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Chapter 12 Gases and Gas Laws

Chapter 12 Gases and Gas Laws. The Nature of Gases. Gases expand to fill their containers Gases are fluid – they flow Gases have low density 1/1000 the density of the equivalent liquid or solid Gases are compressible Gases effuse and diffuse. Gases Are Fluids.

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Chapter 12 Gases and Gas Laws

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  1. Chapter 12 Gases and Gas Laws

  2. The Nature of Gases Gases expand to fill their containers Gases are fluid – they flow Gases have low density 1/1000 the density of the equivalent liquid or solid Gases are compressible Gases effuse and diffuse

  3. Gases Are Fluids • Gases are considered fluids. • The word fluidmeans “any substance that can flow.” • Gas particles can flow because they are relatively far apart and therefore are able to move past each other easily.

  4. Gases Have Low Density • Gases have much lower densities than liquids and solids do - WHY? • Because of the relatively large distances between gas particles, most of the volume occupied by a gas is empty space. • The low density of gases also means that gas particles travel relatively long distances before colliding with each other.

  5. Gases are Highly Compressible • Suppose you completely fill a syringe with liquid and try to push the plunger in when the opening is plugged. • You cannot make the space the liquid takes up become smaller. • The space occupied by the gas particles is very small compared with the total volume of the gas. • Applying a small pressure will move the gas particles closer together and will decrease the volume.

  6. Gases Completely Fill a Container • A solid has a certain shape and volume. • A liquid has a certain volume but takes the shape of the lower part of its container. • In contrast, a gas completely fills its container. • Gas particles are constantly moving at high speeds and are far apart enough that they do not attract each other as much as particles of solids and liquids do. • Therefore, a gas expands to fill the entire volume available.

  7. Gas Pressure • Earth’s atmosphere, commonly known as air, is a mixture of gases: mainly nitrogen and oxygen. • As gas molecules are pulled toward the surface of Earth, they collide with each other and with the surface of Earth more often.Collisions of gas molecules are what cause air pressure.

  8. Measuring Pressure Force Newton (N) Pressure = Area m2, cm2 Units of Pressure 1 atm = 760 torr = 101.3 kPa = 760 mmHg Standard Temperature Pressure (STP) 1 atm, 0°C , 22.4 L , 1 mole

  9. 1. Covert 1.00 atm to mmHg 1.00 atm 760 mmHg = 7.60 x 10^2 mmHg 1 atm 2. Covert 3.00 atm to kPa. 3.00atm 101.3 kPa = 304 kPa 1 atm 3. What is 100.0 KPa in atm? 1 atm 100.0 kPa = 0.9872 atm 101.3 kPa

  10. Measuring Pressure Using Barometer • Measures atmospheric pressure • The atmosphere exerts pressure on the surface of mercury in the dish. • This pressure goes through the fluid and up the column of mercury. • The mercury settles at a point where the pressure exerted downward by its weight equals the pressure exerted by the atmosphere.

  11. Kinetic Molecular Theory Particles of matter are ALWAYS in motion Volume of individual particles is  zero. Collisions of particles with container walls cause pressure exerted by gas. Particles exert no forces on each other. Average kinetic energy is proportional to Kelvin, temperature of a gas. Ideal gas- imaginary perfect gas fitting the theory

  12. Measurable Properties of Gases • Gases are described by their measurable properties. • P = pressure exerted by the gas • V = total volume occupied by the gas • T= temperature of the gas • n = number of moles of the gas Units atm L K mol

  13. **Gas Laws – ABCGG LAWS** n is proportional to V @ constant T vogadro’s • A • B • C • G • G P is inversely proportional to V @ constant T oyles’s V is proportional to T @ constant P harles’s ay- Lussac’s P is proportional to T @ constant V Rate of effusion is inversely proportional to square root of gas’s molar mass raham’s

  14. Pressure-Volume Relationship : Boyle’s Law • Pressure and Volume are inversely proportional at constant temperature •  Pressure =  Volume •  Volume =  Pressure PV = k P1V 1= P2V2

