Gas Behavior

1
Gas Behavior
Chapter 12
2
The Beginning

• The first gas studied was air.
• The studies were very important to understanding
  gas behavior because
• Air is a mixture of gases.
• It still behaved as one gas.
• Soall gases, under similar conditions, behave
  similarly

3
Kinetic-Molecular Theory of Gases

• Assumes gas particles are in constant motion
• Used to explain gas behavior
• Based on five assumptions

4
Kinetic-Molecular Theory of Gases

• Assumption 1
• Gases consist of large numbers of tiny particles
  whose sizes are negligible in comparison to their
  distance from each other.
• Translation Gas molecules are very small and
  far apart, giving the gas a low density and the
  property of compressibility

5
Kinetic-Molecular Theory of Gases

• Assumption 2
• Collisions between gas particles and collisions
  between gas particles and the walls of the
  container are elastic.
• Translation No kinetic energy is lost in the
  collisions it is transferred

6
Kinetic-Molecular Theory of Gases

• Assumption 3
• Particles of gas are continually, rapidly, and
  randomly moving, thereby possessing kinetic
  energy.
• Translation Gas particles are not attracted to
  each other because their kinetic energy is too
  strong gas particles never stop they are always
  moving

7
Kinetic-Molecular Theory of Gases

• Assumption 4
• Gas particles have no forces of attraction or
  repulsion between them.
• Translation Gas particles will not stick
  together when they collide they bounce off each
  other

8
Kinetic-Molecular Theory of Gases

• Assumption 5
• The temperature of the gas affects the average
  kinetic energy of gas particles.
• Translation Temperature is a measure of average
  kinetic energy higher temp.higher KE

9
Kinetic-Molecular Theory of Gases

• The KMT only applies to ideal gasestheoretical
  gases
• Other gases are called real gases
• Real gases behave almost ideally when the
  pressure is not too high and/or the temperature
  is not too low

10
Kinetic Energy (KE)

• The energy of motion
• KE ½ mv2
• m mass
• v speed

11
Try This

• A 68 kg track runner is running at 10 m/s and a
  136 kg football player is running at 5 m/s.
  Which has more kinetic energy?
• KE ½ (68 kg)(10 m/s)2 3400 J
• KE ½ (136)(5 m/s)2 1700 J
• Track runner

12
Volume

• Volume is measured in
• Liters (L)
• Milliliters (mL)
• Cubic centimeters (cm3 or cc)

13
Pressure

• Pressure is measured in
• Atmospheres (atm)
• Millimeters of mercury (mm Hg)
• Kilopascals (kPa)
• Torricelli (torr)
• Pounds per square inch (psi)

14
Temperature

• Temperature is measured in
• Celsius (C)
• Kelvin (K)
• Fahrenheit (F)

15
Standard Temperature and Pressure

• 0C or 273 K
• 1 atm or 760 mm Hg or 760 torr or 101.3 kPa

16
The Gas Laws

• Developed to help create relationships between
  volume, pressure, temperature, and amount of a
  gas.
• First person to collect and analyze a gasRobert
  Boyle

17
The Gas Laws

• Boyles Law volume of a fixed mass of gas
  varies inversely with the pressure at constant
  temperature
• Translation When pressure (P) increases, volume
  (V) decreases. When P decreases, V increases
• Mathematically P1V1P2V2

18
Important!

• Units must be the same thing. For instance if P1
  is in mm of Hg then P2 must also be measured in
  mm of Hg.
• If V1 is measured in mL then V2 must also be
  measured in mL.

19
The Gas Laws

• Boyles Law Problem
• A gas at a pressure of 608 mm Hg is held in a
  container with a volume of 545 L. If the volume
  of the container is increased to 1065 L, and
  temperature is held constant, what is the new
  pressure of the gas?

