NOTES: Unit 4 - AIR Sections A1 – A8: Behavior of Gases and Gas Laws.
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Transcript of NOTES: Unit 4 - AIR Sections A1 – A8: Behavior of Gases and Gas Laws.
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NOTES: Unit 4 - AIR
Sections A1 – A8: Behavior of Gases and Gas Laws
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BEHAVIOR OF GASES
• Gases have weight
• Gases take up space
• Gases exert pressure
• Gases fill their containers
Gases doing all of these things!
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Kinetic Theory of GasesThe basic assumptions of the kinetic molecular
theory are:
Gases are mostly empty space
The molecules in a gas are separate, very small and very far apart
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Kinetic Theory of GasesThe basic assumptions of the kinetic molecular
theory are:
Gas molecules are in constant, chaotic motion
Collisions between gas molecules are elastic (there is no energy gain or loss)
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Kinetic Theory of GasesThe basic assumptions of the kinetic molecular theory
are: The average kinetic
energy of gas molecules is directly proportional to the absolute temperature
Gas pressure is caused by collisions of molecules with the walls of the container
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Measurements of Gases:
To describe a gas, its volume, amount, temperature, and pressure are measured.• Volume: measured in L, mL, cm3 (1 mL = 1 cm3)
• Amount: measured in moles (mol), grams (g)
• Temperature: measured in KELVIN (K)
K = ºC + 273
• Pressure: measured in mm Hg, torr, atm, etc.
P = F / A (force per unit area)
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Moderate Force (about 100 lbs)
Small Area (0.0625 in2)
Enormous Pressure (1600
psi)P = F /A
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Bed of Nails
Large Surface Area (lots of nails)
Moderate ForceSmall
Pressure
P = F / A
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Units of Pressure:
Units of Pressure: 1 atm = 760 mm Hg 1 atm = 760 torr 1 atm = 1.013 x 105 Pa 1 atm = 101.3 kPa
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A.5 - BOYLE’S LAW:
As P, V and vice versa….
INVERSE RELATIONSHIP
P1V1 = P2V2
For a given number of molecules of gas at a constant temperature, the volume of the gas varies inversely with the pressure.
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Boyle’s Law in action…
• Recall: in the lab, when you added more mass (force), and therefore more pressure, to the syringe the smaller the volume of gas inside the syringe became.
• Example: if the volume of the gas in the syringe were changed to half of its original volume by pushing on the plunger, the pressure of the gas sample would be DOUBLED.
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Boyle’s Law in action…
• Example: If the gas volume in the syringe were reduced to ¼ of its original volume, the gas pressure would be 4 TIMES larger.
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Boyle’s Law and Kinetic Molecular Theory:
• How does kinetic molecular theory explain Boyle’s Law?
gas molecules are in constant, random motion;
gas pressure is the result of molecules colliding with the walls of the container;
as the volume of a container becomes smaller, the collisions over a particular area of container wall increase…the gas pressure INCREASES!
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Pressure-Volume Calculations:
Example: Consider the syringe. Initially, the gas occupies a volume of 8 mL and exerts a pressure of 1 atm.
What would the pressure of the gas become if its volume were increased to 10 mL?
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“Reason and Ratio” method:
• First, reason or predict: If the volume INCREASES from 8 mL to 10 mL, the pressure must DECREASE by a proportional amount.
• To determine the proportional amount, determine the volume ratio (2 possibilities):8 mL OR 10 mL
10 mL 8 mL
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“Reason and Ratio” method:
• Which ratio is correct?
