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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.

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HW Answer

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  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)

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