1. 12 The Gaseous State of Matter
Air in a hot air balloon expands upon heating. Some air escapes from the top, lowering the air density, making the balloon buoyant.
Foundations of College Chemistry, 14th Ed.
Morris Hein and Susan Arena
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2. © 2014 John Wiley & Sons, Inc. All rights reserved.
Chapter Outline
12.1 Properties of Gases
A. Measuring the Pressure of a Gas
B. Pressure Dependence: Number of Molecules and Temperature
12.2 Boyle’s Law
12.3 Charles’ Law
12.4 Avogadro’s Law
A. Mole-Mass-Volume Calculations
12.5 Combined Gas Laws
12.6 Ideal Gas Law
A. Kinetic-Molecular Theory
B. Real Gases
12.7 Dalton’s Law of Partial Pressures
12.8 Density of Gases
12.9 Gas Stoichiometry
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Properties of Gases
Gases:
i) Have indefinite volume
Expand to fill a container
ii) Have indefinite shape
Assume the shape of a container
iii) Have low densities
Volume increase 1200 fold when 1 mole of water goes from a liquid to a gas.
Example
Volume occupied by
1 mol of H2O:
as a liquid (18 mL)
as a gas (22.4 L)
dair = 1.2 g/L at 25 °C
dwater = 1.0 g/mL at 25 °C
iv) Have high velocities and kinetic energies
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4. gas molecule collisions with the container walls.
Measuring Pressure
Pressure: Force per unit area
Pressure =
area
force
Pressure depends on:
1) The number of gas molecules
2) Gas temperature
Pressure results from
gas molecule collisions
with the container walls.
3) Volume occupied by the gas
SI unit of pressure is the pascal (Pa) = 1 newton/meter2
Unit Conversions:
1 atm = 760 mm Hg = 760 torr
= kPa = bar = psi
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5. Practicing Pressure Conversions
Convert 740. mm Hg to a) atm and b) kPa. a)
Use the conversion factor:
1 atm = 760 mm Hg
740. mm Hg
× atm
760 mm Hg
= atm b)
Use the conversion factor:
101.3 kPa = 760 mm Hg
740. mm Hg
× kPa
760 mm Hg
= kPa
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6. Atmospheric Pressure Due to the mass of the atmospheric gases pressing
Definition: total pressure exerted by gases in the atmosphere
Due to the mass of the atmospheric gases pressing
downward on the Earth’s surface.
Major Components of Dry Air
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7. © 2014 John Wiley & Sons, Inc. All rights reserved.
Measuring Pressure
Measuring Pressure
Use a Barometer
1) Invert a long tube of Hg
over an open dish of Hg.
2) Hg will be supported
(pushed up) by the pressure
of the atmosphere.
Barometer invented by E. Torricelli- hence the name Torr
3) Height of Hg column can be
used to measure pressure.
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8. Pressure Dependence 1) On the Number of Molecules
Pressure (P ) is directly proportional to the number
of gas molecules present (n ) at constant
temperature (T ) and volume (V ).
Increasing n creates more frequent collisions with the
container walls, increasing the pressure
V = 22.4 L
T = 25.0 °C
0.5 mol H2
P = 0.5 atm
1 mol H2
P = 1 atm
2 mol H2
P = 2 atm
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9. Pressure Dependence 2) On Temperature
Pressure is directly proportional to temperature when
moles (n ) and volume (V ) are held constant.
T = 0 °C
T = 100 °C
2.24 atm
3.06 atm
Increasing T causes:
a) more frequent and
b) higher energy collisions
0.1 mol of gas
in a 1L container
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10. Boyle’s Law The volume of a fixed quantity of gas is inversely
proportional to the pressure exerted by the gas at
constant mass and temperature.
P a V 1
PV = constant (k ) or
P = k × V 1
Most common form:
P1V1 = P2V2
Graph showing inverse PV relationship
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11. Boyle’s Law Problems
What volume will 3.5 L of a gas occupy if the pressure is
changed from 730. mm Hg to 600. mm Hg?
P1V1 = P2V2
Knowns
V1 = 3.5 L P1 = 730. mm Hg P2 = 600. mm Hg
V2 = P2
P1V1
Solve For V2
730. mm Hg
600. mm Hg ×
Calculate
V2 =
3.5 L
= 4.3 L
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12. Boyle’s Law Problems A sample of Ne gas occupies 250. mL at 880. torr.
Calculate the PNe if the volume is increased to 1.0 L,
assuming constant temperature. (Note: Convert mL to L.)
P1V1 = P2V2
Knowns
V1 = L V2 = 1.0 L P1 = 880. mm Hg
P2 = V2
P1V1
Solving For P2
0.250 L
1.0 L
Calculate
P2 =
880. torr ×
= 220 mm Hg
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13. Boyle’s Law Problems
A sample of gaseous nitrogen in a 65.0 L automobile air
bag has a pressure of 745 mm Hg.
If the sample is transferred to a 25.0 L bag at the same
temperature, what is the pressure in the bag?
a) 2.18 mm Hg
b) 1940 mm Hg
c) 287 mm Hg
d) mm Hg
P2 = V2
P1V1
65.0 L
25.0 L =
745 torr ×
= mm Hg
Sense check: As volume decreases, pressure should increase!
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14. Temperature in Gas Law Problems
Kelvin Temperature Scale
Derived from the relationship between temperature
and volume of a gas.
As a gas is cooled by 1 ºC increments,
the gas volume decreases in increments of 1/273.
All gases are expected to have zero volume if cooled to −273 ºC.
Cannot achieve 0 volume as a real gas condenses before you can reach 0K.
V -T relationship of methane
(CH4) with extrapolation (-----) to absolute zero.
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15. Temperature in Gas Law Problems
This temperature (−273 ºC) is referred to as absolute zero.
Absolute zero is the temperature (0 K) when the
volume of an ideal gas becomes zero.
All gas law problems use the Kelvin temperature scale!
Celsius temperature
TK = T°C + 273
Kelvin temperature
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16. Charles’ Law The volume of a fixed quantity of gas is directly
proportional to the absolute temperature of the gas
at constant pressure.
V = k T or
V a T V T
= k
Most common form:
V1 V2
T1 T2 =
© 2014 John Wiley & Sons, Inc. All rights reserved.