  15. Boyle’s Law Calculation A given sample of gas occupies 523mL at 1.00 atm. The pressure is increased to 1.97 atm while the temperature stays the same. What is the new volume of the gas? P1= 1.00 atm P2= 1.97 atm V1= 523 mL V2= ? mL P1V 1= P2V2 P1V1 (1.00 atm) (523 mL) = V2= P2 (1.97 atm) = 265 mL

  16. 1. A sample of oxygen gas has a volume of 150.0mL at a pressure of 0.947 atm. What will the volume of the gas be at a pressure of 1.00 atm if the temperature remains constant? P1= 0.947 atm P2= 1.00 atm V1= 150.0 mL V2= ? mL P1V 1= P2V2 P1V1 (0.974atm) (150.0 mL) V2= = P2 (1.00atm) = 142mL

  17. 2. If 2.5 L of a gas at 110.0 kPa is expanded to 4.0 L at constant temperature, what will be the new value of pressure? P1=110.0 kPa P2= ? kPa V1= 2.5 L V2= 4.0 L P1V 1= P2V2 P1V1 (110.0 kPa) ( 2.5 L) P2= = V2 (4.0 L) = 69 kPa

  18. V1 V2 = T1 T2 Temeperature-Volume Relationship: Charle’s Law • Volume and temperature areproportional at constant pressure •  volume =  temperature (K) •  Volume =  temperature (K)  KE of the gases,  volume @  temperature V = k T

  19. Charles's Law Calculation A balloon is inflated to 665 mL volume at 27°C. It is immersed in a dry-ice bath at −78.5°C. What is its volume, assuming the pressure remains constant? V1= 665 mL V2= ? mL T1= 27°C + 273 K = 300 K T2= -78.5°C + 273 K = 194.5 K V1 V2 = T1 T2 V1 T2 (665 mL)( 194.5 K) V2 = = T1 (300 K) 4.3 x 10^2 mL =

  20. 1. Helium gas in a balloon occupies 2.5 L at 300.0K. The balloon is dipped into liquid nitrogen that is at a temperature of 80.0K. What will be volume of the helium in the balloon at the lower temperature be? V1= 2.5 L V2= ? mL T1= 300 K T2= 80.0 K V1 V2 = T1 T2 T2 V1 (2.5 L)( 80.0 K) V2 = = T1 (300 K) = 0.67 L

  21. 2. A helium filled balloon has a volume of 2.75 L at 20.0 °C . The volume of the balloon changes to 2.46 L when placed outside on a cold day. What is the temperature outside in °C ? V1= 2.75 L V2= 2.46 L T1= 20 °C + 273 K = 293 K T2= ? °C V1 V2 = T1 T2 V2 T1 (2.46 L)( 293 K) T2 = = V1 (2.75 L) = 262.10 K = -10.89 °C= -10.9 °C

  22. P1 P2 = T1 T2 Temperature-Pressure Relationships: Gay-Lussac’s Law • Pressure and temperature areproportional at constant volume •  pressure =  temperature (K) •  pressure =  temperature (K) P = k T

  23. Gay-Lussac’s Law Calculation 1. An aerosol can containing gas at 101 kPa and 22°C is heated to 55°C. Calculate the pressure in the heated can. P1= 101 kPa P2= ? kPa T1= 22 °C + 273 K = 295 K T2= 55 °C + 273K = 328 K P1 P2 = T1 T2 P1 T2 (101 kPa)( 328 K) P2 = = T1 (295 K) = 1.1 x 10^2 kPa

  24. 2. A sample of helium gas is at 122 kPa and 22°C. Assuming constant volume. What will the temperature be when the pressure is 203 kPa? P1= 122 kPa P2= 203 kPa T1= 22 °C + 273 K = 295 K T2= ? K P1 P2 = T1 T2 P2 T1 (203 kPa)(295K) T2 = = P1 (122 kPa) = 4.9 x 10^2 K or 2.2 x10^2 °C

  25. V1 V2 = n1 n2 Volume-Molar Relationships: Avogadro’s Law • Volume and number of moles (n) areproportional at constant temperature and pressure •  volume =  number of moles •  volume =  number of moles • 22.4 L for 1 mole of a gas @ STP V = k n

  26. Avogadro’s Law • What volume of CO2 contains the same number of molecules as 20.0mL of O2 at the same conditions? 20 mL

  27. Gas Laws Combined Gas Law

  28. Molecular Composition of Gases • No gas perfectly obeys all four of these laws under all conditions. • These assumptions work well for most gases and most conditions. • One way to model a gas’s behavior is to assume that the gas is an ideal gas that perfectly follows these laws. • An ideal gas, unlike a real gas, • does not condense to a liquid at low temperatures, • does not have forces of attraction or repulsion between the particles, and is • composed of particles that have no volume.

