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Energy

Energy and States of Matter *add the following papers to your notes: gizmos, pHet simulation, KMT reading guide. Energy. When particles collide, energy is transferred from one particle to another.

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Energy

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  1. Energy and States of Matter*add the following papers to your notes: gizmos, pHet simulation, KMT reading guide

  2. Energy • When particles collide, energy is transferred from one particle to another. • Law of conservation of energy: energy can be neither created nor destroyed; it can only be converted from one form to the other.

  3. Solids • Descriptors: • Hardness • Shape • Malleable • Ductile • Density • Elasticity • Properties: • Particles are very close together • Strong attractive forces between particles • Particles vibrate but do not move out of position • Fixed shape • Fixed volume

  4. Liquids • Descriptors • Viscosity • Concentration • Fluid • Density • Cohesion (ex. water-water) • Adhesion (ex. Water-leaf surface) • Properties • Particles are close together • Weak attractive forces between particles • Particles slide past each other • Takes the shape of the container • Fixed volume

  5. Gasses • Properties • Particles are far apart • No attractive forces between particles • Takes the shape of the container • Particles spread out to fill the container • Can be identified by “burning splint” test: • O2 gas causes the burning splint to re-light • CO2 gas causes the burning splint to go out quietly (fire extinguisher) • H2 gas causes a popping sound • DIATOMIC MOLECULES: • There are seven elements (all gasses) whose atoms are not stable as individuals. • These atoms will always bond with another atom. • If no other type of atom is available, they bond with another atom of the same type. • These are called DIATOMIC MOLECULES. • They are: H, O, F, Br, I, N, Cl

  6. Temperature • Temperature: measure of the average kinetic energy of each particle within an object. • Gases at the same temperature have the same kinetic energy. • Kinetic energy = ½(mv2)

  7. Temperature • Kelvin scale: zero is the temperature at which no more energy can be removed from matter; all motion stops. • Zero on Kelvin scale is called absolute zero; written as 0K • No negative Kelvin temps

  8. Temperature K = C° + 273 Convert the following temperatures into Kelvin: a. 43 oC b. –135 0C Convert the following temperatures into Celsius: a. 340 K b. 30 K

  9. Kinetic Molecular Theory • Gases consist of tiny particles (atoms or molecules). • These particles are so small, compared with the distances between them that the volume (size) of the individual particles can be assumed to be negligible (zero). • The particles are in constant random motion, colliding with the walls of the container. These collisions with the walls cause the pressure exerted by the gas. • The particles are assumed to not attract nor repel each other. • The average kinetic energy of the gas particles is directly proportional to the Kelvin temperature of the gas.

  10. Pressure The amount of pressurea gas exerts depends upon 2 things: • theforceparticles have as they collide with each other & the objects around them; • the amount ofareathe particles have to move around in Pressure of a gas is due to the collisions of particles with the sides of the container in which it is enclosed.

  11. Pressure The metric unit for measuring pressure is ATMOSPHERE (atm). Other units for measuring pressure: • mm of Hg (millimeters of mercury) 760 mm Hg = 1 atm • kPa (kilopascals) 101.3 kPa = 1 atm • psi (pounds per square inch) 14.7 psi = 1 atm

  12. STPStandard Temperature and Pressure 1 atm 1 atm = Air pressure at sea level O°C

  13. Relationships • Direct Relationship: when changing one variable causes the other variable to change in the same direction • when one goes up, the other goes up; when one goes down, the other goes down • Inverse Relationship: when changing one variable causes the other variable to change in the opposite direction • when one goes up, the other goes down • Another word for inverse is indirect.

  14. Direct Inverse

  15. Boyle’s law • Demo – marshmallows and vacuum pump • Independent variable:_______________________ • Dependent variable: ________________________ • Observations: ____________________________________________________________________________________ • Relationship: ______________________________

  16. Boyle’s Law • The volume of a given amount of gas held at a constant temperature varies inversely with the pressure • Mathematical relationship: V1P1 = V2P2

  17. Boyle’s Law Practice: • If a gas has a volume of 100 ml when the pressure is 1.4 atm, what is the volume, in mL, when the pressure is increased to 1.6 atm and the temperature is held constant?

