States of Matter The Solid State

1. States of Matter • The Solid State • Particles are tightly packed, very close together (strong cohesive forces) • Low kinetic energy (energy of motion) • Fixed shape and volume • Crystalline or amorphous structure

2. The Liquid State • Particles are close to each other (making them mostly incompressible) • Attractive forces keep molecules close, but not so close to restrict movement

3. The Gas State • Gas particles move randomly and rapidly. • Size of gas particles is small compared to the space between the particles. • Gas particles exert no attractive forces on each other. • Kinetic energy of gas particles increases with increasing temperature.

4. a The symbol “~” means approximately.

5. Gases and Pressure • When gas particles collide with the walls of a container, they exert a pressure. • Pressure (P) is the force (F) exerted per unit area (A). Force F Pressure = = Area A 760. mm Hg 760. torr 1 atmosphere (atm) = 14.7 psi 101,325 Pa

6. Gas Laws Mathematical relationships describing the behavior of gases with regard to mixing, diffusion, changes in pressure, changes in temperature Boyle’s Law: Describes the relation between pressure and volume of a gas, under a constant temperature PiVi = PfVf where i = initial condition and f = final condition

7. Boyle’s Law: Inverse relation between Pressure and Volume

8. Example: Freon-12, CCl2F2, is used in refrigeration systems. What is the new volume (L) of a 8 L sample of Freon gas initially at 50 mm Hg after its pressure is changed to 200 mm Hg at constant T? • Set up a data table Conditions 1Conditions 2 P1 = 50 mm Hg P2 = 200 mm Hg V1 = 8 L V2 = ?

9. 2. Solve Boyle’s Law for V2: P1V1 = P2V2 V2 = V1P1 P2 V2 = 8 L x 50 mm Hg = 2 L 200 mm Hg

10. Learning Check A sample of helium gas has a volume of 6.4 L at a pressure of 0.70 atm. What is the new volume when the pressure is increased to 1.40 atm (T constant)?

11. Solution P1V1 = P2V2 Solve for V2: V2 = V1P1 P2 V2 = 6.4 L x 0.70 atm = 3.2 L 1.40 atm Volume decreases when there is an increase in the pressure (Temperature is constant).

12. Learning Check A sample of oxygen gas has a volume of 12.0 L at 600. mm Hg. What is the new pressure when the volume changes to 36.0 L? (T and n constant.)

13. Solution Conditions 1 Conditions 2 P1 = 600. mm Hg P2 = ? V1 = 12.0 L V2 = 36.0 L P2 = P1 V1 V2 600. mm Hg x 12.0 L = 200. mm Hg 36.0 L

14. Charles’ Law: Describes relation between temperature and volume of a gas, under constant pressure Vi/Ti = Vf/Tf Charles’s Law: Direct relationship between Volume and Temperature

15. Charles’ Law

16. Example: A balloon has a volume of 785 mL at 21°C. If the temperature drops to 0°C, what is the new volume of the balloon (P constant)? • Set up data table: Conditions 1Conditions 2 V1 = 785 mL V2 = ? T1 = 21°C = 294 K T2 = 0°C = 273 K Be sure that you always use the Kelvin (K) temperature in gas calculations!

17. 2. Solve Charles’ law for V2 V1 = V2 T1 T2 V2 = V1 T2 T1 V2 = 785 mL x 273 K = 729 mL 294 K

18. 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)?

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

21. Combined Gas Law: Describes relation between pressure, temperature and volume of a gas PiVi/Ti = PfVf/Tf

22. Example: 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 1Conditions 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 = ?

23. 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

24. 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)?

25. 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

26. Gay–Lussac’s Law: Describes the relation between pressure and temperature of a gas, at a constant volume • Pressure and temperature are directly related P Pressure = k = constant Temperature T P1 P2 = T1 T2 Note: Temperature must be expressed in kelvins.

