Transcript The Nature of Gases - Gordon State College

Chapter 7
Gases
1
Kinetic Theory of Gases
Particles of a gas
 Move rapidly in straight lines and are in
constant motion.
 Have kinetic energy that increases with an
increase in temperature.
 Are very far apart.
 Have essentially no attractive (or repulsive)
forces.
 Have very small volumes compared to the
volume of the container they occupy.
2
Properties of Gases

Gases are described in terms of four properties:
pressure (P), volume (V), temperature (T), and
amount (n).
3
Barometer


A barometer
measures the pressure
exerted by the gases
in the atmosphere.
The atmospheric
pressure is measured
as the height in mm
of the mercury
column.
4
Learning Check
A. The downward pressure of the Hg in a
barometer is _____ than/as the weight of the
atmosphere.
1) greater
2) less
3) the same
B. A water barometer is 13.6 times taller than a Hg
barometer (DHg = 13.6 g/mL) because
1) H2O is less dense
2) H2O is heavier
3) air is more dense than H2O
5
Solution
A.The downward pressure of the Hg in a
barometer is 3) the same as the weight of the
atmosphere.
B. A water barometer is 13.6 times taller than a
Hg barometer (DHg = 13.6 g/mL) because
1) H2O is less dense
6
Pressure
 A gas exerts pressure, which is defined as a
force acting on a specific area.
Pressure (P) = Force
Area
 One atmosphere (1 atm) is 760 mm Hg.
 1 mm Hg = 1 torr
1.00 atm = 760 mm Hg = 760 torr
7
Units of Pressure

In science, pressure is stated in atmospheres
(atm), millimeters of mercury (mm Hg), and
Pascals (Pa).
8
Learning Check
A. What is 475 mm Hg expressed in atm?
1) 475 atm
2) 0.638 atm
3) 3.61 x 105 atm
B. The pressure in a tire is 2.00 atm. What is this
pressure in mm Hg?
1) 2.00 mm Hg
2) 1520 mm Hg
3) 22,300 mm Hg
9
Solution
A. What is 475 mm Hg expressed in atm?
2) 0.638 atm
485 mm Hg x
1 atm
= 0.638 atm
760 mm Hg
B. The pressure of a tire is measured as 2.00 atm.
What is this pressure in mm Hg?
2) 1520 mm Hg
2.00 atm x 760 mm Hg = 1520 mm Hg
1 atm
10
Boyle’s Law
 The pressure of a
gas is inversely
related to its
volume when T
and n are
constant.
 If volume
decreases, the
pressure
increases.
11
PV Constant in Boyle’s Law

The product P x V remains constant as long as
T and n do not change.
P1V1 = 8.0 atm x 2.0 L = 16 atm L
P2V2 = 4.0 atm x 4.0 L = 16 atm L
P3V3 = 2.0 atm x 8.0 L = 16 atm L

Boyle’s Law can be stated as
P1V1 = P2V2 (T, n constant)
12
Solving for a Gas Law Factor



The equation for Boyle’s Law can be rearranged to
solve for any factor.
To solve for V2, divide both sides by P2.
P1V1 = P2V2
Boyle’s Law
P2
P2
P1V1 =
V2
P2
Solving for P2
P2 = P1 V1
V2
13
PV in Breathing Mechanics


When the lungs
expand, the
pressure in the
lungs decreases.
Inhalation occurs
as air flows towards
the lower pressure
in the lungs.
14
PV in Breathing Mechanics


When the lung
volume decreases,
pressure within the
lungs increases.
Exhalation occurs
as air flows from
the higher pressure
in the lungs to the
outside.
15
Calculation with Boyle’s Law
Freon-12, CCl2F2, is used in refrigeration
systems. What is the new volume (L) of a 8 L
sample of Freon gas initially at 50 mm Hg
after its pressure is changed to 200 mm Hg at
constant T?
1. Set up a data table
Conditions 1
Conditions 2
P1 = 50 mm Hg
V1 = 8 L
P2
V2
= 200 mm Hg
= ?
16
Calculation with Boyle’s Law
(continued)
2. When pressure increases, volume decreases.
Solve Boyle’s Law for V2:
P1V1 = P2V2
V2
= V1P1
P2
V2 = 8 L x
50 mm Hg
200 mm Hg
=
2L
pressure ratio decreases volume
17
Learning Check
The helium in a cylinder
has a volume of 120 mL
and a pressure of 840 mm
Hg. A change in the volume
results in a lower pressure
inside the cylinder. Does
cylinder A or B represent the final volume?
At a new pressure of 420 mm Hg, what is the new
volume?
1) 60 mL
2) 120 mL
3) 240 mL
18
Solution
The helium in a cylinder has a volume of 120 mL and
a pressure of 840 mm Hg. A change in the volume
results in a lower pressure inside the cylinder. Does
cylinder A or B represent the final volume?
B) If P decreases, V increases.
At a new pressure of 420 mm
Hg, what is the new volume
of the cylinder?
3) 240 mL
19
Learning Check
A sample of helium gas
has a volume of 6.4 L at
a pressure of 0.70 atm.
What is the new volume
when the pressure is
increased to 1.40 atm
(T constant)?
A) 3.2 L
B) 6.4 L
C) 12.8 L
20
Solution
A) 3.2 L
Solve for V2: P1V1 = P2V2
V2 = V1P1
P2
V2
= 6.4 L x 0.70 atm = 3.2 L
1.40 atm
Volume decreases when there is an increase in the
pressure (Temperature is constant).
21
Learning Check
A sample of oxygen gas has a volume of 12.0 L at
600. mm Hg. What is the new pressure when the
volume changes to 36.0 L? (T and n constant.)
1) 200. mm Hg
2) 400. mm Hg
3) 1200 mm Hg
22
Solution
1) 200. mm Hg
Data table
Conditions 1
P1 = 600. mm Hg
V1 = 12.0 L
P2
Conditions 2
P2 = ???
V2 = 36.0 L
P1 V1
V2
600. mm Hg x 12.0 L = 200. mm Hg
36.0 L
=
23
Charles’ Law


