Page Physics 207 – Lecture 22Physics 207: Lecture 22, Pg 1Lecture 22Goals:Goals:••WrapWrap--up chapter 15 (fluids)up chapter 15 (fluids)••Start chapter 16 (thermodynamics) Start chapter 16 (thermodynamics) ••AssignmentAssignment HW-9 due Tuesday, Nov 22 Monday: Read through Chapter 16Physics 207: Lecture 22, Pg 2StreamlineslKeep track of a small portion of the fluid:Page Physics 207 – Lecture 22Physics 207: Lecture 22, Pg 3Continuity equationA1A2v1v2A1v1: units of m2m/s = volume/sA2v2: units of m2m/s = volume/sA1v1=A2v2Physics 207: Lecture 22, Pg 4 Assuming the water moving in the pipe is an ideal fluid, relative to its speed in the 1” diameter pipe, how fast is the water going in the 1/2” pipe? Exercise ContinuitylA housing contractor saves some money by reducing the size of a pipe from 1” diameter to 1/2”diameter at some point in your house. v1v1/2(A) 2 v1(B) 4 v1(C) 1/2 v1(D) 1/4 v1Page Physics 207 – Lecture 22Physics 207: Lecture 22, Pg 5Energy conservation: Bernoulli’s eqnv1F1=P1A1net work=(P1ΔV-P2ΔV)=(P1-P2) ΔVv2work=F1Δx=P1A1Δx=P1ΔVΔxF2=P2A2Physics 207: Lecture 22, Pg 6lWith height, the change in the potential energy must also be taken into accountP1+1/2 ρ v12+ρgy1=P2+1/2 ρv22+ρgy2P+1/2 ρ v2+ρgy= constantP1+1/2 ρ v12=P2+1/2 ρ v22net work=(P1-P2) ΔV=1/2 Δm v22-1/2 Δm v12Page Physics 207 – Lecture 22Physics 207: Lecture 22, Pg 7Toricelli’s LawP + ½ ρ v2 = constA BP0+ ρ g h + 0 = P0+ 0 + ½ ρ v2 2g h = v2P0= 1 atmABPhysics 207: Lecture 22, Pg 8Elastic propertieslYoung’s modulus: measures the resistance of a solid to a change in its length.L0∆LFF/A0= Y (ΔL/L0)lBulk modulus: measures the resistance of solids or liquids to changes in volume.V0V0-∆VFF/A0= -B (ΔV/V0)Page Physics 207 – Lecture 22Physics 207: Lecture 22, Pg 9Carbon nanotube 100 x 1010Physics 207: Lecture 22, Pg 10Thermodynamics: A macroscopic description of matterA)One billionB)1015C)1024D)10100E)None of the abovelRoughly, how many atoms are there in a handfullof stuff?Page Physics 207 – Lecture 22Physics 207: Lecture 22, Pg 11Microscopic to macroscopic connectionlOne mole of substance = 6.02 x 1023basic particlesO2has atomic mass of 321 mole of O2will have 32 grams of massmass of one O2molecule=32 grams/6.02 x 1023Physics 207: Lecture 22, Pg 12TemperaturelThree main scales212Farenheit100Celcius32 0 273.15373.15KelvinWater boilsWater freezes0-273.15-459.67Absolute ZeroPage Physics 207 – Lecture 22Physics 207: Lecture 22, Pg 13Phase-diagramslRecall “3” Phases of matter: Solid, liquid & gas lAll 3 phases exist at different p,T conditionslTriple point of water: p = 0.06 atmT = 0.01°CPhysics 207: Lecture 22, Pg 14Ideal Gas LawlAssumptions 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 constantPage Physics 207 – Lecture 22Physics 207: Lecture 22, Pg 15A)DoubleB)Remain the sameC)HalvedD)None of the abovelSuppose 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 22, Pg 16Boltzmann’s constantP V = n R Tn: # of molesn=N/NAP V = n R T=(N/NA) R T=N (R/NA) TP V= N kBTkB: Boltzmann’s constantPage Physics 207 – Lecture 22Physics 207: Lecture 22, Pg 17PV diagramsVolumePressure1 Liter1 atm3 Liters3 atmPhysics 207: Lecture 22, Pg 18PV diagrams: Important processeslIsochoric process: V = const (aka isovolumetric)lIsobaric process: P = constlIsothermal process: T = constPage Physics 207 – Lecture 22Physics 207: Lecture 22, Pg 19VolumePressure2211 TpTp=12VolumePressure2211 TVTV=12VolumePressure2211VpVp=12lWhich one of the following PV diagrams describe an isobaric process (P=constant) A) B) C)V constant,isochoricPV constant,isothermalP
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