Solids Liquids and Phase Changes

Views:
 
Category: Entertainment
     
 

Presentation Description

No description available.

Comments

Presentation Transcript

Solids, Liquids, and Phase Changes : 

Solids, Liquids, and Phase Changes

Condensed States of Matter : 

Condensed States of Matter In gases, particles move past each other, colliding sometimes.

Condensed States of Matter : 

Condensed States of Matter At lower temperatures, the particles are moving slower, on average. Link to animation

Condensed States of Matter : 

Condensed States of Matter Attractive inter-particle forces overcome the motion of the particles.

Condensed States of Matter : 

Condensed States of Matter The gas becomes a liquid as particles are pulled together.

Liquids : 

Liquids Particles vibrate, rotate, and slide past each other Liquids take the shape of their container Liquids, like gases, are fluid

Liquids : 

Liquids When liquids cool, particles get locked into specific positions (freezing)

Phase Changes : 

Phase Changes Gas Liquid Solid Freezing Melting Condensing Evaporating Sublimation Deposition Or Boiling

Crystals and Amorphous Solids : 

Crystals and Amorphous Solids Crystals Particles vibrate and rotate Particles do not move past each other Particles are arranged in orderly, repeating, three-dimensional pattern Crystal lattice Crystalline substances break into smaller crystals with the same repeating pattern

Slide 10: 

Crystals and Amorphous Solids

Crystals and Amorphous Solids : 

Crystals and Amorphous Solids Amorphous solids Particles vibrate and rotate Particles do not move past each other Particles arranged in random order Glassy Amorphous solids break into smaller pieces of varying shapes and sizes.

Crystals and Amorphous Solids : 

Crystals and Amorphous Solids

Vapor Pressure and Boiling Point : 

Vapor Pressure and Boiling Point Vapor pressure Pressure created by an evaporated liquid Pressure can build up in a closed container

Vapor Pressure and Boiling Point : 

Vapor Pressure and Boiling Point Vapor Pressure is related to Temperature.

If the pressure was lowered to 600 torr, what would the boiling points for the three liquids be? : 

If the pressure was lowered to 600 torr, what would the boiling points for the three liquids be?

Boiling Point : 

Boiling Point Temperature at which vapor pressure equals the atmospheric pressure

Normal Boiling Point : 

Normal Boiling Point Boiling point at 1 atmosphere pressure. Lower the pressure, lower the bp Higher the pressure, higher the bp

Boiling Point of Water in PGH : 

Boiling Point of Water in PGH Pgh is above sea level Air pressure is less at this elevation (1200 ft) Vapor pressure reaches that lower air pressure before the temperature gets to 100 degrees Celsius. Water boils below 100 degrees Celsius in Pittsburgh

Boiling point of water in Denver : 

Boiling point of water in Denver What is your estimate of the boiling temperature (boiling point) of water in Denver? Why would it be difficult to cook spaghetti while hiking in the Rocky Mountains?

Intermolecular Forces : 

Intermolecular Forces Strength of interparticle forces affects vapor pressure Dispersion force Dipole-dipole force Hydrogen bonding Metallic bonding Ionic bonding Stronger interparticle forces lead to lower vapor pressures and higher boiling points

Intermolecular Forces and Temperature : 

Intermolecular Forces and Temperature Speeding up molecules (heating the liquid) makes it easier for them to overcome attractive intermolecular forces to evaporate and increase vapor pressure.

Heating and Cooling Processes : 

Heating and Cooling Processes Heat is absorbed or released when a substance changes state. Temperature does not change during a phase change.

Heating : 

Heating Energy gained is used to rearrange particles, changing from solid to liquid or liquid to gas. Endothermic Particles get farther apart

Cooling : 

Cooling Energy lost is released when particles are rearranged, changing from gas to liquid or liquid to solid. Exothermic Particles get closer together

Heating and Cooling Processes : 

Heating and Cooling Processes When a substance changes state, heat is either absorbed or released Released – exothermic Absorbed – endothermic

q=mCpT : 

q=mCpT Use the formula above to calculate the heat absorbed (+) or released (-) during a temperature change. Temperature does not change during a phase change (melting, freezing, boiling...) T would be 0. q would be 0.

Phase Change : 

Phase Change Heat absorbed or released is used to change particle arrangements, not temperature.

Melting : 

Melting Solid becomes liquid Heat moves particles out of fixed positions

Boiling : 

Boiling Heat overcomes interparticle forces, moving particles away from each other.

Condensation : 

Condensation Heat is released as interparticle forces take over, moving the particles closer together

Freezing : 

Freezing Heat is given off as particles move into fixed positions.

Heating Curve (graph) : 

Heating Curve (graph)

Summary : 

Summary A-B heating a solid B-C melting solid to liquid C-D heating a liquid D-E boiling liquid to gas E-F heating a gas

Cooling Curve : 

Cooling Curve A B C D E

Summary : 

Summary A gas cooling B gas condensing C liquid cooling D liquid freezing E solid cooling

Enthalpy Changes : 

Enthalpy Changes Temperature changes H = q H = mCpT Liquid – Solid state change H = m Hfus Heat of fusion Liquid – Gas state change H = m Hvap Heat of vaporization

Sample Problem : 

Sample Problem 42.3 g of gaseous H2O at 123.0 °C is cooled to solid H2O at -28.0 °C. What is the total amount of heat (in Joules) released in this process? 5 steps: A,B,C,D,E

Sample Problem : 

Sample Problem 42.3 g of gaseous H2O at 123.0 °C is cooled to solid H2O at -28.0 °C. What is the total amount of heat released in this process? H = mCpT H = (42.3 g)(2.02 J/g°C)(100°C – 123.0°C) H = (42.3 g)(2.02 J/g°C)(– 23.0°C) H = - 1 970 J

Sample Problem : 

Sample Problem 42.3 g of gaseous H2O at 123.0 °C is cooled to solid H2O at -28.0 °C. What is the total amount of heat released in this process? H = m Hvap H = (42.3 g)(-2260J/g) H = - 95 600 J

Sample Problem : 

Sample Problem 42.3 g of gaseous H2O at 123.0 °C is cooled to solid H2O at -28.0 °C. What is the total amount of heat released in this process? H = mCpT H = (42.3 g)(4.18 J/g°C)(0°C – 100°C) H = (42.3 g)(4.18 J/g°C)(– 100.0°C) H = - 17 700 J

Sample Problem : 

Sample Problem 42.3 g of gaseous H2O at 123.0 °C is cooled to solid H2O at -28.0 °C. What is the total amount of heat released in this process? H = m Hfus H = (42.3 g)(-334J/g) H = - 14 100 J

Sample Problem : 

Sample Problem 42.3 g of gaseous H2O at 123.0 °C is cooled to solid H2O at -28.0 °C. What is the total amount of heat released in this process? H = mCpT H = (42.3 g)(2.06 J/g°C)(–28.0°C – 0°C) H = (42.3 g)(2.06 J/g°C)(– 28.0°C) H = - 2 440 J

Sample Problem : 

Sample Problem 42.3 g of gaseous H2O at 123.0 °C is cooled to solid H2O at -28.0 °C. What is the total amount of heat released in this process?