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Chapter 8 Gases, Liquids, and Solids

Chapter 8 Gases, Liquids, and Solids. 8.1 State of Matter and Their Changes. Solids. Solids have A definite shape. A definite volume. Particles that are close together in a fixed arrangement. Particles that move very slowly. Liquids. Liquids have

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Chapter 8 Gases, Liquids, and Solids

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  1. Chapter 8Gases, Liquids, and Solids

  2. 8.1 State of Matter and Their Changes Solids • Solids have • A definite shape. • A definite volume. • Particles that are close together in a fixed arrangement. • Particles that move very slowly.

  3. Liquids • Liquids have • An indefinite shape, but a definite volume. • The same shape as their container. • Particles that are close together, but mobile. • Particles that move slowly.

  4. Gases • Gases have • An indefinite shape. • An indefinite volume. • The same shape and volume as their container. • Particles that are far apart. • Particles that move fast.

  5. Learning Check Identify each as: 1) solid 2) liquid or 3) gas. ___ A. It has a definite volume, but takes the shape of the container. __ B. Its particles are moving rapidly. __ C. It fills the volume of a container. __ D. It has particles in a fixed arrangement. __ E. It has particles close together that are mobile.

  6. Solution Identify each as: 1) solid 2) liquid or 3) gas. 2 A. It has a definite volume, but takes the shape of the container. 3 B. Its particles are moving rapidly. 3 C. It fills the volume of a container. 1 D. It has particles in a fixed arrangement. 2 E. It has particles close together that are mobile.

  7. 8.2 Gases and the Kinetic- Molecular Theory

  8. Kinetic Theory of Gases Particles of a gas • Move rapidly in straight lines. • Have kinetic energy that increases with an increase in temperature. • Are very far apart. • Have essentially no attractive (or repulsive) forces. • Have very small individual volume compared to the volume of the container they occupy.

  9. Properties of Gases • Gases are described in terms of four properties: pressure (P), volume (V), temperature (T), and amount (n).

  10. Barometer • A barometer measures the pressure exerted by the gases in the atmosphere. • The atmospheric pressure is measured as the height in mm of the mercury column.

  11. Gas Pressure 02 • Units of pressure: atmosphere (atm) Pa (N/m2, 101,325 Pa = 1 atm) Torr (760 Torr = 1 atm) bar (1.01325 bar = 1 atm) mm Hg (760 mm Hg = 1 atm) lb/in2 (14.696 lb/in2 = 1 atm) in Hg (29.921 in Hg = 1 atm) Chapter 09

  12. Learning Check A. The downward pressure of the Hg in a barometer is _____ than/as the weight of the atmosphere. 1) greater 2) less 3) the same B. A water barometer is 13.6 times taller than a Hg barometer (DHg = 13.6 g/mL) because 1) H2O is less dense 2) H2O is heavier 3) air is more dense than H2O

  13. Solution A.The downward pressure of the Hg in a barometer is 3) the same as the weight of the atmosphere. B. A water barometer is 13.6 times taller than a Hg barometer (DHg = 13.6 g/mL) because 1) H2O is less dense

  14. 8.3 Pressure Pressure • A gas exerts pressure, which is defined as a force acting on a specific area. Pressure (P) = Force Area • One atmosphere (1 atm) is 760 mm Hg. • 1 mm Hg = 1 torr1.00 atm = 760 mm Hg = 760 torr

  15. Units of Pressure • In science, pressure is stated in atmospheres (atm), millimeters of mercury (mm Hg), and Pascals (Pa).

