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Chapter 12: Gas Laws

Chapter 12: Gas Laws. KMT. Objective: To describe the kinetic molecular theory. Properties of Gases. Gases have the following properties: Gases have mass Gases are compressible Gases fill their containers Gases diffuse (spread out) Gases exert pressure. Kinetic Molecular Theory (KMT).

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Chapter 12: Gas Laws

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  1. Chapter 12: Gas Laws

  2. KMT • Objective: To describe the kinetic molecular theory.

  3. Properties of Gases • Gases have the following properties: • Gases have mass • Gases are compressible • Gases fill their containers • Gases diffuse (spread out) • Gases exert pressure

  4. Kinetic Molecular Theory (KMT) • KMT is a theory used to explain the behaviors and characteristics of gases. • The word “kinetic” refers to motion • The work “molecular” refers to molecules

  5. KMT: 3 Basic Assumptions • 1.) A gas is composed of small hard particles with an insignificant volume. • 2.) Gas particles constantly move in straight line paths unless they collide with another particle or the side of the container. • 3.) Gas particles undergo perfectly elastic collisions. • This means that no energy is lost or gained. The total kinetic energy remains constant.

  6. The Variables • Variables necessary to study the behavior of gases include: • Volume • Amount of Gas • Temperature • Pressure

  7. Volume (V) • The volume of the gas is simply the volume of the container it is contained in. • The metric unit of volume is the liter (L)

  8. Amount of Gas (n) • The quantity of as in a given sample expressed in terms of moles of gas.

  9. Temperature (T) • The temperature of a gas is generally measured with a thermometer in degrees Celsius. • All calculations involving gases should be made after converting the degrees Celsius into Kelvin. Kelvin = ⁰C + 273

  10. Temperature Conversions • Convert the following measurements into Kelvin: 1.) 0 ⁰C 2.) 50 ⁰ C • Convert the following measurement into degrees Celsius: 3.) 400 K

  11. Temperature Conversions • Convert the following measurements into Kelvin: 1.) 0 ⁰C = 273 K 2.) 50 ⁰ C = 323 K • Convert the following measurement into degrees Celsius: 3.) 400 K = 127 ⁰C

  12. Pressure (P) • The pressure of a gas is the force exerted on the wall of the container a gas is trapped in. • There are several units for pressure depending on the instrument used to measure it including: • atmospheres (atm) • millimeters of Mercury (mmHG) • kilopascal (kPa) or Pascal (Pa) • torr • psi • bar

  13. Pressure Conversions • We will use the following conversion factors: 1 atm = 101.325 kPa = 101325 Pa = 760 mmHg = 760 torr = 14.7 psi = 1.013 bar

  14. Pressure Conversions 1.) Convert 3 atm into torr.

  15. Pressure Conversions 1.) Convert 3 atm into torr. 3 atm 760 torr = 2280 torr 1 atm

  16. Pressure Conversions 2.) Convert 38 kPa into bar.

  17. Pressure Conversions 2.) Convert 38 kPa into bar. 38 kPa 1.013 bar = 0.38 bar 1-1.325 kPa

  18. Pressure Conversions 3.) Convert 50 mmHg into psi.

  19. Pressure Conversions 3.) Convert 50 mmHg into psi. 50 mmHg 14.7 psi = 0.97 psi 760 mmHg

  20. Standard Temperature and Pressure (STP) • The behavior of a gas depends very strongly on the temperature and the pressure at which the gas is held. • To make it easier to discuss the behavior of gases, it is convenient to designate standard conditions, called STP. • At STP, • Temperature = 273 K • Pressure = 1 atm

  21. Boyle’s Law • Objective: To solve problems using Boyle’s Law.

  22. Boyle’s Law • Robert Boyle discovered the relationship between the pressure and volume of a gas. • He measured the volume of air at different pressures, and observed a pattern which led to his mathematical law. • During his experiments, temperature and amount of gas were held constant. http://en.wikipedia.org/wiki/Robert_Boyle

  23. Boyle’s Law As the pressure increases Volume decreases

  24. Boyle’s Law Volume Pressure  Pressure and Volume are indirectly proportional

  25. Boyle’s Law • Boyle’s Mathematical Law: P1V1 = P2V2

  26. Boyle’s Law: Example • A gas has a volume of 3.0 L at 2 atm. What is its volume at 4 atm?

  27. Boyle’s Law: Example • A gas has a volume of 3.0 L at 2 atm. What is its volume at 4 atm? 1.) Determine which variables you have P1 = 2 atm V1 = 3.0 L P2 = 4 atm V2 = ? 2.) Determine which law is being used P and V = Boyle’s Law P1V1 = P2V2 (2 atm)(3.0L) = (4 atm)V2 3.) Solve for the missing variable V2 = 1.5L

  28. Charles’ Law • Jacques Charles discovered the relationship between temperature and volume. • He measured the volume of air at different temperatures and observed a pattern which led to his mathematical law. • During his experiments, pressure of the system and amount of gas were held constant. http://en.wikipedia.org/wiki/File:Jacques_Alexandre_C%C3%A9sar_Charles.jpg

