Transcript Physics 106P: Lecture 1 Notes

Thermodynamics
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
Increase in internal
energy of system
Heat flow
into system
Work done by system
P
P1
Equivalent ways of writing 1st Law:
Q = DU + W
P3
1
2
3
V1
V2
V
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Work Done by a System
M
Dy
M
The work done by the gas as it contracts is
A) Positive
B) Zero
C) Negative
W = F d cosq
=P A d = P A Dy = P DV
W = p DV :For constant Pressure
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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Thermodynamic Systems and
P-V Diagrams
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:
 which point has highest T?
»
 which point has lowest U?
P
P1
A
P3
C
V1
B
V2
V
»to change the system from C to B,
energy must be added to system
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Special PV Cases
 Constant
1
Pressure (isobaric)
1
W = PDV = 0
2
 Constant
4
2
3
DV > 0
V
3
DV = 0
 Constant
W = PDV (>0)
4
P
Volume
(isochoric)
P
V
Temp DU = 0
 Adiabatic
Q=0
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First Law of Thermodynamics
Isobaric Example
P
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. (R=8.31 J/k mole)
P
1
V1
2
V2
V
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First Law of Thermodynamics
Isochoric Example
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.
P
P2
2
P1
1
V
V
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Example: complete cycle
P
1
Wtot = ??
4
P
1
W = PDV (>0)
P
1
2
V
2
4
3
1
P
1
V
Wtot > 0
W = PDV (<0)
2
4
3
DV < 0
V
1
3
V
DV = 0
P
2
3
4
DV > 0
3
W = PDV = 0
P
4
2
V
W = PDV = 0
2
4
3
DV = 0
V
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WORK Question
If we go the opposite direction for the
cycle (4,3,2,1) the net work done by the
system will be
A) Positive
B) Negative
P
1
4
2
If we go
the other
way
then
3
V
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PV 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?
P(atm)
P(atm)
A. Case 1
B. Case 2
A
B
4
4
C. Same
A
B
2
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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PV Question2
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
4
4
C. Same
A
B
2
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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PV Question3
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
A
B
C. Same
4
4
2
A
B
2
Case 1
Q = DU + W
3
Case 2
9 V(m3)
3
9 V(m3)
W is same for both
DU is larger for Case 1
Therefore, Q is larger for Case 1
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First Law Questions
Q = DU + W
Work done by system
Increase in internal
energy of system
Heat flow
into system
Some questions:
P
P1
P3
1
2
3
V1
V2
 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)
V
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Question
Consider a hypothetical device that takes 1000 J of heat
from a hot reservoir at 300K, ejects 200 J of heat to a cold
reservoir at 100K, and produces 800 J of work.
Does this device violate the first law of thermodynamics ?
1. Yes
2. No
W (800) = Qhot (1000) - Qcold (200)
 Efficiency = W/Qhot = 800/1000 = 80%

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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 cold to hot
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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)
P


V
for a complete cycle DU=0  Q=W
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