Chemistry: The Molecular Nature of Matter, 6E
Jespersen/Brady/Hyslop
Group
Problem
In groups of 2-3 brainstorm how
to describe a gas.
 What are some observable properties?
 What variables would you use to describe a gas?
2
Group
Problem
Describe a gas:
o Will expand to fill a volume
o Mostly empty space so can be compressed
o Can expand & contract with temperature
o Particles constantly in motion & constantly colliding
o Some gases are heavier then others and sink to the floor
rather then rise to the ceiling
3
Properties of Common Gases
 Despite wide differences in chemical properties,
all gases more or less obey the same set of
physical properties.
Four Physical Properties of Gases
 Inter-related
1. Pressure (P )
2. Volume (V )
3. Temperature (T )
4. Amount = moles (n)
4
Review: The Mole

Avagadro’s number (NA) allows us to
measure the number of particles of a gas as
the number of moles:
 NA = 6.02214129 × 1023 particles/mole
 We can measure the number of moles of a
gas by measuring its mass and knowing its
Molar Mass
 Molar Mass = mass / (# of moles)
 M = m/n
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
5
Group
Problem
How many moles of the CFC pollutant
CCl2F2 are in 50.0g?
6
Group
Problem
How many moles of the CFC pollutant CCl2F2
are in 50.0g?
50.0 g CCl2F2(1 mol/121 g CCl2F2) =
0.41 mol CCl2F2
7
Group
Problem
Calculate the mass of 3 moles of nerve agent VX:
CH3CH2
C11H26NO2PS
8
Group
Problem
Calculate the mass of 3 moles of nerve agent VX:
CH3CH2
3 mol C11H26NO2PS ( 267 g C11H26NO2PS /mol C11H26NO2PS)
= 801 g C11H26NO2PS
9
Review: Temperature
 Temperature is measured with a thermometer
usually in °C, °F, or Kelvin
 Fahrenheit

O is freezing point of salt water/96 is temperature of life
 Celsius
 O is freezing point of water/100 is boiling point of water
 Kelvin
 Uses absolute 0 where all motion stops
 O°C = 273 K
 °C = (°F -32) × (5/9)
10
Group
Problem
If room temperature is 25°C, what is
room temperature in Kelvin?
In °F?
11
Group
Problem
If room temperature is 25°C,
what is room temperature in Kelvin?
RT = 25°C + 273 K = 298 K
In °F?
25°C = (°F -32) × (5/9)
°F = [(25°C) 9/5] +32 = 77 °F
12
Pressure: Measurement and Units
force
Pressure 
area
 Pressure is force per unit area
 Earth exerts gravitational force on everything with
mass near it
 Weight
 Measure of gravitational force that earth exerts on
objects with mass
 What we call weight is gravitational force acting on
object (weight ≠ mass)
13
Force vs. Pressure
 Consider someone wearing flat shoes vs. high "spike"
heels
 Weight of person is the same F = 120 lbs
 Pressure on floor differs greatly (F/A)
Shoe
Flat
Spike
Area
Pressure
10 in.  3 in.
120 lbs
P =
= 4 psi
2
= 30 in.2
30 in.
0.4 in  0.4 in
120 lbs
= 750 psi
= 0.16 in.2 P =
2
0.16 in.
This is why snow shoes have a large footprint
14
Pressure
 Atmospheric Pressure
 Resulting force per unit area
 When earth's gravity acts on molecules in
air
 Pressure due to air molecules colliding
with object
 Barometer
 Instrument used to measure atmospheric
pressure
15
Vaccum
A vacuum exerts zero pressure on a containers walls
16
Toricelli Barometer
 Simplest barometer
 Tube that is 80 cm in
length
 Sealed at one end
 Filled with mercury
 In dish filled with
mercury
17
Toricelli Barometer
 Air pressure
 Pushes down on
mercury
 Forces mercury up tube
 Weight of mercury in tube
 Pushes down on
mercury in dish
 When two forces balance
 Mercury level stabilizes
 Read atmospheric
pressure
18
Toricelli Barometer
 If air pressure is high
 Pushes down on mercury
in dish
 Increase in level in tube
 If air pressure is low
 Pressure on mercury in
dish less than pressure
from column
 Decrease level in tube
Result:
 Height of mercury in tube is
the atmospheric pressure
19
Standard Atmospheric Pressure
 Typical range of pressure for most places
where people live
730 to 760 mm Hg
 Top of Mt. Everest
Air pressure = 250 mm Hg
Standard atmosphere (atm)
 Average pressure at sea level
 Pressure needed to support column of
mercury 760 mm high measured at 0 °C
20
Units of Pressure
 SI unit for pressure
 Pascal = Pa = 1 N/m2
 1 atm = 101,325 Pa = 101 kPa
 100 kPa = 0.9868 atm
 Other units of pressure
 1.013 Bar = 1013 mBar = 1 atm
 760 mm Hg = 1 atm
 760 torr = 1 atm
 At sea level 1 torr = 1 mm Hg
21
Group
Problem
Express Pressure in atm and kPa for a
gas at 705 mmHg
22
Group
Problem
Express Pressure in atm and kPa for a gas at 705
mmHg.
705 mmHg(1 atm/760 mmHg) = 0.927 atm
0.927 atm (101 kPa/1 atm) = 93.6 kPa
23
Manometers
 Used to measure pressure inside closed reaction
vessels
 Pressure changes caused by gases produced or
used up during chemical reaction
 Open-end manometer
 U tube partly filled with liquid (usually
mercury)
 One arm open to atmosphere
 One arm exposed to trapped gas in vessel
24
Open Ended Manometer
Pgas = Patm
Pgas > Patm
Gas pushes
mercury up
tube
Pgas < Patm
Atmosphere
pushes
mercury down
tube
25
Ex. Using Open Ended Manometers
A student collected a gas in an
apparatus connected to an openend manometer. The mercury in
the column open to the air was 120
mm higher and the atmospheric
pressure was measured to be 752
torr. What was the pressure of the
gas in the apparatus?
This is a case of Pgas > Patm
Pgas = 752 torr + 120 torr
= 872 torr
26
Ex. Using Open Ended Manometers
In another experiment, it was
found that the mercury level in
the arm of the manometer
attached to the container of gas
was 200 mm higher than in the
arm open to the air. What was
the pressure of the gas?
This is a case of Pgas < Patm
Pgas = 752 torr – 200 torr
= 552 torr
27
Group
Problem
CO2 collected in a monometer
in a lab with a barometric
reading of 97 kPa. What is the
Pressure of CO2?
33
mmHg
28
Group
Problem
CO2 collected in a monometer in a lab
with a barometric reading of 97 kPa.
What is the Pressure of CO2?
Pgas < Patm
33mm Hg (101 kPa/760 mmHg)
=4.4 kPa
Pgas = 97 kPa – 4.4 kPa
= 92.06 kPa
33
mmHg
29
Closed-end Manometer
 Arm farthest from vessel (gas) sealed
 Tube filled with mercury
 Then open system to flask and some mercury drains out
of sealed arm
 Vacuum exists above mercury in sealed arm
30
Closed-end Manometer
 Level of mercury in
arm falls, as not
enough pressure in
the flask to hold up
Hg
 Patm = 0
 Pgas = PHg
 So directly read
pressure
31
Your Turn
Gas pressure is measured using a close-ended
mercury manometer. The height of fluid in the
manometer is 23.7 in. Hg. What is this pressure in
atm?
A. 23.7 atm
B. 0.792 atm
C. 602 atm
D. 1.61 atm
2.54 cm
10 mm
1 atm
23.7 in. Hg ´
´
´
= 0.792 atm
in
cm
760 mm
32
Group
Problem
What is the pressure of an
unknown gas in atm within this
closed monometer?
Closed monometer
437 mm
205 mm
33
Group
Problem
What is the pressure of an unknown gas in
atm within this closed monometer?
PHg = 437mm-205mm = 232mm
Closed monometer
232mmHg (1 atm/160 mmHg)
437 mm
= 0.31 atm
205 mm
34
Comparison of Hg and H2O
 Pressure of 1 mm column of
mercury and 13.6 mm
column of water are the
same
 Mercury is 13.6 times more
dense than water
 Both columns have same
weight and diameter, so they
exert same pressure
d = 13.6 g/mL
d = 1.00 g/mL
35
Using Liquids Other Than Mercury in
Manometers and Barometers
 Simple relationship exists between two systems.
 For example, use water (d = 1.00 g/mL) instead of
mercury (d = 13.6 g/mL) in the tube
In general
hA  d A  hB  d B
For converting from
mm Hg to mm H2O
hH O =
2
hHg ´ d Hg
dH O
2
 Use this relationship to convert pressure change in mm
H2O to pressure change in mm Hg
36
Ex. Converting mm Acetone to mm
Hg - Solution
Acetone has a density of 0.791 g/mL. Acetone is
used in an open-ended manometer to measure a
gas pressure slightly greater than atmospheric
pressure, which is 756 mm Hg at the time of the
measurement. The liquid level is 20.4 mm
higher in the open arm than in the arm nearest
the gas sample. What is the gas pressure in torr?
37
Ex. Converting mm Acetone to mm
Hg - Solution
First convert mm acetone to mm Hg
hHg
20.4 mm acetone ´ 0.791 g/mL
=
= 1.19 mm Hg
13.6 g/mL
Then add PHg to Patm to get Ptotal
 Pgas = Patm + PHg