  16. Learning Check A. What is 475 mm Hg expressed in atm?1) 475 atm2) 0.625 atm3) 3.61 x 105 atm B. The pressure in a tire is 2.00 atm. What is this pressure in mm Hg?1) 2.00 mm Hg2) 1520 mm Hg3) 22,300 mm Hg

  17. Solution A. What is 475 mm Hg expressed in atm? 2) 0.625 atm 475 mm Hg x 1 atm = 0.625atm 760 mm Hg B. The pressure of a tire is measured as 2.00 atm. What is this pressure in mm Hg? 2) 1520 mm Hg 2.00 atm x 760 mm Hg = 1520 mm Hg 1 atm

  18. 8.4 Boyle’s Law: The Relation Between Volume and Pressure

  19. Boyle’s Law • The pressure of a gas is inversely related to its volume when T and n are constant. • If volume decreases, the pressure increases.

  20. PV Constant in Boyle’s Law • The product P x V remains constant as long as T and n do not change.P1V1 = 8.0 atm x 2.0 L = 16 atm LP2V2 = 4.0 atm x 4.0 L = 16 atm LP3V3 = 2.0 atm x 8.0 L = 16 atm L • Boyle’s Law can be stated as P1V1 = P2V2 (T, n constant)

  21. Temperature and Volume (Charles’ Law)

  22. Charles’ Law • The Kelvin temperature of a gas is directly related to the volume (P and n are constant). • When the temperature of a gas increases, its volume increases.

  23. Charles’ Law V and T • For two conditions, Charles’ Law is written V1 = V2 (P and n constant) T1 T2 • Rearranging Charles’ Law to solve for V2 V2 = V1T2 T1

  24. Learning Check Solve Charles’ Law expression for T2. V1 = V2 T1 T2

  25. Solution V1 = V2 T1 T2 Cross multiply to give V1T2 = V2T1 Isolate T2 by dividing through by V1 V1T2 = V2T1 V1 V1 T2 = V2T1 V1

  26. Learning Check A sample of oxygen gas has a volume of 420 mL at a temperature of 18°C. At what temperature (in °C) will the volume of the oxygen be 640 mL (P and n constant)? 1) 443°C 2) 170°C 3) – 82°C

  27. Solution 2) 170°C T2 = T1V2 V1 T2 = 291 K x 640 mL = 443 K 420 mL = 443 K – 273 K = 170°C

  28. 8.6 Gay-Lussac’s Law: The Relation Between Pressure and Temperature • The pressure exerted by a gas is directly related to the Kelvin temperature of the gas at constant V and n. P1 = P2 T1 T2

  29. Calculation with Gay-Lussac’s Law A gas has a pressure at 2.0 atm at 18°C. What is the new pressure when the temperature is 62°C? (V and n constant) 1. Set up a data table. Conditions 1 Conditions 2 P1 = 2.0 atm P2 = T1 = 18°C + 273 T2 = 62°C + 273 = 291 K = 335 K ?

  30. Calculation with Gay-Lussac’s Law (continued) 2. Solve Gay-Lussac’s Law for P2 P1 = P2 T1 T2 P2 = P1 T2 T1 P2 = 2.0 atm x 335 K = 2.3 atm 291 K

  31. Learning Check Use the gas laws to complete with 1) Increases 2) Decreases A. Pressure _______ when V decreases. B. When T decreases, V _______. C. Pressure _______ when V changes from 12.0 L to 24.0 L. D. Volume _______when T changes from 15.0 °C to 45.0°C.

  32. Solution Use the gas laws to complete with 1) Increases 2) Decreases A. Pressure 1) Increases, when V decreases. B. When T decreases, V 2) Decreases. C. Pressure 2) Decreases when V changes from 12.0 L to 24.0 L D. Volume 1) Increases when T changes from 15.0 °C to 45.0°C

  33. 8.7 The Combined gas Law

  34. Boyle’s law: V a (at constant n and T) Va T T T P P P V = constant x = K 1 P Combined Gas Law Charles’ law: VaT(at constant n and P) K is a constant PV / T= K

  35. Combined Gas Law • The combined gas law uses Boyle’s Law, Charles’ Law, and Gay-Lussac’s Law (n is constant). P1 V1 = P2 V2 T1 T2

  36. Combined Gas Law Calculation A sample of helium gas has a volume of 0.180 L, a pressure of 0.800 atm and a temperature of 29°C. At what temperature (°C) will the helium have a volume of 90.0 mL and a pressure of 3.20 atm (n constant)? 1. Set up Data Table Conditions 1 Conditions 2 P1 = 0.800 atm P2 = 3.20 atm V1 = 0.180 L (180 mL) V2 = 90.0 mL T1 = 29°C + 273 = 302 K T2 = ??