17. Charles’ Law Problems
3.0 L of H2 gas at −15 ºC is allowed to warm to 27 ºC at
constant pressure. What is the gas volume at 27 ºC?
V1 V2
T1 T2 =
Knowns
V1 = 3.0 L
T1 = −15 ºC = 258 K T2 = 27 ºC = 300. K
V2 = T1
V1T2
Solving For V2
V2 = T1
V1T2
300. K
Calculate =
3.0 L ×
= 3.5 L
258 K
© 2014 John Wiley & Sons, Inc. All rights reserved.

18. Charles’ Law Problems A gas has a volume of 3.00 L at 10.0 ºC.
What is the temperature of the gas if it
expands to 6.00 L, assuming constant pressure?
V1 V2
T1 T2 =
Knowns
V1 = 3.00 L V2 = 6.00 L
T1 = 10.0 ºC = 283 K
T2 = V1
T1V2
Solving For T2
T2 = V1
T1V2
6.00 L
Calculate =
283 K ×
= 566 K
3.00 L
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19. Charles’ Law Problems At 321 K, a gas occupies 635 mL of volume.
If the temperature is decreased to 216 K,
what is the new gas volume?
a) 916 mL
b) 109 mL
c) 943 mL
d) 427 mL
V2 = T1
V1T2
216 K =
635 mL ×
= 427 mL
321 K
Sense Check: As temperature decreases, volume decreases!
© 2014 John Wiley & Sons, Inc. All rights reserved.