  18. If a gas has a volume of 100 ml when the pressure is 1.4 atm, what is the volume, in mL, when the pressure is increased to 1.6 atm and the temperature is held constant? 1. List variables: • V1 = 100 mL • P1 = 1.4 atm • V2 = ? mL • P2 = 1.6 atm 2. Find formula on EOC chart: • V1P1 = V2P2 3. Substitute in known values: • (100mL)(1.4atm) = (V2)(1.6atm) 4. Solve for unknown: • V2 = 87.5 mL • Very Important: Determine if your answer makes sense… pressure increased by a little, so volume should decrease by a little.

  19. Boyle’s Law practice: • The volume of a quantity of a gas held at constant temperature and 1.00 atm of pressure is 100 mL. What pressure does it take to reduce the volume to 95 mL?

  20. Charles’ law • Demo – balloons • Independent variable:_______________________ • Dependent variable: ________________________ • Observations: ______________________________________________________________________________________________________________________________ • Relationship: ______________________________

  21. Charles’ Law • The volume of a given amount of gas is varies directly to its kelvin temperature when pressure is constant. • Mathematical relationship: V1 = V2 T1 T2

  22. Charles’ Law practice • A balloon inflated in an air conditioned room at 27◦C has a volume of 4.00 L. If it is heated to 57◦C and the pressure remains constant, what is the new volume?

  23. A balloon inflated in an air conditioned room at 27◦C has a volume of 4.00 L. If it is heated to 57◦C and the pressure remains constant, what is the new volume? 1. List variables 2. Convert temp to Kelvin: • T1= 27°C + 273 = 300 K • V1 = 4.00 L • T2 = 57°C + 273 = 330 K • V2 = ? L 3. Find formula on EOC chart: • V1 = V2 T1 T2 4. Substitute in known values: • (4.00L) = (V2)_ (300K) (330K) 5. Solve for unknown (by cross-multiplying): • V2 = 4.4 L 6. Determine if your answer makes sense… • temp went up by 10%, • volume went up by 10%.

  24. Charles’ law practice: • A gas kept at constant pressure has a volume of 10.0 L at 25.0° C. At what Celsius temperature would the gas have a volume of 20.0 L?

  25. Gay-Lussac’s law • Demo – crush the can • Independent variable:_______________________ • Dependent variable: ________________________ • Observations: ____________________________________________________________________________________ • Relationship: ______________________________

  26. Gay Lussacs Law • The pressure of a gas varies directly to the Kelvin temperature of the sample, if the volume remains constant. • Mathematical relationship P1= P2 T1 T2 Directly related: temperature increases pressure increases

  27. Graph of Gay-Lussac’s Law(direct relationship)

  28. Gay-Lussac’s Law practice • A gas in an aerosol can is at a pressure of 1.00 atm and 27.0 oC. If the can is thrown into a fire, what is the internal pressure of the gas when the temperature reaches 927 oC?

  29. A gas in an aerosol can is at a pressure of 1.00 atm and 27.0 oC.If the can is thrown into a fire, what is the internal pressure of the gas when the temperature reaches 927 oC? • List variables • Convert temp to Kelvin: • P1 = 1.00 atm • T1= 27°C + 273 = 300 K • P2 = ? atm • T2 = 927°C + 273 = 1200 K • Find formula on EOC chart: • P1V1 = P2V2 T1n1 T2n2 • Substitute in known values: Ignore variables that don’t change. In this example, volume (v) and number of moles (n) don’t change. • (1.00atm) = (P2) (300K) (1200K) • Solve for unknown: • P2 = 4.00 atm • Determine if your answer makes sense…temp quadrupled, pressure quadrupled.

  30. Gay- Lussac’s law practice: • A sample of a gas has a pressure of 851 mm Hg at 285°C. To what Celsius temperature must the gas be heated to double its pressure if there is no change in the volume of the gas?

