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GAS LAWS I - Unit 10
•
•
•
•
•

5-1

THEIR PROPERTIES
AND
BEHAVIOR
Chapter 9.1-9.3 (McM)
Chapter 5.1-5.3 Silberberg
Chapter 5

Gases and the Kinetic-Molecular Theory

5-2
Gases and the Kinetic Molecular Theory
5.1 An Overview of the Physical States of Matter
5.2 Gas Pressure and Its Measurement

5.3 The Gas Laws and Their Experimental Foundations

5-3
Goals & Objectives
• See the Learning Objectives on
pages 217.
• Understand these Concepts:
• 5.1-4.
• Master these Skills:
• 5.1-3.
5-4
An Overview of the Physical States of Matter
Distinguishing gases from liquids and solids.
• Gas volume changes significantly with pressure.
– Solid and liquid volumes are not greatly affected by pressure.

• Gas volume changes significantly with temperature.
– Gases expand when heated and shrink when cooled.
– The volume change is 50 to 100 times greater for gases than for
liquids and solids.

• Gases flow very freely.
• Gases have relatively low densities.
• Gases form a solution in any proportions.
– Gases are freely miscible with each other.

5-5
Figure 5.1

5-6

The three states of matter.
The Atmosphere and
Some Properties of Gases

5-7

• Gases exert pressure on their
surroundings.
• Gases can be compressed.
• Gases can expand without limit.
• Gases diffuse into each other.
• Gases can be described in terms of
their volume, temperature, pressure
and the number of moles of gas
present.
Volume
• any unit of volume may apply
• L or mL are the most common.
• any unit of length cubed may be
used.
– m3, cm3, etc.

5-8
Temperature
• For all gases, absolute temperature
must be used.
• Kelvin is the absolute temperature
scale.
• K= oC + 273

5-9
Pressure
• defined as the force per unit area
• Average atmospheric pressure at sea
level is 14.7 lbs/in2.
• Standard pressure is defined as
• 760 mm(Hg), or
• 760 torr or
• one atmosphere
5-10
Gas Pressure and its Measurement
force
Pressure =
area
Atmospheric pressure arises from the force exerted
by atmospheric gases on the earth’s surface.

Atmospheric pressure decreases with altitude.

5-11
Figure 5.2

5-12

Effect of atmospheric pressure on a familiar
object.
Figure 5.3

5-13

A mercury barometer.
Table 5.1 Common Units of Pressure

5-14
Sample Problem 5.1

Converting Units of Pressure

PROBLEM: A geochemist heats a limestone (CaCO3) sample and
collects the CO2 released in an evacuated flask attached
to a closed-end manometer. After the system comes to
room temperature, Δh = 291.4 mm Hg. Calculate the CO2
pressure in torrs, atmospheres, and kilopascals.
PLAN:

Construct conversion factors to find the other units of pressure.

SOLUTION:

291.4 mmHg x

1torr
1 mmHg

291.4 torr x 1 atm
760 torr

= 0.3834 atm

0.3834 atm x 101.325 kPa
1 atm

5-15

= 291.4 torr

= 38.85 kPa
The Gas Laws
• The gas laws describe the physical behavior of gases in
terms of 4 variables:
–
–
–
–

pressure (P)
temperature (T)
volume (V)
amount (number of moles, n)

• An ideal gas is a gas that exhibits linear relationships
among these variables.
• No ideal gas actually exists, but most simple gases
behave nearly ideally at ordinary temperatures and
pressures.

5-16
Figure 5.5

5-17

Boyle’s law, the relationship between the volume
and pressure of a gas.
Boyle’s Law
At constant temperature, the volume occupied by a
fixed amount of gas is inversely proportional to the
external pressure.
V

1
P

or PV = constant

At fixed T and n,
P decreases as V increases
P increases as V decreases

5-18
Robert Boyle
• Robert Boyle born (25
January 1627 – 30
December1691) was an
Irish natural philosopher,
chemist, physicist,
inventor, and early
gentleman scientist,
noted for his work in
physics and chemistry.
He is best known for the
formulation of Boyle's
Law.

