Understanding Matter and Its Phases: Gases, Liquids, and Solids

1. Fundamentals of General, Organic, and Biological Chemistry 5th Edition Chapter Eight Gases, Liquids, and Solids James E. Mayhugh Oklahoma City University 2007 Prentice Hall, Inc.

2. Outline • 8.1 States of Matter and Their Changes • 8.2 Gases and the Kinetic–Molecular Theory • 8.3 Pressure • 8.4 Boyle’s Law: The Relation Between Volume and Pressure • 8.5 Charles’s Law: The Relation Between Volume and Temperature • 8.6 Gay-Lussac’s Law: The Relation Between Pressure and Temperature • 8.7 The Combined Gas Law • 8.8 Avogadro’s Law: The Relation Between Volume and Molar Amount • 8.9 The Ideal Gas Law • 8.10 Partial Pressure and Dalton’s Law • 8.11 Intermolecular Forces • 8.12 Liquids • 8.13 Water: A Unique Liquid • 8.14 Solids • 8.15 Changes of State Chapter Eight

3. 8.1 States of Matter and Their Changes • Matter exists in any of three phases, or states—solid, liquid, and gas, depending on the attractive forces between particles, temperature, and pressure. • In a gas, the attractive forces between particles are very weak compared to their kinetic energy, so the particles move about freely, are far apart, and have almost no influence on one another. • In a liquid, the attractive forces between particles are stronger, pulling the particles close together but still allowing them considerable freedom to move about. Chapter Eight

4. In a solid, the attractive forces are much stronger than the kinetic energy of the particles, so the atoms, molecules, or ions are held in a specific arrangement and can only wiggle around in place. Chapter Eight

5. Phase change or change of state: The transformation of a substance from one state to another. • Melting point (mp): The temperature at which solid and liquid are in equilibrium. • Boiling point (bp):The temperature at which liquid and gas are in equilibrium. • Sublimation: A process in which a solid changes directly to a gas. • Melting, boiling, and sublimation all have H > 0, and S > 0. This means they are nonspontaneous below and spontaneous above a certain temperature. Chapter Eight

6. Chapter Eight

7. 8.2 Gases and the Kinetic-Molecular Theory • The behavior of gases can be explained by a group of assumptions known as the kinetic–molecular theory of gases. The following assumptions account for the observable properties of gases: • A gas consists of many particles, either atoms or molecules, moving about at random with no attractive forces between them. Because of this random motion, different gases mix together quickly. Chapter Eight

8. The amount of space occupied by the gas particles themselves is much smaller than the amount of space between particles. Most of the volume taken up by gases is empty space, accounting for the ease of compression and low densities of gases. • The average kinetic energy of gas particles is proportional to the Kelvin temperature. Thus, gas particles have more kinetic energy and move faster as the temperature increases. (In fact, gas particles move much faster than you might suspect. The average speed of a helium atom at room temperature and atmospheric pressure is approximately 1.36 km/s, or 3000 mi/hr, nearly that of a rifle bullet.) Chapter Eight

9. Collisions of gas particles, either with other particles or with the wall of their container, are elastic; that is, the total kinetic energy of the particles is constant. The pressure of a gas against the walls of its container is the result of collisions of the gas particles with the walls. The number and force of collisions determines the pressure. • A gas that obeys all the assumptions of the kinetic–molecular theory is called an ideal gas. All gases behave somewhat differently than predicted by the kinetic–molecular theory at very high pressures or very low temperatures. Most real gases display nearly ideal behavior under normal conditions. Chapter Eight

10. 8.3 Pressure • Pressure (P) is defined as a force (F) per unit area (A) pushing against a surface; P = F/A. • A barometer measures pressure as the height of a mercury column. Atmospheric pressure presses down on mercury in a dish and pushes it up a tube. • Pressure units: 1 atm = 760 mm Hg = 14.7 psi = 101,325 Pa 1 mm Hg = 1 torr = 133.32 Pa Chapter Eight

11. Gas pressure inside a container is often measured using an open-end manometer, a simple instrument similar in principle to the mercury barometer. Chapter Eight

12. 8.4 Boyle’s Law: The Relation Between Volume and Pressure • Boyle’s law: The volume of a gas is inversely proportional to its pressure for a fixed amount of gas at a constant temperature. That is, P times V is constant when the amount of gas n and the temperature T are kept constant. • V 1/P or PV = k if n and T are constant • If: P1V1 = k and P2V2 = k • Then: P1V1 = P2V2 Chapter Eight

13. The volume of a gas decreases proportionately as its pressure increases. If the pressure of a gas sample is doubled, the volume is halved. Chapter Eight

14. Graph (a) demonstrates the decrease in volume as pressure increases, whereas graph (b) shows the linear relationship between V and 1/P. Chapter Eight

15. Problem 8.41 The volume of a balloon is 2.85 L at 1.00 atm. What pressure is required to compress the balloon to a volume of 1.70 L? Use Boyle’s Law: P1V1 = P2V2 now, solve for P2 divide both sides by V2 P1V1/V2 = P2 Chapter Eight

