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Gases Chapter 3

Gases Chapter 3. REVIEW p.92. Ideal Gases. Ideal gases are imaginary gases that perfectly fit all of the assumptions of the kinetic molecular theory. Gases consist of tiny particles that are far apart relative to their size.

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Gases Chapter 3

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  1. Gases Chapter 3

  2. REVIEW p.92 Ideal Gases Ideal gases are imaginary gases that perfectly fit all of the assumptions of the kinetic molecular theory. • Gases consist of tiny particles that are far apart • relative to their size. • Collisions between gas particles and between particles and the walls of the container are elastic • No kinetic energy is lost in elastic collisions

  3. Ideal Gases(continued) • Gas particles are in constant, rapid motion. They • therefore possess KE (energy of motion) • There are no forces of attraction between gas • particles • The average KE of gas particles depends on T, not on the identity of the particle

  4. Physical Characteristics of Gases

  5. Pressure (P) p.104 • Is caused by the collisions of molecules with the walls of a container • P is defined as the force per unit area on a surface • SI units = Newton/meter2(N/m2) = 1 Pascal (Pa) • N is the force that will increase the speed of a one-kg mass by 1 m/s each second that the force is applied • 1 standard atmosphere = 101,325 Pa = 101.3 kPa • 1 atm = 760 mm Hg = 760 torr

  6. Measuring Pressure The first device for measuring atmospheric P was developed by Evangelista Torricelli during the 17th century. The device was called a “barometer” • Baro= weight • Meter= measure

  7. An Early Barometer The normal pressure due to the atmosphere at sea level can support a column of mercury that is 760 mm high.

  8. Converting Celsius to Kelvin Gas law problems involving T (temperature) require that the T be in KELVINS! Kelvins = C + 273 °C = Kelvins - 273

  9. STP • P = 1 atm (atmosphere, 760 torr, 760 mm Hg, 101.3kPa) • T = 273 K (0°C) • Molar volume of an ideal gas is 22.4 L at STP

  10. REVIEW: 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

  11. The Gas Laws – p. 118 • Describe HOW gases behave. • Can be predicted by the theory. • Amount of change can be calculated with mathematical equations.

  12. Boyle’s Law P is inversely proportional to V when T is held constant. • Pressure ´ Volume = a constant • P1V1 = P2V2 (T = constant) Robert Boyle (1627-1691)

  13. Graph of Boyle’s Law Boyle’s Law says the pressure is inverse to the volume. Note that when the volume goes up, the pressure goes down P1V1 = P2V2 (T = constant)

  14. Changing the Container Size • In a smaller container molecules have less room to move. • Thus, hit the sides of the container more often. • As V decreases, P increases.

  15. 1 atm • As the pressure on a gas increases 4 Liters

  16. 2 atm • the volume decreases • Gas particles are closer together, shorter distance to go before they impact the containers wall • More molecule impacts per unit time increase in P • P and V are inversely related 2 Liters

  17. - Page 419

  18. Example: A 250.0 mL sample of Cl2 is collected when the barometric pressure is 105.2 kPa. What is the volume of the sample after the barometer drops to 100.3 kPa? Given: P1 = 105.2 kPa Find: V2 = ? V1 = 250.0 mL P2 = 100.3 kPa Use Boyles Law: P1 V1 = P2 V2 Set up for Unknown: V2 = P1 V1 P2 Solve: V2 = 105.2 kPa x 250.0 mL 100.3 kPa = 262.2 mL

  19. Charles’s Law • The V of a gas is directly proportional to T, and extrapolates to zero at zero Kelvin (-273.15°C). (P = constant). See P. 371, Fig. 8 • Charles’s experiments showed that all gases expand to the same extent when heated. Jacques Charles (1746-1823). Isolated boron and studied gases. Balloonist. Temperature MUST be in KELVINS!

  20. Temperature • Raising the T of a gas increases the P, if the V is held constant. • The molecules hit the walls harder, more often.

  21. 300 K • If you start with 1 liter of gas at 1 atm pressure and 300 K • and heat it to 600 K one of 2 things happens

  22. 600 K 300 K • Either the volume will increase to 2 liters at 1 atm Charles’s Law

  23. 600 K 300 K Gay Lussac’s Law • Or the pressure will increase to 2 atm. • Or someplace in between

  24. T1 = 309 K Example: A balloon is inflated with 6.22 L of helium at a temperature of 36 °C. What is the volume of the balloon when the temperature is 22°C? Given: V1 = 6.22 L Find: V2 = ? T1 = 36 °C T1 = ? K T2 = 22 °C T2 = ? K T2 =295 K T1 = Kelvin = C + 273 = 36 °C + 273 = 309 T2 =Kelvin = C + 273 = 22 °C + 273 = 295 V1 V2 T1 T2 6.22L 309K V1 T1 = 295K V2 = = T2 = 5.94 L

  25. Gay-Lussac’s Law The pressure and temperature of a gas are directly related, provided that the volume remains constant. Temperature MUST be in KELVINS! (1778-1850)

  26. Summary of the Named Gas-Laws:

  27. The Combined Gas Law The combined gas law expresses the relationship between P, V and T of a fixed amount of gas. Temperature MUST be in KELVINS!

