑡𝑚
=
2.5 𝑎𝑡𝑚
2.5 𝑎𝑡𝑚
V2
1.44 L = V2

18. 𝑃1 𝑉1 = 𝑃2𝑉2
Convert 350.0 mL at 740.0 mmHg to its new
volume at standard pressure.
GIVEN:
3.
V1 = 350 mL
P1 = 740 mmHg
P2 = standard pressure which is 760mm hg
V2 = ???

19. 𝑃1 𝑉1 = 𝑃2𝑉2
GIVEN:
3.
V1 = 350 mL
P1 = 740 mmHg
P2 = 760mm hg
V2 = ???
P1V1 = P2V2
(740 mmhg) (350 mL) = (760mmhg) V2
259,000mmhg(mL) = (760mmhg)
V2259,000𝑚𝑚ℎ𝑔 (𝑚𝐿)
760𝑚𝑚ℎ𝑔
=
760𝑚𝑚ℎ𝑔
760𝑚𝑚ℎ𝑔
V2 340.79 mL = V2

20. CHARLES’S
LAW: Volume-
Temperature
Relationship

21. JACQUES ALEXANDRE CHARLES
• A French Physicist
• performed an experiment in a balloon, hot water and cold
water more than 100 years later after Boyle’s experiment
about the volume-pressure relationship.
• He then proposed that “the Kelvin temperature and the
volume of a gas are directly related when there is no
change in pressure of a gas”.

22. Graph that shows the
relationship of Volume and
Temperature
Schematic Illustration of
Charles’ Law

23. From his experiment, Charles has
shown us that as the volume
increases and/or decreases, the
temperature manifests the same
increase and/or decreases, which
pertains to a direct relationship.
This relationship is more commonly
called Charles’ Law.

24. This relationship can be expressed in a mathematical way:
V ∝ T, or more simply,
V
T
= k, where K is proportional
constant.
For a given sample of gas under two different conditions at a
constant pressure, the equation can be written as:
𝑉1
𝑇1
=
𝑉2
𝑇2
where 𝑇1 and 𝑉1 are the initial temperature in Kelvin and
Volume, while, 𝑇2, also in Kelvin, and 𝑉2 are the final
temperature and volume of the same amount of gas at the
same pressure.

26. 𝑽𝟏
𝑻𝟏
=
𝑽𝟐
𝑻𝟐
A syringe contains 56.11 mL of gas at 311 K.
Determine the volume that the gas will occupy if
the temperature is increased to 400 K .
1.
GIVEN:
V1 = 56.11 mL
T1 = 311 K
T2 = 400 K
V2 = ???

27. 𝑽𝟏
𝑻𝟏
=
𝑽𝟐
𝑻𝟐
1. GIVEN:
V1 = 56.11 mL
T1 = 311 K
T2 = 400 K
V2 = ???
𝑉1
𝑇1
=
𝑉2
𝑇2
V2 (311 K) = 56.11 mL (400 K)
V2 (311 K) = 22,444 ml K
311 𝐾(𝑉2)
311 𝐾
=
22,444 𝑚𝐿 𝐾
311 𝐾
V2 = 72.17 mL
V2T1 = V1T2
GIVEN:
V1 = 56.11 mL
T1 = 311 K
T2 = 400 K

28. 𝑽𝟏
𝑻𝟏
=
𝑽𝟐
𝑻𝟐
If 540 mL of nitrogen at 0.00 °C is heated to a
temperature of 100 °C, what will be the new
volume of the gas?
2.
GIVEN:
V1 = 540 mL
T1 = 0.00 °C
T2 = 100 °C
WE HAVE TO
CONVERT
CELCIUS TO
KELVIN FIRST
V2 = ???