20
The Gas Laws

• Boyles Law Solution
• Have P1608 mm Hg
• V1545 L
• V21065 L
• Want P2
• Use Boyles Law Equation P1V1P2V2

21
The Gas Laws

• Boyles Law Solution
• P1V1 P2V2
• (608 mm Hg)(545 L) (P2)(1065 L)
• 311 mm Hg

22
Try this!

• A high- altitude balloon contains 30.0 L of
  helium gas at 103 kPa. What is the volume when
  the balloon rises to an altitude where the
  pressure is only 25.0 kPa? (Assume that the
  temperature remains constant)

23
Answer

• P1V1 P2V2
• (103 kPa)(30.0 L) (25.0 kPa)(V2)
• 124 L

24
Have you ever had the experience of buying a
helium filled balloon and then taking it outside
on a cold day? If you have you noticed that the
balloon shrunk and looked like there was not
enough helium put in it. However if you ever put
a helium balloon in your car on a HOT day you
will return to find that the balloon
hasexploded. Why do these things happen?
25
The Gas Laws

• Charles Lawrelationship between gas temperature
  and volume
• the volume of a fixed mass of gas at constant
  pressure varies directly with the Kelvin
  temperature
• Translation when V increases, T increases, and
  when V decreases, T decreases
• Mathematically

26
Important!

• The Kinetic Theory of Gases states that the
  kinetic energy of a gas is proportional to its
  temperature and using a Celsius scale would cause
  the kinetic energy of a gas to be negativewhich
  is impossible!
• Change the temperature to Kelvin!!!!

27
Rememberall temperatures must be in Kelvin

• K ºC 273

28
The Gas Laws

• Charles Law Problem
• A sample of neon gas occupies a volume of 752 mL
  at 25ºC. What volume will the gas occupy at 50ºC
  if the pressure remains constant?
• Rememberthe gas laws will be applied all
  temperatures must be in Kelvin!

29
The Gas Laws

• Charles Law Solution
• Have V1752 mL
• T125ºC 273298 K
• T250ºC 273323 K
• Want V2 in mL
• Use
• Solve 752 mL V2
• 298 K 323 K
• 815 mL

30
Try This!

• A balloon inflated in a room at 24C has a volume
  of 4.00 L. The balloon is then heated to a
  temperature of 58C. What is the new volume if
  the pressure remains constant?

31
Answer

• V1 V2
• T1 T2
• T1 24C 273 297 K
• T2 58C 273 331 K
• 4.00 L V2
• 297 K 331 K
• 4.46 L

32
The Gas Laws

• Gay-Lussacs Lawrelationship between gas
  pressure and temperature
• pressure of a fixed mass of gas at a constant
  volume varies directly with the Kelvin
  temperature
• Translation When pressure increases, temp.
  increases when pressure decreases, temp.
  decreases
• Mathematically

33
The Gas Laws

• Gay-Lussacs Law Problem
• The gas in an aerosol can is at a pressure of
  3.00 atm at 25ºC. Directions on the can warn the
  user not to keep the can in a place where the
  temperature exceeds 52ºC. What would the gas
  pressure in the can be at 52ºC?
• Remember to convert temperatures to Kelvin!!

34
Answer

• P1 P2
• T1 T2
• T1 25C 273 298 K
• T2 52C 273 325 K
• 3.00 atm P2
• 298 K 325 K
• 3.27 atm

35
Try This!

• A gas has a pressure of 6.58 kPa at 539 K. What
  will be the pressure at 211 K if the volume does
  not change?

36
Answer

• P1 P2
• T1 T2
• 6.58 kPa P2
• 539 K 211 K
• 2.58 kPa

37
The Gas Laws

• Combined Gas Law combination of all three laws
• Mathematically

38
The volume of a gas -filled balloon is 30.0 L at
40C and 153 kPa pressure. What volume will the
balloon have at standard temperature and
pressure?
Standard Pressure 760 torr 1 atm
101.3kPa Standard Temp. 273 K
39
Answer

• P1V1 P2V2
• T1 T2
• T1 40C 273 313 K
• T2 0C 273 273 K
• (153 kPa)(30.0 L) (101.3 kPa)(V2)
• 313 K 273 K
• 39.5 L

40
The Gas Laws

• Daltons Law of Partial Pressures the total
  pressure of a mixture of gases is the sum of the
  pressures of each individual gas
• Translation the sum of the parts equals the
  whole
• Mathematically PT P1 P2 P3

41

• Air contains oxygen, nitrogen , carbon dioxide
  and trace amounts of other gases. What is the
  partial pressure of oxygen (PO2) at 101.3 kPa of
  total pressure if the partial pressure of
  nitrogen,carbon dioxide and other gases are 79.10
  kPa, 0.040 kPa and 0.94 kPa respectively.

