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States of Matter The Solid State  Particles are tightly packed, very close together (strong cohesive forces)  Low kinetic energy (energy of motion) 

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Presentation on theme: "States of Matter The Solid State  Particles are tightly packed, very close together (strong cohesive forces)  Low kinetic energy (energy of motion) "— Presentation transcript:

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). Pressure= Force = F AreaA 1 atmosphere (atm) = 760. mm Hg 760. torr 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 P i V i = P f V f where i = initial condition and f = final condition

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

8 Freon-12, CCl 2 F 2, 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? 1.Set up a data table Conditions 1Conditions 2 P 1 = 50 mm HgP 2 = 200 mm Hg V 1 = 8 LV 2 = ? Example:

9 2. Solve Boyle’s Law for V 2 : P 1 V 1 = P 2 V 2 V 2 = V 1 P 1 P 2 V 2 = 8 L x 50 mm Hg = 2 L 200 mm Hg

10 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)? Learning Check

11 P 1 V 1 = P 2 V 2 Solve for V 2 : V 2 = V 1 P 1 P 2 V 2 = 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). Solution

12 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.) Learning Check

13 Conditions 1Conditions 2 P 1 = 600. mm HgP 2 = ? V 1 = 12.0 LV 2 = 36.0 L P 2 = P 1 V 1 V mm Hg x 12.0 L = 200. mm Hg 36.0 L Solution

14 Charles’ Law: Describes relation between temperature and volume of a gas, under constant pressure V i /T i = V f /T f 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)? 1.Set up data table: Conditions 1Conditions 2 V 1 = 785 mLV 2 = ? T 1 = 21°C = 294 KT 2 = 0°C = 273 K Be sure that you always use the Kelvin (K) temperature in gas calculations!

17 2. Solve Charles’ law for V 2 V 1 = V 2 T 1 T 2 V 2 = V 1 T 2 T 1 V 2 = 785 mL x 273 K = 729 mL 294 K

18 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)? Learning Check

19 T 2 = T 1 V 2 V 1 T 2 = 291 K x 640 mL = 443 K 420 mL = 443 K – 273 K = 170°C Solution

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21 Combined Gas Law: Describes relation between pressure, temperature and volume of a gas P i V i /T i = P f V f /T f

22 A sample of helium gas has a volume of L, a pressure of 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 P 1 = atm P 2 = 3.20 atm V 1 = L (180 mL) V 2 = 90.0 mL T 1 = 29°C = 302 KT 2 = ? Example:

23 2. Solve for T 2 P 1 V 1 =P 2 V 2 T 1 T 2 T 2 = T 1 P 2 V 2 P 1 V 1 T 2 = 302 K x 3.20 atm x 90.0 mL = 604 K atm mL T 2 = 604 K – 273 = 331 °C

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

25 Data Table T 1 = 308 K T 2 = -95°C = 178K V 1 = 675 mL V 2 = ? P 1 = 646 mm Hg P 2 = 802 mm Hg Solve for V 2 V 2 =V 1 P 1 T 2 P 2 T 1 V 2 = 675 mL x 646 mm Hg x 178K = 314 mL 802 mm Hg x 308K Solution

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 Pressure Temperature = constant P T = k Note: Temperature must be expressed in kelvins. P1P1 T1T1 = P2P2 T2T2

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28 Avogadro’s Law: Equal volumes of gases measured at the same temperature and pressure contain equal number of molecules V i /n i = V f /n f 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 V 1 = 1.5 LV 2 = ? n 1 = 0.75 mole Hen 2 = 1.2 moles He V 1 /n 1 =V 2 /n 2 V 2 = V 1 n 2 n 1 V 2 = 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 ( L atm/K mol)

31 A cylinder contains 5.0 L of O 2 at 20.0°C and 0.85 atm. How many grams of oxygen are in the cylinder? Ideal Gas Law Example: 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 O 2 (0.0821atm L)(293 K) = mole O 2 x 32.0 g O 2 = 5.8 g O 2 1 mole O 2

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: P i V i = P f V f Charles: V i /T i = V f /T f Avogradro: V i /n i = V f /n f Gay-Lussac: P i /T i = P f /T f Dalton: P (total) = P (gas 1) + P (gas 2) + P (gas 3) *Combined: P i V i /T i = P f V f /T f *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. All covalent compounds exhibit London dispersion forces. The weak interaction between these temporary dipoles constitutes 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

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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 and Boiling, 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 48 Network solid: composed of a vast number of atoms covalently bonded together (SiO 2 ). Metallic solid: a lattice of metal cations surrounded by a cloud of e − that move freely (Cu).

49 Amorphous Solids 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. Amorphous solid: has no regular arrangement of its closely packed particles.


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