Notes One Unit Five

1
Notes One Unit Five
Characteristics of Gases Pressure of
fluids Standard Temperature and
Pressure Converting Pressures Gas Laws
Pages 422-440
2
Pressure Versus Molecular Collision

• Pressure is caused by molecular collision
• A molecule colliding creates a force.
• Catching a ball creates a force.
• PF/A
• pp 427

3
Pressure viewed as created in a fluid

• Created by the weight
• The deeper you go, the more weight .

pp 427
4
Air is a fluidjust like water
pp 427
5
Torricellian Barometer
pp 427
Air Pressure
780 torr
760 torr
740 torr
Mercury
6
Pop can Demo
7
Magdeburg plates Demo
8
Standard Pressure, Temperature and Volume

• 1 atm 14.7 psi
• 1 atm 29.92 in Hg
• 1 atm 760 mm Hg
• 1 atm 760 torr
• 1 atm 101,325 Pa
• 1 atm 101.325 kPa
• 1 atm 1.01325 bar
• 273K or 0oC KoC273
• 22.4 Liter/mole for any gas at STP
• pp 427

9
Converting Pressures

• Convert 25 lb/in2 to torr
• Convert 75 Kpa to in Hg

10
Starter

• The pressure of a gas is measured as 49 torr.
  Convert this pressure to atmospheres,
  kiloPascals, and mmHg.

11
Starter Pressure Conversions

• The pressure of a gas is measured as 49 torr.
  Represent this pressure in atmospheres, Pascals,
  and mmHg.

12
Physical Properties of Gases

• Gas Laws
• Relationships between volume, temperature,
  pressure, and amount of gas.

13
Boyles Law P and V

• as one increases, the
• other decreases
• inversely proportional
• pressure is caused by moving molecules hitting
  container walls
• If V is decreased and the of molecules stays
  constant, there will be more molecules hitting
  the walls per unit

14
Boyles Law P and V

• Boyles Law the V of fixed mass of gas varies
  inversely with P at a constant T.
• PV k
• k is a constant for a certain sample of gas that
  depends on the mass of gas and T
• What kind of graph is V vs. P?
• If we have a set of new conditions for the same
  sample of gas, they will have same k so

15
Boyles Law
16
Boyles Law P and V

• Discovered by Irish chemist, Robert Boyle
• Used a J-shaped tube to experiment with varying
  pressures in multistory home and effects on
  volume of enclosed gas

17
Example Boyles Law

• Consider a 1.53-L sample of gaseous SO2 at a
  pressure of 5.6 x 103 Pa. If the pressure is
  changed to 1.5 x 104 Pa at constant temperature,
  what will be the new volume of the gas?

18
Charles Law V and T

• if P is constant, gases expand when heated
• when T increases, gas molecules move faster and
  collide with the walls more often and with
  greater force
• to keep the P constant, the V must increase

19
Charles Law V and T

• Problem if we use Celsius, we could end up with
  negative values from calculations in gas laws for
  volumes
• we need a T system with no negative values
  Kelvin Temperature Scale
• starts at -273.15 C absolute zero 0 K
• lowest possible temperature

balloon going into liquid nitrogen
20
Charles Law V and T

• Charles Law the V of fixed mass of gas at
  constant P varies directly with Kelvin T.
• V kT
• k is a constant for a certain sample of gas that
  depends on the mass of gas and P
• What kind of graph is V vs. T?
• If we have a set of new conditions for the same
  sample of gas, they will have same k so

21
Charles Law

• discovered by French physicist, Jacques Charles
  in 1787
• first person to fill balloon with hydrogen gas
  and make solo balloon flight

22
Example Charles Law Temp.

• A sample of gas at 15C and 1 atm has a volume
  of 2.58 L. What volume will this gas occupy at
  38C and 1 atm?

23
Charles and Boyles Demo
Pressure is constant.
Temperature is constant.
Moles are constant.
Moles are constant.
P1
1.0atm
P2
2.0atm
1.0atm
P1
1.0atm
P2
V1
1.0L
V2
0.50L
V1
1.0L
2.0L
V2
T1
273K
T2
546K
T1

273K
T2
273K
V1
V2

P1 x
V1
P2 x
V2

T1
T2
1.0a x
1.0L
2.0a x
0.50L

1.0L
2.0L

pp 433-440
273K
546K
24
The CombinedGas Law
25
Combining the gas laws

• So far we have seen two gas laws

Jacques Charles
Robert Boyle
Joseph Louis Gay-Lussac
These are all subsets of a more encompassing law
the combined gas law
Read pages 437, 438. Do Q 26 33 (skip 31)
26
Combined Gas Law Equations
27
Combined Gas Law Equation

• Is..?

pp 433-440
28
Combined gas Law Problem One

• A gas occupies 2.0 m3 at 121.2 K, exerting a
  pressure of 100.0 kPa. What volume will the gas
  occupy at 410.0 K if the pressure is increased to
  220.0 kPa?

