1
GASES and their PROPERTIES
Exercises, Examples, and
BOLD numbered problems
2
Chapter 12 Outline
• Pressure formulas
–Combined
–PV=nRT
• Know Key Equations 12.1-8
3
BEHAVIOR OF GASES
4
Importance of Gases
• Airbags fill with N2 gas in an accident.
• Gas is generated by the decomposition
of sodium azide, NaN3.
• 2 NaN3 ---> 2 Na + 3 N2
5
THREE STATES OF
MATTER
General Properties of
Gases
• There is a lot of “free”
space in a gas.
• Gases can be expanded
infinitely.
• Gases occupy containers
uniformly and completely.
• Gases diffuse and mix
rapidly.
6
7
Properties of Gases
Gas properties can be
modeled using mathematics.
Model depends on—
• V = volume of the gas (L)
• T = temperature (K)
• n = amount (moles)
• P = pressure (atmospheres)
Pressure of air is
measured with a
BAROMETER
(developed by Torricelli
in 1643)
Pressure
Hg rises in tube until force
of Hg (down) balances
the force of atmosphere
(pushing up).
P of Hg pushing down
related to
• Hg density
• column height
8
Pressure
Column height measures
P of atmosphere, where -
1 standard atm
= 760 mm Hg
= 29.9 inches Hg
= about 34 feet of water
SI unit is PASCAL, Pa, where
1 atm = 101.325 kPa
1 atm = 1.01325 barr
9
10
Gas Pressure
• Pressure is defined as the force per unit area, and is
usually measured in Pascals, which are N/m2.
• We measure pressure in mmHg or torr.
• These units of pressure are equivalent come from
measurements using a Torricellian barometer.
F
g
d  cm
d  h  cm
P=
=
=
=
= dh
2
2
2
A
cm
cm
cm
3
2
P = pressure, F = Force ~ mass (g),
A (area) = cm2, d = density (g/cm3),
h = height
P is proportional to the height of the liquid mercury.
11
IDEAL GAS LAW
PV=nRT
Brings together gas
properties.
Can be derived from
experiment and
theory.
12
Boyle’s Law
If n and T are
constant, then
PV = (nRT) = k
This means, for
example, that P
goes up as V goes
down.
Robert Boyle
(1627-1691).
Son of Early of
Cork, Ireland.
13
Boyle’s Law
• Boyle's law states that the pressure is
inversely proportional to volume, ---• at constant n and T.
P a 1/V
As P increases, V decreases
PV = Cb
P1V1 = P2V2
14
Boyle’s Law
A bicycle pump is
a good example
of Boyle’s law.
As the volume of
the air trapped in
the pump is
reduced, its
pressure goes
up, and air is
forced into the
tire.
See Figures
12.3 and 12.4
page 543.
Charles’s
Law
15
If n and P are
constant, then
V = (nR/P)T = kT
V and T are directly
proportional.
Jacques Charles
(1746-1823). Isolated
boron and studied
gases. Balloonist.
Charles’s Law
16
• As temperature increases, volume increases,
at constant n and P.
• If the Kelvin scale is used, V a T.
V = CcT or
V1 V2

T1 T2
17
Charles’s Law
18
Gay-Lussac and Avogadro
Equal volumes of gases contain equal
numbers of molecules at constant T and P.
V a n
or V = Ca n
V1 V2

n1 n 2
19
Gay-Lussac’s Law
Avogadro’s Hypothesis
Equal volumes of gases at the
same T and P have the same
number of molecules.
V = n (RT/P) = kn
V and n are directly proportional.
twice as many
molecules
20
21
Avogadro’s Hypothesis
The gases in this experiment are all
measured at the same T and P.
22
THE IDEAL GAS LAW
• Combining the previous gas laws,
produces the Ideal Gas Law.
• PV = nRT
R = 0.082057 L atm/K mol
• Standard temperature and pressure, STP,
are defined as:
0oC or 273.15 K
and exactly one atmosphere pressure.
