Chapter 10 1

Homework:
10.12, 10.28, 10.42, 10.48, 10.54, 10.66, 10.72
Chapter 10 2

Characteristics of Gases
Expand to fill a volume (expandability)
Compressible
Readily forms homogeneous mixtures with other gases
These behaviors are due to large distances between the gas molecules.
Chapter 10 3

Pressure
Pressure - force acting on an object per unit area.
Chapter 10 4

Pressure
Pressure - force acting on an object per unit area.
P

F
A
Chapter 10 5

Pressure
Pressure - force acting on an object per unit area.
P

F
A
Atmospheric pressure is measured with a barometer.
Standard atmospheric pressure is the pressure required to support 760 mm of Hg in a column.
There are several units used for pressure:
Pascal (Pa), N/m 2
Millimeters of Mercury (mmHg)
Atmospheres (atm)
Chapter 10 6

Pressure
Conversion Factors
1 atm = 760 mmHg
1 atm = 760 torr
1 atm = 1.01325

10 5 Pa
1 atm = 101.325 kPa.
Chapter 10 7

The Gas Laws
There are four variables required to describe a gas:
Amount of substance: moles
Volume of substance: volume
Pressures of substance: pressure
Temperature of substance: temperature
The gas laws will hold two of the variables constant and see how the other two vary.
Chapter 10 8

The Gas Laws
The Pressure-Volume Relationship: Boyle’s Law
Boyle’s Law - The volume of a fixed quantity of gas is inversely proportional to its pressure.
Chapter 10 9

The Gas Laws
The Pressure-Volume Relationship: Boyle’s Law
Boyle’s Law - The volume of a fixed quantity of gas is inversely proportional to its pressure.
P

1
V
(constant n and T )
Chapter 10 10

The Gas Laws
The Pressure-Volume Relationship: Boyle’s Law
Boyle’s Law - The volume of a fixed quantity of gas is inversely proportional to its pressure.

1
P
P
1
V
1
V

P
2
V
2
(constant n and T )
Chapter 10 11

The Gas Laws
The Pressure-Volume Relationship: Boyle’s Law
Chapter 10 12

The Gas Laws
The Temperature-Volume Relationship:
Charles’s Law
Charles’s Law - the volume of a fixed quantity of gas at constant pressure increases as the temperature increases.
Chapter 10 13

The Gas Laws
The Temperature-Volume Relationship:
Charles’s Law
Charles’s Law - the volume of a fixed quantity of gas at constant pressure increases as the temperature increases.
V

T
n
P
Chapter 10 14

The Gas Laws
The Temperature-Volume Relationship:
Charles’s Law
Charles’s Law - the volume of a fixed quantity of gas at constant pressure increases as the temperature increases.
V

T
n
P
V
1
T
1

V
2
T
2
Chapter 10 15

The Gas Laws
The Temperature-Volume Relationship:
Charles’s Law
Chapter 10 16

The Gas Laws
The Quantity-Volume Relationship:
Avogadro’s Law
Avogadro’s Law - The volume of gas at a given temperature and pressure is directly proportional to the number of moles of gas.
Chapter 10 17

The Gas Laws
The Quantity-Volume Relationship:
Avogadro’s Law
Avogadro’s Law - The volume of gas at a given temperature and pressure is directly proportional to the number of moles of gas.
V
 n
P
T
Chapter 10 18

The Ideal Gas Equation
Combine the gas laws (Boyle, Charles, Avogadro) yields a new law or equation.
Chapter 10 19

The Ideal Gas Equation
Combine the gas laws (Boyle, Charles, Avogadro) yields a new law or equation.
Ideal gas equation:
PV = nRT
R = gas constant = 0.08206 L.atm/mol-K
P = pressure (atm) V = volume (L) n = moles T = temperature (K)
Chapter 10 20

The Ideal Gas Equation
We define STP (standard temperature and pressure) as 0

C, 273.15 K, 1 atm.
Volume of 1 mol of gas at STP is 22.4 L.
Chapter 10 21

Further Applications of The Ideal-Gas
Equation
Gas Densities and Molar Mass
Rearranging the ideal-gas equation with
M as molar mass yields d

P
M
RT
Chapter 10 22

Gas Mixtures and Partial Pressures
Dalton’s Law
Dalton’s Law - In a gas mixture the total pressure is given by the sum of partial pressures of each component:
P t
= P
1
+ P
2
+ P
3
+ …
- The pressure due to an individual gas is called a partial pressure.
Chapter 10 23

Gas Mixtures and Partial Pressures
Partial Pressures and Mole Fractions
The partial pressure of a gas can be determined if you know the mole fraction of the gas of interest and the total pressure of the system.
Let n i be the number of moles of gas i exerting a partial pressure P i
, then
P i
=
 i
P t where
 i is the mole fraction (n i
/n t
).
Chapter 10 24

