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NOTES: Unit 4 - AIR Sections A1 – A8: Behavior of Gases and Gas Laws

NOTES: Unit 4 - AIR Sections A1 – A8: Behavior of Gases and Gas Laws

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Page 1: NOTES: Unit 4 - AIR Sections A1 – A8: Behavior of Gases and Gas Laws

NOTES: Unit 4 - AIR

Sections A1 – A8: Behavior of Gases and Gas Laws

Page 2: NOTES: Unit 4 - AIR Sections A1 – A8: Behavior of Gases and Gas Laws

BEHAVIOR OF GASES

• Gases have weight

• Gases take up space

• Gases exert pressure

• Gases fill their containers

Gases doing all of these things!

Page 3: NOTES: Unit 4 - AIR Sections A1 – A8: Behavior of Gases and Gas Laws

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

Page 4: NOTES: Unit 4 - AIR Sections A1 – A8: Behavior of Gases and Gas Laws

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)

Page 5: NOTES: Unit 4 - AIR Sections A1 – A8: Behavior of Gases and Gas Laws

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

Page 6: NOTES: Unit 4 - AIR Sections A1 – A8: Behavior of Gases and Gas Laws

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)

Page 7: NOTES: Unit 4 - AIR Sections A1 – A8: Behavior of Gases and Gas Laws

Moderate Force (about 100 lbs)

Small Area (0.0625 in2)

Enormous Pressure (1600

psi)P = F /A

Page 8: NOTES: Unit 4 - AIR Sections A1 – A8: Behavior of Gases and Gas Laws

Bed of Nails

Large Surface Area (lots of nails)

Moderate ForceSmall

Pressure

P = F / A

Page 9: NOTES: Unit 4 - AIR Sections A1 – A8: Behavior of Gases and Gas Laws

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

Page 10: NOTES: Unit 4 - AIR Sections A1 – A8: Behavior of Gases and Gas Laws

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.

Page 11: NOTES: Unit 4 - AIR Sections A1 – A8: Behavior of Gases and Gas Laws

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.

Page 12: NOTES: Unit 4 - AIR Sections A1 – A8: Behavior of Gases and Gas Laws

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.

Page 13: NOTES: Unit 4 - AIR Sections A1 – A8: Behavior of Gases and Gas Laws

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!

Page 14: NOTES: Unit 4 - AIR Sections A1 – A8: Behavior of Gases and Gas Laws
Page 15: NOTES: Unit 4 - AIR Sections A1 – A8: Behavior of Gases and Gas Laws

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?

Page 16: NOTES: Unit 4 - AIR Sections A1 – A8: Behavior of Gases and Gas Laws

“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

Page 17: NOTES: Unit 4 - AIR Sections A1 – A8: Behavior of Gases and Gas Laws

“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

Page 18: NOTES: Unit 4 - AIR Sections A1 – A8: Behavior of Gases and Gas Laws

Equation for Boyle’s Law:

• P1V1 = P2V2

• where: P1 = initial pressure

V1 = initial volume

P2 = final pressure

V2 = final volume

Page 19: NOTES: Unit 4 - AIR Sections A1 – A8: Behavior of Gases and Gas Laws

P1V1 = P2V2

• Using the same syringe example, just “plug in” the values:

P1V1 = P2V2

(1 atm) (8 mL) = (P2) (10 mL)

Page 20: NOTES: Unit 4 - AIR Sections A1 – A8: Behavior of Gases and Gas Laws

P1V1 = P2V2

(1 atm) (8 mL) = P2

(10 mL)

P2 = 0.8 atm

Page 21: NOTES: Unit 4 - AIR Sections A1 – A8: Behavior of Gases and Gas Laws

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

Page 22: NOTES: Unit 4 - AIR Sections A1 – A8: Behavior of Gases and Gas Laws

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

Page 23: NOTES: Unit 4 - AIR Sections A1 – A8: Behavior of Gases and Gas Laws

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.

Page 24: NOTES: Unit 4 - AIR Sections A1 – A8: Behavior of Gases and Gas Laws

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

Page 25: NOTES: Unit 4 - AIR Sections A1 – A8: Behavior of Gases and Gas Laws
Page 26: NOTES: Unit 4 - AIR Sections A1 – A8: Behavior of Gases and Gas Laws
Page 27: NOTES: Unit 4 - AIR Sections A1 – A8: Behavior of Gases and Gas Laws

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

Page 28: NOTES: Unit 4 - AIR Sections A1 – A8: Behavior of Gases and Gas Laws
Page 29: NOTES: Unit 4 - AIR Sections A1 – A8: Behavior of Gases and Gas Laws

Converting Kelvin:

• To convert from Celsius to Kelvin:

add 273.

Example: What is 110 ºC in Kelvin?

110 ºC+ 273 = 383 K

Page 30: NOTES: Unit 4 - AIR Sections A1 – A8: Behavior of Gases and Gas Laws

Converting Kelvin:

• To convert from Kelvin to Celsius:

subtract 273.

Example: 555 K in Celsius?

555 K - 273 = 282 ºC

Page 31: NOTES: Unit 4 - AIR Sections A1 – A8: Behavior of Gases and Gas Laws

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

Page 32: NOTES: Unit 4 - AIR Sections A1 – A8: Behavior of Gases and Gas Laws

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

Page 33: NOTES: Unit 4 - AIR Sections A1 – A8: Behavior of Gases and Gas Laws

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?

Page 34: NOTES: Unit 4 - AIR Sections A1 – A8: Behavior of Gases and Gas Laws

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…

Page 36: NOTES: Unit 4 - AIR Sections A1 – A8: Behavior of Gases and Gas Laws

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

Page 37: NOTES: Unit 4 - AIR Sections A1 – A8: Behavior of Gases and Gas Laws

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