5-19
Boyle’s Law
• the relationship between volume and
pressure
• At constant temperature, the volume
occupied by a given quantity of gas is
inversely proportional to the pressure
applied to the gas.
• PV=k
• P1V1 = P2V2
5-20
Figure 5.16

A molecular view of Boyle’s law.

Pext increases,
T and n fixed

At any T, Pgas = Pext as
particles hit the walls
from an average
distance, d1.

5-21

Higher Pext causes lower V, which
results in more collisions, because
particles hit the walls from a shorter
average distance (d2 < d1). As a
result, Pgas = Pext again.
Sample Problem 5.2

Applying the Volume-Pressure Relationship

PROBLEM: Boyle’s apprentice finds that the air trapped in a J tube
occupies 24.8 cm3 at 1.12 atm. By adding mercury to the
tube, he increases the pressure on the trapped air to 2.64
atm. Assuming constant temperature, what is the new
volume of air (in L)?
PLAN:

The temperature and amount of gas are fixed, so this
problem involves a change in pressure and volume only.
V1 (cm3)
unit conversions

V1 (L)
multiply by P1/P2

V2 (L)

5-22
Sample Problem 5.2
SOLUTION:
P1 = 1.12 atm
V1 = 24.8 cm3

P2 = 2.64 atm
V2 = unknown

24.8 cm3 x 1 mL
1 cm3
P1V1
n1T1

=

P2V2

L
103 mL

= 0.0248 L

P1V1 = P2V2

n2T2

V2 = V1 x P1
P2

5-23

x

n and T are constant

= 0.0248 L x 1.12 atm
2.46 atm

= 0.0105 L
Boyle’s Law
• At 25oC a sample of helium gas
occupies 400 mL under a pressure of
760 torr. Determine the volume of
this gas at a pressure of 2.00
atmospheres at the same
temperature.

5-24
5-25
5-26
Charles’s Law
At constant pressure, the volume occupied by a
fixed amount of gas is directly proportional to its
absolute (Kelvin) temperature.
V

T

V
= constant
T

At fixed T and n,
P decreases as V increases
P increases as V decreases

5-27
Charles’ Law
• the volume-temperature relationship
• At a constant pressure, the volume
occupied by a definite mass of a gas
is directly proportional to the absolute
temperature.
• V/T = k
• V1/T1 = V2/T2
5-28
Figure 5.6 AB

5-29

Charles’s law, the relationship between the
volume and temperature of a gas.
Jacques Alexandre Cesar
Charles
• Jacques Alexandre
César Charles
(November 12, 1746 –
April 7, 1823) was a
French inventor, scientist,
mathematician, and
balloonist. He is best
know for Charles’ Law.

5-30
Charles’ Law

5-31
Figure 5.18

At T1, Pgas = Patm.

5-32

A molecular view of Charles’s law

Higher T increases
collision frequency,
so Pgas > Patm.

Thus, V increases
until Pgas = Patm at T2.
Sample Problem 5.3

Applying the Pressure-Temperature
Relationship

PROBLEM: A steel tank used for fuel delivery is fitted with a safety
valve that opens when the internal pressure exceeds
1.00x103 torr. It is filled with methane at 23 C and 0.991
atm and placed in boiling water at exactly 100 C. Will the
safety valve open?
PLAN:

We must determine if the pressure will exceed 1.00x103 torr
at the new temperature. Since the gas is in a steel tank, the
volume remains constant.
P1 (atm)

T1 and T2 ( C)
1 atm = 760 torr

T1 and T2 (K)

P1 (torr)
multiply by T2/T1

P2 (torr)

5-33

K = C + 273.15
Sample Problem 5.3
SOLUTION:
P1 = 0.991 atm
T1 = 23 C

P2 = unknown
T2 = 100. C

0.991 atm x 760 torr
1 atm
P1V1
n1T1

=

P2 = P1 x T2
T1

n and V are constant
T1 = 23 + 273.15 = 296 K
T2 = 100. + 273.15 = 373 K

= 753 torr
P2V2

P1

n2T2

T1

= 753 torr x 373 K
296 K

=

P2
T2

= 949 torr

The safety valve will not open, since P2
is less than 1.00 x 103 torr.