Assign variables and calculate V2.
(2.0m3 )
(100.0KPa)
(220.0KPa )
(V2 )

(121.2K )
(410.0K )
(2.0m3 )
(100.0KPa)
(410.0K )
V2

(121.2K )
(220.0KPa )
pp 433-440
V2
3.1m3
29
Combined gas Law Problem Two
10.0 gram

• A 10.0 gram sample of ethane(C2H6) gas is at STP.
  If the volume is changed to 26.0 liters, what is
  the new Kelvin temperature of the gas?

1) Calculate Formula mass.
Mass
E

24.0
C
2x
12.0
6.0
H
6x
1.0
30.0g/m
2) Calculate V1.
10.0g
V1

V1
7.46L
30.0g/mol
22.4L
3) Assign variables and calculate T2.
(101.325KPa)
(26.0L)
(101.325KPa)
(7.46L)

(T2)
(273K)
(273K)
(101.325KPa)
(26.0L)
T2
T2
951K

(101.325KPa)
(7.46L)
pp 433-440
30
Notes Two Unit Five

• Grahams Law Calculation
• Review Mass-Mass Calculation
• Mass-Volume Calculation _at_STP
• Volume-Mass Calculation _at_STP

Pages 441-450
31
Grahams Law Demo
17.0g/m
36.5g/m
pp 442
32
Grahams Law

• Describes how speed compares between gas
  molecules with different masses.
• Two different gases
• 1)Same Temperature
• 2)Different Masses
• Kinetic energy ? ½ M1V12 ½ M2V22

pp 442
33
Grahams Equation
½ M2V22 ½ M1V12
M2V22 M1V12
by M1

M1
M1
M2V22
V12
by V22
V12

M1
V22
V22
Square root of both sides
pp 442
34
Grahams Law Problem One

• At a high temperature molecules of chlorine gas
  travel 15.90cm. What is the mass of vaporized
  metal (gas) under the same conditions, if the
  metal travels 8.97cm?

E

Mass
Cl
2x
35.5
71.0g/m
(15.90cm)
14.9
2

8.97cm
M2
223g/m
pp 442
35
Grahams Law Problem Two

• At a certain temperature molecules of chlorine
  gas travel at 0.450 km/s. What is the speed of
  sulfur dioxide gas under the same conditions?
  

Mass
E

Cl
2x
35.5
71.0g/m
E

Mass
32.1
S
1x
32.1
32.0
O
2x
16.0
64.1g/m
V2

V2
0.474Km/s
pp 442
36
Review Mass-Mass Calculation
How many grams of oxygen will react with 5.00
grams of hydrogen to make water?
___g
42
5.0g
2.0g/m
2
1
2
H2(g) O2(g) ? H2O(l)
X32.0g/m
___m
1.3
___m
2.5

• 1) grams H2 to moles H2
• 5.00g
• 2) moles H2 to moles O2
• 2.5m H2x
• 3) moles O2 to grams O2
• 1.3mO2x

2.0gH2 /m
2.5m H2

Mass
E
H
2.0g/m
(1mO2)
2 x
1.0
1.3mO2
(2mH2)

Mass
E
O
2x
16.0
32.0g/m
32.0g/m
42gO2
37
Mass-Volume Calculation _at_STP
How many liters of oxygen will react with 5.00
grams of hydrogen to make water?
___L
29
5.0g
2.0g/m
pp 449
2
1
2
H2(g) O2(g) ? H2O(l)
X22.4g/m
___m
1.3
___m
2.5

• 1) grams H2 to moles H2
• 5.00g
• 2) moles H2 to moles O2
• 2.5m H2x
• 3) moles O2 to liters O2
• 1.3mO2x

2.0gH2 /m
2.5m H2

Mass
E
H
2.0g/m
(1mO2)
2 x
1.0
1.3mO2
(2mH2)