42
Answer

• PT P1 P2 P3
• 101.3 kPa 79.10 kPa 0.040 kPa 0.94 kPa
  PO2
• PO2 21.2 kPa

43
Ideal Gas Law

• Gases behave differently under different
  circumstances (each gas has a different molar
  mass)
• Use term ideal gas to describe gas behavior
  under all circumstances
• No such thing as ideal gasthey are real gases
• In reality gases can be liquefied and sometimes
  solidified by cooling and by applying pressure
  whereas ideal gases cannot be. So real gases do
  not behave like ideal gases under high pressures
  and at low temperatures.

44
Ideal Gas Law

• PV nRT
• P is pressure may be labeled kPa, atm, mm Hg, or
  torr
• V is volume must be labeled L
• n is moles

45
Ideal Gas Law Continued

• R is a constant whose value is determined by P.
• If P is labeled kPa ? R 8.314
• If P is labeled atm ? R 0.0821
• If P is labeled mm Hg or torr ? R 62.4
• T is temperature must be labeled K

46
Try This!

• You fill a rigid steel cylinder that has a volume
  of 20.0 L with nitrogen gas to a final pressure
  of 150049 torrs at 28C. How many moles of
  nitrogen gas does the cylinder contain?
• Convert Temperature to Kelvin!

47
Answer

• PV nRT
• 28C 273 301 K
• (150049 torr)(20.0 L) n (62.4)(301 K)
• 160. moles of N2

48
Try This!

• What volume is occupied by 5.03 g of hydrogen gas
  at 28C and a pressure of 2.0 atm?
• Hint Convert grams to moles and C to K!

49
Answer

• PV nRT
• 5.03 g 2.0 g/mol 2.515 mol H2
• 28C 273 301 K
• (2 atm)(V) (2.515 mol)(0.0821)(301 K)
• 31 L

50
Ideal Gas Law

• Finding the molar mass
• M mRT
• PV
• M molar mass and m grams
• What is the molar mass of a gas if 372 ml have a
  mass of 0.920 grams at 100ºC and 108 kPa of
  pressure?

51
Answer

• M mRT
• PV
• What is the molar mass of a gas if 372 ml have a
  mass of 0.920 grams at 100ºC and 108 kPa of
  pressure?
• 372 mL 1000 0.372 L
• 100C 273 373 K
• M (0.920 g)(8.314)(373 K)
• (108 kPa)(0.372 L)
• 71.0 g/mol

52
Try This!

• A container holds 2240 L of methane gas (CH4) at
  a pressure of 1.50 kPa and a temperature of 42C.
  How many grams of CH4 does this container hold?

53
Answer

• M mRT
• PV
• 42C 273 315 K
• CH4 12.0 4.0 16.0 g/mol
• 16.0 g.mol (x g)(8.314)(315 K)
• (1.50 kPa)(2240L)
• 20.5 g

54
Diffusion and Effusion

• Diffusion is the gradual mixing of gases due to
  the random, spontaneous motion of the gas
  particles
• Effusion is the process by which gas molecules
  trapped in a container randomly pass through tiny
  openings in the container
• What are everyday examples of diffusion or
  effusion?

55
Examples

• Diffusion perfume spreading, smelling cooking
  food, and smell something burning
• Effusion tire puncture and a pin hole in a
  balloon

56
Diffusion and Effusion

• Rates of diffusion/effusion depends on the
  velocity of the molecules
• Velocity depends on temperature and mass
• Would hot or cold particles move faster?
• Would heavy particles move slower or faster?

57
Diffusion and Effusion

• Grahams Lawrelationship between rate of
  effusion (diffusion) and molar mass
• the rate of effusion of gases at the same
  temperature and pressure are inversely
  proportional to the square root of the molar mass
  
• Mathematically

58
Graham noticed that gases of lower molar mass
effuse faster than gases of higher molar mass.

• Nitrogen effuses at 535 m/s. How much faster
  will helium gas effuse?
• 535 m/s v4
• x m/s v28
• 1415 m/s 535 m/s
• 2.6 times faster