5-34
Charles’ Law
• A sample of hydrogen gas occupies
100 mL at 25oC and 1.00 atm
pressure. Determine the volume of
this gas at 50oC under the same
pressure.

5-35
5-36
5-37
The Combined Gas Law
• For a given quantity of gas,
• P1V1 = P2V2
• T1
T2

5-38
The Combined Gas Law
• A sample of nitrogen gas occupies
750 mL at 75oC under a pressure of
810 mm(Hg). Determine the volume
of this sample of gas at STP.

5-39
5-40
5-41
The Combined Gas Law
• A sample of methane gas, CH4,
occupies 260 mL at 32oC at a
pressure of 0.500 atm. Determine
the temperature at a volume of 500
mL and a pressure of 1200 torr.

5-42
5-43
5-44
Jean Baptiste Andre Dumas
• Jean Baptiste André
Dumas (July 14, 1800 April 10, 1884), French
chemist, best known for
his works on organic
analysis and synthesis,
as well as the
determination of atomic
weights (relative atomic
masses) by measuring
vapor densities

5-45
Figure 5.11

Determining the molar
mass of an unknown
volatile liquid.
based on the method of
J.B.A. Dumas (1800-1884)

5-46
Sample Problem 5.8
PROBLEM:

Finding the Molar Mass of a Volatile Liquid

An organic chemist isolates a colorless liquid from a
petroleum sample. She places the liquid in a preweighed
flask and puts the flask in boiling water, causing the liquid
to vaporize and fill the flask with gas. She closes the flask
and reweighs it. She obtains the following data:

Volume (V) of flask = 213 mL
mass of flask + gas = 78.416 g

T = 100.0 C
P = 754 torr
mass of flask = 77.834 g

Calculate the molar mass of the liquid.
PLAN: The variables V, T and P are given. We find the mass of the
gas by subtracting the mass of the flask from the mass of
the flask with the gas in it, and use this information to
calculate M.

5-47
Sample Problem 5.8
SOLUTION:

m of gas = (78.416 - 77.834) = 0.582 g

V = 213 mL x 1 L = 0.213 L
103 mL

T = 100.0 C + 273.15 = 373.2 K

P = 754 torr x 1 atm = 0.992 atm
760 torr

M=

5-48

mRT
=
PV

atm·L
x 373 K
mol·K
0.213 L x 0.992 atm

0.582 g x 0.0821

= 84.4 g/mol
Gas Behavior at Standard Conditions
STP or standard temperature and pressure
specifies a pressure of 1 atm (760 torr) and a
temperature of 0 C ( 273.15 K).
The standard molar volume is the volume of 1 mol
of an ideal gas at STP.
Standard molar volume = 22.4141 L or 22.4 L

5-49
Standard Temperature and
Pressure (STP)
• T = 273K (0oC)
• P = 1 atm

5-50
Figure 5.9

5-51

Standard molar volume.
Figure 5.10

5-52

The volume of 1 mol (22.4 L) of an ideal gas and of
some familiar objects: 1 gal of milk (3.79 L), a
basketball (7.50 L) and 2.00 L of a carbonated drink.
Example Problem
• Determine the number of moles of
methane gas contained in 8.96 liters
at STP.

5-53
5-54
The Ideal Gas Law
pV = nRT
PV
R=
nT

=

1 atm x 22.414 L
1 mol x 273.15 K

=

0.0821 atm·L
mol·K

R is the universal gas constant;
the numerical value of R depends on the units used.