Kinetic-Molecular Theory
Theory developed to explain gas behavior
To describe the behavior of a gas, we must first describe what a gas is:
–
Gases consist of a large number of molecules in constant random motion.
–
Volume of individual molecules negligible compared to volume of container.
–
Intermolecular forces (forces between gas molecules) negligible.
–
Energy can be transferred between molecules, but total kinetic energy is constant at constant temperature.
–
Average kinetic energy of molecules is proportional to temperature.
Chapter 10 25

Kinetic-Molecular Theory
Boyle’s Law
• Pressure exerted by a gas is the result of bombardment of the walls of the container by the gas molecules.
•
Pressure varies directly with the number of molecules hitting the wall per unit time.
– If the volume is reduced the number of impacts is increased; therefore, the pressure is increased.
Chapter 10 26

Kinetic-Molecular Theory
Charles’ Law
• The force that a gas molecule strikes the side of a container is directly proportional to the temperature.
•
As the temperature is increased the kinetic energy
(force) of the gas particles increases.
•
For the pressure to be constant, the area the force is applied to must be increased (the volume is increased).
Chapter 10 27

Kinetic-Molecular Theory
Dalton’s Law
• There are large distances between the gas molecules.
•
The components of a mixture will bombard the walls of the container with the same frequency in the presence of a mixture as it would by itself.
Chapter 10 28

Kinetic-Molecular Theory
Molecular Speed
• As kinetic energy increases, the velocity of the gas molecules increases.
•
Root mean square speed, u, is the speed of a gas molecule having average kinetic energy.
• Average kinetic energy,

, is related to root mean square speed:
 = ½mu 2
Chapter 10 29

Kinetic-Molecular Theory
Molecular Speed
Chapter 10 30

Molecular Effusion and Diffusion
Molecular Speed
• Consider two gases at the same temperature: the lighter gas has a higher u than the heavier gas.
•
Mathematically:
Chapter 10 31

Molecular Effusion and Diffusion
Molecular Speed
• Consider two gases at the same temperature: the lighter gas has a higher u than the heavier gas.
•
Mathematically: u

3 RT
M
Chapter 10 32

Molecular Effusion and Diffusion
Molecular Speed
• Consider two gases at the same temperature: the lighter gas has a higher u than the heavier gas.
•
Mathematically: u

3 RT
M
•
The lower the molar mass,
M
, the higher the u for that gas at a constant temperature.
Chapter 10 33

Molecular Effusion and Diffusion
Chapter 10 34

Molecular Effusion and Diffusion
Graham’s Law of Effusion
Effusion – The escape of gas through a small opening.
Diffusion – The spreading of one substance through another.
Chapter 10 35

Molecular Effusion and Diffusion
Graham’s Law of Effusion
Graham’s Law of Effusion – The rate of effusion of a gas is inversely proportional to the square root of its molecular mass.
Chapter 10 36

Molecular Effusion and Diffusion
Graham’s Law of Effusion
Graham’s Law of Effusion – The rate of effusion of a gas is inversely proportional to the square root of its molecular mass.
r
1 r
2

M
M
2
1
Chapter 10 37

Molecular Effusion and Diffusion
Graham’s Law of Effusion
Graham’s Law of Effusion – The rate of effusion of a gas is inversely proportional to the square root of its molecular mass.
r
1 r
2

M
M
2
1
•
Gas escaping from a balloon is a good example.
Chapter 10 38

Molecular Effusion and Diffusion
Graham’s Law of Effusion
Chapter 10 39

Molecular Effusion and Diffusion
Diffusion and Mean Free Path
• Diffusion of a gas is the spread of the gas through space.
•
Diffusion is faster for light gas molecules.
•
Diffusion is slowed by gas molecules colliding with each other.
•
Average distance of a gas molecule between collisions is called mean free path.
Chapter 10 40

Molecular Effusion and Diffusion
Diffusion and Mean Free Path
•
At sea level, mean free path is about 6

10 -6 cm.
Chapter 10 41

Real Gases: Deviations from Ideal
Behavior
• From the ideal gas equation, we have
PV = nRT
•
This equation breaks-down at
–
High pressure
• At high pressure, the attractive and repulsive forces between gas molecules becomes significant.
Chapter 10 42

Real Gases: Deviations from Ideal
Behavior
Chapter 10 43

Real Gases: Deviations from Ideal
Behavior
• From the ideal gas equation, we have
PV = nRT
•
This equation breaks-down at
–
High pressure
• At high pressure, the attractive and repulsive forces between gas molecules becomes significant.
–
Small volume
Chapter 10 44

