1 / 54

HW Answer

HW Answer. What was last night’s answer?. HW Problem. A Helium atom has a mass of 6.64 × 10 -27 kilograms. The sun’s core is made of Helium and has a temperature of 5800 K. What is <v> of the atoms in the sun’s core?. HW Problem. A Helium atom has a mass of 6.64 × 10 -27 kilograms.

barton
Download Presentation

HW Answer

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. HW Answer • What was last night’s answer?

  2. HW Problem • A Helium atom has a mass of 6.64 × 10-27 kilograms. • The sun’s core is made of Helium and has a temperature of 5800 K. • What is <v> of the atoms in the sun’s core?

  3. HW Problem • A Helium atom has a mass of 6.64 × 10-27 kilograms. • The sun’s core is made of Helium and has a temperature of 5800 K. • What is <v> of the atoms in the sun’s core?

  4. HW Problem • A Helium atom has a mass of 6.64 × 10-27 kilograms. • The sun’s core is made of Helium and has a temperature of 5800 K. • What is <v> of the atoms in the sun’s core? • <v> = 6013 m/s

  5. Remember Work • Famous Saying: • “Work = Force times Distance” • Really it’s • Work = Force Cross Product Displacement or • Work = F Δx Cos θ, • when θ = 0, Then • Work = F Δx

  6. Remember Work • Work = FΔx • Multiply by y*z/y*z

  7. Remember Work • Work = FΔx • Multiply by y*z/y*z (= times by 1) • Which is Multiplicative Identity Property • Work =(F/y*z)*y*z*Δx

  8. Remember Work • Work = FΔx • Multiply by y*z/y*z • Work =(F/y*z)*y*z*Δx • y*z = Area.

  9. Remember Work • Work = FΔx • Multiply by y*z/y*z • Work =(F/y*z)*y*z*Δx • y*z = Area. • Work = F/Area * Area*Δx

  10. Remember Work • Work = FΔx • Multiply by y*z/y*z • Work =(F/y*z)*y*z*Δx • y*z = Area. • Work = F/Area * Area*Δx • F/Area = Pressure • Area*Δx = 3 dimensions = volume • Work (by gas)= Pressure * ΔV

  11. Work done by a gas • A force is applied to slowly compress the gas • The compression is slow enough for all the system to remain same temperature. • W = - P ΔV • Work done ON the gas. (compression) • W = + P ΔV • Work done BY the gas is + P ΔV (Expansion)

  12. Steam Engines operate this way. • Heat (Energy) is added to a gas • It expands. • A piston turns that expansion into useful work.

  13. Problem. • An ideal gas: • Starts off at 5 m3, 3 N/m2. • Is heated in a sealed container to 5 m3, 6 N/m2. • Allowed to expand to 8 m3 , 6 N/m2. • Cools over time to 8 m3, 3 N/m2. • Forcibly compressed back to 5 m3, 3 N/m2. • What is the total work done? • 5 stages, 4 works. Work is from stage to stage. Find each work and add.

  14. An ideal gas: • Starts off at 5 m3, 3 N/m2. • Is heated in a sealed container to 5 m3, 6 N/m2. • W = + P ΔV • ΔV=(5-5) = 0 so W = 0

  15. Is heated in a sealed container to 5 m3, 6 N/m2. • Allowed to expand to 8 m3 , 6 N/m2. • W = + P ΔV • W = 6 N/m2 x (5 - 8) m3 • W = -18 J • However, gas expanded, so use positive work done by gas. • W = 18 J

  16. Allowed to expand to 8 m3 , 6 N/m2. • Cools over time to 8 m3, 3 N/m2. • W = + P ΔV • ΔV=8 – 8 = 0 so W = 0

  17. Cools over time to 8 m3, 3 N/m2. • Forcibly compressed back to 5 m3, 3 N/m2. • W = + P ΔV • W = 3 N/m2 x (8 - 5) m3 • W = 9 J • However, gas compressed, so use negative work done by gas. • W = -9 J • Total work done is the sum • W = 0 + 18J + 0 - 9J = +9J of work • Work done by the gas

  18. Starts off at 5 m3, 3 N/m2. • Is heated in a sealed container to 5 m3, 6 N/m2. • Allowed to expand to 8 m3 , 6 N/m2. • Cools over time to 8 m3, 3 N/m2. • Forcibly compressed back to 5 m3, 3 N/m2.

  19. Problem. • An ideal gas: • Starts off at 5 m3, 3 N/m2. • Is heated in a sealed container to 5 m3, 6 N/m2. • Allowed to expand to 8 m3 , 6 N/m2. • Cools over time to 8 m3, 3 N/m2. • Forcibly compressed back to 5 m3, 3 N/m2. • What if we had run through those steps backwards?

  20. 5 m3, 3 N/m2. • Gas expands • to 8 m3, 3 N/m2. • W = + P ΔV • W = 3 N/m2 x (8 - 5) m3 • W = 9 J • Gas expanded, so use positive work done by gas.

