Download Chapter 5 Gases Gases Pushing Measuring Air Pressure

Chapter 5
Helleur
Principles of Chemistry: A Molecular Approach, 1st Ed.
Nivaldo Tro
Chapter 5
Gases
1
Gases Pushing
• Gas molecules are constantly in
motion.
• As they move and strike a surface,
they push on that surface.
9 push = force
• If we could measure the total
amount of force exerted by gas
molecules hitting the entire surface
at any one instant, we would know
the pressure the gas is exerting.
9 pressure = force per unit area
2
Measuring Air Pressure
• Use a barometer.
• column of mercury
•
supported by air
pressure
force of the air on the
surface of the mercury
balanced by the pull of
gravity on the column
of mercury
gravity
3
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Chapter 5
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Common Units of Pressure
Average Air Pressure
at Sea Level
101.325
Unit
kilopascal (kPa)
atmosphere (atm)
1 (exactly)
millimeters of mercury (mmHg)
760 (exactly)
torr (torr)
760 (exactly)
pounds per square inch (psi, lbs./in2)
14.7
4
Example 5.1 A bicycle tire has a pressure of 132 psi.
Convert this pressure to mmHg?
Given:
132 psi
Find:
mmHg
Conceptual
Plan:
psi
atm
mmHg
Relationships: 1 atm = 14.7 psi, 1 atm = 760 mmHg
Solution:
5
The Simple Gas Laws
1) Boyle’s Law
• Pressure of a gas is inversely proportional to its
volume.
9 constant T and amount of gas
9 graph P vs. V is curve
9 graph P vs. 1/V is straight line
• As P increases, V decreases by the same factor.
• P × V = constant
P1 × V1 = P2 × V2 (at constant T and mol of gas)
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Boyle’s Experiment
• added Hg to a J-tube
with air trapped inside
• used length of air
column as a measure of
volume
7
Boyle’s Law: A Molecular View
Pressure is caused by the molecules striking the sides of the
container.
When you decrease the volume of the container with the
same number of molecules in the container, more molecules
will hit the wall at the same instant.
This results in
increasing the
pressure.
8
Example 5.2 A cylinder with a movable piston has a volume
of 7.25 L at 4.52 atm. What is the volume at 1.21 atm?
Given: V1 = 7.25 L, P1 = 4.52 atm, P2 = 1.21 atm
Find: V2, L
Conceptual
Plan:
Relationships: P1
V1, P1, P2
V2
· V1 = P2 · V2
Solution:
9
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Chapter 5
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2) Charles’s Law
• Volume is directly proportional to
temperature.
9constant P and amount of gas
9graph of V vs. T is straight line
• As T increases, V also increases.
• Note: Kelvin T = Celsius T + 273.15
• V = constant × T
9if T measured in Kelvin
T is in units of K
10
Charles Law
A Problem to Consider
• A sample of methane gas that has a volume of 3.8
L at 5.0°C is heated to 86.0°C at constant
pressure. Calculate its new volume.
using
and converting temp to units of K
i.e, T1(K)= 5.0 oC + 273.15 = 278.2 K
V2= V1 x T2
T1
=
(3.8L) (359.2K ) = 4.9 L
278.2 K
Copyright © Houghton Mifflin Company. All rights reserved.
Presentation of Lecture Outlines, 5–11
3) Avogadro’s Law
• volume directly proportional to
the number of gas molecules
9V = constant × n
9while P and T remain constant
9more gas molecules = larger
volume
• count number of gas molecules
by moles
• Equal volumes of gases contain
equal numbers of molecules.
9The identity gas doesn’t
matter.
12
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Chapter 5
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Clicker # 5-1
Equal volumes of propane, C3H8, and carbon
monoxide at the same temperature and pressure
have the same
a. density.
b. number of molecules.
c. number of atoms.