Mass
E
O
2x
16.0
32.0g/m
22.4L/m
29LO2
38
Volume-Mass Calculation _at_STP
How many grams of oxygen will react with 56.0
liters of hydrogen to make water?
____g
40.0
56.0L
22.4L/m
pp 449
2
1
2
H2(g) O2(g) ? H2O(l)
____m
1.25
X32.0g/m
____m
2.50

• 1) Liters H2 to moles H2
• 56.0L
• 2) moles H2 to moles O2
• 2.50m H2x
• 3) moles O2 to grams O2
• 1.25mO2x

22.4LH2 /m
2.50m H2

Mass
E
H
2.0g/m
(1mO2)
2 x
1.0
1.25mO2
(2mH2)

Mass
E
O
2x
16.0
32.0g/m
32.0g/m
40.0gO2
39
Notes Three Unit Five

• Kinetic theory of gases
• Molar volume _at_ Non-STP Conditions
• R is Universal Gas Constant

Pages 452-459
40
THE KINETIC THEORY OF GASES

• Large number of particles

6.022x1023atoms/mole
pp 426
41
THE KINETIC THEORY OF GASES

• Large number of particles
• Elastic collisions

For a collision? KEBeforeKEAfter
pp 426
42
THE KINETIC THEORY OF GASES

• Large number of particles
• Elastic collisions
• No external forces

pp 433-440
43
THE KINETIC THEORY OF GASES

• Large number of particles
• Elastic collisions
• No external forces
• Separated by large distances

1.6x1011 times diameter
pp 426
44
THE KINETIC THEORY OF GASES

• Large number of particles
• Elastic collisions
• No external forces
• Separated by large distances
• No forces between particles

pp 426
45
Finding volumes _at_ Non-STP Conditions
PVnRT
Ideal Gas Equation
What is
P?
V?
moles
n?
T?
R?
Universal Gas Constant
(22.4L)
(101.325kpa)
P
V
R

R

(1 m)
(273.15K)
n
T
-
L
Kpa

(8.314 )
R
-
m
K
pp 446
46
Finding Volume at Non-STP

• Oxygen is made reacting 15.00g water at 209.0Kpa
  and 20.0oC.

____L
9.33
15.00g
18.0g/m
2H2O(l)? 2H2(l) 1O2(l)
______m
0.833
_____m
X22.4L/m
0.417
a) What volume of oxygen would be made at STP?
1) grams H2O to moles H2O
0.833m
15.00g
18.0g/m
2) moles H2O to moles O2
( 1mO2 )
0.833m x
0.417mO2
(2m H2O)
3) moles O2 to liters O2
E

Mass
0.417mO2x
22.4L/m
9.33LO2
H
2 x
1.0
2.0
_at_STP
O
16.0
16.0
1 x
18.0g/m
47
Finding Volume at Non-STP
pp 450
b) What is the volume of gas at reaction
conditions?
PVnRT
(0.417m)
(8.314 LKpam-1K-1)
(293.2K)
V
n
T
R
(209.0KPa)
V

P
4.86L
V
c) How many moles O2 gas are produced?
n
0.417mO2
d) How many grams O2 gas are produced?
n x g/m g

Mass
E
0.417m x
32.0g/m
13.3gO2
O
2x
16.0
32.0g/m
48
end
49
Molar Volume Lab

• Molar Volume Lab Data
• Molar Volume Lab Calculations

50
Molar Volume Lab Data
3.39
3.40
3.41
1.37
1.38
1.39
35.62
35.63
35.64
19.7
19.9
19.9
743.0
743.1
743.2
17.5
17.4
17.3
51
Molar Volume Lab Calculations
1. Calculate the grams of magnesium reacted.
(1.38g Mg)

0.0469g
3.40cm X
(100.00cm)
2. Calculate the moles of magnesium reacted.
0.0469g
24.3g/mol

0.00193moles
3.Calculate the pressure of the dry hydrogen gas.
-
(725.7T)
743.1
17.4

P

PH2 Pair- P
H2
H2O
4. Calculate the volume (V2) at STP from data.
P2
V2
V1
(760T)
V2
P1
(725.7T)
(35.63mL)


T1
(19.9oC273.15)
T2
(273.15K)
V2
31.70mL
5. What is the molar volume of hydrogen at STP.
Volume of H2 at STP
31.70mL


16400mL/m
Moles of H2
0.00193m
52
Mass-Volume Calculation _at_STP
How many liters of CO2 will form when 6.00 g of
Al2(CO3)3 decomposes?
E

