CHAPTER 5: GASES CLASS NOTES FOR CHEMISTRY 5.1

CHAPTER 5: GASES
CLASS NOTES FOR CHEMISTRY
5.1 Breathing: Putting Pressure to Work
Pressure: force per unit area; for gases pressure is caused by the gas molecules striking the
surface around them. Generally pressure decreases if there are fewer gas molecules, such as at
higher altitude. At 30,000 feet above sea level where commercial planes fly, the cabins must be
pressurized or passengers would pass out from lack of oxygen.
5.2 Pressure: The Result of Molecular Collisions
Pressure =
force
area
=
F
A
Pressure units:
1 mm of Hg = 1 torr
1 atm = 760 torr (1 atm = 14.7 psi or pounds per square inch)
Mt. Everest has 0.31 atm pressure. A highly inflated mountain bike tire has a pressure of about
6 atm.
Example 5.1 A high-performance road bicycle tire is inflated to a total pressure of 132 psi.
What is this pressure in mm of Hg?
Plan: psi  atm mm of Hg
132 psi X
1 atm
14.7 psi
X
760 mm
1 atm
= 6820 or 6.82 x 103
mm of Hg
The Manometer: A way to Measure Pressure in the Laboratory
5.3 The Simple Gas Laws: Boyles’ Law, Charles’s Law, and Avagadro’s Law
Variables to consider: pressure, P; volume, V; temperature, T; and number of moles, n.
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Boyles’s Law: Volume and Pressure
If number of moles of the gas is constant and the temperature of the gas is constant, Boyle
found the following relationship:
V is proportional to
1
P
, or V is inversely proportional to pressure.
As pressure increases,
volume decreases.
We can change this to an equation by using a proportionality constant.
V = (constant) X
1
P
P1V1 = constant
Rearrange to obtain: PV = constant
and P2V2 = constant Then P1V1 = P2V2 This is Boyle’s Law.
Example 5.2 Boyle’s Law A woman has an initial lung volume of 2.75 L, which is filled with
air at atmospheric pressure of 1.02 atm. If she increases her lung volume to 3.25 L without
inhaling any additional air, what is the pressure in her lungs?
Plan: Use P1V1 = P2V2 P1 = 1.02 atm; V1 = 2.75 L; V2 = 3.25 L P2 = ?
P2 =
P1 V1
V2
=
1.02 atm x 2.75 L
3.25 L
= 0.863 atm
Charles’s Law: Volume and Temperature
If number of moles of the gas is constant and pressure is constant, Charles found that as
temperature increases the volume increases. The gas that fills a hot-air balloon is warmed with
a burner, increasing its volume, lowering its density, and causing it to float in the colder denser
surrounding air.
V is proportional to T (if temperature is in Kelvin). Or V = (constant) X T
Or
V
T
= constant
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V1
= constant,
T1
Charles’s Law.
V2
T2
V1
= constant, therefore:
T1
=
V2
T2
That is a statement of
Example 5.3 A sample of gas has a volume of 2.80 L at an unknown temperature. When the
sample is submerged in ice water at T = 0.00 o C, its volume decreases to 2.57 L. What was its
initial temperature (in K and in o C) ?
V1
T1
=
V2
T2
K = o C + 273
K = 0 + 273 = 273
V1 T1
Rearrange Charles’s Law to solve for T1. T1 =
T1 = 298 K
o
V2
=
2.80 L x 273 K
2.57 L
We change temperature to oC.
C = 298 – 273 = 24 o C
Avogadro’s Law: Volume and Amount (In Moles)
If pressure of the gas is constant and the temperature of the gas is constant, Avogadro found
the following relationship:
V is proportional to n or V = (constant) X n
Rearranged:
V1
n1
=
V2
n2
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Example 5.4 A male athlete in a kinesiology research study has a lung volume of 6.15 L during a
deep inhalation. At that volume, his lungs contain 0.254 moles of air. During the exhalation, his
lung volume decreases to 2.55 L. How many moles of gas did the athlete exhale? Assume
constant temperature and pressure.