The Kelvin
temperature of a
gas is directly
related to the
volume (P and n are
constant).
When the
temperature of a
gas increases, its
volume increases.
24
Charles’ Law V and T

For two conditions, Charles’ Law is written
V1 = V2 (P and n constant)
T1
T2

Rearranging Charles’ Law to solve for V2
V2 = V1T2
T1
25
Learning Check
Solve Charles’ Law expression for T2.
V1 = V2
T1
T2
26
Solution
V1
T1
= V2
T2
Cross multiply to give
V1T2 = V2T1
Isolate T2 by dividing through by V1
V1T2
V1
T2 =
=
V1
V2T1
V2T1
V1
27
Calculations Using Charles’ Law
A balloon has a volume of 785 mL at 21°C. If
The temperature drops to 0°C, what is the new
volume of the balloon (P constant)?
1. Set up data table:
Conditions 1
Conditions 2
V1 = 785 mL
V2 = ?
T1 = 21°C = 294 K T2 = 0°C = 273 K
Be sure that you always use the Kelvin (K)
temperature in gas calculations.
28
Calculations Using Charles’ Law
(continued)
2. Solve Charles’ law for V2
V1 = V2
T1
T2
V2 = V1 T2
T1
V2 = 785 mL x
273 K = 729 mL
294 K
29
Learning Check
A sample of oxygen gas has a volume of 420
mL at a temperature of 18°C. At what
temperature (in °C) will the volume of the
oxygen be 640 mL (P and n constant)?
1) 443°C
2) 170°C
3) – 82°C
30
Solution
170°C
T2 = T1V2
V1
T2 = 291 K x
640 mL
420 mL
= 443 K – 273 K
= 443 K
= 170°C
31
Gay-Lussac’s Law: P and T

The pressure
exerted by a gas is
directly related to
the Kelvin
temperature of
the gas at constant
V and n.
P1 = P2
T1
T2
32
Calculation with Gay-Lussac’s Law
A gas has a pressure at 2.0 atm at 18°C. What
is the new pressure when the temperature is
62°C? (V and n constant)
1.
Set up a data table.
Conditions 1
Conditions 2
?
P1
= 2.0 atm
P2
=
T1
= 18°C + 273
T2
= 62°C + 273
= 291 K
= 335 K
33
Calculation with Gay-Lussac’s Law
(continued)
2. Solve Gay-Lussac’s Law for P2
P1 = P2
T1
T2
P2 = P1 T2
T1
P2 = 2.0 atm x 335 K = 2.3 atm
291 K
34
Learning Check
Use the gas laws to complete with
1)
Increases 2) Decreases
2)
A. Pressure _______ when V decreases.
B. When T decreases, V _______.
C. Pressure _______ when V changes
from 12.0 L to 24.0 L.
D. Volume _______when T changes from
15.0 °C to 45.0°C.
35
Solution
Use the gas laws to complete with
1) Increases 2) Decreases
A. Pressure 1) Increases, when V decreases.
B. When T decreases, V 2) Decreases.
C. Pressure 2) Decreases when V changes
from 12.0 L to 24.0 L
D. Volume 1) Increases when T changes from
15.0 °C to 45.0°C
36
Next Time

