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Thermal Expansion

Thermal Expansion. D L = a L o D T D L = change in_______ a = coefficient of _______expansion L o = original ________ D T = change in o C or___. Thermal Expansion. D L = a L o D T D L = change in_______ a = coefficient of _______expansion L o = original ________

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Thermal Expansion

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  1. Thermal Expansion • D L = a LoD T • D L = change in_______ • a = coefficient of _______expansion • Lo = original ________ • D T = change in oC or___

  2. Thermal Expansion • D L = a LoD T • D L = change in_______ • a = coefficient of _______expansion • Lo = original ________ • D T = change in oC or___

  3. Thermal Expansion • D L = a LoD T • D L = change in length • a = coefficient of linear expansion • Lo = original length • D T = change in oC or K

  4. Thermal Expansion • D L = a LoD T • D L = change in length • a = coefficient of linearexpansion • Lo = original length • D T = change in oC or K

  5. Thermal Expansion • D L = a LoD T • D L = change in length • a = coefficient of linear expansion • Lo = original length • D T = change in oC or K

  6. Thermal Expansion • D L = a LoD T • D L = change in length • a = coefficient of linear expansion • Lo = original length • D T = change in oC or K

  7. Thermal Expansion • D L = a LoD T • D L = change in length • a = coefficient of linear expansion • Lo = original length • D T = change in oC or K

  8. Problem Example • A structure steel beam holds up a scoreboard in an open-air stadium. If the beam is 12 m long when it is put into place on a winter day at 00C how much longer will it be in the summer at 320C. • a = 12x10-6 m-1K-1 • D L = a LoD T= 12x10-6 m-1K-1(12m)(32K) • D L = .0046 m

  9. Problem Example • A structure steel beam holds up a scoreboard in an open-air stadium. If the beam is 12 m long when it is put into place on a winter day at 00C how much longer will it be in the summer at 320C. • a = 12x10-6 m-1K-1 • D L = a LoD T= 12x10-6 m-1K-1(12m)(32K) • D L = .0046 m

  10. Problem Example • A structure steel beam holds up a scoreboard in an open-air stadium. If the beam is 12 m long when it is put into place on a winter day at 00C how much longer will it be in the summer at 320C. • a = 12x10-6 m-1K-1 • D L = a LoD T= 12x10-6 m-1K-1(12m)(32K) • D L = .0046 m

  11. Problem Example • A structure steel beam holds up a scoreboard in an open-air stadium. If the beam is 12 m long when it is put into place on a winter day at 00C how much longer will it be in the summer at 320C. • a = 12x10-6 m-1K-1 • D L = a LoD T= 12x10-6 m-1K-1(12m)(32K) • D L = .0046 m

  12. Thermo Equations • Rate of heat transfer is equal to the ratio of the product of the coefficient of thermal conductivity, Area, temperature difference and the Length of the path H = kADT l -10oC H 25oC 20oC H 25oC Lower heat transfer, A is smaller, DT Is smaller and window thickness is Greater (l) Higher heat transfer, A is larger, DT is larger and the window is not Thick (l)

  13. Thermo Equations • Rate of heat transfer is equal to the ratio of the product of the coefficient of thermal conductivity, Area, temperature difference and the Length of the path H = kADT l -10oC H 25oC 20oC H 25oC Lower heat transfer, A is smaller, DT Is smaller and window thickness is Greater (l) Higher heat transfer, A is larger, DT is larger and the window is not Thick (l)

  14. Thermo Equations • Rate of heat transfer is equal to the ratio of the product of the coefficient of thermal conductivity, Area, temperature difference and the Length of the path H = kADT l -10oC H 25oC 20oC H 25oC Lower heat transfer, A is smaller, DT Is smaller and window thickness is Greater (l) Higher heat transfer, A is larger, DT is larger and the window is not Thick (l)