20. Avogadro’s Law Equal volumes of different gases at constant T and P
contain the same number of molecules.
Avogadro’s Law provided proof for the concept of diatomic molecules for hydrogen and chlorine.
1 volume unit
4 molecules
1 volume unit
4 molecules
2 volume units
8 molecules
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21. Avogadro’s Law Given the following gas phase reaction: N2 + 3 H2 2 NH3
If 12.0 L of H2 gas are present, what volume of N2 gas is
required for complete reaction? T and P are held constant.
By Avogadro’s Law, we can use the reaction stoichiometry
to predict the N2 gas needed.
Knowns
VH2 = 12.0 L
Solving For VN2
Calculate
12.0 L H2
× 1 L N2
= L N2 required
3 L H2
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22. Avogadro’s Law Given the following gas phase reaction: 2 H2 + O2 2 H2O
At constant T and P, how many liters of O2 are required
to make 45.6 L of H2O?
a) 11.4 L
b) 45.6 L
c) 22.8 L
d) 91.2 L
45.6 L H2O
× 1 L O2
= L O2 required
2 L H2O
Sense Check: Less moles of O2 equal less L of O2!
© 2014 John Wiley & Sons, Inc. All rights reserved.

23. Mole/Mass/Volume Relationships
Molar Volume: volume 1 mol of gas occupies at STP
molar volume = 22.4 L/mol at STP
Molar volume can be used as a conversion factor if the
mass and volume occupied by a gas are known.
Example:
1.0 L of O2 at STP has a mass of g. Show that the molar mass of O2 is 32.0 g/mol.
1.429 g O2
22.4 L O2
= g/mol O2 ×
1.0 L O2
1 mol O2
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24. Mole/Mass/Volume Relationships
If 3.00 L of a gas measured at STP has a mass of 5.35 g,
calculate the molar mass.
a) 39.9 g/mol
b) 79.6 g/mol
c) 12.6 g/mol
d) g/mol
5.35 g gas
22.4 L O2
= g/mol ×
3.00 L gas
1 mol gas
Unit Check: Molar mass has units of g/mol,
so use dimensional analysis when setting up the problem!
© 2014 John Wiley & Sons, Inc. All rights reserved.

25. Combined Gas Laws A combination of Boyle’s and Charles’ Laws.
Used in problems involving changes in P, T, and V
with a constant amount of gas.
P1V P2V2 T1 T2 =
The volume of a fixed quantity of gas depends on the
temperature and pressure.
It is not possible to state the volume of gas without
stating the temperature and pressure.
P and V units, as long as consistent, are ok. T must always be in K.
Standard Temperature and Pressure (STP):
0.00 °C ( K) and 1 atm (760 torr)
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26. Combined Gas Law Problems
A sample of gas occupies 125 mL at STP.
What is the volume of the gas at 65 ºC and 320. torr?
P1V P2V2 T1 T2 =
Knowns
V1 = L P1 = 760 torr T1 = 273 K
P2 = 320 torr T2 = 65 ºC = 338 K
V2 =
T1P2
P1V1T2
Solving For V2
Calculate P2
V2 = V1 × P1 ×T2 T1
0.125 L =
× torr
× 338 K
= L
320. torr
273 K
© 2014 John Wiley & Sons, Inc. All rights reserved.

27. Combined Gas Law Problems
What is the volume at STP for a gas that occupies
1.62 L at 616 torr and 42 °C?
P1V P2V2 T1 T2 =
Knowns
V1 = 1.62 L P1 = 616 torr T1 = 42 °C = 315 K
P2 = 760. torr T2 = 273 K
V2 =
T1P2
P1V1T2
Solving For V2
Calculate
V2 = V1 ×P1 × T2 P2
1.62 L = T1
× 616 torr
× 273 K
= L
760. torr
315 K
© 2014 John Wiley & Sons, Inc. All rights reserved.