We complete Chapter 7
37
Combined Gas Law

The combined gas law uses Boyle’s Law, Charles’
Law, and Gay-Lussac’s Law (n is constant).
P1 V1
T1
=
P2 V2
T2
38
Combined Gas Law Calculation
A sample of helium gas has a volume of 0.180 L, a
pressure of 0.800 atm and a temperature of 29°C.
At what temperature (°C) will the helium have a
volume of 90.0 mL and a pressure of 3.20 atm (n
constant)?
1. Set up Data Table
Conditions 1
Conditions 2
P1 = 0.800 atm
P2 = 3.20 atm
V1 = 0.180 L (180 mL) V2 = 90.0 mL
T1 = 29°C + 273 = 302 K T2 = ??
39
Combined Gas Law Calculation
(continued)
2. Solve for T2
P1 V1
T1
=
P2 V2
T2
T2 = T1 P2V2
P1V1
T2 = 302 K x 3.20 atm x 90.0 mL = 604 K
0.800 atm 180.0 mL
T2 = 604 K – 273 = 331 °C
40
Learning Check
A gas has a volume of 675 mL at 35°C and
0.850 atm pressure. What is the volume(mL)
of the gas at –95°C and a pressure of 802 mm
Hg (n constant)?
41
Solution
Data Table
T1 = 308 K
T2 = -95°C + 273 = 178K
V1 = 675 mL
V2 = ???
P1 = 646 mm Hg
P2 = 802 mm Hg
Solve for T2
V2 = V1 P1 T2
P2T1
V2 = 675 mL x 646 mm Hg x 178K = 314 mL
802 mm Hg x 308K
42
Avogadro's Law: Volume and Moles

The volume of a
gas is directly
related to the
number of moles
of gas when T
and P are
constant.
V1 = V2
n1
n2
43
Learning Check
If 0.75 mole of helium gas occupies a
volume of 1.5 L, what volume will 1.2 moles
of helium occupy at the same temperature
and pressure?
1) 0.94 L
2) 1.8 L
3) 2.4 L
44
Solution
3) 2.4 L
Conditions 1
V1 = 1.5 L
n1 = 0.75 mole He
V2 = V1n2
n1
Conditions 2
V2 = ???
n2 = 1.2 moles He
V2 = 1.5 L x 1.2 moles He
0.75 mole He
= 2.4 L
45
STP

The volumes of gases can be compared when they
have the same conditions of temperature and
pressure (STP).
Standard temperature (T)
0°C or 273 K
Standard pressure (P)
1 atm (760 mm Hg)
46
Molar Volume


At STP, 1 mole of
a gas occupies a
volume of 22.4 L.
The volume of
one mole of a gas
is called the molar
volume.
47
Molar Volume as a Conversion
Factor

The molar volume at STP can be used to
form conversion factors.
22.4 L
1 mole
and
1 mole
22.4 L
48
Learning Check
A. What is the volume at STP of 4.00 g of CH4?
1) 5.60 L 2) 11.2 L
3) 44.8 L
B. How many grams of He are present in 8.00 L
of gas at STP?
1) 25.6 g
2) 0.357 g
3) 1.43 g
49
Solution
A.
1) 5.60 L
4.00 g CH4 x 1 mole CH4 x 22.4 L (STP) = 5.60 L
16.0 g CH4
1 mole CH4
B. 3) 1.43 g
8.00 L x 1 mole He
22.4 L
x
4.00 g He = 1.43 g He
1 mole He
50
Ideal Gas Law


The relationship between the four
properties (P, V, n, and T) of gases can be
written equal to a constant R.
PV = R
nT
Rearranging this expression gives the
expression called the ideal gas law.
PV = nRT
51
Universal Gas Constant, R

The universal gas constant, R, can be calculated
using the molar volume of a gas at STP.

At STP (273 K and 1.00 atm), 1 mole of a gas
occupies 22.4 L.
P
V
R =
PV =
(1.00 atm)(22.4 L)
nT
(1 mole) (273K)
n
T
=
0.0821 L atm
mole K

Note there are four units associated with R.
52
Learning Check
Another value for the universal gas constant is
obtained using mm Hg for the STP pressure. What is
the value of R when a pressure of 760 mm Hg is
placed in the R value expression?
53
Solution
What is the value of R when the STP value
for P is 760 mmHg?
R =
PV
nT
= (760 mm Hg) (22.4 L)
(1 mole)
(273K)
= 62.4 L mm Hg
mole K
54
Learning Check
Dinitrogen oxide (N2O), laughing gas, is used
by dentists as an anesthetic. If a 20.0 L tank
of laughing gas contains 2.8 moles N2O at
23°C, what is the pressure (mm Hg) in the
tank?
55
Solution
1. Adjust the units of the given properties to
match the units of R.
V = 20.0 L, T = 296 K, n = 2.8 moles, P = ?
2. Rearrange the ideal gas law for P.
P = nRT
V
P = (2.8 moles)(62.4 L mm Hg)(296 K)
(20.0 L)
(mole K)
= 2.6 x 103 mm Hg
56
Learning Check
A cylinder contains 5.0 L of O2 at
20.0°C and 0.85 atm. How many
grams of oxygen are in the
cylinder?
57
Solution
1. Determine the given properties.
P = 0.85 atm, V = 5.0 L, T = 293 K, n (or g =?)
2. Rearrange the ideal gas law for n (moles).
n = PV
RT
= (0.85 atm)(5.0 L)(mole K) = 0.18 mole O2
(0.0821atm L)(293 K)
3. Convert moles to grams using molar mass.
= 0. 18 mole O2 x 32.0 g O2 = 5.8 g O2
1 mole O2
58
Molar Mass of a Gas
What is the molar mass of a gas if 0.250 g of the gas
occupy 215 mL at 0.813 atm and 30.0°C?
1. Solve for the moles (n) of gas.
n = PV =
(0.813 atm) (0.215 L)
RT (0.0821 L atm/mole K)(303K)
= 0.00703 mole
2. Set up the molar mass relationship.
Molar mass =
g =
0.250 g
= 35.6g/mole
mole
0.00703 mole
59
Gases in Equations