  29. Charles’ Law

  30. Charles’ Law Volume Temperature  Temperature and Volume are directly proportional

  31. Charles’ Law • Charles’ mathematical law: V1 = V2 T1 T2

  32. Charles’ Law: Example • A gas has a volume of 3.0 L at 127⁰C. What is its volume at 2727⁰C?

  33. Charles’ Law: Example • A gas has a volume of 3.0 L at 127⁰C. What is its volume at 2727⁰C? 1.) Determine which variables you have: T1 = 127⁰C + 273 = 400 K V1 = 3.0L T2 = 227⁰C + 273 = 500 K V2 = ? 2.) Determine which law is being used: V1 = V2 T and V = Charles’ Law T1 T2 3.0 L = V2 3.) Solve for the unknown variable 400K 500 K V2 = 3.8 L

  34. Gay-Lussac’s Law • Gay Lussac discovered the relationship between temperature and pressure of a gas. • He measured the temperature of air at varying pressures and observed a pattern which led to his mathematical law. • During his experiments, volume of the system and amount of gas were held constant. http://en.wikipedia.org/wiki/File:Gaylussac.jpg

  35. Gay-Lussac’s Law Pressure Temperature  Temperature and Pressure are directly proportional

  36. Gay-Lussac’s Law • Gay-Lussac’s mathematical law: P1 = P2 T1 T2

  37. Gay-Lussac’s Law: Example • A gas has a pressure of 3.0 atm at 127⁰C. What is its pressure at 227⁰C?

  38. Gay-Lussac’s Law: Example • A gas has a pressure of 3.0 atm at 127⁰C. What is its pressure at 227⁰C? 1.) Determine which variables you have: T1 = 127⁰C + 273 = 400 K P 3.0 atm T2 = 227⁰C + 273 = 500 K P2 = ? 2.) Determine which law is being used: P1 = P2 T and P = Gay-Lussac’s Law T1 T2 3.0 atm = P2 3.) Solve for the unknown variable 400K 500 K P2 = 3.8 L

  39. Combined Gas Law • Combined Gas Law mathematical equation: P1V1 = P2V2 T1 T2

  40. Combined Gas Law: Example #1 • A sample of helium gas has a volume of 0.180 L, a pressure of 0.800 atm, and a temperature of 302 K. What is the new temperature (K) of the gas at a volume of 90.0 mL and a pressure of 3.20 atm?

  41. Combined Gas Law: Example #1 • A sample of helium gas has a volume of 0.180 L, a pressure of 0.800 atm, and a temperature of 302 K. What is the new temperature (K) of the gas at a volume of 900 mL and a pressure of 3.20 atm? 1.) Determine which variables you have: P1 = 0.800 atm P2 = 3.20 atm V1 = 0.180 L V2 = 90.0 mL = 0.090 L T1 = 302 K T2 = ? 2.) Determine which law is being used: P1V1 = P2V2 P, V, and T = Combined Gas Law T1 T2 (0.800atm)(0.180L) = (3.20atm)(0.090L) (302 K) T2 3.) Solve for the unknown variable T2 = 604 K

  42. Combined Gas Law: Example #2 • A gas has a volume of 0.675 L at 308 K and 0.850 atm pressure. What is the temperature in K when the gas has a volume of 0.315 L and a pressure of 1.055 atm?

  43. Combined Gas Law: Example #2 • A gas has a volume of 0.675 L at 308 K and 0.850 atm pressure. What is the temperature in K when the gas has a volume of 0.315 L and a pressure of 1.055 atm? 1.) Determine which variables you have: P1 = 0.850 atm P2 = 1.055 atm V1 = 0.675 L V2 = 0.315 L T1 = 308 K T2 = ? 2.) Determine which law is being used: P1V1 = P2V2 P, V, and T = Combined Gas Law T1 T2 (0.850 atm)(0.675L) = (1.055m)(0.315L) (308 K) T2 3.) Solve for the unknown variable T2 = 178 K

  44. Avogadro’s Law • Avogadro described the relationship between the volume and amount of a gas. http://www.google.com/imgres

  45. Avogadro’s Law • How does adding more molecules of gas change the volume of the air in a tire? • If a tire has a leak, how does the loss of air (gas) molecules change the volume?

  46. Avogadro’s Law • How does adding more molecules of gas change the volume of the air in a tire? The volume will increase. • If a tire has a leak, how does the loss of air (gas) molecules change the volume? The volume will decrease.

  47. Avogadro’s Law As the amount of gas increases, the volume increases.

  48. Avogadro’s Law Volume Amount of Gas  Volume and Amount of Gas are directly proportional

  49. Avogadro’s Law • Avogadro’s mathematical law: V1 = V2 n1 n2

  50. Avogadro’s Law: Example • A 0.15 mole sample of neon gas has a volume of 3.0 L. What happens to the volume if the amount of neon gas is increased to 3.0 moles?

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