28. Combined Gas Law Problems
A balloon is filled with 266 L of He gas, measured at 38 °C
and atm. What will its volume be when the
temperature is lowered to −76 ° C and the pressure
is atm?
a) 299 L
b) 95.0 L
c) 745 L
d) 237 L
V2 = V1 × P1 × T2 P2
266 L = T1
× atm
× 197 K
= 299 L
0.561 atm
311 K
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29. Ideal Gas Law A single equation relating all properties of a gas.
PV = nRT
where R is the universal gas constant
Constant n and T
Constant n and P
Constant P and T
V a 1/P
Boyle’s Law
V a T
Charles’ Law
V a n
Avogadro’s Law
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30. Ideal Gas Constant R is derived from conditions at STP. Calculate R.
PV = nRT
Knowns
P = 1.00 atm V = 22.4 L T = 273 K n = 1.00 mol R PV nT =
Solving For R
Calculate
R = P × V =
1.00 atm
× L
= L . atm
n × T
1.00 mol × 273 K
mol . K
Units are critical in ideal gas problems!
© 2014 John Wiley & Sons, Inc. All rights reserved.

31. How many moles of He are contained in a 0.900 L
Ideal Gas Law Practice
How many moles of He are contained in a L
container at 30. ºC and atm?
PV = nRT
Knowns
P = atm V = L T = 30. ºC = 303 K n PV RT =
Solving For n
Calculate
n = P × V =
0.800 atm
× L
= mol
R × T
L . atm
mol . K
× 303 K
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32. What volume will be occupied by 0.393 mol of N2
Ideal Gas Law Practice
What volume will be occupied by mol of N2
at atm and 24 °C?
PV = nRT
Knowns
P = atm n = mol T = 24 ºC = 297 K V
nRT P =
Solving For V
Calculate
V = nRT =
0.393 mol ×
L . atm
mol . K
× 297 K
= L P
0.971 atm
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33. The ideal gas law can also be written in terms of
Ideal Gas Law Practice
The ideal gas law can also be written in terms of
molar mass of a gas.
PV = nRT
n =
mass in grams (g)
molar mass (M )
PV = M
gRT
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34. Ideal Gas Law Practice A 0.210 g gas sample has a pressure of 432 torr
in a 333 mL container at 23 ºC.
What is the molar mass of the gas?
PV = M
gRT
Knowns
P = 432 torr = atm V = L
T =296 K mass = g
Solving For M
Calculate
M = gRT
= g × L atm/mol K × 296 K
= 27.0 g/mol PV
0.568 atm × L
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35. Ideal Gas Law Practice Calculate the molar mass (M ) of an unknown gas
if g occupies a volume of 754 mL at 30. ºC
and 342 torr.
a) 35.4 g/mol
b) 21.9 g/mol
c) 87.3 g/mol
d) 55.0 g/mol
M = gRT
= g × L atm/mol K × 303 K
= 56.3 g/mol PV
0.450 atm × L
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36. Kinetic Molecular Theory
A general theory developed to explain the behavior
and theory of gases, based on the motion of particles.
Assumptions of Kinetic Molecular Theory (KMT):
1) Gases consist of tiny particles.
2) The distance between particles is large when compared
to particle size. The volume occupied by a gas is
mostly empty space.
3) Gas particles have no attraction for one another.
4) Gas particles move linearly in all directions, frequently
colliding with the container walls or other particles.
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37. Kinetic Molecular Theory
Assumptions of KMT (continued):
5) Collisions are perfectly elastic.
No energy is lost during collisions.
6) The average kinetic energy for particles is the same for
all gases (regardless of molar mass) at the same temperature.
KE = 1/2mv 2
where m is the mass and v is the velocity of the particle
The average kinetic energy is directly proportional to
temperature (in K).
Gases which behave under these assumptions are
know as ideal gases.
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38. Real Gases Real gases typically behave like ideal gases over
a fairly wide range of temperatures and pressures.
Conditions where real gases deviate from ideal gases:
1) At high pressure (small volumes)
Distance between particles is small and the particles
do not behave independently.
2) At low temperature
Particles experience intermolecular interactions.
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39. Dalton’s Law of Partial Pressures
The total pressure of a mixture is the sum of the partial
pressures of the different gases in the mixture.
Ptotal = P1 + P2 + P3…
Each gas behaves independently in the mixture.
Application of Dalton’s Law
Gases collected over H2O contain
both the gas and H2O vapor.
Vapor pressure of H2O is constant
at a given T.
Pbottle is equalized so that Pbottle = Patm thus
Collecting a gas over water
Patm = Pgas + PH2O
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40. Partial Pressures Problems
A sample of O2 gas is collected over water at 22 ºC and
662 torr. What is the partial pressure of O2 gas? The
vapor pressure of water is 19.8 torr at 22 ºC.
Knowns
Patm = 662 torr PH2O = 19.8 torr
Solving For PO2
PO2 = Patm – PH2O
Calculate
PO2 = 662 torr – 19.8 torr
= 642 torr
© 2014 John Wiley & Sons, Inc. All rights reserved.