23
THE IDEAL GAS LAW
• One mole of gas occupies 22.414 L at STP,
and is called the standard molar volume or
the molar volume at standard conditions.
• The gas law can also be written in another
form:
P1 V1
----------n1 T1
P2 V2
= -----------n2 T2
24
Sample Problems
• Calculate the new temperature if a
sample of gas at 25.0oC has a volume
of 475 liters at a pressure of 3.50 atm
and has a new volume of 825 liters at
a pressure of 2.85 atm.
• Calculate the volume of the original
sample at STP.
Using PV = nRT
How much N2 is req’d to fill a small room
with a volume of 960 cubic feet (27,000 L)
to P = 745 mm Hg at 25 oC?
R = 0.0821 L•atm/K•mol
Solution
1. Convert all data into proper units
V = 27,000 L
T = 25 oC + 273 = 298 K
P = 745 mm Hg (1 atm/760 mm Hg)
= 0.980 atm
25
Using PV = nRT
How much N2 is req’d to fill a small room
with a volume of 960 cubic feet (27,000 L)
to P = 745 mm Hg at 25 oC?
R = 0.0821 L•atm/K•mol
Solution
2. Now calc. n = PV / RT
4
(0.980 atm)(2.7 x 10 L)
n =
(0.0821 L • atm/K • mol)(298 K)
n = 1.1 x 103 mol (or about 30 kg of gas)
26
27
Gases and Stoichiometry
Figure 12.9
28
Gases and Stoichiometry
2 H2O2(liq) ---> 2 H2O(g) + O2(g)
Decompose 1.1 g of H2O2 in a flask with a
volume of 2.50 L. What is the pressure of
O2 at 25 oC? Of H2O?
Solution
Strategy:
• Calculate moles of H2O2 and then moles
of O2 and H2O.
• Finally, calculate P from n, R, T, and V.
Gases and Stoichiometry
29
2 H2O2(liq) ---> 2 H2O(g) + O2(g)
Decompose 1.1 g of H2O2 in a flask with a
volume of 2.50 L. What is the pressure of O2
at 25 oC? Of H2O?
Solution
1 mol
1.1 g H2 O2 •
 0.032 mol
34.0 g
1 mol O2
0.032 mol H2O2 •
= 0.016 mol O2
2 mol H2O 2
30
Gases and Stoichiometry
2 H2O2(liq) ---> 2 H2O(g) + O2(g)
Decompose 1.1 g of H2O2 in a flask with a
volume of 2.50 L. What is the pressure of O2
at 25 oC? Of H2O?
Solution
P of O2 = nRT/V
(0.016 mol)(0.0821 L• atm/K •mol)(298 K)
=
2.50 L
P of O2 = 0.16 atm
31
Gases and Stoichiometry
2 H2O2(liq) ---> 2 H2O(g) + O2(g)
Solution
What is P of H2O? Could calculate as
above. But recall Avogadro’s hypothesis.
V
P


n at same T and P
n at same T and V
There are 2 times as many moles of H2O a
moles of O2. P is proportional to n.
Therefore, P of H2O is twice that of O2.
P of H2O = 0.32 atm
Dalton’s Law of Partial Pressures
1. Each component in a gas mixture exerts
a pressure independent of the other
components in the mixture.
2. This pressure is called the partial
pressure.
3. Dalton's law of partial pressures
explains the relationship between these
pressures and the other variables.
4. The mole fraction, XA, is defined as:
moles of A
A 
moles total
32
1. Each component in a gas mixture exerts a
pressure independent of the other
components in the mixture.
2. This pressure is called the partial
pressure.
33
Dalton’s Law of Partial Pressures
34
3. Dalton's law of partial pressures explains the
relationship between these pressures and the
other variables.
Ideal Gas Law
PV = nRT
Ideal Gas Law Relationships
PV = nRT
PV
R
nT
PA VA
R
n A TA
PB VB
R
n B TB
35
Ideal Gas Law Relationships
Rearrangement of Terms
PA VA
PB VB PC VC

R=
n A TA
n BTB n C TC
Combined Gas Law
36
Dalton’s Law of Partial Pressures
37
3. Dalton's law of partial pressures explains the
relationship between these pressures and the
other variables.