Real Gases: Deviations from Ideal
Behavior
• From the ideal gas equation, we have
PV = nRT
•
This equation breaks-down at
–
High pressure
• At high pressure, the attractive and repulsive forces between gas molecules becomes significant.
–
Small volume
• At small volumes, the volume due to the gas molecules is a source of error.
Chapter 10 45

Real Gases: Deviations from Ideal
Behavior
The Van der Waals Equation
• Two terms are added to the ideal gas equation to correct for volume of molecules and one to correct for intermolecular attractions.
Chapter 10 46

Real Gases: Deviations from Ideal
Behavior
The Van der Waals Equation
• Two terms are added to the ideal gas equation to correct for volume of molecules and one to correct for intermolecular attractions.
P

V nRT
 nb
 n
2 a
V
2
Chapter 10 47

Real Gases: Deviations from Ideal
Behavior
The Van der Waals Equation
• Two terms are added to the ideal gas equation to correct for volume of molecules and one to correct for intermolecular attractions.
P

V nRT
 nb
 n
2 a
V
2
•
a and b are constants, determined by the particular gas.
Chapter 10 48

Homework:
10.12, 10.28, 10.42, 10.48, 10.54, 10.66, 10.72
Chapter 10 49

Problem
A sample of gas occupies a volume of 7.50L at
0.988 atm and 28.0
o C. Calculate the pressure of the gas if its volume is decreased to 4.89L while its temperature is held constant.
Chapter 10 50

Problem
A sample of gas occupies a volume of 7.50L at 0.988 atm and 28.0
o C. Calculate the pressure of the gas if its volume is decreased to 4.89L while its temperature is held constant.
--BOYLES LAW
Chapter 10 51

Problem
A sample of gas occupies a volume of 7.50L at 0.988 atm and 28.0
o C. Calculate the pressure of the gas if its volume is decreased to 4.89L while its temperature is held constant.
--BOYLES LAW
P

1
V
1
P
2
V
2
Chapter 10 52

Problem
A sample of gas occupies a volume of 7.50L at 0.988 atm and 28.0
o C. Calculate the pressure of the gas if its volume is decreased to 4.89L while its temperature is held constant.
--BOYLES LAW
0 .
988 atm ( 7 .
P
1
V
2

P
2
V
2
50 L )

P
2
( 4 .
89 L )
Chapter 10 53

Problem
A sample of gas occupies a volume of 7.50L at 0.988 atm and 28.0
o C. Calculate the pressure of the gas if its volume is decreased to 4.89L while its temperature is held constant.
0 .
988
--BOYLES LAW
P
1
V
2

P
2
V
2 atm ( 7 .
50 L )

P
2
( 4 .
89 L )
P
2

0 .
988 atm ( 7 .
50
4 .
89 L
L )

1 .
52 atm
Chapter 10 54

Problem
A sample of gas occupies a volume of 7.50L at 0.988 atm and 28.0
o C. At what temperature in degrees
Celsius is the volume of the gas 4.00L if the pressure is kept constant.
Chapter 10 55

Problem
A sample of gas occupies a volume of 7.50L at 0.988 atm and 28.0
o C. At what temperature in degrees
Celsius is the volume of the gas 4.00L if the pressure is kept constant.
--Charles’ Law
Chapter 10 56

Problem
A sample of gas occupies a volume of 7.50L at 0.988 atm and 28.0
o C. At what temperature in degrees
Celsius is the volume of the gas 4.00L if the pressure is kept constant.
--Charles’ Law
V
1
T
1

V
2
T
2
Chapter 10 57

Problem
A sample of gas occupies a volume of 7.50L at 0.988 atm and 28.0
o C. At what temperature in degrees
Celsius is the volume of the gas 4.00L if the pressure is kept constant.
--Charles’ Law
P
V
2

1

P
V
2
2 convert temperatur e to kelvin
K

273

28

301 K
Chapter 10 58

Problem
A sample of gas occupies a volume of 7.50L at 0.988 atm and 28.0
o C. At what temperature in degrees
Celsius is the volume of the gas 4.00L if the pressure is kept constant.
--Charles’ Law
P
1
V
2

P
V
2
2
7 .
50 L
301 K

4 .
00 L x
Chapter 10 59

Problem
A sample of gas occupies a volume of 7.50L at 0.988 atm and 28.0
o C. At what temperature in degrees
Celsius is the volume of the gas 4.00L if the pressure is kept constant.
--Charles’ Law
P
V
2
1

P
V
2
2
7 .
50 L

301 K
4 .
00 L x x

160 .
5 K
   to o   
112 .
5
Chapter 10 60

Problem
Calcium hydride, CaH
2
, reacts with water to form hydrogen gas:
CaH
2
(s) + 2 H
2
O(l)