V1
n1
=
V2
Rearrange to solve for n2.
n2
n2 =
n 1 V2
V1
=
0.254 mol x 2.55 L
6.15 L
n2 = 0.105 mol
mol exhaled = 0.254 mol- 0.105 mol = 0.149 mol [What he started with minus what he had left.}
Combined Gas Law (THIS IS FOR CHANGING CONDITIONS!)
If we combine the three laws above, we get this. This is very helpful, so that you don’t have to
remember three separate laws. If the temperature is constant, then the T’s cancel. If the
pressure is constant, then the P’s cancel. If the amount of gas, n, is constant, then the n’s
cancel.
V1 𝑃1
𝑛1 T1
=
V2 𝑃2
𝑛2 T2
5.5 The Ideal Gas Law (THIS IS FOR NO CHANGING CONDITIONS)
Volume is proportional to 1/P
(Boyle’s Law)
Volume is proportional to T
(Charles’s Law)
Volume is proportional to n
(Avogadro’s Law)
nT
Therefore V is proportional to P
Or using a proportionality constant R
V =
Rn T
P
Then we rearrange and get: PV = nRT [This is the ideal gas law and a gas that follows this law
exactly is called an ideal gas.]
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If P is in atm; V is in L; n is in mol; T is in K, R = 0.08206
L ∙ atm
mol ∙ K
Example 5.5 Calculate the volume occupied by 0.845 mol of nitrogen gas at a pressure of 1.37
atm and a temperature of 315 K.
Plan: : PV = nRT
or V =
nRT
P
=
0.845x 0.08206 x 315
1.37
Notice all units are correct for the
value of R. V = 15.9 L
Example 5.6 Calculate the number of moles of gas in a 3.24 L basketball inflated to a total
pressure of 24.3 psi at 25 o C.
PV = nRT
We have some bad units. Rearrange to get n. n =
PV
RT
K = 25 + 273 = 298 K
1 atm = 14.7 psi
n=
1.6531 x 3.24
0.08206 x 298
24.3 psi X
1 atm
14.7 psi
= 1.6531 atm
= 0.219 mol
5.5 Applications of the Ideal Gas Law: Molar Volume. Density, and Molar Mass of a Gas
Molar Volume at Standard Temperature and Pressure
The volume occupied by one mole of a substance is called its molar volume.
STP: Standard temperature and pressure are: T = 0 oC or 273 K
V =
nRT
P
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P = 1.00 atm or 760 mmHg
What is the volume of 1.00 mol of gas at STP? V =
1.00 x 0.08206 x 273
1.00
=
22.4 L
Density
Rearrange the equation above:
PV = nRT
Substitute for n since n = mass X
PV= mass X
1 mole
molar mass in g
X RT
1 mole
molar mass in g
or PV =
mass in grams
or molar mass in g/mole
mass in grams
molar mass in g/mole
x RT
Divide both sides by V and group as below:
P=
mass in g
V
or P = D X
X
1 mole
molar mass
1 mole
molar mass in g
X RT
mass
V
X RT
Multiply both sides by molar mass:
D=
Density =
P(molar mass) = DRT
P x molar mass
RT
Example 5.7 Calculate the density of nitrogen gas at 125 o C and a pressure of 755 mmHg.
We have some bad units for this equation. 755 mm of Hg X
K = 125 + 273 = 398 K
D
=
P x molar mass
RT
=
0.993 x 28.02
0.08206 x 398
= 0.852 g/L
Molar Mass of a Gas
PV= mass X
1 mole
molar mass
Rearrange to:
X RT
molar mass =
Mass is in grams.
mass x RT
PV
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1.00 atm
760 mmHg
= 0.993 atm
Example 5.8 A sample of gas has a mass of 0.311 g. Its volume is 0.225 L at a temperature of 55 o C and
a pressure of 886 mmHg. Find its molar mass. Let’s first get the correct units:
P in atm = 886 mmHg X =
K = 55 + 273 = 328 K
molar mass =
760 mmHg
= 1.1658 atm
mass x RT
PV
0.311 x 0.08206 x 328
molar mass =
1 atm
1.1658 x 0.225
= 31.9 g/mol
5.6 Mixtures of Gases and Partial Pressures
The pressure due to any individual component In a gas mixture is its partial pressure, P n .
