Physics 1210/1310 Mechanics&Thermodynamics Lecture 39~40 Thermodynamics.
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Transcript of Physics 1210/1310 Mechanics&Thermodynamics Lecture 39~40 Thermodynamics.
![Page 1: Physics 1210/1310 Mechanics&Thermodynamics Lecture 39~40 Thermodynamics.](https://reader036.fdocuments.net/reader036/viewer/2022082217/5a4d1b357f8b9ab05999cb83/html5/thumbnails/1.jpg)
Physics 1210/1310
Mechanics &
Thermodynamics
Lecture 39~40 Thermodynamics
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Phase Diagrams
For a substance which expands on melting
For an ideal gas
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First and Second Law ofThermodynamics
And Types of Thermodynamic Processes
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Thermodynamic systems
Isolated systems can exchange neither energy nor matter with the environment.
Closed systems exchange energy but not matter with the environment.
Heat
Work
reservoir
Open systems can exchange both matter and energy with the environment.
Heat
Work
reservoir
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Quasi-static processes
Quasi-static processes: near equilibrium
States, initial state, final state, intermediate state: p, V & Twell defined
• Sufficiently slow processes = any intermediate state can be considered as at thermal
equilibrium. • Criterion: It makes sense to define a temperature (ie)
there is a statistical average
• Examples of quasi-static processes: - iso-thermal: T = constant - iso-volumetric: V = constant - iso-baric: P = constant - adiabatic: Q = 0
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Work done equals area under curve in pV diagram
Expansion: work on piston positive, work on gas negativeCompression: work on piston negative, work on gas positive
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1st Law: The change of the total energy of athermodynamic system depends on change ofheat in system and work done.
More generally DE = DK + DUg +DU
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State functions
)( iff VVPW
a. isovolumetricb. isobaric
a. isobaricb. isovolumetric
)( ifi VVPW f
i
V
VPdVW
isothermal
• the work done by a system depends on the initial and final states and on the path it is not a state function.
• energy transfer by heat also depends on the initial, final, and intermediate states it is not a state function either.
a b c
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Kinds of Thermodynamic Processes:
Adiabatic – no heat transfer (by insulation or by very fast process)
U2 – U1 = -W
Isochoric – constant volume process (no work done on surroundings)
U2 – U1 = Q
Isobaric – constant pressure process
W = p (V2 – V1)
Isothermal – constant temperature process (heat may flow but very slowly so that thermal equilibrium is not disturbed)
DU, Q , W not zero: any energy entering as heat must leave as work
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Ideal gas – iso-volumetric process
Isovolumetric process: V = constant
V
P
V1,2
1
202
1
V
VPdVW
TCTTCQ
V
V
D )( 12
Heat
reservoir
During an iso-volumetric process, heat enters (leaves) the system and increases (decreases) the internal energy.
(CV: heat capacity at constant volume)
TCQUWQU
VDDD
22 TNkVP B
11 TNkVP B
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Ideal gas - isobaric process
Isobaric process: P = constant
V
P
V1
1 2 VPVVPPdVWV
VD )( 12
2
1
V2 TC
TTCQ
p
p
D
)( 12(CP: heat capacity at constant pressure)
VPTCWQU
P DDD
During an isobaric expansion process, heat enters the system. Part of the heat is used by the system to do work on the environment; the rest of the heat is used to increase the internal energy.
Heat
Work
reservoir
11 TNkPV B
22 TNkPV B
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Ideal gas - isothermal process
Isothermal process: T = constant
V
P
V1
1
2
TNkVP B22
1
2ln
2
1
2
1
2
1
VVTNk
VdVTNk
dVVTNkPdVW
B
V
VB
V
VB
V
V
V2
0DU
1
2lnVVTNkWQ B
Expansion: heat enters the system all of the heat is used by the system to do work on the environment.
Compression: energy enters the system by the work done on the system, all of the energy leaves the system at the same time as the heat is removed.