Ideal Gas Law
PV = nRT
PA VA
R
n A TA
Dalton’s Law of Partial Pressures
38
3. Dalton's law of partial pressures explains the
relationship between these pressures and the
other variables.
VA
PA
n A TA
VT
VA
VB
PT
 PA
 PB
...
n T TT
n A TA
n BTB
39
These terms can be regrouped
VT
VA
VB
PT
 PA
 PB
...
n T TT
n A TA
n BTB
PT = PA + PB + PC etc.
Same holds for moles since PT = nT
nT = nA + nB + nC etc.
Dalton’s Law of Partial Pressures
4. The mole fraction, XA, is defined as:
moles of A/moles total
PT VT PA VA PB VB PC VC



...
n T TT n A TA n BTB n C TC
For conditions of constant V & T
PT PA PB PC

  ...
nT nA nB nC
40
Dalton’s Law of Partial Pressures
4. The mole fraction, XA, is defined as:
moles of A/moles total
PT PA

 ...
nT nA
nA
PT
 PA
nT
nA
PT
 PT  A =PA
nT
41
Dalton’s Law of Partial Pressures
3. Relationship between these pressures and
the other variables.
PT = PA + PB + PC etc. And ---
nT = nA + nB + nC etc.
4. Mole fraction = XA
PA = XAPTotal
42
Dalton’s Law of Partial Pressures
2 H2O2(liq) ---> 2 H2O(g) + O2(g)
0.32 atm
0.16 atm
What is the total pressure in the flask?
Ptotal in gas mixture = PA + PB + ...
Therefore:
Ptotal = P(H2O) + P(O2) = 0.48 atm
Dalton’s Law: total P is sum of
PARTIAL pressures.
Figures 12.12 and 12.14
43
44
Sample Problems
1. A mixture of 2.50 moles neon, 1.45 moles
helium, and 2.80 moles argon has a
pressure of 1.45 atm. What are the partial
pressure of all the gases in this system?
PA = XAPTotal
nA
PT  A  PT
 PA
nT
45
Sample Problems
A mixture of 2.50 moles neon, 1.45 moles
helium, and 2.80 moles argon has a
pressure of 1.45 atm. What are the partial
pressure of all the gases in this system?
2.50 mol Ne + 1.45 mol He + 2.8 mol Ar = 6.75 mole
total
46
Sample Problems
A mixture of 2.50 moles neon, 1.45 moles
helium, and 2.80 moles argon has a
pressure of 1.45 atm. What are the partial
pressure of all the gases in this system?
PNe =
PHe =
2.50 mole Ne
 1.45 atm = 0.537 atm
6.75 mole total
1.45 mole Ne
6.75 mole total
 1.45 atm = 0.311 atm
47
Sample Problems
A mixture of 2.50 moles neon, 1.45 moles
helium, and 2.80 moles argon has a
pressure of 1.45 atm. What are the partial
pressure of all the gases in this system?
PAr =
2.80 mole Ne
 1.45 atm = 0.601 atm
6.75 mole total
Check Your Answer
PNe +PHe +PAr = .537 atm + .311 atm + .601 atm =
1.45 atm total
48
Sample Problems
2. If 425 mL of hydrogen is collected over
water at 25oC and 755.0 torr, calculate
the volume of “dry” hydrogen at STP.
“wet” H2
“dry” H2 at STP
V=
425 mL
? mL
T=
298 K
273 K
P=
731.2 torr*
*(P
atm
= PH2 + PH2O
760 torr
PH2 = 755.0 - 23.8 torr)
49
Sample Problems
2.
