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Transcript
Physics150
Thermodynamics
Chapter15
TheFirstLawofThermodynamics
•  Let’sconsideranidealgasconfinedinachamberwithamoveablepiston
•  Ifwepressthepistonèthegasinthechambercompressesèpistonis
doingworkèinternalenergyofthegaschanges
ΔU = Q + W
gas-filled
chamber
Physics140,Prof.M.Nikolic
moveable
pistonof
massm
2
Thefirstlawofthermodynamics
ΔU = Q + W
Thechangeininternalenergyofasystemisequaltotheheatflowintothesystem
plustheworkdoneonthesystem(conserva]onofenergy).
Qistheheat
•  IfHEATISADDEDtothesystem(gas)èQ>0
•  IfHEATISRELEASEDbythesystem(gas)èQ<0
Wisthework
•  IfWORKISDONEONthesystem(gas)ègascompressesèW>0
•  IfWORKISDONEBYthesystem(gas)ègasexpandsèW<0
Physics140,Prof.M.Nikolic
3
Conceptualques]on–Thefirstlawof
thermodynamics
Q1
Anamountofheatequalto2500Jisaddedtoasystem,and1800J
ofworkisdoneonthesystem.Whatisthechangeininternal
energyofthesystem?
A. 
B. 
C. 
D. 
2500J
700J
1800J
4300J
ΔU = Q + W
IfheatisaddedèQ>0èQ=2500J
IftheworkisdoneonèW>0èW=1800J
ΔU = 2500J +1800J = 4300J
30
Physics140,Prof.M.Nikolic
4
Thermodynamicprocesses
Gasisexpandingètheworkisdoneby
thegasètheworkonthegasisnega]ve
W = −Fd
W = −PAd
W = −PΔV
Ifworkistobenon-zero,thegasmustexpandorcontract.
•  WhenVf<VièthegashascontractedandW>0
•  WhenVf>VièthegashasexpandedandW<0.
Physics140,Prof.M.Nikolic
5
PVDiagrams
Pressure
Pressure
Pi
Pi
Pf
Vi
Vf Volume
•  Thepressureisstayingconstant.
•  Themagnitudeoftheworkdone
istheareaunderneaththe
“curve.”
Physics140,Prof.M.Nikolic
Vi
Vf Volume
•  Boththepressureandthevolume
change
•  Themagnitudeofthetotalworkis
s]lltheareaunderthecurve.
6
PVDiagrams
isobaricprocess
Pressure
•  Constantpressure(isobaricprocess)
isochoricprocess
Pi
isothermalprocess
W = −PΔV = −P(V f −Vi )
•  Constantvolume(isochoricprocess)
W = −PΔV = 0
Pf
•  Constanttemperature(isothermal
process)
Vi
V f
adiaba]cprocess
(steepercurvethanfor
isothermalprocess)
Physics140,Prof.M.Nikolic
Volume
•  Noheattransfer(adiaba]c
process)
Q=0
7
Thermodynamicprocessesforanidealgas
PV = nR T
•  Constantvolume(isochoricprocess)
W = −PΔV = 0
Q = nCV ΔT
ΔU = Q + W = nCV ΔT
•  Constantpressure(isobaricprocess)
W = −PΔV = −nRΔT
ΔU = Q + W = nCP ΔT − nRΔT
Q = nCP ΔT
CV = CP − R
•  Constanttemperature
(isothermalprocess)
Physics140,Prof.M.Nikolic
ΔU = 0
" Vf %
−Q = W = −nRT ln $ '
# Vi &
8
Exercise:Thermodynamicprocesses
Howmuchheatisrequiredinanisochoricprocesstoraisethetemperatureof
5molesofanideal,monatomicgasfrom-50oCto50oC?
"3 %
Q = nCV ΔT = n $ R ' (Tf − Ti )
#2 &
3
Q = 5 8.315(50 − (−50)) = 6236J
2
Howmuchheatisneededtoisothermallychangethevolumeof20molesofanideal
gasfrom1.0m3to0.4m3ifthetemperatureremainsat-23.0oC?
! Vf $
Q = nRT ln # &
" Vi %
First,convertthetemperaturetoKelvins:273+(–23)=250K.
" 0.4 %
Q = 20 ⋅ 8.315⋅ 250 ⋅ ln $
' = −38095J
# 1 &
Physics140,Prof.M.Nikolic
9
Reversibilityvs.Irreversibility
Areversibleprocessisaprocessthatcanhappenequallywellfrom
stateAtostateBorstateBtostateA.
•  Perfectlyelas]ccollisionisareversibleprocess
AnIrreversibleProcess
Physics140,Prof.M.Nikolic
10
TheSecondLawofThermodynamics
•  Anyprocessthatinvolvesdissipa]onofenergyisnotreversible.
•  Anyprocessthatinvolvesheattransferfromaholerobjectto
acolderobjectisnotreversible.
Thesecondlawofthermodynamics(ClausiusStatement):
Heatneverflowsspontaneouslyfromacolderbodytoaholerbody.
Physics140,Prof.M.Nikolic
11
Heatengines
Aheatengineisadevicedesignedtotakeheatandconvertitto
mechanicalwork.
•  takeacertainamountofenergyQHfroma
“hotreservoir”
•  convertsomeofthatenergytoworkW
•  expeltheremainingenergyQCintoa“cold
reservoir.”