= 756.0 torr + 1.19 torr
 Pgas = 757.2 torr
38
Boyle’s Law
 Studied relationship
between P and V
 Work done at
constant T as well as
constant number of
moles (n)
 T1 = T2
 As V decreases, P
increases
39
Ideal
Gas Law
Charles Law
If Pressure is constant but freeze a balloon, it decreases in V
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
40
Charles’s Law
 Charles worked on
relationship of how V
changes with T
 Kept P and n constant
 Demonstrated V increases
as T increases
41
Gay-Lussac’s Law
 Worked on relationship between pressure and
temperature
 Volume (V ) and number of moles (n) are constant
 P increases as T increases
 This is why we don’t heat canned foods on a campfire
without opening them
 Showed that gas pressure
is directly proportional
to absolute temperature
P µT
Low T, Low P
P
High T, High P
T (K)
42
Group
Problem
Force of Collisions
P
Area
What happens to gas pressure when you raise the
temperature?
If the container can expand in
response to the force
In a rigid walled container
43
Group
Problem
Force of Collisions
P
Area
What happens to gas pressure when you raise the
temperature?
If the container can expand in
response to the force
No change in pressure is
observed because the
area increased.
In a rigid walled container
Pressure increases
because the faster
moving molecules hit
the walls of the
container with greater
force
44
Combined Gas Law
1
o Boyle’s law: P 
V
o Charles Law:
T V
o Guy-Lussac’s Law:
o
T
P
V
T P
is equivalent to
o For any two conditions:
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
T
P
V
PV
=C
T
P1V1 P2V2