29. 𝑽𝟏
𝑻𝟏
=
𝑽𝟐
𝑻𝟐
If 540 mL of nitrogen at 0.00 °C is heated to a
temperature of 100 °C, what will be the new
volume of the gas?
2.
GIVEN:
V1 = 540 mL
T1 = 0.00 °C
T2 = 100 °C
V2 = ???
T1= 0 C + 273 = 273 K
T2= 100 C + 273 =
373 K

30. 𝑽𝟏
𝑻𝟏
=
𝑽𝟐
𝑻𝟐
2. GIVEN:
V1 = 540 mL
T1 = 273 K
T2 = 373 K
V2 = ???
𝑉1
𝑇1
=
𝑉2
𝑇2
V2 (273 K) = 540 mL (373 K)
V2 (272 K) = 201,420 ml K
273 𝐾(𝑉2)
273 𝐾
=
201,420 𝑚𝐿 𝐾
273 𝐾
V2 = 737.8 mL
V2T1 = V1T2

31. 𝑽𝟏
𝑻𝟏
=
𝑽𝟐
𝑻𝟐
At constant pressure, Neon gas contracts from
2.00 L to 0.75 L. The initial temperature is 24 C.
Find the final temperature
3.
GIVEN:
V1 = 2 L
T1 = 24 C
T2 = ???
WE HAVE TO
CONVERT
CELCIUS TO
KELVIN FIRST
V2 = 0.75 L

32. 𝑽𝟏
𝑻𝟏
=
𝑽𝟐
𝑻𝟐
At constant pressure, Neon gas contracts from
2.00 L to 0.75 L. The initial temperature is 24 C.
Find the final temperature
3.
GIVEN:
V1 = 2 L
T1 = 24 C
T2 = ???
V2 = 0.75 L
T1= 24 C + 273 = 297 K
GIVEN:
V1 = 2 L
GIVEN:

33. 𝑽𝟏
𝑻𝟏
=
𝑽𝟐
𝑻𝟐
3.
𝑉1
𝑇1
=
𝑉2
𝑇2
0.75 L (297 K) = 2 L (T2)
222.75 L K = 2 L (T2)
222.75 𝐿𝐾
2𝐿
=
2𝐿 (𝑇2)
2𝐿
111.38 K = T2
V2T1 = V1T2
T1 = 297 K
T2 = ???
V2 = 0.75 L
V1 = 2 L
GIVEN:

34. GAY-LUSSAC’S
LAW:
Temperature-Pressure
Relationship

35. JOSEPH-LOUIS GAY-LUSSAC
- A French Chemist and Physicist.
- He proved that one of the postulates in Kinetic Molecular
Theory
is the effect of temperature on the motion of gas
particles.
- Year 1802, he conducted and experiment and discovered
the
relationship between the Temperature and Pressure.
He found out that the pressure of the pressure of a gas increased
or decreased proportionally with a change in temperature. He
summarized his findings in his proposed Gay-Lussac’s Law which
states that “the pressure of a fixed amount of a gas is directly
proportional to the absolute temperature (Kelvin).

36. Graph that shows the
relationship of Temperature
and Pressure
Schematic Illustration of
Gay-Lussac’s Law

37. It can be mathematically expressed as follows P ∝ T
(n, V constant). Or dividing both sides of the equation
by T, hence, P, = k or
P
T
= K.
For any two sets of pressure and temperature, at
constant volume, the equation can be stated as:
𝑃1
𝑇1
=
𝑃2
𝑇2

38. 𝑃1
𝑇1
=
𝑃2
𝑇2
Determine the pressure change when a constant
volume of gas at 1.00 atm is heated from 20.0 °C
to 30.0 °C.
1.
GIVEN:
P1 = 1 atm
T1 = 20 C
T2 = 30 C
P2 = ???
WE HAVE TO
CONVERT
CELCIUS TO
KELVIN FIRST

39. 𝑃1
𝑇1
=
𝑃2
𝑇2
Determine the pressure change when a constant
volume of gas at 1.00 atm is heated from 20.0 °C
to 30.0 °C.
1.
GIVEN:
P1 = 1 atm
T1 = 20 C
T2 = 30 C
P2 = ???
T1= 20 C + 273 = 293 K
T2= 30 C + 273 = 303 K

