CHAPTER 8
{
GASES
1
Comparison of Solids, Liquids,
and Gases
{
The density of gases is much less than that
of solids or liquids.
Densities
(g/mL)
H2O
Solid
Liquid
Gas
0.917
0.998
0.000588
CCl4
1.70
1.59
0.00503
Gas molecules must be very far apart
compared to liquids and solids.
{
2
The Kinetic-Molecular Theory
{
{
The basic assumptions of kineticmolecular theory are:
Postulate 1
z
z
z
{
Gases consist of molecules that are relatively
far apart.
Gases have few intermolecular attractions.
The volume of individual molecules is very
small compared to the gas’s volume.
Proof - Gases are easily compressible.
3
The Kinetic-Molecular Theory
{
Postulate 2
z
{
Gas molecules are in constant, random,
straight line motion with varying
velocities.
Postulate 3
z
z
Gas molecules have elastic collisions
with themselves and the container.
Total energy is conserved during a
collision.
4
The Kinetic-Molecular Theory
{
Postulate 4
z
z
{
The kinetic energy of the molecules is
proportional to the Kelvin temperature.
The average kinetic energies of
molecules of different gases are equal at
a given temperature.
Proof - Motion increases as
temperature increases.
5
The Kinetic-Molecular Theory
• The kinetic energy of the molecules is
proportional to the absolute
temperature.
6
Pressure
{
Pressure is force per unit area.
z
z
{
lb/in2
N/m2
Atmospheric pressure is measured using a
barometer.
Hg density = 13.6 g/mL
7
Pressure
8
Boyle’s Law:
The Volume-Pressure Relationship
{
{
Volume of a gas decreases, pressure
increases.
Volume of a gas increases, pressure
decreases.
z
z
{
{
NO CHANGE IN TEMPERATURE
NO CHANGE IN # MOLES OF GAS
Thus we write Boyle’s Law mathematically
as P1V1 = P2V2
We can use any pressure or volume units as
long as we use the same units for both P1
and P2 or V1 and V2.
9
Boyle’s Law:
The Volume-Pressure Relationship
10
Boyle’s Law:
The Volume-Pressure Relationship
{
Example 12-1: At 25oC a sample of He has a
volume of 4.00 x 102 mL under a pressure
of 7.60 x 102 torr. What volume would it
occupy under a pressure of 2.00 atm at the
same T?
P1 V1 = P2 V2
P1 V1
V2 =
P2
(
760 torr )(400 mL )
=
1520 torr
= 2.00 × 10 2 mL
11
Boyle’s Law:
The Volume-Pressure Relationship
Inhale—decrease pressure in lungs
Breathe in until pressure in lungs equals atmospheric pressure
Exhale—decrease volume in lungs
Air released to decrease the pressure
12
Charles’ Law:
The Volume-Temperature Relationship;
The Absolute Temperature Scale
{
Charles’s law states that the volume
of a gas is directly proportional to
the absolute temperature at
constant pressure.
z
{
Gas laws must use the Kelvin scale to
be correct.
Relationship between Kelvin and
centigrade.
K = o C + 273
13
Charles’ Law:
The Volume-Temperature Relationship;
The Absolute Temperature Scale
V1 V2
=
T1 T2
14
Charles’ Law:
The Volume-Temperature Relationship;
The Absolute Temperature Scale
{
Example 12-2: A sample of hydrogen, H2,
occupies 1.00 x 102 mL at 25.0oC and 1.00
atm. What volume would it occupy at
50.0oC under the same pressure?
T1 = 25 + 273 = 298
T2 = 50 + 273 = 323
V1 V2
V1T2
=
∴ V2 =
T1 T2
T1
1.00 × 10 mL × 323 K
V2 =
298 K
= 108 mL
2
15
Gay-Lussac’s Law
Pressure-Temperature Relationship
{
Pressure of a gas is directly related to its
temperature
z
As long as the volume and the number of moles
or gas do not change
P1 P2
=
T1 T2
16
Gay-Lussac’s Law
Pressure-Temperature Relationship
{
{
{
In an open container, molecules with
enough kinetic energy leave the surface
and evaporate
In a closed container, vapor accumulates
and creates a pressure called “vapor
pressure”
A liquid’s boiling point is the point where
the vapor pressure is equal to the outside
pressure
17
Gay-Lussac’s Law
Pressure-Temperature Relationship
18
Gay-Lussac’s Law
Pressure-Temperature Relationship
19
Gay-Lussac’s Law
Pressure-Temperature Relationship
20
Summary of Gas Laws
21
Standard Temperature and
Pressure
{
Standard temperature and pressure is
given the symbol STP.
z
{
{
It is a reference point for some gas
calculations.
Standard P ≡ 1.00000 atm or 101.3 kPa
Standard T ≡ 273.15 K or 0.00o
22
The Combined Gas Law Equation
{
The gas laws presented so far can be
combined into one statement is called the
combined gas law equation.
z
Useful when the V, T, and P of a gas are
changing.
P1 V1 P2 V2
=
T1
T2
23
The Combined Gas Law Equation
{
Example 12-3: A sample of nitrogen gas, N2,
occupies 7.50 x 102 mL at 75.00C under a
pressure of 8.10 x 102 torr. What volume
would it occupy at STP?
V 1 = 750 mL
V2 = ?