T1
T2
45
Combined Gas Law
P1V1 P2V2
=
T1
T2
 All T 's must be in K
 Value of P and V can be any units as long as they are the
same on both sides
 Can derive Boyle’s Law, Charle’s Law, and Guy Lussac’s
Law from this equation
 Gives all relationships needed for fixed amount of gas
under two sets of conditions
46
How Other Laws Fit into Combined Gas Law
P1V1
T1
=
Boyle’s Law
T1 = T2
Charles’ Law
P1 = P2
Gay-Lussac’s V1 = V2
Law
P2V2
T2
P1V1 = P2V2
V1
T1
P1
T1
=
=
V2
T2
P2
T2
47
Combined Gas Law
P1V1
T1
=
P2V2
T2
Used for calculating effects of changing conditions
 T in Kelvin
 P and V any units, as long as units cancel
Example: If a sample of air occupies 500. mL at 273.15 K
and 1 atm, what is the volume at 85 °C and 560 torr?
760 torr ´ 500. mL
273.15 K
=
560 torr ´ V2
358 K
V2 = 890. mL
48
Ex. Using Combined Gas Law
 What will be the final pressure of a sample of
nitrogen gas with a volume of 950. m3 at 745 torr and
25.0 °C if it is heated to 60.0 °C and given a final
volume of 1150 m3?
 First, number of moles is constant even though
actual number is not given
 You are given V, P and T for initial state of system as
well as T and V for final state of system and must find
Pfinal
 This is a clear case for combined gas law
49
Ex. Using Combined Gas Law
 List what you know and what you don’t know
 Convert all temperatures to Kelvin
 Then solve for unknown—here P2
P1 = 745 torr
P2 = ?
V1 = 950 m3
V2 = 1150 m3
T1 = 25.0 °C + 273.15
T2 = 60.0 °C + 273.15
= 298.15 K
P2 =
P1V1T2
T1V2
=
= 333.15 K
745 torr ´ 950 m3 ´ 333.15 K
P2 = 688 torr
3
298.15 K ´ 1150 m
50
Ex. Combined Gas Law
 Anesthetic gas is normally given to a patient when
the room temperature is 20.0 °C and the patient's
body temperature is 37.0 °C. What would this
temperature change do to 1.60 L of gas if the pressure
and mass stay the same?
V1
T1
=
 What do we know?
 P and n are constant
 So combined gas law simplifies to
V2
T2
51
Ex. Combined Gas Law
V1 = 1.60 L
V2 = ?
T1 = 20.0 °C + 273.15
T2 = 37.0 °C + 273.15
= 293.15 K
= 310.15 K
 List what you know and what you don’t know
 Convert all temperatures to Kelvin
 Then solve for unknown—here V2
V2 =
V1T2
T1
1.60 L ´ 310.15 K
=
293.15 K
V2 = 1.69 L
52
Your Turn
Which units must be used in all gas law
calculations?
A. K for temperature
B. atm for pressure
C. L for volume
D. no specific units as long as they cancel
53
Relationships between Gas Volumes
 In reactions in which products and reactants are
gases:
 If T and P are constant
 Simple relationship among volumes
hydrogen + chlorine  hydrogen chloride
1 vol
1 vol
2 vol
hydrogen + oxygen  water (gas)
2 vol
1 vol
2 vol
 Ratios of simple, whole numbers
54
Avogadro’s Principle
 When measured at same T and P, equal V 's of gas contain
equal number of moles
 Volume of a gas is directly proportional to its number of
moles, n
 V is proportional to n (at constant P and T )
Coefficients
Volumes
Molecules
Moles
H2(g) + Cl2(g)  2 HCl(g)
1
1
2
1
1
2
1
1
2 (Avogadro's Principle)
1
1
2
55
Standard Molar Volume
 Volume of 1 mole gas must be identical for all gases
under same P and T
 Standard conditions of temperature and pressure — STP
 STP = 1 atm and 273.15 K (0.0 °C)
 Under these conditions
 1 mole gas occupies V = 22.4 L
 22.4 L  standard molar volume
56
Learning Check:
Calculate the volume of ammonia formed by the
reaction of 25 L of hydrogen with excess nitrogen.
N2(g) + 3H2(g)  2NH3(g)
25 L H2 2 L NH3
´
= 17 L NH3
1
3 L H2
57
Learning Check:
N2(g) + 3H2(g)  2NH3(g)
If 125 L H2 react with 50 L N2, what volume of NH3 can be
expected?
125 L H2 2 L NH3
´
= 83.3 L NH3
1
3 L H2
50 L N2 2 L NH3
´
= 100 L NH3
1
1 L N2
H2 is limiting reagent 83.3 L
58
Learning Check:
How many liters of N2(g) at 1.00 atm and 25.0 °C are produced
by the decomposition of 150. g of NaN3?
2NaN3(s)  2Na(s) + 3N2(g)
150. g NaN 3 1 mol NaN 3
3 mol N2
´
´
= 3.461 mol N2
1
65.0099 g 2 mol NaN 3
3.461 mol N2
22.4 L
´
= 77.53 L
1
1 mol at STP
V 1 V2
V 1T2
=
; V2 =
T 1 T2
T1
77.53 L ´ 298.15 K
V2 =
= 84.6 L
273.15 K
59
Your Turn
How many liters of SO3 will be produced when 25 L of
sulfur dioxide reacts with 75 L of oxygen ? All gases are at
STP.
A. 25 L
B. 50 L
C. 100 L
D. 150 L
E. 75 L
2SO2(g)
+ O2(g)
25 L SO2 ´
75 L O2 ´
2SO3(g)
2 L SO3
2 L SO2
2 L SO3
1 L O2
= 25 L SO3
= 150 L SO3
60
Ideal Gas Law
 With Combined Gas Law we saw that
PV
=C
T
 With Avogadro’s results we see that this is modified to
PV
= n ´R
T
 Where R = a new constant = Universal Gas constant
PV = nRT
61
Ideal Gas Law
PV = nRT
 Equation of state of a gas:
 If we know three of these variables, then we can calculate
the fourth
 Can define state of the gas by defining three of these
values
Ideal Gas
 Hypothetical gas that obeys ideal gas law relationship
over all ranges of T, V, n and P
 As T increases and P decreases, real gases act as ideal
gases
62
What is the value of R?
 Plug in values of T, V, n and P for 1 mole of gas at STP
(1 atm and 0.0 °C)
 T = 0.0 °C = 273.15 K
 P = 1 atm
 V = 22.4 L
 n = 1 mol
PV
1 atm ´ 22.4 L
R=
=
nT 1 mol ´ 273.15 K
R = 0.082057 L atm mol–1 K–1
63
Learning Check: PV = nRT
How many liters of N2(g) at 1.00 atm and 25.0 °C are produced
by the decomposition of 150. g of NaN3?
2NaN3(s)  2Na(s) + 3N2(g)
V=?
V = nRT/P
P = 1 atm
T = 25C + 273.15 = 298.15 K
150. g NaN 3 1 mol NaN 3
3 mol N2
n = mol N2 =
´
´
1
65.01 g
2 mol NaN 3
n = 3.461 mol N2
3.461 mol N ) ( 0.082057
(
V =
2
1.00 atm
L×atm
mol×K
) (298.15 K )
V = 84.6 L
Ex. Ideal Gas Law Problem
 What volume in milliliters does a sample of
nitrogen with a mass of 0.245 g occupy at 21 °C
and 750 torr?
 What do I know?
 Mass and identity (with the MM) of substance –
can find moles
 Temperature
 Pressure
 What do I need to find?
 Volume in mL
65
Ex. Ideal Gas Law Problem Solution
V = ? (mL)
mass = 0.245 g MM = 2  14.0 = 28.0 g/mol
 Convert temperature from °C to K
T = 21°C + 273.15 K = 294 K
 Convert pressure from torr to atm
æ 1 atm ö
÷÷ =0.987 atm
P = 750 torr çç
è 760 torr ø
 Convert mass to moles
m
0.245 g
–3
n=
=
=8.75
´
10
mol
MM 28.0 g mol–1
66
Ex. Ideal Gas Law Problem Solution
nRT
V =
P