• Your “reasoning” indicates the pressure should DECREASE, so use the ratio that is less than 1: 8 mL / 10 mL
• Now, multiply the original pressure by this volume ratio:
1 atm x (8 mL / 10 mL) = 0.8 atm
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Equation for Boyle’s Law:
• P1V1 = P2V2
• where: P1 = initial pressure
V1 = initial volume
P2 = final pressure
V2 = final volume
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P1V1 = P2V2
• Using the same syringe example, just “plug in” the values:
P1V1 = P2V2
(1 atm) (8 mL) = (P2) (10 mL)
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P1V1 = P2V2
(1 atm) (8 mL) = P2
(10 mL)
P2 = 0.8 atm
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Example: A sample of gas occupies 12 L under a pressure of 1.2 atm. What would its volume be if the pressure were increased to 3.6 atm? (assume temp
is constant)
P1V1 = P2V2
(1.2 atm)(12 L) = (3.6 atm)V2
V2 = 4.0 L
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Example: A sample of gas occupies 28 L under a pressure of 200 kPa. If the volume is decreased to 17 L, what be the new pressure? (assume temp is
constant)
P1V1 = P2V2
(200 kPa)(28 L) = (P2)(17 L)
P2 = 329 kPa
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A.7: Temperature – Volume Relationships
• What happens to matter when it is heated?
• It EXPANDS.
• What happens to matter when it is cooled?
• It CONTRACTS.
• Gas samples expand and shrink to a much greater extent than either solids or liquids.
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Charles’ Law:Jacques Charles (1746-1828)
The volume of a given number of molecules
is directly proportional to the
Kelvin temperature.
As T , V (when P and amt. of gas are constant) and vice versa…. DIRECT RELATIONSHIP
Equation:2
2
1
1
T
V
T
V
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Temperature – Volume Relationship:
• So, doubling the Kelvin temperature of a gas doubles its volume;
• reducing the Kelvin temperature by one half causes the gas volume to decrease by one half…
• WHY KELVIN?
the Kelvin scale never reaches “zero” or has negative values
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Converting Kelvin:
• To convert from Celsius to Kelvin:
add 273.
Example: What is 110 ºC in Kelvin?
110 ºC+ 273 = 383 K
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Converting Kelvin:
• To convert from Kelvin to Celsius:
subtract 273.
Example: 555 K in Celsius?
555 K - 273 = 282 ºC
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Example: A sample of nitrogen gas occupies 117 mL at 100.°C. At what temperature would it occupy 234 mL if the pressure does not change? (express
answer in K and °C)
V1 = 117 mL; T1 = 100 + 273 = 373 K V2 = 234 mL; T2 = ???
V1 / T1= V2 / T2
(117 mL) / (373 K) = (234 mL) / T2
T2 = 746 K (now subtract 273 to get ºC) T2 = 473 ºC
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Example: A sample of oxygen gas occupies 65 mL at 28.8°C. If the temperature is raised to 72.2°C,
what will the new volume of the gas?
V1 = 65 mL; T1 = 28.8 + 273 = 301.8 K V2 = ??? mL; T2 = 72.2 + 273 = 345.2 K
V1 / T1= V2 / T2
(65 mL) / (301.8 K) = (V2) / 345.2 K V2 = (65 mL) (345.2 K) / (301.8 K) V2 = 74.3 mL
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A.8: Temperature – Pressure Relationships
• Picture a closed, rigid container of gas (such as a scuba tank) – the volume is CONSTANT.
• So, what would happen to the
kinetic energy of the gas
molecules in the container if
you were to heat it up?
• How would this affect pressure?
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Temperature – Pressure Relationships:
• Raising the Kelvin temperature of the gas will cause an INCREASE in the gas pressure.
• WHY? • With increasing temperature, the K.E. of
the gas particles increases – they move faster!
• They collide more often and with more energy with the walls of the container…
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Temperature – Pressure Relationships:
• The pressure increases!
• So, as temperature INCREASES, pressure INCREASES, and;
• As temperature DECREASES, pressure DECREASES.
• DIRECT RELATIONSHIP!• Equation:
P1 = P2
T1 T2
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Example: A sample of oxygen gas is in a rigid steel container. The pressure inside the container is 2 atm and the
temperature is 45ºC. If the temperature is cooled to 32ºC, what will be the new pressure inside the container?
P1 = 2 atm; T1 = 45 + 273 = 318 K P2 = ??? mL; T2 = 32 + 273 = 305 K
P1 / T1= P2 / T2
(2 atm) / (318 K) = (P2) / (305 K) V2 = (2 atm) (305 K) / (318 K) V2 = 1.9 atm