Mass
54.0
Al
2 x
27.0
____L
1.72
6.00g
234.0g/m
36.0
C
3 x
12.0
144.0
O
9 x
16.0
3
Al2(CO3)3(s)? Al2O3(s) CO2(g)
1
1
234.0g/m
X22.4g/m
______m
0.0769
______m
0.0256

• 1) grams Al2(CO3)3 to moles Al2(CO3)3
• 6.00g
• 2) moles Al2(CO3)3 to moles CO2
• 0.0256mx
• 3) moles CO2 to liters CO2
• 0.0769mCO2x

234.0g/m
0.0256m
(3 m CO2)
0.0769m
(1 Al2(CO3)3 )
22.4L/m
1.72LCO2
pp 449
53
Unit Five

• Combined Quiz Review Unit Eight

54
Combined gas Law Quiz Example One

• A gas occupies 4.0 m3 at 135.1 K, exerting a
  pressure of 101.3 kPa. What volume will the gas
  occupy at 390.0 K if the pressure is increased to
  150.0 kPa?

Assign variables and calculate V2.
(4.0m3 )
(101.3KPa)
(150.0KPa )
(V2 )

(135.1K )
(390.0K )
(4.0m3 )
(101.3KPa)
(390.0K )
V2

(135.1K )
(150.0KPa )
V2
7.8M3
55
Combined gas Law Quiz Example Two

• A 53.0 gram sample of ethyne(C2H2) gas is at
  STP. If the volume is changed to 41.0 liters,
  what is the new Kelvin temperature of the gas?

53.0 gram
1) Calculate Formula mass.
Mass
E

24.0
C
2x
12.0
2.0
H
2x
1.0
26.0g/m
2) Calculate V1.
53.0g
V1

V1
45.6L
26.0g/mol
22.4L
3) Assign variables and calculate T2.
(101.325KPa)
(41.0L)
(101.325KPa)
(45.6L)

(T2)
(273K)
(273K)
(101.325KPa)
(41.0L)
T2
T2
245K

(101.325KPa)
(45.6L)
56
Seven Rows
Eight Rows
57
Seven Rows
Eight Rows
58
Seven Rows
Eight Rows
59
Seven Rows
Eight Rows
60
Seven Rows
Eight Rows
61
Seven Rows
Eight Rows
62
Quiz Two Notes

• Mass-Volume Calculation _at_STP
• Grahams Law Problem

63
Mass-Volume Calculation _at_STP
How many liters of ozone will react with 4.00
grams of nitrogen to make dinitrogen pentoxide?
____L
5.33
4.00g
28.0g/m
pp 449
3
5
3
N2(g) O3(g) ? N2O5(l)
____m
0.238
X22.4g/m
_____m
0.143

• 1) grams N2 to moles N2
• 4.00g
• 2) moles N2 to moles O3
• 0.143m N2x
• 3) moles O3 to liters O3
• 0.238mO3x

28.0gN2 /m
0.143m N2

Mass
E
N
28.0g/m
(5mO3)
2 x
14.0
0.238mO3
(3mN2)

Mass
E
O
3x
16.0
48.0g/m
22.4L/m
5.33L
64
Volume-Mass Calculation _at_STP
How many grams of ozone will react with 56.0
liters of nitrogen to make dinitrogen pentoxide?
____g
200.
56.0L
22.4L/m
pp 449
3
5
3
N2(g) O3(g) ? N2O5(l)
____m
4.17
X48.0g/m
____m
2.50

• 1) Liters N2 to moles N2
• 56.0L
• 2) moles N2 to moles O3
• 2.50m N2x
• 3) moles O3 to grams O3
• 4.17mO3x

22.4LN2 /m
2.50m N2

Mass
E
N
28.0g/m
(5mO3)
2 x
14.0
4.17mO3
(3mN2)

Mass
E
O
3x
16.0
48.0g/m
48.0g/m
200.g
65
Grahams Law Problem Three
If a sulfur hexafluoride molecule has a speed of
128m/s, what would heliums speed be at the same
temperature?
E

Mass
32.1
S
1x
32.1
114.0
F
6x
19.0
M2
V1
146.1g/m

V2
M1

Mass
E
He
1x
4.0
4.0g/m
4.0g/m
128m/s

V2
146.1g/m
146.1g/m
x(128m/s)
V2

770m/s
4.0g/m
pp 442
66
Seven/Eight Rows
67
Seven/Eight Rows
68
Final Quiz Examples

• Fill in Word Bank
• Combined gas Law Problems11 and 12
• Volume at Non-STP Items 13.1 and 13.2
• Grahams Law Item 14
• Written Response Item 15

69
Word Bank Items 1 through 10
70
Combined Gas Law Problems 11 and 12

• A gas occupies 7.1 m3 at 162.3 K, exerting a
  pressure of 121.5 kPa. What volume will the gas
  occupy at 250.0 K if the pressure is increased to
  172.1 kPa?