P1 = n 1 X
RT
P2 = n 2 X
V
Or Pt = P1 + P2 + P3 + ˑˑˑˑˑ
RT
P3 = = n 3 X
V
RT
V
This is called Dalton’s Law of Partial Pressure
Example 5.9 A 1.00 L mixture of helium, neon, and argon has a total pressure of 662 mmHg at 298
K. If the partial pressure of helium is 341 mmHg and the partial pressure of neon is 112 mmHg, what
mass of argon is present in the mixture.
Plan: Use Dalton’s Law to get partial pressure of argon. Use ideal gas law to get mass of argon.
662 mmHg = 341 mmHg + 112 mmHg + PAr
PAr = 209 mmHg
molar mass of Ar = 39.95 g/mol
209 mmHg X
PV = nRT
1 atm
760 mmHg
Solve for n:
0.01125 mol X
39.95 g Ar
1 mol
= 0.275 atm
n=
𝑃𝑉
𝑅𝑇
n=
0.275 x 1
0.08206 𝑥 298
= 0.449 g of argon
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= 0.01125 mol Ar
When the desired product of a chemical reaction is a gas, the gas is often collected by the
displacement of water. Look at the experimental set up Figure 5.14 on page 218. The gas is
always mixed with water vapor by this method because some water molecules evaporate and
mix with the gas. Look at Table 5.4 on page 218 for the vapor pressure of water at various
temperatures.
Example 5.11 In order to determine the rate of photosynthesis, the oxygen gas emitted by an
aquatic plant is collected over water at a temperature of 293 K and a total pressure of 755.2 mmHg.
Over a specific time period, a total of 1.02 L of gas is collected. What mass of oxygen gas in grams is
formed?
Plan: Use Dalton’s Law of Partial Pressure and the ideal gas law, then find grams.
K = C + 273
oC
293 = C + 273
= 20
Temperature in oC is 20.0 o C. From the Table on page 218, the partial pressure of the water is 17.55
torr.
Pt = Poxygen + Pwater
737.65 mmHg X
n=
PV
RT
n=
0.041175 mol X
755.2 = Poxygen + 17.55
1 atm
760 mmHg
Poxygen = 737.65 mmHg
= or 0.97059 atm
0.97059 x 1.02
0.08206 𝑥 293
= 0.041175
32.00 g diatomic oxygen
1 mol
or 4.1175 x 10-2 mol
= 1.32 g O2
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5.8 Kinetic Molecular Theory: A model for Gases
1. The size of a gas particle is negligibly small.
2. The average kinetic energy of a gas particle is proportional to the temperature in kelvins.
3. The collision of one gas particle with another (or with the walls of its container) is completely
elastic. This means that when two particles collide, they may exchange energy, but there is no
overall loss of energy.
Diffusion: The process by which gas molecules spread out in response to a concentration
gradient.
Effusion: The process by which gas escapes from a container into a vacuum through a small
hole.
Graham’s Law of Effusion
rate A
rate B
=√
molar mass B
molar mass A
Where rate is the effusion rate of molecule A and effusion rate of molecule B.
5.10 Real Gases: The Effects of Size and Intermolecular Forces
Under conditions of high pressure the volume of the gas molecules becomes more important
and the gas exhibits nonideal behavior. The ideal gas law: PV = nRT is not valid.
Under conditions of low temperature the attraction among the gas molecules becomes more
important and the gas exhibits nonideal behavior. The ideal gas law is not valid.
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