16. P2 = P1V1/V2 P2 = (1.00 atm)(2.85 L)/(1.70 L) = 1.68 atm Chapter Eight

17. 8.5 Charles’ Law: The Relation Between Volume and Temperature • Charles’s law: The volume of a gas is directly proportional to its Kelvin temperature for a fixed amount of gas at a constant pressure. That is, V divided by T is constant when n and P are held constant. • V T or V/T = k if n and P are constant • If: V1/T1 = k and V2/T2 = k • Then: V1/T1 = V2/T2 Chapter Eight

18. If the Kelvin temperature of a gas is doubled, its volume doubles. Chapter Eight

19. As the temperature goes up, the volume also goes up. Chapter Eight

20. Problem 8.47 A hot air balloon has a volume of 875 L. What is the original temperature of the balloon if its volume changes to 955 L when heated to 56oC? Charles’ Law: V1/T1 = V2/T2now, solve for T1 divide both sides by V1 1/T1 = V2/V1T2 take reciprocal of both sides T1 = V1T2/V2 Chapter Eight

21. Make sure to use Kelvin degrees T1 = V1T2/V2 T2 = 56oC + 273 = 329 K T1 = (875 L)(329K)/(955 L) = 301K convert to oC (subtract 273) 28oC Chapter Eight

22. 8.6 Gay-Lussac’s Law: The Relation Between Pressure and Temperature • Gay-Lussac’s law: The pressure of a gas is directly proportional to its Kelvin temperature for a fixed amount of gas at a constant volume. That is, P divided by T is constant when n and V are held constant. • P T or P/T = k if n and V are constant • If: P1/T1 = k and P2/T2 = k • Then: P1/T1 = P2/T2 Chapter Eight

23. As the temperature goes up, the pressure also goes up. Chapter Eight

24. Problem 8.52 An aerosol can has an internal pressure of 3.85 atm at 25oC. What temperature is required to raise the pressure to 18.0 atm? Use Gay-Lussac’s law P1/T1 = P2/T2 Now, solve for unknown variable. divide both sides by P2 P1/T1P2 = 1/T2 take reciprocal of both sides T2 = T1P2/P1 Chapter Eight

25. T2 = T1P2/P1 T1 = 25oC + 273 = 298 K T2 = (298 K)(18.0 atm)/(3.85 atm) = 1390 K (1117 oC) Chapter Eight

26. 8.7 The Combined Gas Law • Since PV, V/T, and P/T all have constant values for a fixed amount of gas, these relationships can be merged into a combined gas law for a fixed amount of gas. • Combined gas law: PV/T = k if n constant • P1V1/T1 = P2V2/T2 • If any five of the six quantities in this equation are known, the sixth can be calculated. • See problems 8.8, 9, 54, 57, 60 Chapter Eight

27. 8.8 Avogadro’s Law: The Relation Between Volume and Molar Amount • Avogadro’s law:The volume of a gas is directly proportional to its molar amount at a constant pressure and temperature. That is, V divided by n is constant when P and T are held constant. • V n or V/n = k if P and T are constant • If: V1/n1 = k and V2/n2 = k • Then: V1/n1 = V2/n2 Chapter Eight

28. The molar amounts of anytwo gases with the same volume are the same at a given T and P. • Standard temperature and pressure: (STP) = 0C (273.15 K) and 1 atm (760 mm Hg) • Standard molar volume of a gas at STP = 22.4 L/mol Chapter Eight

29. Problem 8.72 What is the mass of CH4 in a sample that occupies a volume of 16.5 L at STP? 1 mol of any gas at STP = 22.4 L 16.5 L x 1 mol/22.4 L x 16.0 g CH4/1 mol CH4 = 11.8 g CH4 Chapter Eight

30. Density of gases g/L Density of Helium: 1 mol/22.4 L x 4.00g He/mol He = 0.178 g/L Density of Nitrogen: 1 mol/22.4 L x 28.0g N2/mol N2 = 1.25 g/L Density of a gas is proportional to its Mol. Wt. So… H2 < He < CH4 < CO = N2 < CO2 Chapter Eight

31. 8.9 The Ideal Gas Law • Ideal gas law: The relationships among the four variables P, V, T, and n for gases can be combined into a single expression called the ideal gas law. • PV/nT = R (a constant value) or PV = nRT • If the values of three of the four variables in the ideal gas law are known, the fourth can be calculated. • Values of the gas constant R: For P in atm: R = 0.0821 L·atm/mol·K For P in mm Hg: R = 62.4 L·mm Hg/mol·K Chapter Eight

32. Problem 8.11 An aerosol spray can of deodorant with a volume of 350 mL contains 3.2 g of propane gas (C3H8) as a propellant. What is the pressure in the can at 20oC? PV = nRT solve for P P = nRT/V Chapter Eight

33. Convert units P = nRT/V R= 0.0821 L.atm/mol.K n = moles propane (C3H8) = 3.2 g C3H8 x 1 mol C3H8/ 44 g C3H8 = 0.073 mol T= K degrees = 20oC + 273 = 293 K V = Liters = 350 mL x 1 L/1000 mL = 0.350 L Chapter Eight