41. Partial Pressures Problems
A 250. mL sample of O2 was collected over water at
23 ºC and 760 torr. What volume will the O2 occupy at
23 ºC when PO2 is 760. torr?
The vapor pressure of water at 23 ºC is 21.2 torr.
Knowns
VO2 + H2O = 250 mL Patm = PO2 + PH2O = 760. torr
PH2O= 21.2 torr
Solving For VO2
Solve for PO2 using Dalton’s Law
Solve for VO2 using Boyle’s Law
Calculate
PO2 = Ptotal – PH2O
= 760 torr – 21.2 torr
= 739 torr
© 2014 John Wiley & Sons, Inc. All rights reserved.

42. Partial Pressures Problems
(continued)
A 250. mL sample of O2 was collected over water at
23 ºC and 760 torr. What volume will the O2 occupy at
23 ºC when PO2 is 760 torr?
The vapor pressure of water at 23 ºC is 21.2 torr.
Calculate
Solve for VO2 with Boyle’s Law
V2 = P2
P1V1
P1V1 = P2V2
V2 = P2
P1V1
739 mm Hg
760 mm Hg
0.250 L = ×
= L O2
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43. Gas Density
Density of a liquid or solid is expressed in g/mL, but
gas density is very low, so the standard units are g/L.
mass L
volume
density (d ) = = g
The density of a gas at STP can also be related to the
compound’s molar mass. ( ) ) g (
1 mol
22.4 L g
dstp = molar mass × =
mol L
Note: gas densities must be cited at a specific
temperature as volume changes as a function of
temperature (Charles’ Law).
© 2014 John Wiley & Sons, Inc. All rights reserved.

44. ( ) ( ) Gas Density Practice Calculate the density of Cl2 at STP. L
1 mol
22.4 L
d = molar mass = g ×
mol ) ( )
molar mass Cl2 = 70.9 g/mol
d =
70.9 g Cl2
× 1 mol Cl2
= g/L
1 mol Cl2
22.4 L Cl2
Sense Check: Gas densities are expected to be low.
© 2014 John Wiley & Sons, Inc. All rights reserved.

45. Gas Stoichiometry At STP: the molar volume can be used as a conversion
factor to convert between moles and volume.
Non STP Conditions: use the ideal gas law to convert
between moles and volume.
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46. Gas Stoichiometry Practice at STP
For the following reaction:
Calculate the number of moles of phosphorus needed to
react with 4.0 L of H2 gas at 273 K and 1.0 atm.
P4 (s) + 6 H2 (g) PH3 (g)
Knowns
V =4.0 L T = 273 K P = 1.0 atm
Solution Map
L H mol H mol P4
mol P4 =
Calculate
4.0 L H2 ×
22.4 L H2
1 mol H2
1 mol P4
6 mol H2 ×
= mol P4
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47. Gas Stoichiometry Volume Practice
Calculate the volume of N2 necessary to react with
9.0 L of H2 gas at 450 K and 5.00 atm.
N2 (g) + 3 H2 (g) NH3 (g)
a) 9.0 L
b) 3.0 L
c) 27.0 L
d) 1.0 L
3 L H2
9.0 L H2
1 L N2 ×
= 3.0 L N2
At constant T and P, the volume ratio can be used
in place of the mole ratio!
© 2014 John Wiley & Sons, Inc. All rights reserved.