Ca(OH)
2
(aq) + 2 H
2
(g)
How many grams of CaH
2
10.0L of H
23 o C?
2 are needed to generate gas if the pressure of H
2 is 740 torr at
Chapter 10 61

Problem
CaH
2
(s) + 2 H
2
O(l)

Ca(OH)
2
(aq) + 2 H
2
(g)
Calculate moles of H
2 formed
Calculate moles of CaH
2 needed
Convert moles CaH
2 to grams
Chapter 10 62

Problem
CaH
2
(s) + 2 H
2
O(l)

Ca(OH)
2
(aq) + 2 H
2
(g)
P

740 torr / 760

0 .
974
V

10 .
0 L
T

23 o
C

273

296 K r

0 .
08206 L ( atm ) / mol ( K )
Chapter 10 63

Problem
CaH
2
(s) + 2 H
2
O(l)

Ca(OH)
2
(aq) + 2 H
2
(g)
P

740 torr / 760

0 .
974
V

10 .
0 L
T r

23 o
C

273

296 K

0 .
08206 L ( atm ) / mol ( K ) n
H
2

0 .
974 atm ( 10 .
0 L )
0 .
08206 ( 296 K )
Chapter 10 64

Problem
CaH
2
(s) + 2 H
2
O(l)

Ca(OH)
2
(aq) + 2 H
2
(g)
P

740 torr / 760

0 .
974
V

10 .
0 L
T r

23 o
C

273

296 K

0 .
08206 L ( atm ) / mol ( K ) n
H
2

0 .
974 atm ( 10 .
0 L )
0 .
08206 ( 296 K ) n
H
2

0 .
401 mol
Chapter 10 65

Problem
CaH
2
(s) + 2 H
P

2
O(l)

Ca(OH)
2
740 torr / 760

0 .
974
(aq) + 2 H
2
(g)
V

10 .
0 L
T

23 o
C

273

296 K r

0 .
08206 L ( atm ) / mol ( K ) n
H
2 n
H
2

0 .
974 atm ( 10 .
0 L )
0 .
08206 ( 296 K )

0 .
401 mol
1 CaH
2
2 H
2
 x
0 .
401 mol
Chapter 10 66

Problem
CaH
2
(s) + 2 H
P

2
O(l)

Ca(OH)
2
740 torr / 760

0 .
974
(aq) + 2 H
2
(g)
V

10 .
0 L
T r

23 o
C

273

296 K

0 .
08206 L ( atm ) / mol ( K ) n
H
2 n
H
2

0 .
974 atm ( 10 .
0 L )
0 .
08206 ( 296 K )

0 .
401 mol
1 CaH
2 H
2
2

0 .
x
401 x

0 .
2005 mol mol
Chapter 10 67

Problem
CaH
2
(s) + 2 H
P

2
O(l)

Ca(OH)
2
740 torr / 760

0 .
974
(aq) + 2 H
2
(g)
V

10 .
0 L
T r

23 o
C

273

296 K

0 .
08206 L ( atm ) / mol ( K ) n
H
2 n
H
2

0 .
974 atm ( 10 .
0 L )
0 .
08206 ( 296 K )

0 .
401 mol
1 x
CaH
2 H
2
2

0 .
x
401

0 .
2005 mol mol gCaH
2

0 .
2005 mol ( 42 .
10 g / mol )
Chapter 10 68

Problem
CaH
2
(s) + 2 H
P

2
O(l)

Ca(OH)
2
740 torr / 760

0 .
974
(aq) + 2 H
2
(g)
V

10 .
0 L
T r

23 o
C

273

296 K

0 .
08206 L ( atm ) / mol ( K ) n
H
2

0 .
974 atm ( 10 .
0 L )
0 .
08206 ( 296 K ) n
H
2

0 .
401 mol
1 CaH
2 H
2
2

0 .
x
401 x

0 .
2005 mol mol gCaH
2

8 .
44 g

0 .
2005 mol ( 42 .
10 g / mol )
Chapter 10 69

Problem
CaH
2
(s) + 2 H
P

2
O(l)

Ca(OH)
2
740 torr / 760

0 .
974
(aq) + 2 H
2
(g)
V

10 .
0 L
T r

23 o
C

273

296 K

0 .
08206 L ( atm ) / mol ( K ) n
H
2

0 .
974 atm ( 10 .
0 L )
0 .
08206 ( 296 K ) n
H
2

0 .
401 mol
1 CaH
2 H
2
2

0 .
x
401 x

0 .
2005 mol mol gCaH
2

8 .
44 g

0 .
2005 mol ( 42 .
10 g / mol )
Chapter 10 70