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Chapter 14(page 440)
Why is it important to know the volume - mass relationship of gases, the Idea Gas law, and the stoichiometry of gases??
Kinetic molecular theory Brownian motion Pressure Pascal Barometer Diffusion Boltzmann’s constant
Gases consist of widely separated atoms or molecules in constant, random motion
No interaction between atoms or molecules, except during collisions
Mostly empty spaceStraight trajectory until a collision occurs
Properties of gases
Gases have a unique set of physical properties:
1. Gases are translucent or transparent.
2. Gases have very low densities when
compared to liquids or solids.
3. Gases are highly compressible
compared to liquids and solids.
4. Gases can expand or contract to fill
any container. Mostly empty space
Properties of gases
These can be explained by the kinetic molecular theory
kinetic molecular theory: the theory that explains the observed thermal and physical properties of matter in terms of the average behavior of a collection of atoms and molecules.
Gases consist of atoms or molecules with
a lot of space in between,
that are in constant, random motion
If the liquid and gas are both made from the same molecules (H2O),
you can explain the “disappearance” by assuming that
the molecules are much more spread out in the gas phase.
Properties of gases
Evidence for the atomic / molecular nature of matter:
Brownian motion can be seen by
magnifying diluted milk and observing
tiny fat globules getting knocked around by the
surrounding water molecules
Brownian motion
What Brownian motion tells us:1. Matter consists of discrete particles (molecules or atoms)
2. Molecules (or atoms) are in constant, vigorous motion as a result of temperature
Brownian motion
Brownian motion provides a peek into the microscopic world
of atoms to see details that are normally hidden by the law of averages,
and the enormous number of incredibly small atoms.
In the early 1800’s Joseph Gay-Lussac studied gas volume relationships involving a chemical reaction between hydrogen and oxygen and observed that 2 L of hydrogen would react with 1 L of oxygen to form 2 L of water vapor at constant temperature and pressure
Hydrogen gas + oxygen gas ---- water vapor 2 L 1 L ---- 2 L 2 volumes 1 volume ---- 2 volumes
Take a new sheet of paper and fold it into three sections
Write your name, the title of the chapter and the number
On the first section from the sheet of paper, please write six things that you learned from your notes so far that could appear on your test.
In Gay-Lussac experiment it was found that the reaction took place in a 2:1:2 relationship between hydrogen, oxygen and water vapor
If you had 600 L H2, 300 L O2 you would get 600 L H2O formed
With hydrogen and chlorine combining then:
Hydrogen gas + chlorine gas ---- hydrogen chloride gas 1 L 1 L ---- 2 L 1 volumes 1 volume ---- 2 volumes
The relationship that he found between gas volumes is now known to be a law
Gay-Lussac’s law of combining volumes of gases states – at constant temperature and pressure, the volumes of gaseous reactants and products can be expressed as ratios of small whole numbers
Remember from Dalton’s atomic theory that atoms are indivisible
Also remember that Dalton believed that one atom of one reactant always combined with one atom of the other reactant (which caused questions when forming water vapor)
Gay-Lussac disproved the second theory of Dalton, but a scientist by the name of Avogadro formulated an explanation of the problem
Avogadro reasoned that molecules contained more than one atom and formulated his theory which later became a law
Avogadro’s law: equal volumes of gases at the same temperature and pressure contain equal numbers of molecules
Remember that one mole is equal to the atomic mass of that atom
Well, one mole of a molecule or a compound is equal to the combined molecular masses of the atoms
Avogadro suggested that one mole of an element, molecule or an compound contained the same number of particles which is 6.022 x 1023 particles
So 1 mole of H2 (2.015g) contained the same number of particles as 1 mole O2 (32g) at standard pressure (1 atm) and temperature (273 K) STP
Standard volume at STP is 22.414 L
Write a detailed three dollar summary of what you learned (a paragraph, with a topic sentence and three supporting sentences)
Turn to page 468 and complete # 1 – 4 then turn them in
Honors chemistry homework Page 468 # 7 - 16
Molar Volume Ideal gas
Gases consist of widely separated atoms or molecules in constant, random motion
Mostly empty space
No interaction between gas atoms or molecules except in collisions
Straight trajectories until collision occurs
Gas pressure is increased by more frequent and/or harder collisions
1. the density
More molecules means more impacts and a higher pressure.