The ideal gas law can also be expressed by the
combined equation:

P1V1
P2V2
=
T1
T2
5-55
Figure 5.11

5-56

The individual gas laws as special cases of the
ideal gas law.
The Ideal Gas Equation
• Determine the volume occupied by
50.0 g of ethane gas, C2H6, at 140oC
under a pressure of 1820 torr. (At. wt.
C = 12.0, H = 1.0 amu)

5-57
5-58
5-59
Example Problems
• Calculate the pressure exerted by
50.0 g of ethane, C2H6, gas in a
25.0L container at 25oC

5-60
5-61
5-62

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  • 1. GAS LAWS I - Unit 10 • • • • • 5-1 THEIR PROPERTIES AND BEHAVIOR Chapter 9.1-9.3 (McM) Chapter 5.1-5.3 Silberberg
  • 2. Chapter 5 Gases and the Kinetic-Molecular Theory 5-2
  • 3. Gases and the Kinetic Molecular Theory 5.1 An Overview of the Physical States of Matter 5.2 Gas Pressure and Its Measurement 5.3 The Gas Laws and Their Experimental Foundations 5-3
  • 4. Goals & Objectives • See the Learning Objectives on pages 217. • Understand these Concepts: • 5.1-4. • Master these Skills: • 5.1-3. 5-4
  • 5. An Overview of the Physical States of Matter Distinguishing gases from liquids and solids. • Gas volume changes significantly with pressure. – Solid and liquid volumes are not greatly affected by pressure. • Gas volume changes significantly with temperature. – Gases expand when heated and shrink when cooled. – The volume change is 50 to 100 times greater for gases than for liquids and solids. • Gases flow very freely. • Gases have relatively low densities. • Gases form a solution in any proportions. – Gases are freely miscible with each other. 5-5
  • 6. Figure 5.1 5-6 The three states of matter.
  • 7. The Atmosphere and Some Properties of Gases 5-7 • Gases exert pressure on their surroundings. • Gases can be compressed. • Gases can expand without limit. • Gases diffuse into each other. • Gases can be described in terms of their volume, temperature, pressure and the number of moles of gas present.
  • 8. Volume • any unit of volume may apply • L or mL are the most common. • any unit of length cubed may be used. – m3, cm3, etc. 5-8
  • 9. Temperature • For all gases, absolute temperature must be used. • Kelvin is the absolute temperature scale. • K= oC + 273 5-9
  • 10. Pressure • defined as the force per unit area • Average atmospheric pressure at sea level is 14.7 lbs/in2. • Standard pressure is defined as • 760 mm(Hg), or • 760 torr or • one atmosphere 5-10
  • 11. Gas Pressure and its Measurement force Pressure = area Atmospheric pressure arises from the force exerted by atmospheric gases on the earth’s surface. Atmospheric pressure decreases with altitude. 5-11
  • 12. Figure 5.2 5-12 Effect of atmospheric pressure on a familiar object.
  • 14. Table 5.1 Common Units of Pressure 5-14
  • 15. Sample Problem 5.1 Converting Units of Pressure PROBLEM: A geochemist heats a limestone (CaCO3) sample and collects the CO2 released in an evacuated flask attached to a closed-end manometer. After the system comes to room temperature, Δh = 291.4 mm Hg. Calculate the CO2 pressure in torrs, atmospheres, and kilopascals. PLAN: Construct conversion factors to find the other units of pressure. SOLUTION: 291.4 mmHg x 1torr 1 mmHg 291.4 torr x 1 atm 760 torr = 0.3834 atm 0.3834 atm x 101.325 kPa 1 atm 5-15 = 291.4 torr = 38.85 kPa
  • 16. The Gas Laws • The gas laws describe the physical behavior of gases in terms of 4 variables: – – – – pressure (P) temperature (T) volume (V) amount (number of moles, n) • An ideal gas is a gas that exhibits linear relationships among these variables. • No ideal gas actually exists, but most simple gases behave nearly ideally at ordinary temperatures and pressures. 5-16
  • 17. Figure 5.5 5-17 Boyle’s law, the relationship between the volume and pressure of a gas.