  29. Ideal Gas Law PV = nRT P = pressure in atm V = volume in liters n = moles R = proportionality constant = 0.0821 L atm/ mol·K T = temperature in Kelvins The combined gas law expresses the relationship between pressure, volume and temperature of a fixed amount of gas.

  30. Ideal Gas Law Calculation How many moles of gas are contained in 22.4 L liter at 100 atm and 283K? PV = nRT P = 100 atm V = 22.4 L n = ? Moles R = 0.0821 L·atm/mol·K T = 283 K PV n = RT (100 atm)(22.4L) = (0.0821 L·atm/mol· K) ( 283 K) =96.4 moles

  31. Calculate the pressure exerted by 43 mol of nitrogen in a 65L of cylinder at 5.0°C. P = ? atm V = 65 L n = 43 mol R = 0.0821 L·atm/mol·K T = 5°C + 273K = 278 K PV = nRT nRT P = V (43 mol)(0.0821 L·atm/mol· K) ( 278 K) = (65 L) =15 atm

  32. What will be the volume of 111 mol of nitrogen where the temperature is -57°C and pressure is 250 atm? P = 250 atm V = ? L n = 111 mol R = 0.0821 L·atm/mol·K T = -57°C + 273K = 216 K PV = nRT nRT V = P (111 mol)(0.0821 L·atm/mol· K) ( 216 K) = (250 atm) =7.9 L

  33. Hot Balloon • Before putting on the balloon, what can you say about the pressure inside the flask and outside the flask? • What happens to the liquid particles when temperature is increased? • What happens to the gas particles when temperature is increased? • What is pressure? • What is the connection between increasing temperature of the gas particles and pressure (inside the flask)? • After liquid boils and evaporates into gas, what can you say about the pressure inside the flask and outside the flask? • What happens when one side has greater pressure than the other? • Now, explain the reaction of the balloon. • Discuss with your partner and write the detailed explanation for extra credit on the test (up to 5pts)

  34. Gas Behavior – Diffusion/Effusion • Diffusion is the movement of particles from regions of higher density to regions of lower density. • The passage of gas particles through a small opening is called effusion.

  35. Effusion

  36. Graham’s Law • The molecular speeds, vAand vB, of gases A and B can be compared according to Graham’s law of diffusion shown below. • Graham’s law of diffusion states that the rate of diffusion of a gas is inversely proportional to the square root of the gas’s molar mass. • Particles of low molar mass travel faster than heavier particles.

  37. Graham’s Law Calculation • At the same temperature, which molecule travels faster O2 or H2? = 3.98 Hydrogen travels 3.98 times faster than oxygen.

  38. Graham’s Law Calculation Oxygen molecules have a rate of about 480 m/s at room temperature. At the same temperature, what is the rate of molecules of sulfur hexafluoride, SF6? rO2 = 480 m/s MO2 = 32g rSF6= ? m/s MSF6= 146g = 230 m/s

  39. Dalton’s Law • The pressure of each gas in a mixture is called the partial pressure. • The total pressure of a mixture of gases is the sum of the partial pressures of the gases. This principle is known as Dalton’s law of partial pressure. • Ptotal = PA+ PB+ PC

  40. Dalton’s Law Calculation • What is the total pressure in a balloon filled with air (O2 & N2) if the pressure of the oxygen is 170 mm Hg and the pressure of nitrogen is 620 mm Hg? • Ptotal = PA + PB + PC….. • Ptotal = POxygen + Pnitrogen • = 170 mmHg + 620 mmHg • = 790 mmHg

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