Paperwork Next Week: Today ?’s 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? 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

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)

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

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?

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…

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

Parameters: Gas Stuff pV = nRT Q = 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?

Parameters: Gas Stuff pV = nRT Q = 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

Parameters: Gas Stuff pV = nRT Q = 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?)

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

Friday Monday: Guest Lecture Finish Ch. 20