C
QH = W
C
QH = W + QC
Physics140,Prof.M.Nikolic
Nolosses!
100%efficiency!
12
Efficiencyofanengine
Theefficiencyofaheatengineisequaltohowmuchenergyit
convertstoworkfromtheenergyistakesinfromthehotreservoir.
net work output Wnet
e=
=
heat input
QH
QH − QC
QC
e=
= 1−
QH
QH
Efficiency(e)isaunitlessquan]tythatgoesfrom0(W=0)to1(QC=0).
Physics140,Prof.M.Nikolic
13
Refrigeratorsandheatpumps
HeatpumpsandrefrigeratorsbothoperateundertheprincipleofremovingheatQC
fromoneenvironmentandexpellingheatQHintoanotherenvironmentby
performingworkW.
QC + W + QH = 0 ⇒ W + QC = −QH
•  QC>0becauseisbeingaddedtothesystem
•  W>0becauseisbeingaddedtothesystem
•  QH<0becauseisbeingremovedfromthesystem
PerformanceofheatpumpsismeasuredwiththecoefficientofperformanceK:
QH
heat delivered
•  Heatpumps K P = net work input = W
net
•  Refrigerators K R =
Physics140,Prof.M.Nikolic
Q
heat removed
= C
net work input Wnet
14
Exercise:Heatengine
Wnet
e=
QH
a)  Howmuchheatdoesanenginewithefficiencyof33.3%absorb
inordertodeliver1.00kJofwork?
Wnet
e=
QH
Wnet
QH =
e
1000J
QH =
= 3030.3J
0.33
b)Howmuchheatisexhaustedbytheengine?
QH = W + QC
QC = QH − W
QC = 3030.3J −1000J = 2030.3J
Physics140,Prof.M.Nikolic
15
ReversibleEnginesandHeatPumps
Let’simagineacombina]onofaheatengineandaheatpump.Theworkgenerated
bytheheatengine,WH.E.,canbeusedastheworkweputintotheheatpump,
WH.P..Theyoperatebetweenthesametemperatureextremes.
Noenginecanhaveanefficiencygreaterthanthatofanidealreversibleengine
thatusesthesametworeservoirs.Theefficiencyofthisidealreversibleengineis
TC
er = 1 − .
TH
HotReservoir,TH
QH,HE
QH,HP
WHE
WHP
QC,HE
Physics140,Prof.M.Nikolic
QC,HP
ColdReservoir,TC
16
Carnotcycle
Theidealreversibleenginewiththehighestefficiency.
Aheatengineiscyclic–itreturnstothesamepressure,volume
andtemperaturethatithadaperthecyclehasbeencompleted
1.  Theenginesucksinheatfromthehotreservoirbut,byexpanding,
keepsthetemperatureconstant
2.  Thegascon]nuesexpandingadiaba]cally,loweringthetemperature
3.  Thegasexhaustsheatandcompresses,keepingthetemperature
constant
4.  Thegascon]nuescompressingadiaba]callyandreturnstoitsoriginal
configura]on
Physics140,Prof.M.Nikolic
17
TheCarnotcycle
illustrated
Physics140,Prof.M.Nikolic
18
Exercise:Engineefficiency
er = 1−
TC
TH
Anengineoperatesbetweentemperatures650Kand350Kat65.0%ofits
maximumefficiency.
a)Whatistheefficiencyofthisengine?
Themaximumpossibleefficiencyis
er = 1−
TC
TH
350 K
er = 1−
= 0.462
650K
Theengineoperatesate=0.65er=0.30or30%efficiency.
Physics140,Prof.M.Nikolic
19
Exercise:Engineefficiency
e=
QH − QC
Q
= 1− C
QH
QH
Anengineoperatesbetweentemperatures650Kand350Kat65.0%ofits
maximumefficiency.
b)If6.3×103Jisexhaustedtothelowtemperaturereservoir,howmuchworkdoes
theenginedo?
QH − QC
QC
e=
= 1−
QH
QH
QC
(1− e) =
QH
QC
QH =
1− e
QH =
Wnet = QH − QC
Physics140,Prof.M.Nikolic
6300J
= 9000J
1− 0.3
Wnet = 9000 − 6300 = 2700J
20
Entropy
Heatflowsfromobjectsofhightemperaturetoobjectsatlowtemperaturebecause
thisprocessincreasesthedisorderofthesystem.
Entropyisameasureofasystem’sdisorder.
IfanamountofheatQflowsintoasystematconstanttemperature,thenthe
changeinentropyis
Q
ΔS =
T
Everyirreversibleprocessincreasesthetotalentropyoftheuniverse.
Reversibleprocessesdonotincreasethetotalentropyoftheuniverse.
Physics140,Prof.M.Nikolic
21
Entropyandthesecondlawof
thermodynamics
Athermallyisolatedsystemwillneverseeitsentropydecrease,i.e.,
ΔS≥0foranyprocessinathermallyisolatedsystem.
Physics140,Prof.M.Nikolic
22
TheThirdLawofThermodynamics
Itisimpossibletocoolasystemtoabsolutezero.
•  Allgasesliquefyatlowtemperaturessonoexperimentaldataavailable
forlowtemperatures
•  Extrapolate!
Physics140,Prof.M.Nikolic
23