Chem 1151: Ch. 6 States of Matter. Physical States of Matter Matter can exist as:  Solid  Liquid  Gas Temperature Dependent States

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Presentation transcript:

Chem 1151: Ch. 6 States of Matter

Physical States of Matter Matter can exist as:  Solid  Liquid  Gas Temperature Dependent States

Physical Properties States can be distinguished by different properties:  Density: m/V  Shape: Physical dimensions  Compressibility: Volume change due to pressure  Thermal Expansion: Volume change due to temperature change Seager SL, Slabaugh MR, Chemistry for Today: General, Organic and Biochemistry, 7 th Edition, 2011;

Kinetic Molecular Theory of Matter Theory to explain the behavior of matter in different states 1.Matter is composed of tiny particles (molecules) 2.These particles are in constant motion and have kinetic energy (KE) 3.The particles possess potential energy (PE) as a result of attracting or repelling each other. 4.The average particle speed increases as the temperature increases 5.The particles transfer energy from one to another during collisions in which no net energy is lost from the system. m = mass (g, Kg) v = velocity = (Distance (m) /Time (s))

Kinetic Energy 1.Particles are in constant motion and have kinetic energy (KE) m = mass (g, Kg) v (nu)= velocity = (Distance/Time) Calculate KE for two particles with masses of 2.00 g and 4.00 g if they are both moving with a velocity of 15 m/s.

Kinetic Energy How Bruce Lee kicked out of Chuck Norris

Potential Energy Potential energy results from attractions or repulsions of particles.  Gravity  Electrostatic (charge)

Cohesive and Disruptive Forces in Matter Cohesive forces: Associated with PE. Tend to attract particles towards each other. Temperature-independent Disruptive forces: Associated with KE. Tend to scatter particles away from each other. Temperature-dependent State of a substance depends on relative strengths of these forces

Solid State Graphite: Each Carbon is covalently bonded to 3 other carbons in ring Diamond: Each carbon is bonded to 4 other carbons Copper Characteristics of solids: Cohesive forces stronger than disruptive forces High Density Definite Shape (strong cohesive forces) Small Compressibility Very small Thermal Expansion (particles vibrate but volume increases limited due to cohesive forces) Bridge expansion joint

Liquid State Characteristics of liquids: Particles packed randomly and close together Particles in constant motion Particles slide over each other but lack enough KE to separate completely High Density (particles not widely separated) Indefinite Shape (expand to shape of container) Small Compressibility (very little space between molecules) Small Thermal Expansion (particles vibrate, push away from each other, but volume increases limited due to cohesive forces) Seager SL, Slabaugh MR, Chemistry for Today: General, Organic and Biochemistry, 7 th Edition, 2011

Gaseous State Seager SL, Slabaugh MR, Chemistry for Today: General, Organic and Biochemistry, 7 th Edition, 2011; Characteristics of gases: Disruptive forces stronger than cohesive forces between particles Particles in constant random motion Particles far apart, travel in straight lines, collide frequently Low Density (particles widely separated) Indefinite Shape (little cohesion, particles expand to shape of container) Large Compressibility (gas is mostly empty space) Moderate Thermal Expansion (increase in temperature causes particles to collide with more energy, increases volume)

Gas Laws Seager SL, Slabaugh MR, Chemistry for Today: General, Organic and Biochemistry, 7 th Edition, 2011 Describe behavior of gases when mixed, subjected to pressure or temperature changes, or allowed to diffuse Laws describes relationships between temperature (T), volume (V), pressure (P) and mass Pressure (P) = Force/Area Boyle’s law Charles’s law Combined gas law Avogadro’s law Ideal gas law

P, V, T Relationships Boyle’s Law A constant relationship exists between pressure (P) and volume (V) If pressure increases, volume occupied by the gas decreases If volume increases, pressure created by the gas decreases Charles’s Law At constant pressure, the volume of a gas sample is directly proportional to the temperature (expressed in kelvins) If temperature increases, volume increases at constant pressure

P, V, T Relationships Charles’s Law At constant pressure, the volume of a gas sample is directly proportional to the temperature (expressed in kelvins) If temperature increases, volume increases at constant pressure Seager SL, Slabaugh MR, Chemistry for Today: General, Organic and Biochemistry, 7 th Edition, 2011

P, V, T Relationships Combined gas Law Boyle’s law and Charles’s law can be combined to relate P, V and T Because k’’ is a constant, we can use this equation to evaluate changes in these variables over time (between some initial state and a final state)

Ideal Gas Law The combined gas law applicable when mass of gas remains constant during changes in P, V and T What happens when mass changes? Avogadro’s law Two different gases of equal volume measured at same T and P contain equal numbers of molecules Mass would not be identical due to different MW’s ideal gas law Combines Boyle’s law, Charles’s law and Avogadro’s law P = Pressure V = Volume n = number of moles T = Temperature R = Universal Gas Constant