  15. Thermo Equations • Rate of heat transfer is equal to the ratio of the product of the coefficient of thermal conductivity, Area, temperature difference and the Length of the path H = kADT l -10oC H 25oC 20oC H 25oC Lower heat transfer, A is smaller, DT Is smaller and window thickness is Greater (l) Higher heat transfer, A is larger, DT is larger and the window is not Thick (l)

  16. Chapter 13 Temperature • 1717 Fahrenheit – Instrument maker 0 Lowest temperature he could achieve with water,ice, and sea _____ 96 body temperature By chance water froze at ___Fand boiled at____F

  17. Chapter 13 Temperature • 1717 Fahrenheit – Instrument maker 0 Lowest temperature he could achieve with water,ice, and sea salt 96 body temperature By chance water froze at 32 and boiled at 212.

  18. Chapter 13 Temperature • 1742 Anders Celsius used the freezing and boiling points of water for reference points and then divided them into 100 equal parts. • 100 = freezing pt of ______ • 0 = boiling pt of ______ at standard pressure • Later changed to 0 C = Fpt and 100 C as Bpt • First known as centigrade scale • 1954 recognized as _______scale

  19. Chapter 13 Temperature • 1742 Anders Celsius used the freezing and boiling points of water for reference points and then divided them into 100 equal parts. • 100 = freezing pt of water • 0 = boiling pt of water at standard pressure • Later changed to 0 C = Fpt and 100 C as Bpt • First known as centigrade scale • 1954 recognized as Celsius scale

  20. Gas Laws • Vol OoC 100oC Temperature 0C

  21. Gas Laws Volume is directly related to temperature • Vol • V = constant • T OoC 100oC Temperature 0C

  22. Gas Laws • Vol -273 oC OoC 100oC Temperature 0C

  23. Gas Laws • Pressure OoC 100oC Temperature 0C

  24. Gas Laws Pressure is directly related to temperature • Pressure • P = constant • T OoC 100oC Temperature 0C

  25. Gas Laws Pressure -273 oC OoC 100oC Temperature 0C

  26. Absolute Zero • The temperature at which the volume of of an ideal gas is zero and the pressure is zero due to the lack of _______of the particles • 1848 – William Thomson – latter to be known as Lord Kelvin formalized the concept theoretically. • -273.150C is considered absolute ______or 273.16 degrees below the _____point of water (ice, water and water vapor exists at .010C at a pressure of 610 Pa)

  27. Absolute Zero • The temperature at which the volume of of an ideal gas is zero and the pressure is zero due to the lack of motion of the particles • 1848 – William Thomson – latter to be known as Lord Kelvin formalized the concept theoretically. • -273.150C is considered absolute zero or 273.16 degrees below the triple point of water (ice water and water vapor exists at .010C at a pressure of 610 Pa)

  28. Absolute Zero 0C to Kelvin = 0C +_____= Kelvin Kelvin to oC = K - _____= oC

  29. Absolute Zero 0C to Kelvins = 0C + 273 = Kelvins Kelvins to oC = K - 273 = oC

  30. Absolute Zero 0C to Kelvins = 0C + 273 = Kelvins Kelvins to oC = K - 273 = oC

  31. Gas laws continued • P Volume

  32. Gas laws continued • The pressure of a gas sample is inversely proportional to its volume • P PV = constant Volume

  33. Gas law problem • A tank having a volume of 1.00 m3 is filled with air at 00C to 20 times atmospheric pressure. How much volume will that gas occupy at 1.00 atm and 200C?

  34. Gas law problem • A tank having a volume of 1.00 m3 is filled with air at 00C to 20 times atmospheric pressure. How much volume will that gas occupy at 1.00 atm and 200C? • P1V1 = P2V2 • T1 T2 • 1.00 m3(20 atm) = V2(1.00 atm) 273 K 293 K V2 = 21.46 m3

  35. Ideal Gas Equation • PV = nRT • P = 1.013 x 105 Pa • V =.0224 m3 • n = ________ • T = 273 Kelvins • R = 8.31 m3 Pa / mol K

  36. Ideal Gas Equation • PV = nRT • P = 1.013 x 105 Pa • V =.0224 m3 • n = 1.0 moles • T = 273 Kelvins • R = 8.31 m3 Pa / mol K