28. Avogadro’s Law: Equal volumes of gases measured at the same temperature and pressure contain equal number of molecules Vi/ni = Vf/nf where n = number of moles

29. Avogadro’s Law Example: 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? Conditions 1Conditions 2 V1 = 1.5 L V2 = ? n1 = 0.75 mole He n2 = 1.2 moles He V1/n1 =V2/n2 V2 = V1n2 n1V2 = 1.5 L x 1.2 moles He = 2.4 L 0.75 mole He

30. Ideal Gas Law: Describes relation between pressure, volume, temperature and the number of molecules in an ideal gas sample PV = nRT where R = universal gas constant (0.0821 L atm/K mol)

31. Ideal Gas Law Example: A cylinder contains 5.0 L of O2 at 20.0°C and 0.85 atm. How many grams of oxygen are in the cylinder? P = 0.85 atm, V = 5.0 L, T = 293 K, n (or g =?) PV = nRT  n = PV RT = (0.85 atm)(5.0 L)(mole K) = 0.18 mole O2 (0.0821atm L)(293 K) = 0. 18 mole O2 x 32.0 g O2 = 5.8 g O2 1 mole O2

32. Partial Pressure: Pressure an individual gas in a mixture would exert were it alone in the same container Dalton’s Law: Total pressure exerted by a mixture of gases equals the sum of the partial pressures P(total) = P(gas 1) + P(gas 2) + P(gas 3) etc.

33. Dalton’s Law of Partial Pressures

34. Summary of Gas Laws Boyle: PiVi = PfVfCharles: Vi/Ti = Vf/Tf Avogradro: Vi/ni = Vf/nf Gay-Lussac: Pi/Ti = Pf/Tf Dalton: P(total) = P(gas 1) + P(gas 2) + P(gas 3) *Combined: PiVi/Ti = PfVf/Tf *Ideal: PV = nRT * Memorize for exam

35. Intermolecular Forces Intermolecular forces: attractive forces that exist between molecules. In order of increasing strength, these are: • London dispersion forces • dipole–dipole interactions • hydrogen bonding

36. London Dispersion Forces London dispersion forces: weak interactions due to the momentary changes in electron density in a molecule. • Change in electron density creates a temporary dipole. • The weak interaction between these temporary dipoles constitutes London dispersion forces. • All covalent compounds exhibit London dispersion forces. • The larger the molecule, the larger the attractive force, and the stronger the intermolecular forces.

37. Dipole-dipole interaction: attraction between positive end of one polar molecule and negative end of a different polar molecule

38. Hydrogen bonding: Specific type of dipole-dipole force, between the partial positive charge on H and partial negative charge on an electronegative element such as O, N, F

40. Intermolecular Forces:Boiling Point and Melting Point Boiling point: temperature at which a liquid is converted to a gas Melting point: temperature at which a solid is converted to a liquid The stronger the intermolecular forces on a substance, the higher its boiling point and melting point are.

41. Examples of Intermolecular Forces andBoiling, Melting Points:

42. Both molecules have London dispersion forces and nonpolar bonds. • In this case, the larger molecule will have stronger attractive forces.

43. Vapor Pressure Evaporation: the conversion of liquids into the gas phase. • Evaporation is endothermic—it absorbs heat from • the surroundings. Condensation: the conversion of gases into the liquid phase. • Condensation is exothermic—it gives off heat to • the surroundings.

44. Viscosity and Surface Tension Viscosity: a measure of a fluid’s resistance to flow freely • Compounds with strong intermolecular forces • tend to be more viscous than compounds with weaker forces. • Substances composed of large molecules tend to be more viscous, too, because large molecules do not slide past each other as freely.

45. Surface tension: a measure of the resistance of a liquid to spread out. • Interior molecules in a • liquid are surrounded by • intermolecular forces on • all sides. • Surface molecules only • experience intermolecular • forces from the sides and • from below.

46. The Solid State: Types of Solids • Crystalline solid: has a regular arrangement of particles—atoms, molecules, or ions—with a repeating structure. • There are four different types of crystalline solids— ionic, molecular, network, and metallic.

47. Crystalline Solids Ionic solid: composed of oppositely charged ions Molecular solid: composed of individual molecules arranged regularly

48. Network solid: composed of a vast number of atoms covalently bonded together (SiO2). Metallic solid: a lattice of metal cations surrounded by a cloud of e− that move freely (Cu). 48

49. Amorphous Solids Amorphous solid: has no regular arrangement of its closely packed particles. • They can be formed when liquids cool too quickly • for regular crystal formation. • Very large covalent molecules tend to form • amorphous solids, because they can become folded and intertwined. Examples: rubber, glass, and plastic.