2. the volume of the container
With less space to move around, there are more collisions and a higher pressure.
You can affect the gas pressure by changing:
1. the density
More molecules means more impacts and a higher pressure.
2. the volume of the container
With less space to move around, there are more collisions and a higher pressure.
3. the temperature
With more kinetic energy, the molecules move faster. The collisions are harder and more frequent.
You can affect the gas pressure by changing:
Boyle’s law: V versus P
Robert Boyle’s experiment:
- Mercury (Hg) was poured down a tube shaped like the letter “J.”
- The tube was closed on the lower end.
- The gas inside the tube took up space (volume).
- The temperature and number of gas molecules inside the tube stayed constant.
- Boyle observed the change in pressure in mmHg as a function of volume. He then graphed the relationship between pressure and volume.
Boyle’s law: V versus P
Pressure versus volume
• Temperature• Number of moles constant
Boyle’s law: V versus P
Pressure versus volume
• Temperature• Number of moles constant
Boyle’s law: V versus P
If a weather balloon is released on the ground with a volume of 3.0 m3
and a pressure of 1.00 atm, how large will it get when it reaches an
altitude of 100,000 ft, where the pressure is 0.0100 atm?
Boyle’s law: V versus P
On the second section of that sheet of paper, please write six things that you learned from your notes so far that could appear on your test.
Boyle’s law: V versus P
If a weather balloon is released on the ground with a volume of 3.0 m3 and a pressure of 1.00 atm, how large will it get when it reaches an altitude of 100,000 ft, where the pressure is 0.0100 atm?
Asked: Volume of the balloon when it reaches 100,000 ft
Given:
Relationships:
Solve:
Answer: The balloon will have a volume of 300 m3.
31 1 2
32
3
2
1 1 2 2
3
1.00 3.0 0.0100
1.00 3.0
1.00 , 3.0 , 0.0
0.0010030
0
0
10
atm m atm P
atm m
P atm V m P atm
PV PV
Pat
mm
Charles’s law: V versus T
The volume increases as
the temperature increases
Charles’s law: V versus T
Volume versus temperature
• Pressure• Number of moles constant
Charles’s law: V versus T
Doubling the Kelvin temperature will double the volume of a gas
Kelvin temperatures simplify the V versus T relationship
Charles’s law: V versus T
If you inflate a balloon to a size of 8.0 L inside where the temperature is 23oC, what will be the new size of the balloon when you go outside where it is 3oC?
Charles’s law: V vs. T
If you inflate a balloon to a size of 8.0 L inside where the temperature is 23oC, what will be the new size of the balloon when you go outside where it is 3oC?
Charles’s law: V vs. T
Asked: Volume of the balloon when the temperature drops to 3oC
Given:
Relationships:
Solve:
1 1 2
1 2
1 2
1
2
22
8.0 8.0 276
296 276 29
8.0 , 23 , 3
273
:
23 273 296
3 27
7.5
2
6
3 76
o o
Kelvin Celsius
o
o
V L T C T C
V Vand T T
T T
Convert temperature to Kelv
LVL L K
so VK K
in
T C
C
K
K
T K
Combined gas law
Imagine you were to hitch a ride on a high-altitude research balloon that reaches and altitude of 100,000 ft. At sea level, where the pressure is 1.00 atm and the temperature is 20oC, you’ll need 18 m3 of helium to fill the balloon. What will be the new volume of the gas when you reach altitude, where the pressure is 0.0100 atm and the temperature is –50oC?