Assign variables and calculate V2.
(7.1m3 )
(121.5KPa)
(172.1KPa )
(V2 )

(162.3K )
(250.0K )
(250.0K )
(7.1m3 )
(121.5KPa)
V2

(162.3K )
(172.1KPa )
V2
7.7M3
71
Volume at Non-STP Items 13.1 and 13.2

• Oxygen is made decomposing12.13g sodium peroxide
  at 129.5Kpa and 22.1oC.

____L
1.74
12.13g
78.0g/m
2Na2O2 (l)? 2Na2O(l) 1O2(g)
_____m
0.156
______m
X22.4L/m
0.0778
a) What volume of oxygen would be made at STP?
1) grams Na2O2 to moles Na2O2
0.156m
12.13g
78.0g/m
2) moles Na2O2 to moles O2
( 1mO2 )
0.156m x
0.0778mO2
(2m Na2O2)
3) moles O2 to liters O2
E

Mass
0.0778mO2x
22.4L/m
1.74LO2
Na
2 x
23.0
46.0
_at_STP
O
16.0
32.0
2 x
78.0g/m
72
Volume at Non-STP Items 13.1 and 13.2
Oxygen is made decomposing12.13g sodium peroxide
at 129.5Kpa and 22.1oC.
b) What is the volume of gas at reaction
conditions?
PVnRT
(0.0778m)
(8.314 LKpam-1K-1)
(295.3K)
V
n
T
R
(129.5KPa)
V

P
1.47L
V
73
Grahams Law Item 14
If oxygen molecules have a speed of 168m/s, what
would the speed of hydrogen molecules be at the
same temperature?
E

Mass
O
2x
16.0
32.0g/m
E

Mass
H
2.0g/m
2 x
1.0
V2

V2
670m/s
pp 442
74
Written Response Item 15
Why is the molar volume quantity, 22.4L/m,
different at non-standard conditions?
Molar volume is different under non-standard
conditions due to gas volume being dependent on
pressure and temperature.

According to Boyles law,
if the volume is decrease, diagrams 1, the
pressure will increase. This increase is due to
the number of collisions occurring at a smaller
volume. More collisions mean higher pressure.
According to Charles Law, if
the temperature is increased at constant
temperature, the volume of a gas will increase.
The increase occurs due the molecules striking
the container walls with more force and causing
the volume to expand. See diagrams 2.
Before After
Before After
diagrams 2
diagrams 1
75
end
76

• Calculate the mean free path of a nitrogen
  molecule at 1 Torr, 298 K. (Nitrogen, which makes
  up 80 of the atmosphere, has a molecular
  diameter of 3.7Å)

77
(No Transcript)
78
Finding Volume at Non-STP

• Oxygen is made decomposing 26.31g iron(III)
  sulfate at 79.0Kpa and 21.1oC. Fe2(SO4)3(s)?
  FeO(s) SO3(g) O2(g)
• a) What volume of oxygen would be made at STP?
• 1) grams Fe2(SO4)3 to moles Fe2(SO4)3
• 26.31g
• 2) moles Fe2(SO4)3 to moles O2
• 0.06579m Fe2(SO4)3 x
• 3) moles O2 to liters O2
• 0.03290mO2x
• b) What is the volume of gas at reaction
  conditions?
• c) How many moles of the gas is produced?
• d) How many grams O2 gas are produced?

2
3
1
0.5
399.9gFe2(SO4)3/m
0.06579m Fe2(SO4)3
( 1mO2 )
0.03290mO2
(2m Fe2(SO4)3
0.737LO2
22.4L/m
(0. 03290m)
(8.314 LKpam-1K-1)
(294.3K)
PVnRT
V
(79.0KPa)
V
1.02L
n
0.03290mO2
n x g/m g
0. 03290m x
32.0g/m
1.05gO2