  • 18. Boyle’s Law At constant temperature, the volume occupied by a fixed amount of gas is inversely proportional to the external pressure. V 1 P or PV = constant At fixed T and n, P decreases as V increases P increases as V decreases 5-18
  • 19. Robert Boyle • Robert Boyle born (25 January 1627 – 30 December1691) was an Irish natural philosopher, chemist, physicist, inventor, and early gentleman scientist, noted for his work in physics and chemistry. He is best known for the formulation of Boyle's Law. 5-19
  • 20. Boyle’s Law • the relationship between volume and pressure • At constant temperature, the volume occupied by a given quantity of gas is inversely proportional to the pressure applied to the gas. • PV=k • P1V1 = P2V2 5-20
  • 21. Figure 5.16 A molecular view of Boyle’s law. Pext increases, T and n fixed At any T, Pgas = Pext as particles hit the walls from an average distance, d1. 5-21 Higher Pext causes lower V, which results in more collisions, because particles hit the walls from a shorter average distance (d2 < d1). As a result, Pgas = Pext again.
  • 22. Sample Problem 5.2 Applying the Volume-Pressure Relationship PROBLEM: Boyle’s apprentice finds that the air trapped in a J tube occupies 24.8 cm3 at 1.12 atm. By adding mercury to the tube, he increases the pressure on the trapped air to 2.64 atm. Assuming constant temperature, what is the new volume of air (in L)? PLAN: The temperature and amount of gas are fixed, so this problem involves a change in pressure and volume only. V1 (cm3) unit conversions V1 (L) multiply by P1/P2 V2 (L) 5-22
  • 23. Sample Problem 5.2 SOLUTION: P1 = 1.12 atm V1 = 24.8 cm3 P2 = 2.64 atm V2 = unknown 24.8 cm3 x 1 mL 1 cm3 P1V1 n1T1 = P2V2 L 103 mL = 0.0248 L P1V1 = P2V2 n2T2 V2 = V1 x P1 P2 5-23 x n and T are constant = 0.0248 L x 1.12 atm 2.46 atm = 0.0105 L
  • 24. Boyle’s Law • At 25oC a sample of helium gas occupies 400 mL under a pressure of 760 torr. Determine the volume of this gas at a pressure of 2.00 atmospheres at the same temperature. 5-24
  • 25. 5-25
  • 26. 5-26
  • 27. Charles’s Law At constant pressure, the volume occupied by a fixed amount of gas is directly proportional to its absolute (Kelvin) temperature. V T V = constant T At fixed T and n, P decreases as V increases P increases as V decreases 5-27
  • 28. Charles’ Law • the volume-temperature relationship • At a constant pressure, the volume occupied by a definite mass of a gas is directly proportional to the absolute temperature. • V/T = k • V1/T1 = V2/T2 5-28
  • 29. Figure 5.6 AB 5-29 Charles’s law, the relationship between the volume and temperature of a gas.
  • 30. Jacques Alexandre Cesar Charles • Jacques Alexandre César Charles (November 12, 1746 – April 7, 1823) was a French inventor, scientist, mathematician, and balloonist. He is best know for Charles’ Law. 5-30
  • 32. Figure 5.18 At T1, Pgas = Patm. 5-32 A molecular view of Charles’s law Higher T increases collision frequency, so Pgas > Patm. Thus, V increases until Pgas = Patm at T2.
  • 33. Sample Problem 5.3 Applying the Pressure-Temperature Relationship PROBLEM: A steel tank used for fuel delivery is fitted with a safety valve that opens when the internal pressure exceeds 1.00x103 torr. It is filled with methane at 23 C and 0.991 atm and placed in boiling water at exactly 100 C. Will the safety valve open? PLAN: We must determine if the pressure will exceed 1.00x103 torr at the new temperature. Since the gas is in a steel tank, the volume remains constant. P1 (atm) T1 and T2 ( C) 1 atm = 760 torr T1 and T2 (K) P1 (torr) multiply by T2/T1 P2 (torr) 5-33 K = C + 273.15