8.75 10
V
3
 

moles  0.082057 L atm mol -1 K -1  294 K 
0.987 atm
1000 mL
V = 0.214 L ´
= 214 mL
1L
67
Your Turn
Dry ice is solid carbon dixoide. What is the pressure, in
atm, of CO2 in a 50.0 L container at 35 °C when 33.0 g
of dry ice becomes a gas?
A. 0.043 atm
B. 0.010 atm
C. 0.38 atm
D. 0.08 atm
æ 1 mol CO ö æ
ö
E. 38 atm
L
atm
2
(33.0 g CO ) ççè 44.01 g CO ÷÷øççè0.0821 K mol ÷÷ø (308 K)
2
Pressure of CO2 =
= 0.38 atm
2
50.0 L
68
Group
Problem
N2 + H2  NH3
How much H2 at 0°C and 0.86 atm do you need
to react completely with 750 mL of N2 at 1.5 atm
and 20°C to form ammonia?
Hint: is this equation balanced?
69
Group
Problem
N2 + 3 H2  2 NH3
How much H2 at 0°C and 0.86 atm do you need to react
completely with 750 mL of N2 at 1.5 atm and 20°C to
form ammonia?
nN2 = PV/RT = (1.5 atm)(0.750L)/[(0.082 L atm mol–1 K–1)(293K)]= 0.047 mol
0.047 mol Nz (3 mol Hz/1 molN2) = 0.14 mol H2
V = nRT/P = [(0.14 mol)(0.082 L atm mol–1 K–1)(273K)]/0.86 atm = 3.66L H2
Or another way:
750 mL N2 *(3 mL H2/1 mLN2) = 2250 mL H2 - at 1.5 atm and 20 °C
(1.5 atm)(2250 mL)/293 K = 0.86 atm(VH2)/273
VH2 = 3656 mL = 3.6 L
70
Group
Problem
A sample of helium gas occupies 500.0 mL
at 1.21 atm Calculate the volume of the gas
if the pressure is reduced to 491 torr
Group
Problem
A sample of helium gas occupies 500.0 mL at 1.21 atm
Calculate the volume of the gas if the pressure is
reduced to 491 torr
Use Boyle’s Law:
P1V1 = P2V2
1.21 atm(500.0 mL) = 491 torr(1atm/760 torr)(V2)
V2 = 936.5 mL
Determining Molecular Mass of Gas
If you know P, T, V and mass of gas
 Use ideal gas law to determine moles (n) of gas
 Then use mass and moles to get MM
If you know T, P, and density (d ) of a gas
 Use density to calculate volume and mass of gas
 Use ideal gas law to determine moles (n) of gas
 Then use mass and moles to get MM
73
Ex. Molecular Mass of a Gas
The label on a cylinder of an inert gas became illegible,
so a student allowed some of the gas to flow into a 300
mL gas bulb until the pressure was 685 torr. The sample
now weighed 1.45 g; its temperature was 27.0 °C. What
is the molecular mass of this gas? Which of the Group
7A gases (inert gases) was it?
What do I know?
 V, mass, T and P
74
Ex. Molar Mass of a Gas
1L
= 0.300 L
 V = 300 mL ´
1000 mL
 Mass = 1.45 g
 Convert T from °C to K
 T = 27.0 °C + 273.15 K = 300.2 K
 Convert P from torr to atm
1 atm
P = 685 torr ´
= 0.901 atm
760 torr
 Use V, P, and T to calculate n
(
)(
)
)(
0.901 atm 0.300 L
PV
n=
=
= 0.01098
RT
0.082057 atm L mol–1 K –1 300.2 K
mole 75
(
)
Ex. Molar Mass of a Gas – Solution
 Now use the mass of the sample and the moles of the
gas (n) to calculate the molar mass (MM)
Molar Mass =
mass
n
1.45 g
=
= 132 g/mol
0.01098 mol
 Gas = Xe (Atomic Mass = 131.29 g/mol)
76
Ex. Molecular Mass and Molecular
Formula of a Gas
A gaseous compound of phosphorus and fluorine with
an empirical formula of PF2 was found to have a density
of 5.60 g/L at 23.0 °C and 750 torr. Calculate its
molecular mass and its molecular formula.
Know
 Density
 Temperature
 Pressure
77
Ex. Molecular Mass and Molecular
Formula Solution
1 L weighs 5.60 g
 So assume you have 1 L of gas
 V = 1.000 L
 Mass = 5.60 g
 Convert T from °C to K
 T = 23.0 °C + 273.15 K = 296.2 K
 Convert P from torr to atm
 d = 5.60 g/L
1 atm
P = 750 torr ´
= 0.9868 atm
760 torr
78
Ex. Molecular Mass and Molecular
Formula Solution
(
)(
)
0.9868 atm 1.000 L
PV
n=
=
=
RT
0.082057 L atm mol–1 K –1 296.2 K
(
)(
)
0.04058 mole
 Use n and mass to calculate molar mass
Molar Mass =
mass
n
5.60 g
=
= 138 g/mol
0.04058 mol
79
Ex. Molecular Mass and Molecular
Formula Solution
 Now to find molecular formula given empirical
formula and MM
 First find mass of empirical formula unit
 1 P = 1  31 g/mol = 31 g/mol
 2 F = 2  19 g/mol = 38 g/mol
 Mass of PF2
= 69 g/mol
molecular mass
138 g/mol
=
=2
empirical mass
69 g/mol
The correct molecular formula is P2F4
80
Which Gas Law to Use?
 Which gas law to use in calculations?
 If you know ideal gas law, you can get all the rest
Amount of gas
given or asked
for in moles or g
Use Ideal Gas
Law
PV = nRT
Amount of gas
remains constant
or not mentioned
Gas
Law
Problems
Use Combined
Gas Law
P1V1 P2V2
=
n1T1 n2T2
81
Your Turn
A 7.52 g sample of a gas with an empirical formula of
NO2 occupies 2.0 L at a pressure of 1.0 atm and 25 °C.
Determine the molar mass and molecular formula of the
compound.
A. 45.0 g/mol, NO2
B. 90.0 g/mol, N2O4
C. 7.72 g/mol, NO
D. 0.0109 g/mol, N2O