1) a only
2) b only
3) c only
4) a and b only
5) a, b, and c
Example 5.4 A 0.225-mol sample of He has a volume of
4.65 L. How many moles must be added to give 6.48 L?
Given: V1 = 4.65 L, V2 = 6.48 L, n1 = 0.225 mol
Find: n2, and added moles
Conceptual
Plan:
V1, V2, n1
n2
Relationships: mol added = n2 – n1,
Solution:
14
Standard Conditions
• Since the volume of a gas varies with
pressure and temperature, chemists have
agreed on a set of conditions to report our
measurements so that comparison is easy.
We call these standard conditions.
Called STP (standard temperature and pressure)
• standard pressure = 1.00 atm
• standard temperature = 273.15 K (or 0.00 oC)
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Chapter 5
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Molar Volume at STP
9The volume of one mole of gas is called the
molar gas volume, Vm.
9Volumes of gases are often compared at
standard temperature and pressure (STP),
chosen to be 0.00 oC and 1.00 atm pressure.
At STP, the molar volume, Vm, that is, the
volume occupied by one mole of any gas, is
22.4 L/mol
or 6.022 × 1023 molecules of gas
Molar Volume is the same for any gas
17
A Problem to Consider
How many moles of gas at STP are in
39.3 L ?
moles of gas =
VSTP
22.4 L/mol
moles of gas =
39.3 L
22.4 L/mol
moles of gas = 1.75 mol
Copyright © Houghton Mifflin Company. All rights reserved.
Presentation of Lecture Outlines, 5–18
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Chapter 5
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Clicker # 5-2
Which of the following will have the greatest
volume at STP?
a.
b.
c.
d.
e.
5.0 g H2
5.0 g N2
5.0 g O2
5.0 g CO
All of the above have
the same volume.
Practice—How many liters of O2 at STP can be made from
the decomposition of 100.0 g of PbO2?
2 PbO2(s) → 2 PbO(s) + O2(g)
(PbO2 = 239.2 g/mole, O2 = 32.00 g/mol)
Answer: 4.68 L O2
20
4) Ideal Gas Law
•
•
•
•
•
By combining the gas laws we can write a general
equation.
R is called the gas constant.
The value of R depends on the units of P and V.
9We will use 0.08206
and convert P to atm and
V to L.
The other gas laws are found in the ideal gas law if two
variables are kept constant.
This allows us to find one of the variables if we know the
other three.
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Example 5.6 How many moles of gas are in a basketball
with total pressure 24.3 psi and volume of 3.24 L at 25 °C?
Given: V = 3.24 L, P = 24.3 psi, t = 25 °C
Find: n, mol
Conceptual P, V, T, R
n
Plan:
Relationships:
Solution:
22
Clicker # 5-3
How many moles of helium are found in a balloon
that contains 5.5 L of helium at a pressure of 1.15
atm and a temperature of 22.0 ºC?
a.
b.
c.
d.
e.
3.5
0.26
2.6
0.35
3.8
Density at Standard Conditions
• Density of a gas is generally given in g/L.
• mass of 1 mole = molar mass
• volume of 1 mole at STP = 22.4 L
Density at STP = Molar mass
VSTP
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Chapter 5
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Practice—Calculate the density of N2(g) at STP.
Given:
Find:
Conceptual
Plan:
N2
dN2, g/L
MM
d
Relationships:
Solution:
25
Application of the gas laws:
1)
Gas Density (when not at STP)
• Density is directly proportional to molar mass.
26
Example 5.7 Calculate the density of N2
at 125 °C and 755 mmHg.
Given:
Find:
Conceptual
Plan:
dN2, g/L
P, MM, T, R
d
Relationships:
Solution:
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Chapter 5
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Clicker 5-4
Which of the following gases has the greatest
density at 2.5 atm and 25°C?