Imagine you were to hitch a ride on a high-altitude research balloon that reaches and altitude of 100,000 ft. At sea level, where the pressure is 1.00 atm and the temperature is 20oC, you’ll need 18 m3 of helium to fill the balloon. What will be the new volume of the gas when you reach altitude, where the pressure is 0.0100 atm and the temperature is –50oC?
Asked: Volume of the balloon when it reaches 100,000 ft
Given:
Relationships:
Solve:
31 1 1
2 2
1.00 , 18.0 , 20 293 ,
0.0100 , 50 223
o
o
P atm V m T C K
P atm T C K
32
3
2
1
1 2
3
1 2 2
0.01001.00 18.0
293 223
1.00 18.0 223
293 0.01001,370
atm Vatm m
K K
atm m KV
PV P
K atm
T
m
V
T
On the third section of that sheet of paper, please write six things that you learned from your notes so far that could appear on your test.
Avogadro’s law: V versus moles
Two equal volumes of hydrogen react with one volume of oxygen to produce two volumes of water vapor.
Gas volumes act like moles because the same size container has the same number of molecules (at the same temperature and pressure).
Avogadro’s law: V versus moles
Moles versus volume
• Temperature• Pressure constant
Avogadro’s law: V versus moles
Moles versus volume
• Temperature• Pressure constant
Twice the volume,
twice the number of molecules
assuming the temperature and
pressure are constant
Moles versus volume
• Temperature• Pressure constant
Avogadro’s law: V versus moles
Restrictions
The ideal gas law
Combining the previous gas laws, we obtain the ideal gas law
In reality, the ideal gas law is an approximation which is accurate for many gases over a wide range of conditions.
The ideal gas law is not accurate at very high density or at very low temperature.
The ideal gas law
The universal gas constant
R is the only constant
Calculating the universal gas constant using various units
Watch out for the units!
The ideal gas law
On the first section of back side on that sheet of paper, please write six things that you learned from your notes so far that could appear on your test.
The ideal gas law
Limitations of the ideal gas law
In an ideal gas, we assume that:
1. individual gas molecules take up
no space
2. gas molecules do not interact
with each other
For very small volumes or very low temperatures, gas atoms and molecules are very close together, and van der Waals attractions are no longer negligible
PV = nRT
What would be the pressure inside of a 50.0 L tank containing 1,252 g of helium at 20oC?
PV = nRT
What would be the pressure inside of a 50.0 L tank containing 1,252 g of helium at 20oC?
PV = nRT
Asked: Tank pressure
Given:
Relationships:
Solve:
50.0 , 20 , 1,252 , : 4.003oV L T C mass g He g mole
, 0.08206atm L
PV nRT Rmole K
:
20.0 273
50.0
293
11,25 312.82
4.003
o
Convert to units of R
V
T C
molen g
g
L
K
moles
What would be the pressure inside of a 50.0 L tank containing 1,252 g of helium at 20oC?
PV = nRT
Asked: Tank pressure
Given:
Relationships:
Solve:
50.0 , 20 , 1,252 , : 4.003oV L T C mass g He g mole
, 0.08206atm L
PV nRT Rmole K
50.0 , 293 , 312.8V L T K n moles
50.0 312.8 0.08206 293
312.8 0.0820
15
6 29
0
3
50.0
atm LP L mole
atm
s Kmole K
atm Lmoles K
mole KP
L
P
What would be the new volume of a bubble in a bread dough once it goes from a room temperature (20oC) volume of 0.050 cm3 to a 191oC oven?
What would be the new volume of a bubble in a bread dough once it goes from a room temperature (20oC) volume of 0.050 cm3 to a 191oC oven?
Asked: Find the volume of a bubble under changing temperature conditions
Given:
Relationships:
Solve: Pressure and moles stay constant, so they cancel out.
1 1 2 2
1 1 2 2
1 2
1 2
PV PV
nT n
V V
T T
T
3
33
2
2
0.050 4640.079
293
0.050
293 464
cm
Vcm
K K
KV cm
K
31 1 20.050 , 20 293 , 191 464o oV cm T C K T C K
1 1 2 2
1 1 2 2
PV PV
nT n T
What would be the new volume of a bubble in a bread dough once it goes from a room temperature (20oC) volume of 0.050 cm3 to a 191oC oven?
