Paperwork
• HMWK deadline off by one hour
– Everyone get some bonus?
• Guest Instructors
– Monday – Chapter 20.1-20.3
– Week After Mon & Fri
• That’s when the exam is ( 2 weeks)
• Move exam back one week, skip one lab?
Schedule Short Term
• Today – Chapter 19
• Next Week
– Monday –Chapter 20.1-20.3 (Guest)
– Tuesday – Lab #2
• Quiz#2 [Chapter 18, Labs]
– Wed – Ch. 20.4-20.5
– Friday – Practice Problems
• Week After
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Monday – Ch. 20.6-20.7 (Guest)
Tuesday – Lab 3 & Quiz 3
Wed (Was exam) Review 17-20
Friday 21.1-21.3 (Guest)
• Then -
Chapter 19
• 1st Law of Thermo
– Q=DU+W (eq. 19.5)
– Q = Heat
– W = Work
– U = Internal Energy [Any Guesses]
• Internal Energy
– Sum of all KE [Thermo]
– Plus Sum of all interactions [bonds]
– Not U as in grav. potential energy
Signs & Such
• Q=DU+W (eq. 19.5)
• Talks about how heat affects a system
• What does Q being + mean?
– Heat added to system
• What happens if DU is positive?
– Raise temperature, change state…
– Chemical bonds have negative energy
– Solid to liquid means DU increases, less neg.
• What happens if W is positive?
– System does work on its surroundings
– Maybe heats up a container, etc…
Signs & Such
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Q=DU+W (eq. 19.5)
Talks about how heat affects a system
Q+  heat enters system
DU+  Internal energy raised
– Bonds broken, temperature increased
• W+  Work done on outside world
• W-  Work done on system from outside
Isolated System
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Q=DU+W
System completely isolated from outside
What is Q? (say as a function of time)
What is W? (time dependence as well)
Implications?
Isolation can be attained by expansion…
Discussion Q18.10
Start
Gas
# molecules = n0
Temperature = T0
pressure = p0
Volume = V0
Vacuum
Discussion Q18.10
“Sudden” Hole in wall
Gas Initial State
# molecules = n0
Temperature = T0
pressure = p0
Volume = V0
What Happens here?
Gas Final State
# molecules = ?
Temperature = ?
pressure = ?
Volume = ?
Changes in System
• Follow equation: Q=DU+W
– Look at small changes
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dQ = dU + dW
dU = dQ – dW
dW = pdV [gaseous systems]
dU = dQ – pdV [1st law thermo for gas]
Types of Changes
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Adiabatic (Constant Heat)
No heat transfer to/from system
Q=0
dU = dQ – dW
dU = -dW
As a whole: DU = -W
For a gas: dU = -pdV
Types of Changes
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Isochoric (Constant Volume)
No change in volume [Stiff container]
dV = 0
pdV = 0 = W
dU = dQ
As a whole: DU = Q
For a gas: DU = Q
Usually implies no work done that changes
volume
• Example: Stirring liquid usually still “isochoric”
Types of Changes
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Isobaric (Constant Pressure)
No change in volume [Stiff container]
p = constant
dQ = dU + dW
dQ = dU + pdV
p is constant of integration, no V dependence
Integrate both sides  Q = DU + p(DV)
Example: Boiling water in an open pot
Types of Changes
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Isothermal (Constant Temperature)
No change in Temp [Could add heat though…]
T = constant
dQ = dU + dW
dQ = dU + pdV
Complicated  pV = nRT
p = nRT/V (Ideal Gas)
So integration not trivial even for ideal gas
Example: Icewater mixture, while both exist
Internal Energy of Ideal Gas
• Q=DU+W
• dQ = dU + dW
• Gas: dW = pdV
• What does U depend on?
– Reminder about U
– Measure of internal KE & PE between
particles
Discussion Q18.10
Start
Gas
# molecules = n0
Temperature = T0
pressure = p0
Volume = V0
Vacuum
Discussion Q18.10
Final State (Again)
Q=DU+W
Is there work done on gas?
Is there heat input?
Gas Initial State
# molecules = n0
Temperature = T0
pressure = p0
Volume = V0
So what is DU?
Gas Final State
# molecules = ?
Temperature = ?
pressure = ?
Volume = ?
Internal Energy of Ideal Gas
• Q=DU+W
• What does U depend on?
– Temperature
– Gas doesn’t change phase (or its not a gas)
– Ideal Gas: No interactions between particles
– No potential energy (bonding) between
particles
– Diatomic molecules?
Ideal Gas Heat Capacities
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Different for different conditions
dQ = nCdT [molar heat capacity]
Constant Pressure: CP
Constant Volume: CV
Constant Temperature: CT
– Wouldn’t that mean dQ = 0?
• Constant n: Cn
– n is not in C, other part of Equation
Relationship
• CP = CV + R
• Derivation in text
• Heat capacity larger for isobaric process
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Ratio of heat capacities
g = CP/CV = 1.67 (monotomic)
g = CP/CV = 1.4 (diatomic)
g = CP/CV = 1.3 (“triatomic”)
Look at this closer later next week…
Reading & Assignments
• Chapter 18 assignment
– Up today, Due next Friday
• Chapter 19 Assignment
– Up today, Due in 1.5 weeks
• Put up practice problems
– Hopefully answers, etc…
• Read chapter 19 & sect. 20.1 to 20.3 for
Monday
Schedule Short Term
• Today – Chapter 19
• Next Week
– Monday –Chapter 20.1-20.3 (Guest)
– Tuesday – Lab #2
• Quiz#2 [Chapter 18, Labs]
– Wed – Ch. 20.4-20.5
– Friday – Practice Problems
• Week After
–
–
–
–
Monday – Ch. 20.6-20.7 (Guest)
Tuesday – Lab 3 & Quiz 3
Wed (Was exam) Review 17-20
Friday 21.1-21.3 (Guest)
• Then -