  37. Ideal Gas Equation Problem • PV = nRT • What is the volume of 1.00 mole of any gas at 1.013 x105 Pa and 273 K? • V = n R T • P

  38. Ideal Gas Equation Problem • PV = nRT • What is the volume of 1.00 mole of any gas at 1.013 x105 Pa and 273 K? • V = n R T = 1.00 mol (8.31 J / mol k) (273 K) • P 1.013 x105 Pa

  39. Ideal Gas Equation Problem • PV = nRT • What is the volume of 1.00 mole of any gas at 1.013 x105 Pa and 273 K? • V = n R T = 1.00 mol (8.31 J / mol k) (273 K) • P 1.013 x105 Pa • V= .0224 m3

  40. Ideal Gas Equation • PV = nRT • n = N • NA=6.02x1023 • PV= N kB T • kB = 1.38 x 10-23 J/K

  41. Kinetic Theory • How does the microscopic movement of particles determine the macroscopic properties of matter? • James Clerk Maxwell and Ludwig Boltzman • Gas molecules are extremely small, hard, and perfectly _______-

  42. Kinetic Theory • How does the microscopic movement of particles determine the macroscopic properties of matter? • James Clerk Maxwell and Ludwig Boltzman • Gas molecules are extremely small, hard, and perfectly elastic-

  43. Kinetic Theory Assumptions • Ideal gas molecules – have no ______of their own and have ___attraction for other gas molecules. • Theoretical calculations based on the conservation of momentum associated with the _______of gas molecules on a surface and the gas laws have created an equation that relates the temperature of a gas sample to the average kinetic energy of the molecules. • K.E.avg = 3/2 KBT T is in _____Kb=____________

  44. Kinetic Theory Assumptions • Ideal gas molecules – have no volume of their own and haveNO attraction for other gas molecules. • Theoretical calculations based on the conservation of momentum associated with the collision of gas molecules on a surface and the gas laws have created an equation that relates the temperature of a gas sample to the average kinetic energy of the molecules. • K.E.avg = 3/2 KBT T is in Kelvin Kb1.38x10-23 J/K

  45. Kinetic Theory Assumptions • Ideal gas molecules – have no volume of their own and haveNO attraction for other gas molecules. • Theoretical calculations based on the conservation of momentum associated with the collision of gas molecules on a surface and the gas laws have created an equation that relates the temperature of a gas sample to the average kinetic energy of the molecules. • K.E.avg = 3/2 KBT T is in Kelvin Kb1.38x10-23 J/K

  46. Kinetic Theory Assumptions • Ideal gas molecules – have no volume of their own and haveNO attraction for other gas molecules. • Theoretical calculations based on the conservation of momentum associated with the collision of gas molecules on a surface and the gas laws have created an equation that relates the temperature of a gas sample to the average kinetic energy of the molecules. • K.E.avg = 3/2 KBT T is in Kelvin Kb1.38x10-23 J/K

  47. Average speed of gas molecules • K.E.avg = • ½ mv2avg= • (v2)avg= Root mean square velocity =

  48. Average speed of gas molecules • K.E.avg = 3/2 KBT • ½ mv2avg=3/2 KBT • (v2)avg= 3kBT m Root mean square velocity = 3kBT m

  49. Example Problem • What is the root mean square velocity of hydrogen molecules and oxygen molecules at STP? H2 = 2.0g / mole 2.0 amu • O2 = 32.0g / mole 32.0 amu Root mean square velocity= 3kBT m vrms= 3 (1.38x10-23 J / k) (273 K) =1839 m/s • 2.0 amu(1.67x10-27kg)

  50. Example Problem • What is the root mean square velocity of hydrogen molecules and oxygen molecules at STP? • O2 = 32.0g / mole 32.0 amu Root mean square velocity= 3RT M vrms= 3 (1.38x10-23 J / k) (273 K) =460 m/s • 32.0 g(1 kg/100g)

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