“wet” H2
“dry” H2 at STP
V=
425 mL
? mL
T=
298 K
273 K
P=
731.2 torr*
*(P
760 torr
= 755.0 - 23.8 torr)
Using the Combined Gas Law = 375 mL H2
50
GAS DENSITY
Low
density
High
density
51
GAS DENSITY
PV = nRT
n
P
=
V
RT
m
P
=
M• V
RT
m
M
n
where M = molar mass
m
PM
d =
=
V
RT
52
USING GAS DENSITY
The density of air at 15 oC and 1.00 atm is
1.23 g/L. What is the molar mass of air?
Solve the equation for M.
M = dRT/P
M = 29.1 g/mol
Note: Volume not needed!
KINETIC MOLECULAR THEORY
(KMT)
Theory used to explain gas laws.
KMT assumptions are:
• Gases consist of molecules in
constant, random motion.
• P arises from collisions with container
walls.
• No attractive or repulsive forces
between molecules. Collisions elastic.
• Volume of molecules is negligible.
53
54
Kinetic Molecular Theory
Because we assume molecules are in
motion, they have a kinetic energy.
Ave. KE = (1/2)(mass)(ave. speed)2
At the same T, all gases have the same
average KE.
As T goes up, KE also increases and so
does average speed.
KE a T,
1/2 mm2 = CT
(C is a nonspecific constant)
55
Kinetic Molecular Theory
At the same T, all gases have
the same average KE.
As T goes up, KE also
increases and so does speed.
56
Distribution of Molecular Speeds
• The equation: 1/2 mm2 = CT becomes
1/2 Mm2 = 3/2RT and Mm2 = 3RT
R = 8.31 J/K mol and M the molar mass kg /mole.
• The average speed of molecules, m, can be
calculated if the temperature and molar mass are
know.
• For two gases at the same temperature, the
following is true:
MAmA2 = MBmB2
Kinetic Molecular Theory
Expressed by Maxwell’s equation
m2
3RT
M
root mean square speed
where m is the speed and M is the
molar mass.
• speed INCREASES with T
• speed DECREASES with M
57
Velocity of Gas Molecules
Molecules of a given gas have a range of
speeds.
Figure 12.18
58
Distribution of Gas
Molecule Speeds
59
60
GAS DIFFUSION
AND EFFUSION
An application of KMT
• Diffusion is the
gradual mixing of
molecules of
different gases.
• Effusion is the
movement of
molecules through a
small hole into an
empty container.
61
GAS DIFFUSION AND
EFFUSION
Molecules effuse thru holes in
a rubber balloon, for example,
at a rate (= moles/time) that is:
• proportional to T
• inversely proportional to M.
Therefore, He effuses more
rapidly than O2 at same T.
He
62
GAS DIFFUSION AND EFFUSION
Graham’s law
governs effusion
and diffusion of
gas molecules.
Rate for A
Rate for B
M of B
M of A
Rate of effusion is
inversely proportional
to its molar mass.
Thomas Graham, 18051869. Professor in
Glasgow and London.
Gas Diffusion
relation of mass to rate of diffusion
• HCl and NH3 diffuse
from opposite ends
of tube.
• Gases meet to form
NH4Cl.
MHCl > MNH
3
• Therefore, NH4Cl
forms closer to HCl
end of tube.
63
Using KMT to Understand
Gas Laws
64
Recall that KMT assumptions are:
• Gases consist of molecules in
constant, random motion.
• P arises from collisions with container
walls.
• No attractive or repulsive forces
between molecules. Collisions elastic.
• Volume of molecules is negligible.
Avogadro’s Hypothesis and
Kinetic Molecular Theory
65
Gas Pressure,
Temperature, and Kinetic
Molecular Theory
66
Boyle’s Law and
Kinetic Molecular
Theory
67
Deviations from
Ideal Gas Law
• Real molecules
have volume.
• There are
intermolecular
forces.
–Otherwise a gas
could not
become a liquid.
Fig. 12.22
68
69
Deviations from Ideal Gas Law
Account for volume of
molecules and intermolecular
forces with VAN DER
WAAL’S EQUATION.
Measured P
P
n2 a
+ ----V2
Measured V = V(ideal)
V
- nb
nRT
vol. correction
intermol. forces
J. van der Waals,
1837-1923,
Professor of
Physics,
Amsterdam.
