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Paperwork

Paperwork. Next Week: Mon: Guest Lecture – Finish Ch. 20 Tues: Lab3 Wed: Problem Solving Friday: Guest Lecture – Begin Ch. 21 Today Problems 18&19 (maybe20) ?’s. Problem Ch. 18. Extension of 18.66 At what pressures does an ideal gas no longer behave like an ideal gas?

macy-rutledge
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Paperwork

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  1. Paperwork • Next Week: • Mon: Guest Lecture – Finish Ch. 20 • Tues: Lab3 • Wed: Problem Solving • Friday: Guest Lecture – Begin Ch. 21 • Today • Problems 18&19 (maybe20) • ?’s

  2. Problem Ch. 18 • Extension of 18.66 • At what pressures does an ideal gas no longer behave like an ideal gas? • Consider Neutral O2 molecules • Size of 1 molecule ~ 2x10-10m • T = 300K • Container is sphere with diameter 1 m • Chemically Inert • Discuss: What makes an ideal gas ideal? • Look at pressure range from high to low

  3. High Pressure Side • Looking for breakdown of pV=nRT • pV = NkBT • So edge of where this applies • When pressure gets too high, what happens? • Volume of one molecule ~ (2x10-10m)3 • p = (NkBT)/V • kB = 1.38x10-23 J/K • p ~ 714MPa ~ 7,000 atm ~ << Jupiter • At 1K P ~ 23 atm (300psi) << welding tank • Low T physics is fun (Shand)

  4. Low Pressure Side • Ideal gas – gas bounces off each other! • When gas bounces off container more than gas – completely different physics • Look at mean free path

  5. Low Pressure Side • Ideal gas – gas bounces off each other! • When gas bounces off container more than gas – completely different physics • Look at mean free path • What defines l? • When M.F.P. ~ l things go funny! No longer “fluid” Implications?

  6. Low Pressure Side • Consider Neutral O2 molecules • Size of 1 molecule ~ r=2x10-10m • T = 300K • Container is sphere with diameter 1 m • Chemically Inert p ~ 6 mPa ~ 6x10-8 atm ~ 4.5x10-5 Torr Good vacuum pump or cruddy outer space vacuum (near earth?) Notice what happens when you reduce container size (or particle #) If l = 1nm  p = 60 atm (low side) so at room pressure gas non-ideal Why is nanoscience cool? Why are “semi” metals fun? Also interesting aspects of different dimensions Surface area of sphere  4pr2 (see in above eq.) “surface area” of circle (2D) is 2pr…

  7. Chapter 19 • 19.36: A player bounces a basketball on the floor, compressing it to 80.0 % of its original volume. The air (assume it is essentially N2 gas) inside the ball is originally at a temperature of 20.0oC and a pressure of 2.00 atm. The ball's diameter is 23.9 cm. • A) What is T at max compression? • B) What is change in U (of air inside ball) from original state to maximum compression? • How to solve: First set up knowns & unknowns & tools we’ll need. • Tools = Equations/Conservation Laws/Constraints

  8. Parameters: Gas StuffpV = nRTQ = DU + W • 19.36: A player bounces a basketball on the floor, compressing it to 80.0 % of its original volume. The air (assume it is essentially N2 gas) inside the ball is originally at a temperature of 20.0oC and a pressure of 2.00 atm. The ball's diameter is 23.9 cm. • First find T at max compression (part A, why not?) • Original • p0 = 2 atm ~ 203 kPa • T0 = 20oC ~ 293 K • V0 = (4/3)pr3 = 5.72x10-2 m3 • Universal • M = 28 g/mol [Always] • R = 8.31 • n = (p0V0)/(RT0) = 4.77 mol • Final • VF = 0.8 V0 = 4.57x10-2 m3 • Now what? • Type of process? • What is constant? • T? V? p? Q? W? n? m?

  9. Parameters: Gas StuffpV = nRTQ = DU + W • 19.36: A player bounces a basketball on the floor, compressing it to 80.0 % of its original volume. The air (assume it is essentially N2 gas) inside the ball is originally at a temperature of 20.0oC and a pressure of 2.00 atm. The ball's diameter is 23.9 cm. • First find T at max compression (part A, why not?) • Original • p0 = 2 atm ~ 203 kPa • T0 = 20oC ~ 293 K • V0 = (4/3)pr3 = 7.15x10-3 m3 • Universal • M = 28 g/mol [Always] • R = 8.31 • n = (p0V0)/(RT0) = 4.77 mol • Final • VF = 0.8 V0 = 5.67x10-3 m3 • Q constant  Adiabatic • T0(V0)g-1 = TF(VF)g-1 • g = 1.4 (diatomic) • TF = T0 (V0/VF)g-1 • TF = T0 (1/.8).4 = 320 K

  10. Parameters: Gas StuffpV = nRTQ = DU + W • 19.36: A player bounces a basketball on the floor, compressing it to 80.0 % of its original volume. The air (assume it is essentially N2 gas) inside the ball is originally at a temperature of 20.0oC and a pressure of 2.00 atm. The ball's diameter is 23.9 cm. • First find DU. • Original • p0 = 2 atm ~ 203 kPa • T0 = 20oC ~ 293 K • V0 = (4/3)pr3 = 7.15x10-3 m3 • Universal • M = 28 g/mol [Always] • R = 8.31 • n = (p0V0)/(RT0) = 0.59 mol • Final • VF = 0.8 V0 = 5.67x10-3 m3 • Q constant  Adiabatic • T0(V0)g-1 = TF(VF)g-1, g=1.4 • TF = 320 K • Ideal Gas: DU = CVnDT • Air: CV ~ 20 J/(mol K) • DU = 345 J (Why CV?)

  11. Why is DU dependent on CV? • Know: DU depends ONLY on T • Examine constant V case • dQ = dU + dW • dW = 0 if dV = 0 • dQ = nCVdT (heat for constant volume) • dU = nCVdT  DU = nCVDT • So what happens for adiabatic? • Same thing, if true for one, true for all • Depends ONLY ON T, not PATH

  12. Friday Monday: Guest Lecture Finish Ch. 20

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