  21. An ideal gas: • Is cooled in a sealed container to 5 m3, 6 N/m2 to • 5 m3, 3 N/m2. • W = + P ΔV • ΔV=(5-5) = 0 so W = 0

  22. 8 m3 , 6 N/m2. • Compress in a sealed container • to 5 m3, 6 N/m2. • W = + P ΔV • W = 6 N/m2 x (5 - 8) m3 • W = -18 J • Gas compressed, so use negative work done by gas.

  23. Start with 8 m3, 3 N/m2. • Heat to 8 m3 , 6 N/m2. • W = + P ΔV • ΔV=8 – 8 = 0 so W = 0 • Total work done is the sum • W = 9J + 0J + -18J + 0J = -9J of work • Work done on the gas • So doing the steps in reverse order gives the same magnitude of work with an opposite sign. “Opposite Work”.

  24. How can we apply this? • If an engine burns gas, the expanding gas can do work on the wheels and move the car. • Q: But in reverse, if the wheels move the engine backwards, is there energy that is created that can be stored? • Ans: Yes • What do we call it when this engine can go both directions? Can an engine create gas? How do we store the energy?

  25. Thermal expansion • Heat causes material to expand. • (Heat = adding thermal energy) • Cold causes material to contract. • (Cold = subtracting thermal energy Cold = negative heat)

  26. Thermal Expansion • As temperature increases, the amplitude of molecular vibration increases. • This causes the overall object as a whole to expand

  27. This is why most thermometers work

  28. Expansion of Gas • For a gas, use ideal gas law. • PV =NkbT

  29. Expansion of Gas • For a gas, use ideal gas law. • PV =NkbT • N will always be constant as long as no new matter joins the cloud. • kb will always be constant.

  30. Expansion of Gas • So, • PV/T =Nkb • PV /T = “constant” • So: P1V1/T1 = P2V2/T2

  31. P V T Equation • P1V1/T1 = P2V2/T2 • Works only if T is in Kelvin.

  32. P V T Equation • P1V1/T1 = P2V2/T2 • Works only if T is in Kelvin. • Tip: It isn’t always necessary to convert to Kelvin, but it is never wrong to convert to Kelvin.

  33. HW Problem • A cylinder of gas occupies 750 mL at 10.0 C and at a pressure of 1.50 atm. What will its volume be at 1.20 atm and 80.0 C?

  34. HW Problem • A cylinder of gas occupies 750 mL at 10.0 C and at a pressure of 1.50 atm. What will its volume be at 1.20 atm and 80.0 C? Using P1V1/T1 = P2V2/T2 Gives (1.50atm*750mL/283K)*(353K/1.20atm) = V2 (in mL)

  35. Expansion of Solids • Solids also expand and contract with temperature. There are three ways to look at it. • Linear Expansion (1-D) • Area Expansion (2-D) • Volume Expansion(3-D)

  36. New Symbol • Another alpha • Coefficient of linear expansion: • How “expandable” the length of a material is. How much it expands in the X direction. • Depends on the material. Look it up in the book, or wikipedia.

  37. Linear Expansion • For small changes in temperature

  38. Linear Expansion Add equation below to WOD.

  39. Problem • For Aluminum a = .0000231 • A strip of aluminum 1 meter long at 20 C is heated from 20 C to 100 C. Find the new length. L = Lo + Δ L

  40. Area Expansion • Coefficient of area expansion: = g g = gamma • How “expandable” the area of a material is. How much it expands in the X and Y direction or area. • Depends on the material. Look it up in the book, or wikipedia. • g = 2a (usually)

  41. Original Size, Unheated Twist bar into a circle for the next slide. Heated Size, Expands in all Directions Δarea = gareaoΔT

  42. Question • A metal ring is heated. • What will happen to b? • What will happen to a?

  43. .A .A Original Size, Unheated Twist bar into a circle for the next slide. Heated Size, Expands in all Directions Δarea = gareaoΔT

  44. Area Expansion • Area expansion is “photographic.” • Gaps increase in size. • The metal will not expand “into the hole” • Radius a gets bigger

  45. Area Expansion • Before: • After:

  46. Expansion of gaps • A gap inside a material will expand as if it were made of the surrounding material.

  47. I punched a hole in a penny • The diameter of the hole is .3 cm. • Pennies are made of copper, so g = 3.4 * 10-5 • I increase the temperature from 20 C to 175 C. • What is the area of the hole now?

  48. I punched a hole in a penny • The diameter of the hole is .3 cm. • Pennies are made of copper, so g = 3.4 * 10-5 • I increase the temperature from 20 C to 175 C. • What is the area of the hole now? • Δarea = gareaoΔTareanew = areao+Δarea

  49. Area Expansion. Add equation below to WOD.

  50. New Symbol • Volume Expansion • = b= “beta” • How much it expands in the X, Y, and Z directions or volume. • How “expandable” a material is in 3 dimensions. • Depends on the material. Look it up in the book, or wikipedia. • b = 3a (usually)

More Related