1) N2O
2) F2
3) HNF2
4) C3H8
5) NF3
Application of the gas laws:
2)
Molar Mass of a Gas
• One of the methods chemists use to
determine the molar mass of an unknown
substance is to heat a weighed sample
(ms) until it becomes a gas, measure the
temperature, pressure, and volume, and
use the ideal gas law to calculate moles
29
Example 5.8 Calculate the molar mass of a gas with mass of
0.311 g that has a volume of 0.225 L at 55 °C and 886 mmHg.
Given:
Find: molar mass, g/mol
Conceptual
P, V, T, R
n
Plan:
n, m
MM
Relationships:
Step 2
Step 1
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Chapter 5
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Practice—What is the molar mass of a gas if 12.0
g occupies 197 L at 3.80 × 102 torr and 127 °C?
Identify the gas from the periodic table.
Answer: 4.00 g/mol
31
Our air
Mixtures of Gases
• When gases are mixed together, their
molecules behave independently of each other.
9 All the gases in the mixture have the same volume.
¾ All completely fill the container ∴ each gas’s volume = the
volume of the container.
9 All gases in the mixture are at the same temperature.
¾ Therefore, they have the same average kinetic energy.
• Therefore, in certain applications, the mixture
can be thought of as one gas.
9 Even though air is a mixture, we can measure the
pressure, volume, and temperature of air as if it were a
pure substance.
9 We can calculate the total moles of molecules in an air
sample, knowing P, V, and T, even though they are
different molecules.
32
Partial Pressure
• The pressure of a single gas in a mixture of gases
is called its partial pressure.
• We can calculate the partial pressure of a gas (Px)
if
9 we know what mole fraction of the mixture it composes
and the total pressure (Ptot)
9 or, we know the number of moles of the gas (nx) in a
container of known volume and temperature
• The sum of the partial pressures of all the gases in
the mixture equals the total pressure.
9 Dalton’s Law of Partial Pressures
9 because the gases behave independently
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Chapter 5
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The partial pressure of each gas in a
mixture can be calculated using the ideal
gas law.
Eq #1
Eq #2
34
Example 5.9
Determine the mass of Ar (mAr) in the mixture of Ar, He
and Ne given:
PHe = 341 mmHg, PNe = 112 mmHg, Ptot = 662 mmHg
and, at a V = 1.00 L, T = 298 K
Solution on next page
Conceptual
Plan
Ptot, PHe, PNe
PAr
PAr, V, T
nAr
mAr
PAr = Ptot – (PHe + PNe)
Relationships:
step 1
step3
step 2
Step 4
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Chapter 5
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Clicker 5-5
A gas phase mixture of H2 and N2 has a total pressure
of 784 torr with an H2 partial pressure of 124 torr. What
mass of N2 gas is present in 2.00 L of the mixture at
298 K?
a. 0.0133 g
b. 0.0710 g
c. 0.374 g
d. 0.994 g
e. 1.99 g
Mole Fraction
The fraction of the total pressure that a
single gas contributes is equal to the
fraction of the total number of moles that
a single gas contributes.
The ratio of the moles of a single
Eq #3
component to the total number of
moles in the mixture is called the mole
fraction, χ.
Eq #4
The partial pressure of a gas is equal to
the mole fraction of that gas times the
total pressure.
38
Example 5.10 Find the mole fractions and partial pressures in a
12.5-L tank with 24.2 g He and 4.32 g O2 at 298 K.
Conceptual mgas
Plan:
ngas
χgas, Ptotal
χgas
ntot, V, T, R
Ptot
Pgas
Step 1
Step 3
Step 2
Step 4
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Chapter 5
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Collecting Gases over water
• Gases are often collected by having them
displace water from a container.
• The problem is that since water evaporates,
there is also water vapor in the collected gas.
• The partial pressure of the water vapor,
called the vapor pressure, depends only on
the temperature.
9 so you can use a table to find out the partial
pressure of the water vapor in the gas you collect
• If you collect a gas sample with a total
pressure of 758.2 mmHg* at 25 °C, the partial
pressure of the water vapor will be 23.78
mmHg—so the partial pressure of the dry gas
will be 734.4 mmHg.