T1 = 348 K
T 2 = 273 K
P1 = 810 torr
P 2 = 760 torr
Solve for V 2 =
=
(810
P1 V 1 T 2
P 2 T1
torr )(750 mL )(273 K
(760 torr )(348 K )
= 627 mL
)
24
The Combined Gas Law Equation
{
Example 12-4 : A sample of methane,
CH4, occupies 2.60 x 102 mL at 32oC
under a pressure of 0.500 atm. At
what temperature would it occupy 5.00
x 102 mL under a pressure of 1.20 x
103 torr?
You do it!
25
The Combined Gas Law Equation
V1 = 260 mL
V2 = 500 mL
P1 = 0.500 atm
P2 = 1200 torr
= 380 torr
T1 = 305 K
T2 = ?
T1 P2 V2 (305 K )(1200 torr )(500 mL )
T2 =
=
(380 torr )(260 mL )
P1 V1
= 1852 K ≈ 1580 C
o
26
Avogadro’s Law and the
Standard Molar Volume
{
Example 12-5: One mole of a gas
occupies 36.5 L and its density is 1.36 g/L
at a given temperature and pressure. (a)
What is its molar mass? (b) What is its
density at STP?
? g 365
. L 136
. g
=
×
= 49.6 g / mol
mol mol
L
?g
49 . 6 g 1 mol
=
×
= 2 . 21 g/L
L STP
mol
22 . 4 L
27
Avogadro’s Law and the
Standard Molar Volume
{
{
{
Avogadro’s Law states that at the same
temperature and pressure, equal volumes of two
gases contain the same number of molecules (or
moles) of gas.
If we set the temperature and pressure for any
gas to be STP, then one mole of that gas has a
volume called the standard molar volume.
volume
The standard molar volume is 22.4 L at STP.
z
z
{
This is another way to measure moles.
For gases, the volume is proportional to the
number of moles.
11.2 L of a gas at STP = 0.500 mole
z
44.8 L = ? moles
28
Avogadro’s Law and the
Standard Molar Volume
29
Summary of Gas Laws:
The Ideal Gas Law
{
{
All of the gas laws can be combined to
give a single Law called the Ideal Gas
Law.
PV = nRT
R is a proportionality constant called the
universal gas constant.
30
Summary of Gas Laws:
The Ideal Gas Law
{
We must determine the value of R.
z
z
Recognize that for one mole of a gas at 1.00
atm, and 273 K (STP), the volume is 22.4 L.
Use these values in the ideal gas law.
PV (1.00 atm )( 22.4 L )
R =
=
nT (1.00 m ol )( 273 K )
L atm
= 0.0821
m ol K
31
Summary of Gas Laws:
The Ideal Gas Law
{
Example 12-6: What volume would
50.0 g of ethane, C2H6, occupy at 1.40
x 102 oC under a pressure of 1.82 x 103
torr?
To use the ideal gas law correctly, it is very
important that all of your values be in the
correct units!
T = 140 + 273 = 413 K
P = 1820 torr (1 atm/760 torr) = 2.39 atm
50 g (1 mol/30 g) = 1.67 mol
z
1.
2.
3.
32
Summary of Gas Laws:
The Ideal Gas Law
n R T
V =
P
L atm ⎞
⎛
(1 . 67 mol )⎜ 0 . 0821
⎟ (413 K
mol K ⎠
⎝
=
2 . 39 atm
= 23 . 6 L
)
33
Summary of Gas Laws:
The Ideal Gas Law
{
Example 12-7: Calculate the
number of moles in, and the mass
of, an 8.96 L sample of methane,
CH4, measured at standard
conditions.
You do it!
34
Summary of Gas Laws:
The Ideal Gas Law
(1.00 atm)( 8.96 L)
PV
n =
=
= 0.400 mol CH 4
L atm ⎞
RT ⎛
⎜ 0.0821
⎟ ( 273 K)
⎝
mol K⎠
16.0 g
? g CH 4 = 0.400 mol ×
= 6.40 g
mol
35
Summary of Gas Laws:
The Ideal Gas Law
{
Example 12-8: Calculate the pressure
exerted by 50.0 g of ethane, C2H6, in a
25.0 L container at 25.0oC.
You do it!
n = 1.67 mol and T = 298 K
n R T
P =
V
L atm ⎞
⎛
(1.67 mol )⎜ 0.0821
⎟ (298 K )
mol K ⎠
⎝
P =
25.0 L
P = 1 . 63 atm
36
Dalton’s Law of Partial Pressures
{
Dalton’s law states that the
pressure exerted by a mixture of
gases is the sum of the partial
pressures of the individual gases.
Ptotal = PA + PB + PC + .....
37
Composition of the Atmosphere and
Some Common Properties of Gases
Composition of Dry Air
Gas
% by Volume
N2
78.09
O2
20.94
Ar
0.93
CO2
0.03
He, Ne, Kr, Xe
0.002
CH4
0.00015
H2
0.00005
38
Dalton’s Law of Partial Pressures
39
Dalton’s Law of Partial Pressures
{
Example 12-11: If 1.00 x 102 mL of
hydrogen, measured at 25.0 oC and 3.00
atm pressure, and 1.00 x 102 mL of
oxygen, measured at 25.0 oC and 2.00
atm pressure, were forced into one of the
containers at 25.0 oC, what would be the
pressure of the mixture of gases?
PT otal = P H 2 + PO 2
= 3.00 atm + 2.00 atm
= 5.00 atm
40
Dalton’s Law of Partial Pressures
{
Blood Gasses
41
Dalton’s Law of Partial Pressures
{
Blood Gasses
42