Nobel Prize 1910.
70
Deviations from Ideal Gas Law
Cl2 gas has a = 6.49, b = 0.0562
For 8.0 mol Cl2 in a 4.0 L tank at 27 oC.
P (ideal) = nRT/V = 49.3 atm
P (van der Waals) = 29.5 atm
Measured P
P
n2 a
+ ----V2
Measured V = V(ideal)
V
- nb
nRT
vol. correction
intermol. forces
71
Practice Problems
1. A sample of a gas exerts a pressure of
625 torr in a 300. mL vessel at 25oC. What
pressure would this gas sample exert if it
were placed in a 500. mL container at 25oC?
2. What would the volume of a gas be at STP
if it was found to occupy a volume of 255
mL at 25oC and 650 torr?
3. What volume will 25.0 g of oxygen occupy
at 20.oC and a pressure of 0.880 atm?
72
Practice Problems
4. If 2.5 L of hydrogen at STP is reacted with
chlorine, how many liters of hydrogen
chloride will be produced at 25oC and 1.25
atm?
5. If 45 mL of hydrogen at 25oC and 1.25 atm
is reacted with oxygen, how many liters of
water will be produced at STP?
6. If 45 mL of oxygen at 25oC and 1.25 atm is
reacted with nitrogen oxide, how many
grams of nitrogen dioxide will be
produced?
73
Practice Problems
7. At 29 oC and 745 mm, 325 mL of oxygen
are collected by water displacement. What
would be the volume of dry oxygen at STP?
8. The rate of effusion of an unknown gas
was determined to be 2.92 times greater
than that of ammonia. What is the
molecular mass of the unknown gas?
9. A sample of an unknown gas weighing
7.10 g at 741 torr and 44oC occupies a
volume of 5.40 L. What is the molar mass
of the gas?
Practice Problems
74
For problems 10-12
2 NO + O2 --> 2 NO2
10. How many liters of NO2 at 25oC and 1.0 atm
can be produced from 24 g of oxygen?
11. How many liters of NO2 at 20.oC and 1.4 atm
can be produced from 6 L of oxygen at STP?
12. How many liters of NO2 at 25oC and 600. mm
can be produced from 1.86 L of NO at 50.oC
and 720. mm?
75
Practice Problems
13. 2.5 L of a gas at 1.4 atm is changed to
2.1 atm. What is the new volume?
14. 500 mL of nitrogen at 720 torr is
compressed to 125 mL. What is the new
pressure?
15. 12.5 L of nitrous oxide at 25oC is
compressed to 2900 mL. What is the new
temperature in oC?
16. A sample of carbon monoxide at 640 torr
and 30.oC is heated to 90.oC. What is the
new pressure in atm?
76
Practice Problems
17. A 1.25 L expandable container of
hydrogen at 25oC and 4.5 atm is changed to
75oC and 3.6 atm. What is the new volume?
18. A balloon containing 45 mL of carbon
dioxide at 40oC and 140 torr is heated to
60.oC and the pressure becomes 1.00 atm.
What is the new volume?
19. At 25oC and 600. torr a 1.2 L balloon of
oxygen is changed to 1500 mL and 1.4 atm.
What is the new temperature in oC?
77
Practice Problems Answers
1. 375 torr
3. 21.3 L
5. .052 L
7. 276 mL
9. 35.1 g/mole
11. 9 L
13. 1.7 L
15. -204 oC
17. 1.8 L
19. 390 oC
2. 2.0 x 102 mL
4. 4.4 L
6. .21 g
8. 1.99 g/mole
10. 37 L
12. 2.06 L
14. 3000 torr
16. 1.0 atm
18. 8.8 mL
78
Sample Problems
1. The pressure of 500. mL of a gas at
2.00 atm is increased to 4.00 atm.
What is the new volume?