Ideal Gas Law P = Pressure V = Volume n = number of moles T = Temperature R = Universal Gas Constant Also, because m = mass MW = molecular weight We can also express the ideal gas law as STP (Standard Temperature and Pressure) T = 0 °C P = 1.0 atm V of 1 mol gas (any gas) = 22.4 L at STP

PROBLEMS Example 6.6, 6.7, 6.8, 6.9

Changes in State  Transition of matter from one state to another (solid  liquid  gas)  Temperature-related  Exothermic process: Heat released  Particles move closer together  Stronger cohesive forces  Endothermic process: Heat absorbed  Particles move farther apart  Stronger disruptive forces Seager SL, Slabaugh MR, Chemistry for Today: General, Organic and Biochemistry, 7 th Edition, 2011

Evaporation and Condensation Evaporation (vaporization): Molecules leave the surface of a liquid  Endothermic process  Rate depends on temperature and surface area of liquid  Temperature relates to speed and KE of molecules and their ability to escape cohesive forces at liquid surface  Evaporating molecules carry KE away from water  removes heat from remaining liquid  This is how sweating cools the body Seager SL, Slabaugh MR, Chemistry for Today: General, Organic and Biochemistry, 7 th Edition, 2011 Condensation: Gas molecules converted to liquid or solid state  Exothermic process

Evaporation and Vapor Pressure Evaporation (vaporization): Molecules leave the surface of a liquid Condensation: Gas molecules converted to liquid or solid state  In open system, liquid evaporates into atmosphere  In a closed system, evaporation and condensation reach an equilibrium Seager SL, Slabaugh MR, Chemistry for Today: General, Organic and Biochemistry, 7 th Edition, 2011 Vapor pressure: Pressure exerted by vapor in equilibrium with a liquid  Pressure is due to constant number of molecules exerting force on liquid and walls of container  For water, increasing T increases vapor pressure (higher KE)  Compounds that mix with water have lower vapor pressure than nonpolar compounds

Boiling and the Boiling Point  Vaporization occurs at surface of liquid  As liquid heated, vapor pressure increases  Boiling: When vapor pressure equals atmospheric pressure, vaporization begins to occur beneath surface of liquid  Boiling Point: Temperature when vapor pressure equals atmospheric pressure  If you decrease atmospheric pressure, boiling point decreases Seager SL, Slabaugh MR, Chemistry for Today: General, Organic and Biochemistry, 7 th Edition, 2011

Sublimation and Melting  Solids have low vapor pressures due to strong cohesive forces  Vapor pressures increase with temperature  Sublimation: Vapor pressure of solid high enough for molecules to transition from solid directly to gas  Ex. Freeze drying  Melting: Breakdown of solid into liquid state  Melting Point: Temperature where solid and liquid have same vapor pressure  KE of solid particles large enough to overcome strong cohesive forces holding particles together Seager SL, Slabaugh MR, Chemistry for Today: General, Organic and Biochemistry, 7 th Edition, 2011

Energy and the States of Matter  KE (associated with particle motion) is related to heat  PE is associated with particle separation distances, not motion  Increase in T on adding heat increases KE of particles  Adding heat with no increase in T increases PE of particles  Adding heat may or may not result in T increase AB Solid heated from -20 to 0 °C KE increases BC Temp constant while solid melts PE increases CD Adding more heat increases temp KE increases DE liquid  vapor at 100 °C PE increases EF Temp increases with heat of steam KE increases Phase change Phase change

Energy and the States of Matter  Specific Heat: Amount of heat (calories or joules) required to change the temperature of a specified amount of substance (1 g) by 1 °C.  1 cal = J  Substance with high specific heat can absorb more heat with small temp. change  Heat of fusion: Amount of heat (calories or joules) required to melt 1 g of substance at constant temperature  Heat of vaporization: Amount of heat (calories or joules) required to boil 1 g of substance at constant temperature  Ex: Heats of fusion and vaporization for water are 80 and 540 cal/g.  This is why a steam burn is worse than burn by boiling water: higher energy of steam that is released when steam condenses on skin.

Heat Calculations Heat = (sample mass)(specific heat)(temp. change)  Specific Heat: Amount of heat (calories or joules) required to change the temperature of a specified amount of substance (1 g) by 1 °C.  1 cal = J  Ex. 1. How much heat (in J) absorbed by g of ethylene glycol if temperature changes from 30.0 °C to 85.0 °C?

Heat Calculations Heat released = (sample mass)(specific heat)(temp. change)  Heat of vaporization: Amount of heat (calories or joules) required to boil 1 g of substance at constant temperature  1 cal = J  Ex. 2. Calculate the heat released when 5.00 × 10 3 g of steam at 120 °C condenses to water at 100 °C. Part 01 Heat associated with temp. change Heat released = (sample mass)(heat of vaporization) Part 02 Heat associated with phase change