  • 34. Sample Problem 5.3 SOLUTION: P1 = 0.991 atm T1 = 23 C P2 = unknown T2 = 100. C 0.991 atm x 760 torr 1 atm P1V1 n1T1 = P2 = P1 x T2 T1 n and V are constant T1 = 23 + 273.15 = 296 K T2 = 100. + 273.15 = 373 K = 753 torr P2V2 P1 n2T2 T1 = 753 torr x 373 K 296 K = P2 T2 = 949 torr The safety valve will not open, since P2 is less than 1.00 x 103 torr. 5-34
  • 35. Charles’ Law • A sample of hydrogen gas occupies 100 mL at 25oC and 1.00 atm pressure. Determine the volume of this gas at 50oC under the same pressure. 5-35
  • 36. 5-36
  • 37. 5-37
  • 38. The Combined Gas Law • For a given quantity of gas, • P1V1 = P2V2 • T1 T2 5-38
  • 39. The Combined Gas Law • A sample of nitrogen gas occupies 750 mL at 75oC under a pressure of 810 mm(Hg). Determine the volume of this sample of gas at STP. 5-39
  • 40. 5-40
  • 41. 5-41
  • 42. The Combined Gas Law • A sample of methane gas, CH4, occupies 260 mL at 32oC at a pressure of 0.500 atm. Determine the temperature at a volume of 500 mL and a pressure of 1200 torr. 5-42
  • 43. 5-43
  • 44. 5-44
  • 45. Jean Baptiste Andre Dumas • Jean Baptiste André Dumas (July 14, 1800 April 10, 1884), French chemist, best known for his works on organic analysis and synthesis, as well as the determination of atomic weights (relative atomic masses) by measuring vapor densities 5-45
  • 46. Figure 5.11 Determining the molar mass of an unknown volatile liquid. based on the method of J.B.A. Dumas (1800-1884) 5-46
  • 47. Sample Problem 5.8 PROBLEM: Finding the Molar Mass of a Volatile Liquid An organic chemist isolates a colorless liquid from a petroleum sample. She places the liquid in a preweighed flask and puts the flask in boiling water, causing the liquid to vaporize and fill the flask with gas. She closes the flask and reweighs it. She obtains the following data: Volume (V) of flask = 213 mL mass of flask + gas = 78.416 g T = 100.0 C P = 754 torr mass of flask = 77.834 g Calculate the molar mass of the liquid. PLAN: The variables V, T and P are given. We find the mass of the gas by subtracting the mass of the flask from the mass of the flask with the gas in it, and use this information to calculate M. 5-47
  • 48. Sample Problem 5.8 SOLUTION: m of gas = (78.416 - 77.834) = 0.582 g V = 213 mL x 1 L = 0.213 L 103 mL T = 100.0 C + 273.15 = 373.2 K P = 754 torr x 1 atm = 0.992 atm 760 torr M= 5-48 mRT = PV atm·L x 373 K mol·K 0.213 L x 0.992 atm 0.582 g x 0.0821 = 84.4 g/mol
  • 49. Gas Behavior at Standard Conditions STP or standard temperature and pressure specifies a pressure of 1 atm (760 torr) and a temperature of 0 C ( 273.15 K). The standard molar volume is the volume of 1 mol of an ideal gas at STP. Standard molar volume = 22.4141 L or 22.4 L 5-49
  • 50. Standard Temperature and Pressure (STP) • T = 273K (0oC) • P = 1 atm 5-50
  • 52. Figure 5.10 5-52 The volume of 1 mol (22.4 L) of an ideal gas and of some familiar objects: 1 gal of milk (3.79 L), a basketball (7.50 L) and 2.00 L of a carbonated drink.
  • 53. Example Problem • Determine the number of moles of methane gas contained in 8.96 liters at STP. 5-53
  • 54. 5-54
  • 55. The Ideal Gas Law pV = nRT PV R= nT = 1 atm x 22.414 L 1 mol x 273.15 K = 0.0821 atm·L mol·K R is the universal gas constant; the numerical value of R depends on the units used. The ideal gas law can also be expressed by the combined equation: P1V1 P2V2 = T1 T2 5-55
  • 56. Figure 5.11 5-56 The individual gas laws as special cases of the ideal gas law.
  • 57. The Ideal Gas Equation • Determine the volume occupied by 50.0 g of ethane gas, C2H6, at 140oC under a pressure of 1820 torr. (At. wt. C = 12.0, H = 1.0 amu) 5-57
  • 58. 5-58
  • 59. 5-59
  • 60. Example Problems • Calculate the pressure exerted by 50.0 g of ethane, C2H6, gas in a 25.0L container at 25oC 5-60
  • 61. 5-61
  • 62. 5-62