TNkVP B11
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Ideal gas - adiabatic process
V
P
V1
1
2
V2
Adiabatic process: Q = 0
dTNkfdU
TNkfU
B
B
2
2
2
1
2
1
),(V
V
V
VdVTVPPdVW
TNkPV B
dTNkVdPPdVTNkdPVd
B
B
)()(
PdVdWdQdU
Ideal gas:
Adiabatic process: PdVf
dTNk
PdVdTNkf
B
B
22
PdVf
VdPPdV2
f is degree of freedom
111 TNkVP B
222 TNkVP B
T)P(V,P
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V
P
V1
1
2
V2
),( TVPP
0)21(
2
PdVf
VdP
PdVf
VdPPdV
0VdV
PdP
constantVPPV
VPPV
VV
PP
VdV
PdP V
V
P
P
11
11
11
0ln
0lnln
011
Ideal gas - adiabatic process (contd)
let , and divided by PV)21(f
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V
P
V1
1
2
V2
constantPV
2
1
2
111
V
V
V
V VdVVPPdVW
)11()1(
1
1211
VVVPW
Ideal gas - adiabatic process (contd)
constantVPPV 11
constant
For monatomic gas,
7.1321
3
f
For diatomic gas, = 1+2/5 = 1.4 f =5 at normal T (threshold not met) f=7 at very high T
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V
P
V1
1
2
V2
constantPV
2211 VPVP
222
111
TNkVPTNkVP
B
B
2
1
2
1
2
1
TT
VV
PP
1
2
2
1
VV
PP
2
11
1
12
TT
VV
or
Ideal gas - adiabatic process (contd)
constantVPPV 11
constantVTVT 122
111
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V
P
V1
1
2
V2
constantPV constantVPPV 11
constantVTVT 122
111
Ideal gas - adiabatic process (contd)
TNkPV B
During an adiabatic expansion process, the reduction of the internal energy is used by the system to do work on the environment.
During an adiabatic compression process, the environment does work on the system and increases the internal energy.
)11()1(
1
1211
VVVPW
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SummaryQuasi-static
process Character UD WQ
adiabatic 0Q WU D
isothermal T = constant 0DU
isovolumetric
isobaric
V = constant
P = constant
QU D TCQ VD 0W
VPW DWQU D TCQ PD
1
2lnVVTNkW BWQ
)11()1(
1
1211
VVVPW
0Q
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Carnot CycleThe best-e cycle
2 reversible isothermaland 2 reversible adiabaticprocesses.
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4 stroke or Otto engine intake stroke compression stroke ignition power stroke exhaust stroke
Bottom right Ta, bottom left Tb
Use T*V-1 = const. lawfor the two adiabatic processes
For r=8, j = 1.4 e=56%
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Diesel Cycle
Typical rexp ~ 15, rcomp ~ 5
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Second Law of Thermodynamics
No system can undergo a process where heatis absorbed and convert the heat into workwith the system ending in the state where itbegan: No perpetuum mobile.
b/c heat cannotflow from a colder toa hotter body w/oa cost (work).
iow e=100% is notpossible!
Reversible vs irreversibleprocesses.
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Second Law of Thermodynamics:
gives direction to processes
No system can undergo a process where heatis absorbed and convert the heat into workwith the system ending in the state where itbegan: No perpetuum mobile.
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There are many other useful state functions:the thermodynamic potentials
Enthalpy H = U + PV
Free energy at const P, T G = U + PV - TS
Free energy at const. TF = U – TS
Entropy Setc.
The first law revised (for a p-V-T system):DU= TdS - pdV
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Entropy: macroscopic interpretation:
Cost of Order – reversible and irreversible processes
Total entropy change zero:reversible
Use dQ = T dS in first law!
DU = TDS - pDV
What is this entropy?No easy answer, dep. on the case
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Entropy: microscopic meaning
Total entropy change zero:reversible
w no of possible states
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Loose ends: The 3rd Law of Thermodynamics
3rd Law: It is impossible to reach absolute zero.
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‘Fun’ mnemonic about thermodynamic laws:
1st Law ‘You can’t win, you can only break even’
2nd Law ‘You can break even only at absolute zero’
3rd Law ‘You cannot reach absolute zero’
Moral: one can neither win nor break evenThe American Scientist , March 1964, page 40A
The laws of thermodynamics give ALL processes a direction,even the one’s in mechanics, E&M, etc.