40. 𝑃1
𝑇1
=
𝑃2
𝑇2
1.
𝑃1
𝑇1
=
𝑃2
𝑇2
P2(293 K) = 1 atm (303 K)
P2 (293 K)= 303 atm K
293 𝐾 𝑃2
293 𝐾
=
303 𝑎𝑡𝑚 𝐾
293 𝐾
P2 = 1.03 atm
P2T1 = P1T2
GIVEN:
P1 = 1 atm
T1 = 293 K
T2 = 303 K
P2 = ???

41. 𝑃1
𝑇1
=
𝑃2
𝑇2
A gas has a pressure of 699.0 mmHg at 40.0 °C.
What is the temperature at standard pressure?
2.
GIVEN:
P1 = 699.0 mmhg
T1 = 40 C
T2 = ?
P2 = standard pressure
WE HAVE TO
CONVERT
CELCIUS TO
KELVIN FIRST

42. 𝑃1
𝑇1
=
𝑃2
𝑇2
A gas has a pressure of 699.0 mmHg at 40.0 °C.
What is the temperature at standard pressure?
2.
GIVEN:
P1 = 699.0 mmhg
T1 = 40 C
T2 = ?
P2 = standard pressure
T1= 40 C + 273 = 313 K
Standard pressure is 760
mmhg

43. 𝑃1
𝑇1
=
𝑃2
𝑇2
2.
𝑃1
𝑇1
=
𝑃2
𝑇2
760(313 K) = 699 (T2)
237,880 = 699 (T2)
237,880
699
=
699 (𝑇2)
699
340.31 K = T2
P2T1 = P1T2
GIVEN:
P1 = 699.0 mmhg
T1 = 313K
T2 = ?
P2 = 760 mmhg

44. COMBINED GAS
LAWS:
Volume-Pressure-Temperature
Relationship

45. The three properties of a gas (pressure, volume
and temperature) usually change at once under
experimental condition. The four possible
variations are as follows:
1. Both T and P cause an increase in V.
2. Both T and P cause a decrease in V.
3. T causes an increase in V and P cause a
decrease in V.
4. T causes a decrease in V and P cause an
increase in V.

46. The three gas laws (Boyle’s, Charles’ and Gay-Lussac’s laws)
discussed earlier can be combined and may be expressed into one
equation,
PV
T
= K, and expressed as the Combined Gas Law or the
general gas law. Combined Gas Laws allows you to directly solve
for the changes in pressure, volume, or temperature.
Scientists generally follow the customary reference point of gases
which is 0˚C and 1 atm pressure by international agreement. This
reference point refers to the Standard Temperature and Pressure
or (STP). The temperature can also be expressed as 273 K and
standard pressure as 760 torr,760 mmHg and 76 cmHg.

47. The equation can be expressed as follows:
BOYLE’S LAW: V ∝
1
𝑝
CHARLES’S LAW: V ∝ T
COMBINED GAS LAW: V ∝
𝑇
𝑝
or PV ∝ T or PV = constant
Thus,
PV
T
= k
In a constant n, or mole, this can be derived as:
𝑃1𝑉1
𝑇1
=
𝑃2𝑉2
𝑇2

48. 𝑃1𝑉1
𝑇1
=
𝑃2𝑉2
𝑇2
A gas has a volume of 800.0 mL at −23.0 °C and
300.0 torr. What would the volume of the gas be
at 227.0 °C and 600.0 torr of pressure?
1.
GIVEN:
P1 = 300 torr
T1 = -23 C
V1 = 800 mL
P2 = 600 torr
T2 = 227 C
V2 = ?
WE HAVE TO
CONVERT
CELCIUS TO
KELVIN FIRST