P1V1 = P2V2
V2 =
P1V1
P2
(2.00 atm)(500. mL)
=
4.00 atm
V2 = 250. mL
79
Sample Problems
2. 355 L of a gas at 850 torr is expanded to
652 L. What is the new pressure in atm?
P1V1 = P2V2
P2 =
P1V1
(855 torr)(355 L) 1 atm
=
V2
P2 =
652 L
0.61 atm
760 torr
80
Sample Problems
1. The temperature of 462 mL of hydrogen
at 45oC is decreased to 25oC. Calculate
the new volume.
V1
V2
=
T1
T2
V2 =
T2V1
=
T1
V2 =
(298 K)(462 mL)
318 K
433 mL
81
Sample Problems
2. Calculate the temperature in oC of 25.6 mL
of oxygen at 148 K if the volume is changed
to 1.42 L.
V1
V2
=
T1
T2
T2 =
T1V2
=
V1
T2 = 8210 K =
(148 K)(1.42 L)
0.0256 L
7940 oC
82
P and T Law
• As temperature increases, pressure
increases, at constant V and n.
• If the Kelvin scale is used, P a T.
P = CdT or
P1 P 2

T1 T2
83
Sample Problems
1. A sample of nitrogen at 1.45 atm and 25oC
is compressed to a pressure of 2.14 atm,
what is the new temperature in oC?
P1
T1
T2 =
T1P2
P2
=
=
P1
T2 = 440. K =
T2
(298 K)(2.14 atm)
1.45 atm
167 oC
84
Sample Problems
2. A sample of helium at 145 K and 0.780 atm.
is cooled to 125 K, what is the new
pressure in mm?
P1
P2
=
T1
T2
P2 =
T2P1
=
T1
P2 =
(125 K)(0.780 atm) 760 mm
145 K
511 mm
1 atm
85
THE COMBINED GAS LAW
Combining the previous gas laws, produces
the Combined Gas Law.
P1 V1
----------n1 T1
P2 V2
= -----------n2 T2
86
Sample Problems
1. A 25.6 L sample of gas at 25oC and 1.0 atm
is changed to 125oC and 0.50 atm.
Calculate the new volume.
P1V1
T1
V2 =
P1V1T2
=
T1P2
V2 =
P2V2
=
T2
(1.0 atm)(25.6 L)(398 K)
(298 K)(0.50 atm)
68 L
87
Sample Problems
2. Calculate the new pressure in atm if 25 L of
hydrogen at 35oC and 742 torr is compressed
to 1200 mL at 45oC.
P1V1
T1
P2 =
P1V1T2
T1V2
=
P2V2
=
T2
(742 torr)(25 L)(318 K) 1 atm
(308 K)(1.2 L)
P2 =
21 atm
760 torr
88
THE IDEAL GAS LAW
Combining the previous gas laws,
produces the Ideal Gas Law.
PV = nRT
where R = 0.0821 L atm/K mol
89
Sample Problems
1. Calculate the pressure of 4.4 mole of
oxygen contained in a 42.8 L sample at
25oC?
PV = nRT
P=
nRT
V
=
(4.4 mol)(0.0821 L atm/mole K)(298 K)
42.8 L
P =
2.5 atm
90
Sample Problems
2. Calculate the volume of 2.14 moles of
neon at 35oC and 425 mm.
PV = nRT
V=
nRT
=
(2.14 mol)(0.0821 Latm/mol K)(308 K)
P
(425/760) atm
V =
96. 8 L
91
MOLAR VOLUME at STP
One mole of gas occupies 22.4 L at STP,
and is called the standard molar volume
or the molar volume at standard
conditions (SC or STP).
92
Sample Problems
1. Calculate the volume of 12.0 g of fluorine
at STP?
12.0 g
Mole
22.4 L
38.0 g
mole
=
7.07 L
93
Sample Problems
2. What is the mass of 14,500 mL of
hydrogen at STP?
14.5 L
mole
2.0 g
22.4 L
mole
=
1.3 g
COMBINED vs. IDEAL
GAS LAW
94
A sample of 6.9 moles
of carbon monoxide
gas is present in a
container which has
a volume of 30.4 L.
What is the pressure
of the gas if the
temperature is 62oC?