E. Not enough data to determine molar mass
82
Your Turn - Solution
 Molar Mass = g/mol
 Know mass = 7.52 g
 Moles = n= PV/RT
 Therefore MM = 7.52g /[PV/RT] = 7.52g * (RT/PV)
7.52 g) ( 0.0821 L atm K mol ) (298 K )
(
Molar Mass =
(1.0 atm) (2.0 L )
–1
–1
92.0 g/mol
= 90.0
æ 90
92 g mol–1 ö
çç
÷ =2
–1 ÷
è 45.0
46 g mol ø
Molecular formula is N2O 4
83
Stoichiometry of Reactions Between Gases
 Can use stoichiometric coefficients in equations to
relate volumes of gases
 Provided T and P are constant
 Volume is proportional to moles (V  n)
84
Ex. Stoichiometry of Gases
Methane burns with the following equation:
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
1 vol
2 vol
1 vol
2 vol
 The combustion of 4.50 L of CH4 consumes how
many liters of O2? (Both volumes measured at STP.)
 P and T are all constant so just look at ratio of
stoichiometric coefficients
Volume of O2= 4.50 L ´
= 9.00 L O2
2 L O2
1 L CH4
85
Ex. Ideal Gas Law
In one lab, the gas collecting apparatus used a gas bulb with
a volume of 250 mL. How many grams of Na2CO3(s) would
be needed to prepare enough CO2(g) to fill this bulb when
the pressure is at 738 torr and the temperature is 23 °C?
The equation is:
Na2CO3(s) + 2 HCl(aq)  2 NaCl(aq) + CO2(g) + H2O
86
Ex. Ideal Gas Law – Solution
 What do I know?
 T, P, V and MM of Na2CO3
 What do I need to find?
 Mass of Na2CO3
 How do I find this?
 Use ideal gas law to calculate moles CO2
 Convert moles CO2 to moles Na2CO3
 Convert moles Na2CO3 to grams Na2CO3
87
Ex. Ideal Gas Law – Solution
1.
Use ideal gas law to calculate moles CO2
a. First convert mL to L
1L
V = 250 mL ´
= 0.250 L
1000 mL
a. Convert torr to atm
1 atm
P = 738 torr ´
= 0.971 atm
760 torr
b. Convert °C to K
T = 23.0 °C + 273.15 K = 296.2 K
88
Ex. Ideal Gas Law – Solution
1.
Use ideal gas law to calculate moles CO2
PV
0.971 atm ´ 0.250 L
n=
=
RT 0.082057 atm L mol–1 K –1 ´ 296.2 K
= 9.989 × 10–3 mole CO2
2. Convert moles CO2 to moles Na2CO3
1 mol Na 2 CO3
9.989 10 mol CO 2 
1 mol CO 2
–3
= 9.989 × 10–3 mol Na2CO3
89
Ex. Ideal Gas Law – Solution
3. Convert moles Na2CO3 to grams Na2CO3
9.989 ´ 10-3
æ 106 g Na CO ö
2
3÷
mol Na2CO3 çç
÷
1
mol
Na
CO
è
2
3 ø
= 1.06 g Na2CO3
90
Your Turn
2Na(s) + 2H2O(l ) → 2NaOH(aq) + H2(g)
How many grams of sodium are required to produce 20.0
L of hydrogen gas at 25.0 °C, and 750 torr?
A. 18.6 g
B. 57.0 g
C. 61.3 g
D. 9.62 g
E. 37.1 g
91
Your Turn - Solution
 Moles of H2 produced:
n=
æ 1 atm ö
750 torr çç
÷÷ 20.0 L
è 760 torr ø
(
(
)(
)
0.0821 L atm K –1 mol–1 298 K
)
= 0.807 mol H2
 Grams of sodium required:
æ 2 mol Na ö æ 23.0 g ö
÷ çç
÷÷ = 37.1 g
mass Na = 0.807 mol H2 çç
÷
è 1 mol H2 ø è 1 mol Na ø
92
Group
Problem
At what temperature will 1.50 moles
of CH4 occupy a 1 L container at 10
atm?
93
Group
Problem
At what temperature will 1.50 moles of CH4
occupy a 1 L container at 10 atm?
T= PV/nR =
(10 atm)(1L)
(1.50 mol)(0.082 L atm mol–1 K–1)
= 81.3 K
94
Group
Problem
o PV = nRT
od = m / V
oM = m / n
Write out the ideal gas law in
terms of density & then in
terms of molar mass
95
Ideal
Gas Law
Considering Density & Molar Mass
o PV = nRT
od=m/V
oM=m/n
P (m / d) = nRT  d = Pm/nRT
PV = (m / M) RT  M = mRT/PV
M / d = RT / P
Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E
96
Dalton's Law of Partial Pressure
 For mixture of non-reacting gases in container
 Total pressure exerted is sum of the individual
partial pressures that each gas would exert
alone
 Ptotal = Pa + Pb + Pc + ···
 Where Pa, Pb, and Pc are the partial pressures
 Partial pressure
 Pressure that particular gas would exert if it
were alone in container
97
Dalton’s Law of Partial Pressures
 Assuming each gas behaves ideally
 Partial pressure of each gas can be calculated
from ideal gas law
Pa =
naRT
Pb =
V
nbRT
Pc =
V
 So total pressure is
ncRT
V
Ptotal = Pa + Pb + Pc + × × ×
=
naRT
V
+
nbRT
V
+
ncRT
V
+×××
98
Dalton’s Law of Partial Pressures
 Rearranging
Ptotal
æ RT
= na + nb + nc + × × × çç
èV
Ptotal
æ RT
= ntotal çç
èV
 Or
(
)
ö
÷÷
ø
ö
÷÷
ø
 Where ntotal = na + nb + nc + ···
ntotal = sum of number moles of various gases in
mixture
99
Dalton’s Law of Partial Pressures
Means for mixture of ideal gases
 Total number of moles of particles is important
 Not composition or identity of involved particles
 Pressure exerted by ideal gas not affected by identity of gas
particles
 Reveals two important facts about ideal gases
1. Volume of individual gas particles must be important
2. Forces among particles must not be important

