1. 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

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 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?

9. 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

10. 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?)

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