Eq #5
P =P -P
9 *Table 5.3 (next slide)
gas
tot
water vap.
40
Clicker # 5-6
The partial pressures of CH4, N2, and O2 in a sample
of gas were found to be 100., 450., and 200. mmHg,
respectively. Calculate the mole fraction of nitrogen.
1) 0.133
2) 0.267
3) 0.333
4) 0.600
5) 0.733
Vapor Pressure of Water
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Chapter 5
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Collecting Gas by Water Displacement
43
Example 5.11 1.02 L of O2 is collected over water at 293
K with a total pressure of 755.2 mmHg. Find mass O2.
Given: V = 1.02 L, P = 755.2 mmHg, T = 293 K
or T= 20oC
Find: mass O2, g
Conceptual Ptot, PH2O
Plan:
PO2
PO2,V,T
nO2
gO2
Step3
Step1 Solution:
.3)
Step 2
Step 4
O2
44
Practice—0.120 moles of H2 is collected over water
in a 10.0 L container at 323 K. Find the total
pressure (in mmHg). [Need to use Table 5.3]
Answer: 334 mm Hg
45
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Chapter 5
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Reactions Involving Gases
• The principles of reaction stoichiometry from
•
Chapter 4 can be combined with the gas laws for
reactions involving gases.
In reactions of gases, the amount of a gas is often
given as a volume.
9 instead of moles
9 As we’ve seen, you must state pressure and temperature.
• The ideal gas law allows us to convert from the
•
volume of the gas to moles; then we can use the
coefficients in the equation as a mole ratio.
When gases are at STP, use 1 mol = 22.4 L.
P, V, T of Gas A
mole A
mole B
P, V, T of Gas B
46
Example 5.12 What volume of H2 is needed to make 35.7 g
of CH3OH at 0.971 atm and 355 K?
CO(g) + 2 H2(g) → CH3OH(g)
Given: mCH3OH = 37.5 g, P = 0.971 atm, T = 355 K
Find: VH2, L
Conceptual g CH3OH
Plan:
mol CH3OH
mol H2
P, n, T, R
V
Step 1
Step2
H2
47
Example 5.13 How many grams of H2O form when 1.24 L H2
reacts completely with O2 at STP?
O2(g) + 2 H2(g) → 2 H2O(g)
Answer: 0.998 g H2O
16
Chapter 5
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Properties of Gases
• 1) Expand to completely fill their container
• 2) Take the shape of their container
• 3) Low density
9much less than solid or liquid state
• 4) Compressible
• 5) Mixtures of gases are always homogeneous
• 6) Fluid
49
Kinetic Molecular Theory
• 1) The particles of the gas
(either atoms or molecules) are
constantly moving.
• 2) The attraction between
particles is negligible.
• 3) When the moving gas
particles hit another gas particle
or the container, but they
bounce off [ elastic collision].
9like billiard balls
50
Kinetic Molecular Theory
• 4) There is a lot of empty space
between the gas particles.
9compared to the size of the particles
• 5) The average kinetic energy of
the gas particles is directly
proportional to the Kelvin
temperature.
9As you raise the temperature of the
gas, the average speed of the
particles increases.
¾But don’t be fooled into thinking all
the gas particles are moving at the
same speed!!
51
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Chapter 5
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Gas Properties Explained—
Pressure
• Because the gas particles are constantly
moving, they strike the sides of the
container with a force.
• The result of many particles in a gas sample
exerting forces on the surfaces around them
is a constant pressure.
52
Gas Properties Explained—
Compressibility
Because there is a lot of unoccupied space in the structure
of a gas, the gas molecules can be squeezed closer together.
53
Gas Laws Explained—
Boyle’s Law
• Boyle’s law states that the volume of a gas is
inversely proportional to the pressure.
• Decreasing the volume forces the molecules into
a smaller space.
• More molecules will collide with the container at
any one instant, increasing the pressure.
54
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