If they were important, P would be dependent on identity of gas
100
Ex. Partial Pressure
Mixtures of helium and oxygen are used in scuba diving
tanks to help prevent “the bends.” For a particular dive, 46
L He at 25 °C and 1.0 atm and 12 L O2 at 25 °C and 1.0
atm were pumped into a tank with a volume of 5.0 L.
Calculate the partial pressure of each gas and the total
pressure in the tank at 25 °C.
101
Ex. Partial Pressure – Solution
 Have two sets of conditions
 Before and after being put into the tank
He
O2
Pi = 1.0 atm Pf = PHe
Pi = 1.0 atm Pf = PO2
Vi = 46 L
Vi = 12 L
Vf = 5.0 L
Vf = 5.0 L
102
Ex. Partial Pressure – Solution
 First calculate pressure of each gas in 5 L tank (Pf )
using combined gas law
PiVi
1 atm ´ 46 L
PHe =
=
= 9.2 atm
Vf
5L
PiVi
1 atm ´ 12 L
PO =
=
= 2.4 atm
2
Vf
5L
 Then use these partial pressures to calculate total
pressure
Ptotal = PHe + PO = 9.2 atm + 2.4 atm = 11.6 atm
2
103
Your Turn
A mixture of 250 mL of methane, CH4, at 35 ˚C and 0.55
atm and 750 mL of propane, C3H8, at 35˚ C and 1.5 atm,
were introduced into a 10.0 L container. What is the final
pressure, in torr, of the mixture?
A. 95.6 torr
B. 6.20 × 104 torr
C. 3.4 × 103 torr
D. 760 torr
E. 59.8 torr
104
Your Turn - Solution
(0.55 atm) (0.250 L)
PCH =
4
PC H =
3 8
10.0 L
1.5 atm 0.750 L
(
)(
10.0 L
)
= 0.0138 atm
= 0.112 atm
æ 760 torr ö
÷÷ = 95.6 torr
PT = 0.0138 atm + 0.112 atm çç
è atm ø
(
)
105
Mole Fractions and Mole Percents
Mole Fraction (χ)
 Ratio of number moles of given component in mixture
to total number moles in mixture
cA =
nA
n A + nB + nC + × × × + nZ
=
nA
n total
Mole Percent (mol%)
Mole % = c A ´ 100%
106
Mole Fractions of Gases from Partial Pressures
æV
n A = PA çç
è RT
 If V and T are constant then,
ö
÷÷
ø
= constant
 For mixture of gases in one container
æV ö
PA çç
÷÷
è RT ø
XA =
æV ö
æV ö
æV
PA çç
÷÷ + PB çç
÷÷ + PC çç
è RT ø
è RT ø
è RT
V
RT
ö
æV
÷÷ + × × × + PZ çç
ø
è RT
ö
÷÷
ø
107
Mole Fractions of Gases from Partial Pressures
V cancels, leaving
RT
PA
cA =
PA + PB + PC + × × × + PZ
or
cA =
PA
Ptotal
=
nA
n total
108
Ex. Partial Pressures
 The partial pressure of oxygen was observed to be 156 torr
in air with a total atmospheric pressure of 743 torr.
Calculate the mole fraction of O2 present
 Use
cA =
c O2
PA
Ptotal
156 torr
=
= 0.210
743 torr
109
Partial Pressures and Mole Fractions
 Partial pressure of particular component of gaseous
mixture
 Equals mole fraction of that component times total
pressure
PA = c A ´ Ptotal
110
Ex. Partial Pressure
The mole fraction of nitrogen in the air is 0.7808.
Calculate the partial pressure of N2 in air when the
atmospheric pressure is 760. torr.
PN =cN ´ Ptotal
2
2
PN = 0.7808 ´ 760 torr = 593 torr
2
111
Your Turn
A mixture of 250 mL of methane, CH4, at 35˚ C
and 0.55 atm and 750 mL of propane, C3H8, at
35˚ C and 1.5 atm was introduced into a
10.0 L container. What is the mole fraction of
methane in the mixture?
A. 0.50
B. 0.11
C. 0.89
D. 0.25
E. 0.33
112
Your Turn - Solution
0.55 atm ´ 0.250 L
PCH4 =
= 0.0138 atm
10.0 L
1.5 atm ´ 0.750 L
PC 3H8 =
= 0.112 atm
10.0 L
0.0138 atm
c CH =
= 0.110
4
0.0138 atm + 0.112 atm
113
Group
Problem
A mixture of 4.00 g of hydrogen and
10.0 g of helium are in a 4.30-L flask
at 0°C. What is the total pressure of
the container and the partial
pressures of each gas?
114
Group
Problem
A mixture of 4.00 g of hydrogen and 10.0 g of
helium are in a 4.30-L flask at 0°C. What is
the total pressure of the container and the
partial pressures of each gas?
mol H2 = 4.00 g (1 mol H2/2.00 g H2)= 2.00 mol
mol He = 10.0 g He (1 mol He/4.00 g He) = 2.50 mol
Total mol = 2.00 + 2.50 = 4.50 mol
P total = nRT = (4.50 mol)(0.0821 L atm/(mol K))(273 K ) = 23.5 atm
V
4.30 L
XH2 = 2.00 mol = 0.444
4.5 mol
XHe = 2.5 mol = 0.556
4.5 mol
PH2 = XH2 Ptotal = (0.444)(23.5 atm) = 10.4 atm
PHe= XHe Ptotal = (0.556)(23.5 atm) = 13.1 atm
115
Collecting Gases over Water
 Application of Dalton’s Law of Partial Pressures
 Gases that don’t react with water can be trapped over
water
 Whenever gas is collected by displacement of water,
mixture of gases results
 Gas in bottle is mixture of water vapor and gas being collected
116
Collecting Gases over Water
 Water vapor is present because molecules of water
escape from surface of liquid and collect in space above
liquid
 Molecules of water return to liquid
 When rate of escape = rate of return
 Number of water molecules in vapor state remains constant
 Gas saturated with water vapor = “Wet” gas
117
Vapor Pressure
 Pressure exerted by vapor present in space
above any liquid
 Constant at constant T
 When wet gas collected over water, we
usually want to know how much “dry” gas
this corresponds to
 Ptotal = Pgas + Pwater
 Rearranging
 Pgas = Ptotal – Pwater
118
Ex. Collecting Gas over Water
A sample of oxygen is collected over water at 20.0 ˚
C and a pressure of 738 torr. Its volume is 310 mL.
The vapor pressure of water at 20 ˚ C is 17.54 torr.
a. What is the partial pressure of O2?
b. What would the volume be when dry at STP?
a. PO2 = Ptotal – Pwater
= 738 torr – 17.5 torr = 720 torr
119
Ex. Collecting Gas – (Soln.)
b. Use the combined gas law to calculate PO2 at STP
P1 = 720 torr
P2 = 760 torr
V1 = 310 mL
V2 = ?
T1 = 20.0 + 273.12 = 293 K
T2 = 0.0 + 273 K = 273 K
P1V1 P2V 2