The amounts of gases reacted or produced in
a chemical reaction can be calculated using
the ideal gas law and mole factors.
Problem:
What volume (L) of Cl2 gas at 1.2 atm and
27°C is needed to completely react with 1.5 g
of aluminum?
2Al(s) + 3Cl2 (g)
2AlCl3(s)
60
Gases in Equations (continued)
2Al(s) + 3Cl2 (g)
1.5 g
? L 1.2 atm, 300K
1.
2AlCl3(s)
Calculate the moles of Cl2 needed.
1.5 g Al x 1 mole Al x 3 moles Cl2 = 0.083 mole Cl2
27.0 g Al
2 moles Al
2. Place the moles Cl2 in the ideal gas equation.
V = nRT = (0.083 mole Cl2)(0.0821 Latm/moleK)(300K)
P
1.2 atm
= 1.7 L Cl2
61
Learning Check
What volume (L) of O2 at 24°C and 0.950 atm are
needed to react with 28.0 g NH3?
4NH3(g) + 5O2(g)
4NO(g) + 6H2O(g)
62
Solution
1. Calculate the moles of O2 needed.
28.0 g NH3 x 1 mole NH3 x 5 mole O2
17.0 g NH3
4 mole NH3
= 2.06 mole O2
2. Place the moles O2 in the ideal gas equation.
V = nRT = (2.06 moles)(0.0821 L atm/moleK)(297K)
P
0.950 atm
= 52.9 L O2
63
Partial Pressure

In a mixture of gases, the partial
pressure of each gas is the pressure
that gas would exert if it were by itself
in the container.
64
Dalton’s Law of Partial Pressures

The total pressure
exerted by a gas
mixture is the sum
of the partial
pressures of the
gases in that
mixture.
PT = P1 + P2 + .....
65
Partial Pressures

The total pressure of a gas mixture
depends on the total number of gas
particles, not on the types of
particles.
66
Total Pressure

For example, at STP, one mole of gas particles in a
volume of 22.4 L will exert the same pressure as
one mole of a mixture of gas particles in 22.4 L.
V = 22.4 L
1.0 mole N2
1.0 atm
0.4 mole O2
0.6 mole He
1.0 mole
1.0 atm
0.5 mole O2
0.3 mole He
0.2 mole Ar
1.0 mole
1.0 atm
67
Learning Check
A scuba tank contains O2 with a pressure of 0.450
atm and He at 855 mm Hg. What is the total
pressure in mm Hg in the tank?
68
Solution
1. Convert the pressure in atm to mm Hg
0.450 atm x 760 mm Hg = 342 mm Hg = PO
2
1 atm
2. Calculate the sum of the partial pressures.
Ptotal = PO + PHe
2
Ptotal = 342 mm Hg + 855 mm Hg
= 1197 mm Hg
69
Gases We Breathe
70
Health Note: Scuba Diving



When a scuba diver makes a deep dive, the
increased pressure causes more N2 (g) to dissolve
in the blood.
If a diver rises too fast, the dissolved N2 will form
bubbles in the blood, a dangerous and painful
condition called "the bends."
Helium, which does not dissolve in the blood, is
mixed with O2 to prepare breathing mixtures for
deep descents.
71
Learning Check
For a deep dive, some scuba divers are using
a mixture of helium and oxygen gases with a
pressure of 8.00 atm. If the oxygen has a
partial pressure of 1280 mm Hg, what is the
partial pressure of the helium?
1) 520 mm Hg
2) 2040 mm Hg
3) 4800 mm Hg
72
Solution
3)
4800 mm Hg
PTotal = 8.00 atm x 760 mm Hg = 6080 mm Hg
1 atm
PTotal = PO + PHe
2
PHe = PTotal - PO2
PHe = 6080 mm Hg - 1280 mm Hg
= 4800 mm Hg
73