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UW-Madison PHYSICS 207 - Lecture notes

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Lecture 23Thermodynamics: A macroscopic description of matterModern Definition of Kelvin ScaleTransferring energy to a solid (ice)Temperature scalesSome interesting factsIdeal gas: Macroscopic descriptionBoltzmann’s constantThe Ideal Gas LawExampleExample problem: Air bubble risingSlide 15Buoyancy and the Ideal Gas LawBuoyancy and the Ideal Gas LawPV diagrams: Important processesWork and Energy Transfer (Ch. 16)Work and Energy Transfer (Ch. 17)1st Law of ThermodynamicsSlide 221st Law: Work & HeatSlide 241st Law: Work (“Area” under the curve)RecapPhysics 207: Lecture 23, Pg 1Lecture 23Goals:Goals:•Chapter 16Chapter 16 Use the ideal-gas law. Use pV diagrams for ideal-gas processes.•Chapter 17Chapter 17 Employ energy conservation in terms of 1st law of TD Understand the concept of heat. Relate heat to temperature change Apply heat and energy transfer processes in real situations  Recognize adiabatic processes.•AssignmentAssignment HW9, Due Wednesday, Apr. 15th  HW10, Due Wednesday, Apr. 22nd (9 AM)Physics 207: Lecture 23, Pg 2Thermodynamics: A macroscopic description of matterRecall “3” Phases of matter: Solid, liquid & gas All 3 phases exist at different p,T conditions Triple point of water: p = 0.06 atm T = 0.01°CTriple point of CO2: p = 5 atm T = -56°CPhysics 207: Lecture 23, Pg 3Modern Definition of Kelvin ScaleWater’s triple point on the Kelvin scale is 273.16 KOne degrees Kelvin is defined to be 1/273.16 of the temperature at the triple point of waterTriple pointAccurate water phase diagramPhysics 207: Lecture 23, Pg 4Transferring energy to a solid (ice)1. Temperature increase or2. State ChangeIf a gas, then V, p and T are interrelated….equation of statePhysics 207: Lecture 23, Pg 7Temperature scalesThree main scales212Farenheit100Celcius32 0 273.15373.15KelvinWater boilsWater freezes0-273.15-459.67Absolute ZeroFTTCFo3259 FTTFCo3295K 15.273TTCK 15.273CTTPhysics 207: Lecture 23, Pg 8Some interesting factsIn 1724, Gabriel Fahrenheit made thermometers using mercury. The zero point of his scale is attained by mixing equal parts of water, ice, and salt. A second point was obtained when pure water froze (originally set at 30oF), and a third (set at 96°F) “when placing the thermometer in the mouth of a healthy man”.  On that scale, water boiled at 212. Later, Fahrenheit moved the freezing point of water to 32 (so that the scale had 180 increments).In 1745, Carolus Linnaeus of Upsula, Sweden, described a scale in which the freezing point of water was zero, and the boiling point 100, making it a centigrade (one hundred steps) scale. Anders Celsius (1701-1744) used the reverse scale in which 100 represented the freezing point and zero the boiling point of water, still, of course, with 100 degrees between the two defining points.T (K)1081071061051041031001010.1Hydrogen bombSun’s interiorSolar coronaSun’s surfaceCopper meltsWater freezesLiquid nitrogenLiquid hydrogenLiquid heliumLowest T~ 10-9KPhysics 207: Lecture 23, Pg 9Ideal gas: Macroscopic descriptionConsider a gas in a container of volume V, at pressure P, and at temperature TEquation of state Links these quantities Generally very complicated: but not for ideal gasPV = nRTR is called the universal gas constantIn SI units, R =8.315 J / mol·Kn = m/M : number of molesEquation of state for an ideal gas Collection of atoms/molecules moving randomly No long-range forces Their size (volume) is negligible Density is low Temperature is well above the condensation pointPhysics 207: Lecture 23, Pg 10Boltzmann’s constantIn terms of the total number of particles NP, V, and T are the thermodynamics variablesPV = nRT = (N/NA ) RTkB is called the Boltzmann’s constantkB = R/NA = 1.38 X 10-23 J/K PV = N kB TNumber of moles: n = m/M One mole contains NA=6.022 X 1023 particles : Avogadro’s number = number of carbon atoms in 12 g of carbon m=mass M=mass of one molePhysics 207: Lecture 23, Pg 11What is the volume of 1 mol of gas at STP ?T = 0 °C = 273 Kp = 1 atm = 1.01 x 105 PanRTpV  4.22m0224.0Pa1001.1K 273Kmol/J31.835PnRTVThe Ideal Gas LawPhysics 207: Lecture 23, Pg 12ExampleA spray can containing a propellant gas at twice atmospheric pressure (202 kPa) and having a volume of 125.00 cm3 is at 27oC. It is then tossed into an open fire. When the temperature of the gas in the can reaches 327oC, what is the pressure inside the can? Assume any change in the volume of the can is negligible.Steps1. Convert to Kelvin (From 300 K to 600 K)2. Use P/T = nR/V = constant  P1/T1 = P2/T23. Solve for final pressure  P2 = P1 T2/T1http://www.thehumorarchives.com/joke/WD40_StupidityPhysics 207: Lecture 23, Pg 13Example problem: Air bubble risingA diver produces an air bubble underwater, where the absolute pressure is p1 = 3.5 atm. The bubble rises to the surface, where the pressure is p2 = 1.0 atm. The water temperatures at the bottom and the surface are, respectively, T1 = 4°C, T2 = 23°CWhat is the ratio of the volume of the bubble as it reaches the surface,V2, to its volume at the bottom, V1? (Ans.V2/V1 = 3.74)Is it safe for the diver to ascend while holding his breath? No! Air in the lungs would expand, and the lung could rupture.Physics 207: Lecture 23, Pg 15Example problem: Air bubble risingA diver produces an air bubble underwater, where the absolute pressure is p1 = 3.5 atm. The bubble rises to the surface, where the pressure is p2 = 1 atm. The water temperatures at the bottom and the surface are, respectively, T1 = 4°C, T2 = 23°CWhat is the ratio of the volume of the bubble as it reaches the surface,V2, to its volume at the bottom, V1? (Ans.V2/V1 = 3.74)pV=nRT  pV/T = const so p1V1/T1 = p2V2/T2 V2/V1 = p1T2/ (T1 p2) = 3.5 296 / (277 1)If thermal transfer is efficient. [More than likely the expansion will be “adiabatic” and, for a diatomic gas, PV = const. where  = 7/5, see Ch. 17 & 18]Physics 207: Lecture 23, Pg 16Buoyancy and the Ideal Gas LawA typical 5 passenger hot air balloon has approximately 700 kg of total mass and the balloon itself can be thought as spherical with a radius of 10.0 m. If the balloon is launched on a day with conditions of 1.0 atm and 273 K, how hot would you have to heat the air inside (assuming the density of the surrounding air is 1.2 kg/m3 and the air behaves and as an


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UW-Madison PHYSICS 207 - Lecture notes

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