T1
T2
P1V1T 2
V2 
T1P2
720 torr ) (310 mL ) (273 K )
(
V=
(293 K ) (760 torr)
2
V2 = 274 mL
120
Your Turn
An unknown gas was collected by water
displacement. The following data was recorded:
T = 27.0 °C;
P = 750 torr;
V = 37.5 mL;
Gas mass = 0.0873 g;
PH2O(vap) = 26.98 torr
Determine the molecular weight of the gas.
A. 5.42 g/mol
B. 30.2 g/mol
C. 60.3 g/mol
D. 58.1 g/mol
E. 5.81 g/mol
121
Your Turn - Solution
n= PV/RT
MM = g/n = g (RT/PV)
gRT
Molar Mass =
=
PV
(
)(
)(
0.0873 g 0.0821 L atm K –1 mol–1 300 K
(750 torr - 26.98 torr) (0.0375 L)
= 60.3 g/mol
122
)
Group
Problem
32.5 mL of Hydrogen gas is collected over water at
25 ºC and 755 torr. What is the pressure of dry
hydrogen gas? (VP25ºC = 23.76 mmHg)
123
Group
Problem
32.5 mL of Hydrogen gas is collected over water at
25 ºC and 755 torr. What is the pressure of dry
hydrogen gas? (VP25ºC = 23.76 mmHg)
Correct Pt to find the Pdry gas:
755 torr - 23.76 torr = 731.24 torr
731 torr = Phydrogen
124
Diffusion
 Complete spreading
out and
intermingling of
molecules of one gas
into and among
those of another gas
 e.g. Perfume in
room
125
Effusion
 Movement of gas
molecules
 Through extremely small
opening into vacuum
Vacuum
 No other gases present in
other half
126
Thomas Graham
 Studied relationship between effusion rates
and molecular masses for series of gases
 Wanted to minimize collisions
 Slow molecules down
 Make molecules bump aside or move to
rear
127
Graham's Law of Effusion
 Rates of effusion of gases are inversely proportional
to square roots of their densities, d, when compared
at identical pressures and temperatures
Effusion Rate 
1
d
Effusion Rate  d  k
(constant P and T)
(constant P and T)
k is virtually identical for all gases
Effusion Rate (A)  d A  Effusion Rate (B)  d B  k
128
Graham's Law of Effusion
 Rearranging
dB
Effusion Rate ( A )
dB


Effusion Rate (B )
dA
dA
 Finally, dA  MM (constant V and n)
Effusion Rate (A)
dB
MB


