1、 2019Central Heating first second laws thermodynamics.adiabatic,isochoric,isothermal,and isobaric ideal efficiency coefficient of performance Usually the internal energy consists of the sum of the potential and kinetic energies of the working gas molecules.hothotIncrease in Internal Energy,U.Initial
2、 State:P1 V1 T1 n1 Final State:P2 V2 T2 n2Decrease in Internal Energy,U.Initial State:P1 V1 T1 n1 Final State:P2 V2 T2 n2Q=U+W final-initial)Q=U+W final-initial)Work BY a gas is positive Work ON a gas is negativeUU Q=U+WApply First Law:U=+280 JW is positive:+120 J(Work OUT)Q is positive:+400 J(Heat
3、IN)U=+280 JThe increase in internal energy is:Energy is conserved:Q=U+W+U-U1=+U-UHeat input increases V with const.P.BAPV1 V2 VA VB TA T B=Work=Area under PV curveWorkPVBAPV1 V2 VA VB TA T B=U=0 U=0 PAVA=PBVBSlow compression at constant temperature:-.BAPAV2 V1PB400 J of energy is absorbed by gas as
4、400 J of work is done on gas.T=U=0U=T=0PAVA VBPB PAVA=PBVBTA=TBlnBAVWnRTVIsothermal Work Work done at EXPENSE of internal energy INPUT Work INCREASES internal energyWork OutWork InU+UQ=0W=-UU=-WInsulated Walls:Q=0BA400 J of WORK is done,DECREASING the internal energy by 400 J:Net heat exchange is ZE
5、RO.Q=0BA=AABBPVPVOPTIONAL TREATMENTCheck with your instructor to see if this more thorough treatment of thermodynamic processes is required.Qcm t Q=nCv TQ=+4220 J .U=nCv TThus,U is determined by the change of temperature and the specific heat at constant volume.Q=U+WQ=4220 J+J SameU=nCvTCpCvA 200 K4
6、00 K800 K3(101,300Pa)(0.002m)0.122 mol(8.314J/mol K)(200K)PVnRTTA T BP B=2 atm or 203 kPa2 L 200 K400 K800 KAnalyze first law for ISOCHORIC process AB.200 K400 K800 KQ=+514 J W=0U=+514 J What is the volume at point C(&D)?200 K400 K800 K V C=V D=4 L Process BC is ISOBARIC.U=+1028 J 200 K400 K800 K Wo
7、rk depends on change in V.P=0 Work=P VW=+405 J2 L 1 atm200 K400 K800 K4 L 2 atmAnalyze first law for BC.Q=1433 J W=+405 J 200 K400 K800 K U=1028 J What is temperature at point D?800 K T D=400 K BAPB2 L 1 atm200 K400 K800 KCDAnalyze first law for ISOCHORIC process CD.U=(0.122 mol)(21.1 J/mol K)(400 K
8、-800 K)Q=-1028 J W=0U=-1028 J PB 1 atm200 K400 K800 K400 KProcess DA is ISOBARIC.U=-514 J 200 K400 K800 K4 L 400 K.W=-203 JAD 200 K400 K800 K 400 KAnalyze first law for DA.Q=-717 J W=-203 JU=-514 J 200 K400 K800 K 400 KProcess Q U W AB 514 J 514 J 0 BC 1433 J 1028 J 405 J CD-1028 J-1028 J 0 DA-717 J
9、-514 J-203 J Totals 202 J 0 202 J C2 L B -202 JArea=(1 atm)(2 L)Net Work=2 atm L=202 J Q=0PAVA PBVB=Example 2:A diatomic gas at 300 K and 1 atm is compressed adiabatically,decreasing its volume by 1/12.(VA=12VB).What is the new pressure and temperature?(=1.4)Q=0PB=32.4 atm or 3284 kPa1.412BBABVPPV1.
10、4(1 atm)(12)BP 12300 KABABVPPVQ=0TB=810 K 300 KAABBABPVPVTTQ=0B300 K810 K 1 atm300 K810 K W=-3.50 JA heat engine is any device which through a cyclic process:Cold Res.TCEngineHot Res.THQhotWoutQcoldIt is impossible to construct an engine that,operating in a cycle,produces no effect other than the ex
11、traction of heat from a reservoir and the performance of an equivalent amount of work.Not only can you not win(1st law);you cant even break even(2nd law)!WoutCold Res.TCEngineHot Res.THQhotQcoldCold Res.TCEngineHot Res.TH400 J300 J100 J A possible engine.An IMPOSSIBLE engine.EngineHot Res.TH400 J400
12、 JEngineQHWQCe=1-QCQHe=WQHQH-QC QH800 JW600 Je=1-600 J800 Je=1-QCQHe=25%Question:How many joules of work is done?For a perfect engine,the quantities Q of heat gained and lost are proportional to the absolute temperatures T.e=1-TCTHe=TH-TC THQHWQCe=1-TCTHe=1-300 K500 Ke=40%Actual e=0.5ei=20%e=WQHW=eQ
13、H=0.20(600 J)Work=120 JWin+Qcold=QhotWIN=Qhot-QcoldEngineQhotQcoldWinIf this were possible,we could establish perpetual motion!Cold Res.TCHot Res.THQhotQcoldHot Res.THQHWQCK=TH TH-TCQC WK=QH QH-QCA Carnot refrigerator operates between 500 K and 400 K.It extracts 800 J from a cold reservoir during ea
14、ch cycle.What is C.O.P.,W and QH?Engine800 JWQH500 K400 KK=TC TH-TC=C.O.P.(K)=4.0 Engine800 JWQH500 K400 K QC QH-QCQH=1000 J=4.0 800 JW1000 J500 K400 KWork=200 JQ=U+W final-initial)U=nCv TThe Molar Specific Heat capacity,C:Units are:Joules per mole per Kelvin degreeThe following are true for ANY process:Q=U+WPV=nRTAABBABPVPVTTQhotQcoldWoutNot only can you not win(1st law);you cant even break even(2nd law)!The efficiency of a heat engine:e=1-QCQHe=1-TCTHThe coefficient of performance of a refrigerator:CCinHCQQKWQQCHCTKTT
