Paperwork
• Today
– Problems ch.20
– ?’s
• Friday
– Guest Lecture to Kick off Ch. 21
• What are on quizzes?
– Seriously – Read chapter (Yellow!)
Lab 3 Thoughts
• Exemplified heat engine processes
• Quantitative analysis of work done
Entropy Question
• A 740 g quantity of an ideal gas undergoes a
reversible isothermal compression at a
temperature of 330 K. The compression reduces
the volume of the gas from 0.40 m3 initially, to
0.32 m3 finally. The molecular mass of the gas is
320 g/mol. The entropy change for the gas, in SI
units, is ?
• Entropy, Ideal Gas, Energy Tools:
2
dQ
S  
T
1
pV  nRT
dQ  dU  dW
First Thought
• Entropy Change positive or negative?
• Is it zero?
• Why do we usually talk about S & not S?
2nd thought
• Parameters
• Initial
• Final
• Tools
2
dQ
S  
T
1
pV  nRT
dQ  dU  dW
What are constants?
• Temperature?
– Yes 330 K
• Mass (Moles)?
– Yes: Mass = 740 g & M.M. = 320 g/mol
– # moles =
– n = 2.31 moles
• Volume?
– No: Goes from 0.4 to 0.32 m3.
• Pressure?
– No: Increases
• Heat?
– No. dQ = dU + dW
– dQ = dW, dW not likely zero with volume change
• Fill in previous page, Follow main entropy Equation Next
Fill in & Follow Thru
• Parameters (SI UNITS): R=8.31
• Initial
– V1 = 0.4
n1 = 2.31
– p1 = (nRT)/V = 15357
T1 =330
• Final
– V2 = 0.32
n2 = 2.31
– p2 = (nRT)/V = 19161
T2 =330
• Tools
2
dQ
S  
T
1
pV  nRT
dQ  dU  dW
Equation Fun
Isothermal Implications
2
dQ
S  
T
1
pV  nRT
dQ  dW [Isothermal]
dQ  pdV
2
S  
1
pdV
T
dQ  dU  dW
Equation Fun
Insert Ideal Gas Constraint
2
S  
1
dQ
T
dQ  dW
pV  nRT
dQ  dU  dW
[Isothermal]
dQ  pdV
2
S  
1
pdV
T
nRT
V
2
2
2
p
 1 nRT 
1
S   dV   
dV

nR

 dV

T
T
V
V

1
1
1
p
S  nR ln V 2   ln V 1   nR ln V 2 / V 1  4.3
Entropy Units?
Why not use:
dQ = nCdT?
Make Random Engine
Make p-V Diagram
Calculate Work Done, Entropy
Constraints
Ideal Gas
cV = 3R/2, CP = 5R/2 (monotomic)
For Each Cycle: Something Constant
Tools
2
dQ
S  
T
1
pV  nRT
dQ  nCdT
dU  nCV dT
dQ  dU  dW
Friday
• Guest Lecture Ch. 21