49. 𝑃1𝑉1
𝑇1
=
𝑃2𝑉2
𝑇2
A gas has a volume of 800.0 mL at −23.0 °C and
300.0 torr. What would the volume of the gas be
at 227.0 °C and 600.0 torr of pressure?
1.
GIVEN:
P1 = 300 torr
T1 = -23 C
V1 = 800 mL
P2 = 600 torr
T2 = 227 C
V2 = ?
T1= -23 C + 273 = 250 K
T2= 227 C + 273 = 500 K

50. 𝑃1𝑉1
𝑇1
=
𝑃2𝑉2
𝑇2
1.
𝑃1𝑉1
𝑇1
=
𝑃2𝑉2
𝑇2
240,000
250
=
600 (V2)
500
120,000,000 = 150,000 (V2)
120000000
150000
=
150000 (𝑉2)
150000
800 mL = V2
300 (800)
250
=
600 (V2)
500
GIVEN:
T2 = 227 C
V2 = ?
P1 = 300 torr
T1 = -23 C
V1 = 800 mL
P2 = 600 torr

51. 𝑃1𝑉1
𝑇1
=
𝑃2𝑉2
𝑇2
A balloon of air now occupies 10.0 L at 25.0 °C and 1.00
atm. What temperature was it initially, if it occupied
9.40 L and was in a freezer with a pressure of 0.939
atm?
2.
GIVEN:
P1 = 0.939 atm
T1 = ?
V1 = 9.40 L
P2 = 1 atm
T2 = 25 C
V2 = 10 L
WE HAVE TO
CONVERT
CELCIUS TO
KELVIN FIRST
T1= 25 C + 273 = 298 K

52. 𝑃1𝑉1
𝑇1
=
𝑃2𝑉2
𝑇2
2.
𝑃1𝑉1
𝑇1
=
𝑃2𝑉2
𝑇2
8.8266
𝑇1
=
10
298
10(𝑇1)= 2,630.3268
10 (𝑇1)
10
=
2,630.3268
10
T1 = 263.03 K or
-10.03 C
0.939 (9.4)
𝑇1
=
1 (10)
298
GIVEN:
P1 = 0.939 atm
T1 = ?
V1 = 9.40 L
P2 = 1 atm
T2 = 298 K
V2 = 10 L

53. 𝑃1𝑉1
𝑇1
=
𝑃2𝑉2
𝑇2
Suppose you have a sample of gas at 303K in a container with
a volume of 2L and pressure of 760mmHg. The sample shifts
to a temperature of 340 K and the volume increases slightly to
2.1L. What is the pressure of the sample now?
3.
GIVEN:
P1 = 760 mmhg
T1 = 303 K
V1 = 2 L
P2 = ?
T2 = 340 K
V2 = 2.1 L

54. 𝑃1𝑉1
𝑇1
=
𝑃2𝑉2
𝑇2
3.
𝑃1𝑉1
𝑇1
=
𝑃2𝑉2
𝑇2
1520
303
=
P2(2.1)
340
636.3 (𝑃2)= 516,800
636.3 (𝑃2)
636.3
=
516,800
636.3
P2 = 812.20
mmhg
760 (2)
303
=
P2 (2.1)
340
GIVEN:
P1 = 760 mmhg
T1 = 303 K
V1 = 2 L
P2 = ?
T2 = 340 K
V2 = 2.1 L

55. Avogadro’s law,
𝑉1
𝑛1
=
𝑉2
𝑛2
Amadeo Avogadro was an Italian physicist who stated, in 1811,
that the volume of any gas is proportional to the number of
molecules of gas (measured in Moles – symbol mol). In other
words if the amount of gas increases, then so does its volume.