Asked: Find the volume of a bubble under changing temperature conditions
Given:
Relationships:
Solve: Pressure and moles stay constant, so they cancel out.
31 1 20.050 , 20 293 , 191 464o oV cm T C K T C K
1 1 2 2
1 1 2 2
PV PV
nT n T
1 1 2 2
1 1 2 2
1 2
1 2
PV PV
nT n T
V V
T T
3
33
2
2
0.050 464
293
0
0.079
.050
293 464
cm
Vcm
K K
cK
VK
m
PV k T and n constant
Vk P and n constant
T
Boyle’s law
Charles’s law
Avogadro’s law
1 1 2 2
1 2
PV P VPVk or
T T T
n constant
Vk T and P constant
n
Combined gas law
1 1 2 2
1 1 2 2
PV P VPV nRT or
nT n T
Ideal gas law
R = universal gas law
Write a three dollar summary of what you learned (a paragraph, with a topic sentence and three supporting sentences)
Turn to page 468 and complete # 5 – 6 then turn them in
Honors chemistry Homework: Page 468 # 17 - 26
Molar Volume Ideal gas
N/A
1 1 2 2
1 1 2 2
PV P VPV nRT or
nT n T
Ideal gas law
R = universal gas law
:
:
:
:
P pressure
V volume
n moles
T temperature
We can now solve
stoichiometry problems
involving gases
We can now solve
stoichiometry problems
involving gases
and
solids
solutions
other gases
Steps for solving stoichiometry problems
2C20H42 + 61O2 → 40CO2 + 42H2O
If you burn a 125 g candle made of paraffin wax, the peak temperature of the flame is about 1,400oC. Assuming the carbon dioxide produced is at that temperature, what volume of CO2 is produced at a pressure of 790 mmHg? The combustion reaction is:
Asked: Volume of CO2 produced
Given: Paraffin: mass of 125 g
CO2: P = 790 mmHg, T = 1,400oC
Relationships: Molar mass of paraffin = 282.6 g/mole
Mole ratio: 2 moles C20H42 ~ 40 moles CO2
PV = nRT
12.011 20 1.0079
282.6
2
/
4
g mole
If you burn a 125 g candle made of paraffin wax, the peak temperature of the flame is about 1,400oC. Assuming the carbon dioxide produced is at that temperature, what volume of CO2 is produced at a pressure of 790 mmHg? The combustion reaction is:
2C20H42 + 61O2 → 40CO2 + 42H2O
Asked: Volume of CO2 produced
Given: Paraffin: mass of 125 g; CO2: P = 790 mmHg, T = 1,400oC
Relationships: Molar mass of paraffin = 282.6 g/mole
Mole ratio: 2 moles C20H42 ~ 40 moles CO2
PV = nRT
Solve:
20 4220 4 20 422
20 42
1125
282.0.442
6mo
mole C Hg C H
g C Hles C H
125 g C20H420.442 moles
C20H42
Asked: Volume of CO2 produced
Given: Paraffin: mass of 125 g; CO2: P = 790 mmHg, T = 1,400oC
Relationships: Molar mass of paraffin = 282.6 g/mole
Mole ratio: 2 moles C20H42 ~ 40 moles CO2
PV = nRT
Solve:
220 42
202
42
400.442
28.85
moles COmoles C H
molesmol
C Hes CO
0.442 moles C20H42
8.85 moles CO2
Asked: Volume of CO2 produced
Given: Paraffin: mass of 125 g; CO2: P = 790 mmHg, T = 1,400oC
Relationships: Molar mass of paraffin = 282.6 g/mole
Mole ratio: 2 moles C20H42 ~ 40 moles CO2
PV = nRT
Solve:
8.85 0.08206 1,6731,170
1.04
atm
nRTPV nRT s
Lmoles K
mole KV L
a
oP
tm
V
8.85 moles CO2
1.04
8.85
1,673
P atm
n moles
T K
Asked: Volume of CO2 produced
Given: Paraffin: mass of 125 g; CO2: P = 790 mmHg, T = 1,400oC
Relationships: Molar mass of paraffin = 282.6 g/mole
Mole ratio: 2 moles C20H42 ~ 40 moles CO2
PV = nRT
Solve:
8.85 moles CO2
1.04
8.85
1,673
P atm
n moles
T K
8.85 0.08206 1,673
1.0,
41170
nRTPV nRT so
Patm L
moles Kmole K
V
V
atmL
1,170 L CO2
A common reaction occurs when an acid reacts with a metal to produce hydrogen gas. Consider a reaction in which 0.050 L of 1.25 M hydrochloric acid reacts with excess magnesium:
If the gas produced is captured in a 1.00 L container that starts out empty, what will be the pressure at the end of the reaction when the gas has cooled to 22oC?