The volume of a
sample of hydrogen
gas is 5.80 L
measured at 1.00 atm
and 25oC. What is
the pressure of the
gas if the volume is
changed to 9.65 L?
(6.2 atm)
(0.601 atm)
95
Gases and Stoichiometry
These problems are worked the
same as any stoichiometry
problem.
For example:
9.82 L of nitrogen at STP will
produce how many liters of
ammonia at STP?
3 H2 + N2 --> 2 NH3
96
1. 9.82 L of nitrogen at STP will produce
how many liters of ammonia at STP?
STEP 1
Write the balanced chemical equation.
3 H2 + N2
--> 2 NH3
97
STEP 2
Write the given and requested
information below the equation.
3 H2
+
N2 -->
9.82 L
2 NH3
?L
98
STEP 3
Calculate using the information.
3 H2
+
N2 -->
9.82 L
9.82 L N2 mole N2
22.4 L N2
2 NH3
?L
2 mole NH3 22.4 L NH3
mole N2
mole NH3
Or
9.82 L N2 2 L NH3
1 L N2
=
19.6 L NH3
0o C
2. 9.82 L of nitrogen at
and 2.0 atm
will produce how many liters of
ammonia at 0oC and 2.0 atm?
3 H2
+
N2 -->
9.82 L
2 NH3
?L
273 K, 2.0 atm 273 K, 2.0 atm
9.82 L N2 mole N2
Or
? L N2
9.82 L N2 2 L NH3
1 L N2
2 mole NH3
mole N2
=
? L NH3
mole NH3
19.6 L NH3
99
0o C
3. 9.82 L of nitrogen at
and 2.0 atm
will produce how many liters of
ammonia at 25oC and 1.0 atm?
3 H2
+
N2 --> 2 NH3
9.82 L
?L
273 K, 2.0 atm 298 K, 1.0 atm
9.82 L N2 2 L NH3
1 L N2
=
19.6 L NH3
at 273 K, 2.0 atm
Using the Combined Gas Law = 43 L NH3
at 298 K, 1.0 atm
100
0o C
4. 9.82 L of nitrogen at
and 1.0 atm
will produce how many grams of
ammonia?
3 H2
+
N2 -->
9.82 L
2 NH3
?g
273 K, 1.0 atm
9.82 L N2 1 mole N2
2 mole NH3 17.0 g NH3
22.4 L N2 1 mole N2
= 14.9 g NH3
mole NH3
101
5. 34.0 g of nitrogen will produce how
many liters of ammonia at 25oC and
2.0 atm?
3 H2
+
N2 -->
34.0 g
102
2 NH3
?L
298 K, 2.0 atm
34.0 g N2 1 mole N2
28.0 N2
2 mole NH3
1 mole N2
= 2.43 mole NH3
Using the Ideal Gas Law = 29.7 L NH3
at 298 K, 2.0 atm
103
Sample Problems
1. A mixture of 2.50 moles neon,
1.45 moles helium, and 2.80 moles
argon has a pressure of 1.45 atm.
What is the partial pressure of neon?
PNe =
2.50 mole Ne
6.75 mole total
X 1.45 atm = 0.537 atm
104
Sample Problems
2. If 425 mL of hydrogen is collected over
water at 25oC and 755.0 torr, calculate
the volume of “dry” hydrogen at STP.
“wet” H2
“dry” H2 at STP
V=
425 mL
? mL
T=
298 K
273 K
P=
731.2 torr*
*(P
= 755.0 - 23.8 torr)
760 torr
105
Sample Problems
2.
“wet” H2
“dry” H2 at STP
V=
425 mL
? mL
T=
298 K
273 K
P=
731.2 torr*
*(P
760 torr
= 755.0 - 23.8 torr)
Using the Combined Gas Law = 375 mL H2
106
Sample Problems
1. Determine the density of carbon dioxide
at 375 K and 5.0 atm.
(5.0 atm)(44.0 g/mole)
d=
(0.0821 L atm/mole K)(375 K)
= 7.1 g/L