  37. Combined Gas Law Calculation (continued) 2. Solve for T2 P1 V1 = P2 V2 T1 T2 T2 = T1 P2V2 P1V1 T2 = 302 K x 3.20 atm x 90.0 mL = 604 K 0.800 atm 180.0 mL T2 = 604 K – 273 = 331 °C

  38. Learning Check A gas has a volume of 675 mL at 35°C and 0.850 atm pressure. What is the volume(mL) of the gas at –95°C and a pressure of 802 mm Hg (n constant)?

  39. Solution Data Table T1 = 308 K T2 = -95°C + 273 = 178K V1 = 675 mL V2 = ??? P1 = 646 mm Hg P2 = 802 mm Hg Solve for V2 V2 = V1 P1 T2 P2T1 V2 = 675 mL x 646 mm Hg x 178K = 314 mL 802 mm Hg x 308K

  40. 8.8 Avogadro’s law: The Relation Between Volume and Molar Amount Avogadro's Law: Volume and Moles • The volume of a gas is directly related to the number of moles of gas when T and P are constant.V1 = V2n1 n2

  41. Learning Check If 0.75 mole of helium gas occupies a volume of 1.5 L, what volume will 1.2 moles of helium occupy at the same temperature and pressure? 1) 0.94 L 2) 1.8 L 3) 2.4 L

  42. Solution 3) 2.4 L Conditions 1 Conditions 2 V1 = 1.5 L V2 = ??? n1 = 0.75 mole He n2 = 1.2 moles He V2 = V1n2n1V2 = 1.5 L x 1.2 moles He = 2.4 L 0.75 mole He

  43. STP • The volumes of gases can be compared when they have the same conditions of temperature and pressure (STP). Standard temperature (T) 0°C or 273 K Standard pressure (P) 1 atm (760 mm Hg)

  44. Molar Volume • At STP, 1 mole of a gas occupies a volume of 22.4 L. • The volume of one mole of a gas is called the molar volume.

  45. Molar Volume as a Conversion Factor • The molar volume at STP can be used to form conversion factors.22.4 L and 1 mole 1 mole 22.4 L

  46. Learning Check A. What is the volume at STP of 4.00 g of CH4? 1) 5.60 L 2) 11.2 L 3) 44.8 L B. How many grams of He are present in 8.00 L of gas at STP? 1) 25.6 g 2) 0.357 g 3) 1.43 g

  47. Solution A. 1) 5.60 L 4.00 g CH4 x 1 mole CH4 x 22.4 L (STP) = 5.60 L 16.0 g CH4 1 mole CH4 B. 3) 1.43 g 8.00 L x 1 mole He x 4.00 g He = 1.43 g He 22.4 L 1 mole He

  48. 8.9 The Ideal Gas Law _McG 3/4

  49. Boyle’s law: V a (at constant n and T) Va nT nT nT P P P V = constant x = R 1 P Ideal Gas Equation Charles’ law: VaT(at constant n and P) Avogadro’s law: V a n(at constant P and T) R is the gas constant PV = nRT

  50. Universal Gas Constant, R • The universal gas constant, R, can be calculated using the molar volume of a gas at STP. • At STP (273 K and 1.00 atm), 1 mole of a gas occupies 22.4 L. P V R = PV = (1.00 atm)(22.4 L) nT (1 mole) (273K) n T = 0.0821 L atm mole K • Note there are four units associated with R. PV = nRT

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