Lecture 23Ideal Gas LawSlide 3Boltzmann’s constantPV diagramsPV diagrams: Important processesSlide 7Work done on a gasSlide 9Slide 10Slide 11Work and Energy TransferSlide 13First Law of ThermodynamicsSlide 15Isochoric process (const V)Isothermal expansion (const T)Physics 207: Lecture 23, Pg 1Lecture 23Goals:Goals:•Wrap up Chapter 16 and start Chapter 17Wrap up Chapter 16 and start Chapter 17•AssignmentAssignment HW-9 due Tuesday, Nov 22 Wednesday: Read Chapter 17 Third test on Thursday, December 1Physics 207: Lecture 23, Pg 2Ideal Gas LawAssumptions that we will make: “hard sphere” model for the atoms density is low temperature not too highP V = n R Tn: # of molesR=8.31 J/mol K: universal gas constantPhysics 207: Lecture 23, Pg 3A)DoubleB)Remain the sameC)HalvedD)None of the aboveSuppose that we have a sealed container at a fixed temperature. If the volume of the container is doubled, what would happen to pressure?P V = n R TP V = constant if n, T fixedPhysics 207: Lecture 23, Pg 4Boltzmann’s constantP V = n R Tn: # of molesn=N/NAP V = n R T=(N/NA) R T =N (R/NA) TP V= N kB TkB: Boltzmann’s constantPhysics 207: Lecture 23, Pg 5PV diagramsVolumePressure1 Liter1 atm3 Liters3 atmPhysics 207: Lecture 23, Pg 6PV diagrams: Important processesIsochoric process: V = const (aka isovolumetric)Isobaric process: P = constIsothermal process: T = constPhysics 207: Lecture 23, Pg 7VolumePressure2211 TpTp12VolumePressure2211 TVTV12VolumePressure2211VpVp 12Which one of the following PV diagrams describe an isobaric process (P=constant) A) B) C)V constant,isochoricPV constant,isothermalP constant,isobaricPhysics 207: Lecture 23, Pg 8Work done on a gasFext=P AΔxΔW = Fext Δx = P A Δx= P ΔVΔV=Vfinal-VinitialΔW=-P ΔVPhysics 207: Lecture 23, Pg 9Work done on a gasΔW=-P ΔVW= - the area under the P-V curveW= the area under the F-x curveVolumePressure12Physics 207: Lecture 23, Pg 10For an isochoric process (V=const), W=0VolumePressure12Physics 207: Lecture 23, Pg 11How does the work done on the gas compare for the below two processes?VolumePressure12A) |W1|>|W2| B) |W1|=|W2|C) |W1|<|W2|Physics 207: Lecture 23, Pg 12Work and Energy TransferWhen you do work on a system, the energy of the system increases:Potential Energy (ΔU)Kinetic Energy (ΔK)Thermal Energy (ΔEthermal)ΔU+ΔK+ΔEthermal=ΔEsystem=WexternalPhysics 207: Lecture 23, Pg 13This description is incomplete. Energy can be transferred without doing any work. Q: Thermal energy transferΔU+ΔK+ΔEthermal=ΔEsystem=Wexternal+QQ>0, environment is at a higher temperatureQ<0, environment is at a lower temperaturePhysics 207: Lecture 23, Pg 14First Law of ThermodynamicsΔU+ΔK+ΔEthermal=ΔEsystem=Wexternal+QFor systems where there is no change in mechanical energy:ΔEthermal =Wexternal+Q1 calorie=4.186 Joules1 food calorie=1000 caloriesPhysics 207: Lecture 23, Pg 15SystemEnvironmentHeat to system, Q>0Work on system, W>0 Work by system, W<0Heat from system, Q<0Energy in Energy outPhysics 207: Lecture 23, Pg 16Isochoric process (const V)VolumePressurePiPfW=0•Work is: -(the area under the curve)Work is: -(the area under the curve)•Using ideal gas law:Using ideal gas law:PV=nRTTf < Ti•From first law of thermodynamics we conclude:From first law of thermodynamics we conclude:ΔEthermal =W+Q<0Q < 0Physics 207: Lecture 23, Pg 17Isothermal expansion (const T)VolumePressurePiPfViVfΔEthermal=0•Temperature does not changeTemperature does not changeW<0•Work done on the gas is:Work done on the gas is:•From first law of thermodynamics we conclude:From first law of thermodynamics we conclude:ΔEthermal =W+Q=0
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