34. Now Solve for P P = nRT/V n R T / V = (0.073 mol)(0.0821 L.atm/mol.K)(293 K)/0.350 L = 5.0 atm Chapter Eight

35. Chapter Eight

36. 8.10 Partial Pressure and Dalton’s law • Dalton’s law: The total pressure exerted by a gas mixture of (Ptotal) is the sum of the partial pressures of the components in the mixture. • Dalton’s law Ptotal = Pgas1 + Pgas2 + Pgas3 + … • Partial pressure: The contribution of a given gas in a mixture to the total pressure. Chapter Eight

37. Dry Air at Sea Level Pressure at sea level is 760 mm Hg (1 atm) Percent composition of dry air is: 78% N2 (78% of the pressure is caused by N2) 21% O2 (21% of the pressure is caused by O2) 1% Ar (1% of the pressure is caused by Ar) Ptotal = PN2 + PO2 + PAr = (0.78)(760) + (0.21)(760) + (0.01)(760) 760= 593 mm +160 mm + 7 mm Chapter Eight

38. Problem 8.14 Deep sea divers breathe a mixture of 98% He and 2% O2 at 9.50 atm of pressure. How does the partial pressure of O2 in diving gas compare with its partial pressure in normal air? Normal air is 21% O2 PPO2 in air: (0.21 O2)(760 mm Hg) = 160 mm Hg Tank pressure is 9.50 atm x 760 mm Hg/1 atm = 7220 mm Hg PPO2 in tank = (0.02)(7220) = 144 mm Hg Chapter Eight

39. 8.11 Intermolecular Forces in Liquids • Intermolecular force: A force that acts between molecules and holds molecules close to one another. There are three major types of intermolecular forces. • Dipole–dipole forces are weak, with strengths on the order of 1 kcal/mol • London dispersion forces are weak, in the range 0.5–2.5 kcal/mol. They increase with molecular weight and molecular surface area. • Hydrogen bonds can be quite strong, with energies up to 10 kcal/mol. Chapter Eight

40. 5.8 Polar Covalent Bonds and Electronegativity • Electrons in a covalent bond occupy the region between the bonded atoms. • If the atoms are identical, as in H2 and Cl2, electrons are attracted equally to both atoms and are shared equally. • If the atoms are not identical, however, as in HCl, the bonding electrons may be attracted more strongly by one atom than by the other and thus shared unequally. Such bonds are known as polar covalent bonds. Chapter Five

41. When charges separate in a neutral molecule, the molecule has a dipole moment and is said to be polar. Chapter Five

42. In HCl, electrons spend more time near the chlorine than the hydrogen. Although the molecule is overall neutral, the chlorine is more negative than the hydrogen, resulting in partial charges on the atoms. • Partial charges are represented by a d- on the more negative atom and d+ on the more positive atom. • The ability of an atom to attract electrons is called the atom’s electronegativity. • Fluorine, the most electronegative element, assigned a value of 4, and less electronegative atoms assigned lower values. Chapter Five

43. Elements at the top right of the periodic table are most electronegative, those at the lower left are least electronegative. Noble gases are not assigned values. Chapter Five

44. As a rule of thumb, electronegativity differences of less than 0.5 result in nonpolar covalent bonds, differences up to 1.9 indicate increasingly polar covalent bonds, and differences of 2 or more indicate ionic bonds. • There is no sharp dividing line between covalent and ionic bonds; most bonds fall somewhere in-between. Chapter Five

45. 5.9 Polar Molecules • Entire molecules can be polar if electrons are attracted more strongly to one part of the molecule than to another. • Molecules polarity is due to the sum of all individual bond polarities and lone-pair contribution in the molecule. • Polarity has a dramatic effect on the physical properties of molecules, particularly on melting points, boiling points, and solubility. Chapter Five

46. Dipoles or polarity can be represented by an arrow pointing to the negative end of the molecule with a cross at the positive end resembling a + sign. Chapter Five

47. Just because a molecule has polar covalent bonds does not mean that the molecule is polar overall. • Carbon dioxide and tetrachloromethane molecules have no net polarity because their symmetrical shapes cause the individual bond polarities to cancel each other out. Chapter Five

48. Dipole–dipole forces: The positive and negative ends of polar molecules are attracted to one another by dipole–dipole forces. As a result, polar molecules have higher boiling points than nonpolar molecules of similar size. Chapter Eight

49. Only polar molecules experience dipole–dipole forces, but all molecules, regardless of structure, experience London dispersion forces. • (a) On average, the electron distribution in a nonpolar molecule is symmetrical. (b) At any instant, it may be unsymmetrical, resulting in a temporary polarity that can attract neighboring molecules. Chapter Eight

50. A hydrogen bond is an attractive interaction between an unshared electron pair on an electronegative O, N, or F atom and a positively polarized hydrogen atom bonded to another electronegative O, N, or F. Hydrogen bonds occur in both water and ammonia. Chapter Eight