  31. Combined Gas Law • Use when p, t, and v all change • Temperature must be in Kelvin • P1V1 = P2V2 n1T1 n2T2

  32. Combined Gas Law practice • A 25.0 ml balloon at 1.20 atm and 45oC, what temperature (in Celsius) would the gas be when the volume changes to 100 mL and the pressure is 0.816 atm?

  33. A 25.0 ml balloon at 1.20 atm and 45oC, what temperature would the gas be when the volume changes to 100 mL and the pressure is 0.816 atm? 1. List variables 2. Convert temp to Kelvin: • V1 = 25.0 mL • P1 = 1.20 atm • T1= 45°C + 273 = 318 K • T2 = ? K • V2 = 100 ml • P2 = 0.816 atm 3. Find formula on EOC chart: • P1V1 = P2V2 n1T1 n2T2 4. Substitute in known values: (1.20atm)(25.0mL) = (0.816atm)(100mL) (318K) (T2) 5. Solve for unknown: Combine variables, then cross-multiply. • T2 = 865 K 6. Determine if your answer makes sense… • volume increase is more than pressure decrease, so net result should be that temp increases

  34. Combined gas law practice: • A sample of gas is stored in a 500.0 mL flask at 1.07 atm and 10.0oC. The gas is transferred to a 750.0 mL flask at 21.0oC. What is the new pressure in the flask?

  35. Combined gas law practice: • A balloon has a volume of 2.1L at 0.998 atm and 36oC. If it rises to an elevation where the pressure is 0.900 atm and the temperature is 28oC, what will be the new volume of the balloon?

  36. IDEAL VS. REAL GASES • Ideal gases don’t really exist, but many gases do behave ideally under certain conditions (far apart and not able to attract each other). • Ideal behavior occurs when the Pressure is ______________ Temperature is __________ Mass is ___________ Volume is ____________ Molecules are nonpolar.

  37. Which would act more ideally? • He(g) • H2O(g) Why? • Does helium act more ideally at: • 800K • 80K Why? • Does helium act more ideally at: • 20.0 atm • 1.00 atm Why?

  38. Ideal Gas Law • relates • Pressure • Volume • Temperature • number of moles(n) • For a gas at STP, moles(n) and volume(v) are DIRECTLYrelated. • 1 mole = 22.4 L at STP • This is called “molar volume”

  39. READ ONLY, DO NOT COPY!!!!!! • For any ideal gas, the ratio VP is constant. nT • We call this ratio R, the ideal gas constant. • Using standard temp and pressure conditions, we can calculate the value of R. • R = (22.4L)(1atm) (1mol)(273K) • R = 0.0821 L atm/mole K • This value is on your EOC reference chart • Because of the units on R, P must be in atm, V must be in L, and T must be in K. • Since R is a constant, we will never be solving for it. Rearrange R = VP nT to PV = nRT PV = nRT(pronounced “pivnert”) is called the ideal gas law equation

  40. Ideal Gas Law practice • What volume would 1.41 moles of oxygen occupy at 351K and 2.30 atm?

  41. What volume would 1.41 moles of oxygen occupy at 351K and 2.30 atm? 1. List variables 2. Convert temp to Kelvin: • V = ? • n = 1.41 moles • T = 351 K • P = 2.30 atm • R = 0.0821 L atm/mol K 3. Find formula on EOC chart: • PV = nRT 4. Substitute in known values: (2.30atm)(V) = (1.41mol)(0.0821Latm/molK)(351K) 5. Solve for unknown: Combine variables, then divide to get v by itself. • V = 17.7 L

  42. Ideal gas law practice: What temperature, in Celsius, would 6.00 moles of Helium occupy in a 25.0 L container at 1.26 atm?

  43. Avogadro’s Principle Equal volumes of gases at the same temperature and pressure contain the same number of molecules. *** The type of gas doesn’t matter.*** V1= V2 n1 n2

  44. Avogadro’s principle example: Relationship between n and V Increased number of gas particles Increased number of collisions with the walls of the container Increased total force of collisions Inside pressure greater than outside pressure Container expands

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