Mg(s) + 2HCl(aq) → H2(g) + MgCl2(aq)
Asked: Pressure of H2 produced
Given: HCl: V = 0.050 L, Molarity = 1.25 moles/L
H2: V = 1.00 L, T = 22oC
Relationships: Mole ratio: 2 moles HCl ~ 1 mole H2
PV = nRT
Mg(s) + 2HCl(aq) → H2(g) + MgCl2(aq)
A common reaction occurs when an acid reacts with a metal to produce hydrogen gas. Consider a reaction in which 0.050 L of 1.25 M hydrochloric acid reacts with excess magnesium:
If the gas produced is captured in a 1.00 L container that starts out empty, what will be the pressure at the end of the reaction when the gas has cooled to 22oC?
Asked: Pressure of H2 produced
Given: HCl: V = 0.050 L, Molarity = 1.25 moles/L
H2: V = 1.00 L, T = 22oC
Relationships: Mole ratio: 2 moles HCl ~ 1 mole H2
PV = nRT
Solve:Volume
0.050 L HCl1.25 M HCl
1.250.05 0.0620
15
moles HClL HCl
L Hmole
lCl
Cs H
0.0625 moles HCl
Asked: Pressure of H2 produced
Given: HCl: V = 0.050 L, Molarity = 1.25 moles/L
H2: V = 1.00 L, T = 22oC
Relationships: Mole ratio: 2 moles HCl ~ 1 mole H2
PV = nRT
Solve:
0.0625 moles HCl
221
0.06252
0.0313mole H
moles HClmoles
molH l
HC
es
0.0313 moles H2
Asked: Pressure of H2 produced
Given: HCl: V = 0.050 L, Molarity = 1.25 moles/L
H2: V = 1.00 L, T = 22oC
Relationships: Mole ratio: 2 moles HCl ~ 1 mole H2
PV = nRT
Solve:
0.0313 moles H2
1.00
0.0313
22 273
295
o
V L
n mole
T
s
T C
K
0.756 atm H2
0.0313 0.08206 295
1.000.756
nRTPV nRT so
Va
P
a
tm Lmoles K
mole KP
Ltm
What volume of butane gas is needed at room temperature (23oC) and typical pressure (0.954 atm) to produce 85.0 L of carbon dioxide at 825oC and a pressure of 1.04 atm? The reaction is:
2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(g)
2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(g)
What volume of butane gas is needed at room temperature (23oC) and typical pressure (0.954 atm) to produce 85.0 L of carbon dioxide at 825oC and a pressure of 1.04 atm? The reaction is:
Asked: Volume of C4H10 needed to produce 85.0 L of CO2
Given: C4H10: P = 0.984 atm, T = 23oC
CO2: P = 1.04 atm, T = 825oC, V = 85.0 L
Relationships: Mole ratio: 2 moles C4H10 ~ 8 moles CO21 1 2 2
1 1 2 2
PV PV
nT n T
Asked: Volume of C4H10 needed to produce 85.0 L of CO2
Given: C4H10: P = 0.984 atm, T = 23oC
CO2: P = 1.04 atm, T = 825oC, V = 85.0 L
