Transcript Lecture 27

PHYSICS 220
Lecture 26
Thermodynamics I
Lecture 26
Purdue University, Physics 220
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Lecture 26
Purdue University, Physics 220
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First Law of Thermodynamics
Energy Conservation
The change in internal energy of a system (DU) is
equal to the heat flow into the system (Q) minus the
work done by the system (W)
DU = Q - W
Work done by system
Increase in internal
energy of system
Heat flow
into system
Equivalent ways of writing 1st Law:
Q = DU + W
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Lecture 26
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Signs Example
You are heating some soup in a pan on the stove.
To keep it from burning, you also stir the soup.
Apply the 1st law of thermodynamics to the soup.
What is the sign of
A) Q
Positive, heat flows into soup
B) W Negative, stirring does work on soup
C) DU Positive, soup gets warmer
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Work Done by a System
M
M
Dy
W = F d cosq = P A d =
P A Dy = P DV
W = P DV
W > 0 if DV > 0
expanding system does positive work
W < 0 if DV < 0
contracting system does negative work
W = 0 if DV = 0
system with constant volume does no work
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PV Diagram
• Ideal gas law: PV = nRT
• For n fixed, P and V determine “state” of system
T = PV/nR
U = (3/2)nRT = (3/2)PV
• Examples:
P
P1
– which point has highest T?
P3
•B
– which point has lowest U?
•C
– to change the system from C to B,
energy must be added to system
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A
B
C
V1
V2
V
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Example (Isobaric)
2 moles of monatomic ideal gas is taken
from state 1 to state 2 at constant
pressure p=1000 Pa, where V1 = 2m3
and V2 = 3m3. Find T1, T2, DU, W, Q.
1. pV1 = nRT1  T1 = pV1/nR = 120K
2. pV2 = nRT2  T2 = pV2/nR = 180K
3. DU = (3/2) nR DT = 1500 J
P
P
1
2
V1
V2
V
J/K
R  N A k B  8.31
m
DU = (3/2) D pV = (3/2) (3000-2000)J =1500 J
(has to be the same)
4. W = p DV = 1000 J
5. Q = DU + W = 1500 + 1000 = 2500 J
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Example (Isochoric)
2 moles of monatomic ideal gas is taken
from state 1 to state 2 at constant volume
V=2m3, where T1 = 120K and T2 = 180K.
Find Q.
1.
2.
3.
4.
5.
pV1 = nRT1  p1 = nRT1/V = 1000 Pa
pV2 = nRT2  p2 = nRT2/V = 1500 Pa
DU = (3/2) nR DT = 1500 J
W = p DV = 0 J
Q = DU + W = 0 + 1500 = 1500 J
P
P2
P1
2
1
V
V
Requires less heat to raise T at const. volume than
at const. pressure CV=CP-R
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Question
Shown in the picture below are the pressure versus volume graphs for two
thermal processes, in each case moving a system from state A to state B
along the straight line shown. In which case is the work done by the system
the biggest?
A) Case 1
P(atm)
P(atm)
B) Case 2
C) Same
A
B
4
4
2
A
B
2
Case 1
3
Case 2
9 V(m3)
3
9 V(m3)
Net Work = area under P-V curve
Area the same in both cases!
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Question
Shown in the picture below are the pressure versus volume graphs for two
thermal processes, in each case moving a system from state A to state B
along the straight line shown. In which case is the change in internal energy
of the system the biggest?
P(atm)
P(atm)
A) Case 1
B) Case 2
A
B
C) Same
4
4
2
A
B
2
Case 1
3
Case 2
9 V(m3)
3
9 V(m3)
DU = 3/2 D(PV)
Case 1: D(PV) = 4x9-2x3=30 atm*m3
Case 2: D(PV) = 2x9-4x3= 6 atm*m3
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iClicker
Shown in the picture below are the pressure versus volume graphs for two
thermal processes, in each case moving a system from state A to state B
along the straight line shown. In which case is the heat added to the system
the biggest?
A) Case 1
P(atm)
P(atm)
B) Case 2
C) Same
B
4
Q = DU + W
2
A
4
A
B
2
Case 1
W is same for both
3
Case 2
9 V(m3)
3
9 V(m3)
DU is larger for Case 1
Therefore, Q is larger for Case 1
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First Law Questions
Q = DU + W
Heat flow
into system
Work done by system
P
P1 1
Increase in internal
energy of system
P3
3
2
Some questions:




V1
V2
V
Which part of cycle has largest change in internal energy, DU ?
2  3 (since U = 3/2 PV)
Which part of cycle involves the least work W ?
3  1 (since W = PDV)
What is change in internal energy for full cycle?
DU = 0 for closed cycle (since both P & V are back where they started)
What is net heat into system for full cycle (positive or negative)?
DU = 0  Q = W = area of triangle (>0)
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Special PV Cases
P
1
• Constant Pressure
– isobaric
– isochoric
1
W = PDV = 0
2
4
• Constant Temp DU = 0
2
4
P
• Constant Volume
W = PDV (>0)
3
DV > 0
V
3
DV = 0 V
– isothermic
• Adiabatic Q=0 (no heat is transferred)
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1st Law Summary
1st Law of Thermodynamics: Energy Conservation
Q = DU + W
Work done by system
Increase in internal
energy of system
Heat flow
into system
Point on P-V plot completely specifies
state of system (PV = nRT)
 Work done is area under curve
 U depends only on T
(U = 3nRT/2 = 3PV/2)
 For a complete cycle DU=0  Q=W
P

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Reversible?
• Most “physics” processes are reversible, you
could play movie backwards and still looks
fine. (drop ball vs throw ball up)
• Exceptions:
– Non-conservative forces (friction)
– Heat Flow
Heat never flows spontaneously from a colder body
to a hotter body.
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Summary of Concepts
• First Law of thermodynamics: Energy Conservation
– Q = DU + W
• Heat Engines
– Efficiency = 1-QC/QH
• Refrigerators
– Coefficient of Performance = QC/(QH – QC)
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