48. Gas Stoichiometry Practice With the Ideal Gas Law
Given the following reaction:
2 NaN3 (s) Na (s) + 3 N2 (g)
If an air bag should be filled with a pressure of 1.09 atm
at 22 ºC, what amount of solid NaN3 is needed to fill
a bag with a volume of 45.5 L?
Knowns
P = 1.09 atm V = 45. 5L T = 295K
Solving for n of N2 then find the mass of NaN3 needed.
Calculate
n = PV RT =
L atm/mol K × 295 K
1.09 atm × 45.5 L
= mol N2
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49. Gas Stoichiometry Practice With the Ideal Gas Law
(continued)
Given the following reaction:
2 NaN3 (s) Na (s) + 3 N2 (g)
If an air bag should be filled with a pressure of 1.09 atm
at 22.0 ºC, what amount of solid NaN3 is needed to fill
a bag with a volume of 45.5 L?
Calculate
Use the reaction stoichiometry!
2.05 mol N2
× 2 mol NaN3
3 mol N2
× g NaN3
1 mol NaN3
= g NaN3
© 2014 John Wiley & Sons, Inc. All rights reserved.

50. Gas Stoichiometry Practice
What volume of O2 at 760. torr and 25 ºC is needed to
react fully with 3.2 g of C2H6 (propane)?
2 C2H6 (g) O2 (g) CO2 (g) + 6 H2O (l)
Knowns
m = 3.2 g T = 298 K P = 1.00 atm
Solution Map
m C2H mol C2H mol O volume O2
Calculate
3.2 g C2H6 ×
30.08 g C2H6
1 mol C2H6
2 mol C2H6
7 mol O2 ×
= mol O2 =
0.37 mol × × 298 K
L . atm
mol . K
1.00 atm
V = nRT P
= 9.1 L
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51. Gas Stoichiometry Practice
What volume of H2 at 739 torr and 21 ºC is liberated
by 42.7 g of Zn when it reacts with HCl?
Zn (s) HCl (g) ZnCl2 (s) + H2 (g)
a) 7.6 L
b) 16.2 L
c) 3.2 L
d) 1.8 L
m Zn mol Zn mol H2 volume H2
42.7 g Zn
× 1 mol Zn
65.38 g Zn
1 mol Zn
× 1 mol H2
= mol H2
V = nRT P
0.972 atm
0.653 mol × L atm/mol K × 294 K =
= L H2
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52. Chemistry in Action What the Nose Knows
Dogs use smell to detect many drugs, explosives, etc.
based on trace amounts of chemical compounds in the air.
Sensing low concentrations of chemicals is useful!
Better Coffee
Better Science
Artificial noses could sniff
out cancer or explosives!
For more information, see:
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53. © 2014 John Wiley & Sons, Inc. All rights reserved.
Learning Objectives
12.1 Properties of Gases
1) Explain atmospheric pressure and how it is measured.
2) Be able to convert between the various units of pressure.
12.2 Boyle’s Law
3) Use Boyle’s Law to calculate changes in pressure or
volume of a gas at constant temperature.
12.3 Charles’ Law
4) Use Charles’ Law to calculate changes in temperature or
volume of a gas at constant pressure.
© 2014 John Wiley & Sons, Inc. All rights reserved.

54. © 2014 John Wiley & Sons, Inc. All rights reserved.
Learning Objectives
12.4 Avogadro’s Law
5) Solve problems using the relationships between moles,
mass, and volume of gases.
12.5 Combined Gas Law
6) Use the combined gas law to calculate changes in
pressure, volume, or temperature of a gas sample.
12.6 Ideal Gas Law
7) Use the ideal gas law to solve problems involving
pressure, volume, temperature, and moles of a gas.
© 2014 John Wiley & Sons, Inc. All rights reserved.

55. © 2014 John Wiley & Sons, Inc. All rights reserved.
Learning Objectives
12.7 Dalton’s Law of Partial Pressures
8) Use Dalton’s Law of Partial Pressures to calculate the
total pressure for a mixture of gases or the pressure of
a single gas in a mixture of gases.
12.8 Density of Gases
9) Calculate the density of a gas. (Pay attention to units!)
12.9 Gas Stoichiometry
10) Solve stoichiometry problems involving gases.
(Pay attention to the states of matter and use gas laws
only for gases!)
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