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?
• 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
V

4 2r 2 N
pV  Nk BT
V k BT

N
p
1

4 2r 2
1
p
4 2r 2
k BT
p
k BT

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
k BT
1
p
4 2r 2 
Implications?
• What defines ?
• When M.F.P. ~  things go funny! No longer “fluid”
Low Pressure Side
k BT
1
p
4 2r 2 
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  = 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  4r2 (see in above eq.)
“surface area” of circle (2D) is 2r…
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)r3 = 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)r3 = 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
• Final
– T0 = 20oC ~ 293 K
•VF = 0.8 V0 = 5.67x10-3 m3
– V0 = (4/3)r3 = 7.15x10-3 m3
Universal
•Q constant  Adiabatic
– M = 28 g/mol [Always]
•T0(V0)g-1 = TF(VF)g-1 , g=1.4
– R = 8.31
•TF = 320 K
– n = (p0V0)/(RT0) = 0.59 mol
•Ideal Gas: DU = CVnDT
• Air: CV ~ 20 J/(mol K)
• DU = 345 J (Why CV?)
Why is DU dependent on CV?
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•
•
•
•
•
•
•
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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