Effusion Rate (B )
dA
MA
 Result: Rate of effusion is inversely
proportional to molecular mass of gas
Effusion Rate  MM  k
(constant P and T )
129
Graham's Law of Effusion
Effusion Rate  MM  k
 Heavier gases effuse more slowly
 Lighter gases effuse more rapidly
Ex. Effusion Calculate the ratio of the
effusion rates of hydrogen gas (H2) and
uranium hexafluoride (UF6) - a gas used in
the enrichment process to produce fuel for
nuclear reactors.
130
Ex. Effusion
131
Your Turn
If it takes methane 3.0 minutes to diffuse 10.0 m, how
long will it take sulfur dioxide to travel the same
distance ?
A. 1.5 min
B. 12.0 min
C. 1.3 min
D. 0.75 min
E. 6.0 min
132
Your Turn - Solution
133
Ex. Effusion
 For the series of gases He, Ne, Ar, H2, and O2
what is the order of increasing rate of
effusion?
Substance He
Ne
Ar
H2
O2
MM
20
40
2
32
4
134
Kinetic Theory and Gas Laws
 So far, considered gases from experimental point of
view
 At P < 1 atm, most gases approach ideal
 Ideal gas law predicts behavior
 Does not explain it
 Recall scientific method
 Law is generalization of many observations
 Laws allow us to predict behavior
 Do not explain why
135
Kinetic Theory and the Gas Law
 To answer WHY it happens—must construct
theory or model
 Models consist of speculations about what
individual atoms or molecules might be doing to
cause observed behavior of macroscopic system
(large number of atoms/molecules)
 For model to be successful:
 Must explain observed behavior in question
 Predict correctly results of future experiments
136
Kinetic Theory and the Gas Law
 Theories can never be proved absolutely true
 Often valid within defined boundaries
 Approximation by its very nature
 Bound to fail at some point
 One example is kinetic theory of gases
 Attempts to explain properties of ideal gases.
 Describes behavior of individual gas particles
137
Postulates of Kinetic Theory of Gases
① Gas particles are tiny, their V is negligible.
② Particles travel in a straight line, in random
directions.
③ 0 intermolecular attraction.
④ Elastic collisions, no Energy is lost.
⑤ If KE α T, then assume average KE α T.
138
Kinetic Theory of Gases
 Kinetic theory of matter and heat transfer (Chapter 7)
 Heat  PV  KEave
 But for constant number of moles of ideal gas
 PV = nRT
 Where nR is proportionality constant
 This means T  KEave
 Specifically
3
KEave = RT
2
 As T increases, KEave increases
 Increase in number collisions with walls, thereby
increasing pressure
139
Real Gases
 Don’t conform to these assumptions
 Have finite volumes
 Do exert forces on each other
 However, kinetic theory of gases does explain
ideal gas behavior
 True test of model is how well its predictions
fit experimental observations
140
Postulates of Kinetic Theory of Gases
 Picture ideal gas consisting of particles having
no volume and no attractions for each other
 Assumes that gas produces pressure on its
container by collisions with walls
141
Kinetic Theory Explains Gas Laws
P and V (Boyle's Law)
 For given sample of ideal gas at given T
(n and T constant)
 If V decreases,
P increases
P = (nRT )
1
V
By kinetic theory of gases
 Decrease in V, means
gas particles hit wall more often
 Increase P
142
P and T (Gay-Lussac's Law)
 For given sample of ideal
gas at constant V (n and V
constant)
 P is directly proportional to
T
æ nR ö
P = çç ÷÷ T
èV ø
143
P and T (Gay-Lussac's Law)
Kinetic theory of gases accounts
for this
 As T increases
 KEave increase
 Speeds of molecules increases
 Gas particles hit wall more often
as V same
 So P increase
144
T and V (Charles' Law)
 For given sample of ideal
gas at constant P (n and P
constant)
 V is directly proportional
to T
æ nR ö
V = çç ÷÷ T
èP ø
145
T and V (Charles' Law)
Kinetic theory of gases
account for this
 As T increases
 KEave increases
 Speeds of molecules
increases
 Gas particles hit wall
more often as pressure
remains the same
 So volume increases
146
V and n (Avogadro's Principle)
 For ideal gas at constant T and P
 V is directly proportional to n
æ RT
V = çç
è P
ö
÷÷ n
ø
 Kinetic Theory of Gases account for this
 As the number of moles of gas particles increase
at same T
 Holding T and P constant
 Must V must increase
147
Dalton's Theory of Partial Pressures
Ptotal   Pindiv idual gases
 Expected from kinetic theory of gases
 All gas particles are independent of each other
 Volume of individual particles is unimportant
 Identities of gases do not matter
 Conversely, can think of Dalton's Law of Partial
Pressures as evidence for kinetic theory of gases
 Gas particles move in straight lines, neither attracting nor
repelling each other
 Particles act independently

Only way for Dalton's Law to be valid
148
Law of Effusion (Graham's Law)
Effusion Rate A
MB

Effusion Rate B
MA
 Key conditions:
 Comparing two gases at same P and T
 Conditions where gases don't hinder each other
 Hence, particles of two gases have same Keave
KE1 = KE2
 Let v = average of velocity squared of molecules of gases
2
 Then
1
2
m1v 12  12 m2v 22
149
Law of Effusion (Graham's Law)
 Rearranging
v 12
m1

v 22 m 2
 Taking square root of both sides
 Since
m1  M1
 Now
v1
m1
M1


Rate of Effusion
M2
v 2  m2
 So
Effusion Rate = k
Effusion Rate of Gas1
Effusion Rate of Gas2
v1
m1

m2
v2
v
v
=
M2
M1
150
Absolute Zero
T  KE av e
1
2
 m( v )
2
 If KEave = 0, then T must = 0.
 Only way for KEave = 0, is if v = 0 since
m  0.
 When gas molecules stop moving, then gas
as cold as it can get
 Absolute zero
151
Real Gases: Deviations from Ideal Gas Law
 Combined Gas Law
 Ideal Gas Law
PV
= constant
T
PV
=R
nT
 Real gases deviate Why?
152
Real Gases Deviate from Ideal Gas Law
Gas molecules have
finite volumes
1.





They take up space
Less space of kinetic
motions
Vmotions < Vcontainer
Particles hit walls of
container more often
Pressure is higher
compared to ideal
153
Real Gases
2. Particles do attract each other


Even weak attractions means they hit walls of container
less often
Therefore, pressure is less than ideal gas
154
Effect of Attractive Forces on Real Gas
155
van der Waal's equation for Real Gases
2 ö
æ
na
ççP +
÷÷ V - nb = nRT
2
V ø
è
(
corrected P
)
corrected V
 a and b are van der Waal's constants
 Obtained by measuring P, V, and T for
real gases over wide range of conditions
156
van der Waal's equation for Real Gases

n 2a 
 P  2   V  nb   nRT
V 

corrected P
 a — Pressure correction
 Indicates some attractions between molecules
 Large a

Means strong attractive forces between molecules
 Small a

Means weak attractive forces between molecules
157
van der Waal's equation for Real Gases
2 ö
æ
na
ççP +
÷÷ V - nb = nRT
2
V ø
è
(
)
corrected V
 b — Volume correction
 Deals with sizes of molecules
 Large b

Means large molecules
 Small b

Means small molecules
 Gases that are most easily liquefied have largest van der Waal's
constants
158