  28. The combined gas law contains all the other gas laws! • If the temperature remains constant... • Meaning T1 = T2 P1 V1 P2 x V2 x = T1 T2 Boyle’s Law

  29. The combined gas law contains all the other gas laws! • If the pressure remains constant... P1 V1 P2 x V2 x = T1 T2 Charles’s Law

  30. The combined gas law contains all the other gas laws! • If the volumeremains constant... P1 V1 P2 x V2 x = T1 T2 Gay-Lussac’s Law

  31. Unknown: P2 = ? T1 = ? K T2 = ? K Example: A student collects 285 mL of O2 gas at a temperature of 15°C and a pressure of 99.3 kPa. The next day, the same sample occupies 292 mL at a temperature of 11°C. What is the new pressure of the gas? Known: V1 = 285 mL T1 = 15°C P1 = 99.3 kPa V2 = 292 mL T2 = 11°C T1 = Kelvin = C + 273 = 15°C + 273 = 288 K T2 = Kelvin = C + 273 = 11°C + 273 = 284 K

  32. Unknown: P2 = ? T1 = 288 K T2 = 284 K Example: A student collects 285 mL of O2 gas at a temperature of 15°C and a pressure of 99.3 kPa. The next day, the same sample occupies 292 mL at a temperature of 11°C. What is the new pressure of the gas? Known: V1 = 285 mL T1 = 15°C P1 = 99.3 kPa V2 = 292 mL T2 = 11°C T2P1V1 T1V2 P2 =

  33. Unknown: P2 = ? T1 = 288 K T2 = 284 K Example: A student collects 285 mL of O2 gas at a temperature of 15°C and a pressure of 99.3 kPa. The next day, the same sample occupies 292 mL at a temperature of 11°C. What is the new pressure of the gas? Known: V1 = 285 mL T1 = 15°C P1 = 99.3 kPa V2 = 292 mL T2 = 11°C T2P1V1 T1V2 (284 K)(99.3 kPa)(285 mL) (288 K)(292 mL) P2 = = = 95.6 kPa

  34. The effect of adding gas. • When we blow up a balloon we are adding gas molecules. • Doubling the number of gas particles doubles the pressure. (of the same V, at the same T)

  35. If you double the number of molecules 1 atm

  36. If you double the number of molecules • You double the pressure. 2 atm

  37. 4 atm • As you remove molecules from a container

  38. 2 atm • As you remove molecules from a container the pressure decreases

  39. 1 atm • As you remove molecules from a container the pressure decreases • Until the pressure inside equals the pressure outside • Molecules naturally move from high to low pressure

  40. Pressure and the number of molecules are directly related. • More molecules means more collisions. • Fewer molecules means fewer collisions. • Gases naturally move from areas of high P to low P (because of empty space to move in).

  41. Think of it terms of pressure. • The same P at the same T should require that there be the same number of particles. • The smaller particles must have a greater average speed to have the same KE.

  42. Measuring & Comparing the Volumes of Reacting Gases Gay-Lussac’s law of combining volumes of gases states that at constant T and P, the volumes of gaseous reactants and products can be expressed as ratios of small whole numbers • Examples: • 1 liter of nitrogen gas reacts with 3 liters of hydrogen gas to produce 2 liters of ammonia gas N2(g) + 3H2(g) -----> 2NH3(g) • Since ALL the reactants and products are gases, the mole ratio of N2(g):H2(g):NH3(g) of 1:3:2 is also the ratio of the volumes of gases • Therefore, 10mL of N2 gas would react with 10 x 3 = 30mL of H2 gas to produce 10 x 2 = 20mL NH3 gas

  43. Avogadro’s Law • For a gas at constant T and P, the volume is directly proportional to the number of moles of gas • or…at the same T and P equal V have equal # of molecules, regardless of what type of gas they contain. V=an a = proportionality constant V = volume of the gas n = number of moles of gas

  44. Avogadro’s Hypothesis 2 Liters of Helium • Has the same number of particles as .. 2 Liters of Oxygen

  45. This is where we get22.4 L =1 mole • Only at STP • 273K or 0ºC • 1 atm • This way we compare gases at the same T and P.

  46. Standard Molar Volume– p. 141 Equal volumes of all gases at the same temperature and pressure contain the same number of molecules. - Amadeo Avogadro Standard Molar Volume – the volume occupied by one mole of a gas at STP

  47. Standard Molar Volume One mole of any gas will occupy the same volume as one mole of any other gas (at the same T and P), despite mass differences!

  48. Practice Problem • At STP, what is the volume of 7.08 mol of N2 gas? 22.4 L N2 7.08 mol N2 = 159 L N2 1 mol N2

  49. Dalton’s Law of Partial Pressures- p. 144 For a mixture of gases in a container, PTotal = P1 + P2 + P3 + . . . This is particularly useful in calculating the pressure of gases collected over water.

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