56. 𝑉1
𝑛1
=
𝑉2
𝑛2
The initial amount of dry ice (gas) is 1 mol, and occupies a
volume of 2 L. What is the volume if we increase the amount
of dry ice (gas) to 2 mol?
1.
GIVEN:
n1 = 1 mol
V1 = 2 L
n2 = 2 mol
V2 = ?
n1 = 1 mol
V1 = 2 L

57. 𝑉1
𝑛1
=
𝑉2
𝑛2
1. GIVEN:
𝑉1
𝑛1
=
𝑉2
𝑛2
V2 (1) = 2 (2)
V2 = 4
V2 = 4 L
V2n1 = V1n2
n2 = 2 mol
V2 = ?
n1 = 1 mol
V1 = 2 L

58. 𝑉1
𝑛1
=
𝑉2
𝑛2
5.00 L of a gas is known to contain 0.965 mol. If the amount of
gas is increased to 1.80 mol, what new volume will result (at
an unchanged temperature and pressure)?
2.
GIVEN:
n1 = 0.965 mol
V1 = 5 L
n2 = 1.80 mol
V2 = ?

59. 𝑉1
𝑛1
=
𝑉2
𝑛2
2. GIVEN:
𝑉1
𝑛1
=
𝑉2
𝑛2
V2 (0.965) = 5 (1.8)
V2 (0.965) = 9
V2n1 = V1n2
n1 = 0.965 mol
V1 = 5 L
n2 = 1.80 mol
V2 = ?
𝑉2 (0.965)
0.965
=
9
0.965
V2 = 9.33 L

60. The Ideal Gas
Law

61. • Aside from pressure, volume and temperature,
moles may also be combined to the three
properties of gas establish the relationship among
the four variables affecting the behavior of gases.
• Boyle’s, Charles’s, Gay-Lussac’s and Avogadro’s
Law when combined constitute the ideal Gas Law
or ideal gas equation.

62. •The ideal Gas is based on the experimental
measurements of the physical properties of gases:
temperature, pressure, volume and number of
moles.
•Ideal Gas equation can be expressed as:
PV = nRT

63. •Where R is the proportionality constant at STP
and is equal to 0.0821 L . atm/mol . K and
referred as the ideal gas constant. This tells that
the volume of a gas varies directly with the
number of moles and absolute temperature and
inversely proportionality with pressure.
PV = nRT

64. 𝑃𝑉 = 𝑛𝑅𝑇
What pressure is required to contain 0.023 moles of nitrogen
gas in a 4.20 L container at a temperature of 20 C.
1.
GIVEN:
P = ?
V = 4.20 L
n = 0.023 moles
T = 20 + 273 = 293 K
R = 0.0821 (L atm/mol K)

65. 1. GIVEN:
P (4.20) = 0.5532719
𝑃 (4.20)
4.20
=
0.5532719
4.20
P = 0.13 atm
𝑃𝑉 = 𝑛𝑅𝑇
𝑃𝑉 = 𝑛𝑅𝑇
P (4.20) = (0.023)(0.0821)(293)
P = ?
V = 4.20 L
n = 0.023 moles
T = 20 + 273 = 293 K
R = 0.0821 (L atm/mol K)

66. 𝑃𝑉 = 𝑛𝑅𝑇
Oxygen gas is collected at a pressure of 123 kPa in a container
which has a volume of 10.0 L. What temperature must be
maintained on 0.500 moles of this gas in order to maintain
this pressure? Express the temperature in degree Celsius.
2.
GIVEN:
P = 123 kPa
V = 10.0 L
n = 0.500 moles
T = ?
R = 0.0821 (L atm/mol K)
1 atm = 760 mm hg = 760 torr = 101.325 kPa

67. 2. GIVEN:
12.1 = 0.04105 (T)
12.1
0.04105
=
0.04105 (𝑇)
0.04105
294.76 K or 21.76 C = T
𝑃𝑉 = 𝑛𝑅𝑇
𝑃𝑉 = 𝑛𝑅𝑇
1.21 (10) = (0.500)(0.0821)(𝑇)
P = 121 atm
V = 10.0 L
n = 0.500 moles
T = ?
R = 0.0821 (L atm/mol K)

69. Thank you! 