Relationships: Mole ratio: 2 moles C4H10 ~ 8 moles CO2
Solve:
1 1 2 2
1 1 2 2
PV PV
nT n T
C4H10 CO2
4 10
4 10
4 10 4 10
4 1
4 10 2
0
4 10
0.984 1
0.984 1.04 85.0
296 1,0
.04 85.0
296 4 1,098
0.00332 0.0201
6. 5
8
0
9
C H
C H C H
C H
C
C O
H
H
C
H C
at
atm V atm L
n K n K
atm a
m atm L
tm LV
V
K K
V L
n K n K
4 10
2
2 4 10
296
1,098
4
C H
CO
CO C H
T K
T K
n n
Asked: Volume of C4H10 needed to produce 85.0 L of CO2
Given: C4H10: P = 0.984 atm, T = 23oC
CO2: P = 1.04 atm, T = 825oC, V = 85.0 L
Relationships: Mole ratio: 2 moles C4H10 ~ 8 moles CO2
Solve:
1 1 2 2
1 1 2 2
PV PV
nT n T
C4H10 CO2
2 4 10
4 10
2
296
1,0
4
98
CO
C H
C
C
O
H
T K
T
n n
K
4 10
4 10
4 10 4 10
4 1
4 10 2
0
4 10
0.984 1
0.984 1.04 85.0
296 1,0
.04 85.0
296 4 1,098
0.00332 0.0201
6. 5
8
0
9
C H
C H C H
C H
C
C O
H
H
C
H C
at
atm V atm L
n K n K
atm a
m atm L
tm LV
V
K K
V L
n K n K
Use the mole ratio to substitute for nCO2
Asked: Volume of C4H10 needed to produce 85.0 L of CO2
Given: C4H10: P = 0.984 atm, T = 23oC
CO2: P = 1.04 atm, T = 825oC, V = 85.0 L
Relationships: Mole ratio: 2 moles C4H10 ~ 8 moles CO2
Solve:
4 10
4
4 10
4
4 10
4 10
1
10 4
0
1
2
0
0.984 1
0.00332
0.984 1.04 85.0
2
0.0201
6.05
.04 85.0
296 4 1,09
96 1,098
8C H C
C H
C
C H
H
C
C H
C H O
H
atm atm L
n K n
atm V atm L
n K n K
atm atm LV
K K
V
L
K
V
2 4 10
4 10
2
296
1,0
4
98
CO
C H
C
C
O
H
T K
T
n n
K
1 1 2 2
1 1 2 2
PV PV
nT n T
Asked: Volume of C4H10 needed to produce 85.0 L of CO2
Given: C4H10: P = 0.984 atm, T = 23oC
CO2: P = 1.04 atm, T = 825oC, V = 85.0 L
Relationships: Mole ratio: 2 moles C4H10 ~ 8 moles CO2
Solve:
4 10
4 10 2
4 10
4 10
4 10
4 10
4 10
0.984 1.04 85.0
296 1,098
0.984 1.04 85.0
296 4 1,098
0
6.0
.00332 0
5
.0201
C H
C H CO
C H
C H
C H C H
C H
atm V atm L
n K n K
atm V atm L
n K n K
at
V
m atm LV
K K
L
4 10
2
2 4 10
296
1,098
4
C H
CO
CO C H
T K
T K
n n
1 1 2 2
1 1 2 2
PV PV
nT n T
M = m R T
PVm = massM = molar mass
D = M P RT
D = density
Write a three dollar summary of what you learned (a paragraph, with a topic sentence and three supporting sentences)
Turn to page 469 and complete # 32 then turn it in.
Honors chemistry Homework: Page 471 # 73 - 81
Homework requirement: Learn all terms and concepts covered on this topic.
Make sure you have all assignments between page 224 and 227 completed and turned in by your test date.