Presentation on theme: "Thermodynamics -chap. 15 note for AP B physics: if it says ON gas, use W= -PDV if it says BY gas, use W= PDV always: U= Q + W."— Presentation transcript:
1Thermodynamics -chap note for AP B physics: if it says ON gas, use W= -PDV if it says BY gas, use W= PDV always: U= Q + W
2The Ideal Gas Law (review chap.14) P V = N kB TN = number of moleculesN = number of moles (n) x NA molecules/molekB = Boltzmann’s constant = 1.38 x J/KP V = n R TR = ideal gas constant = NAkB = 8.31 J/mol/KNA = Avogrado’s number = 6.02 x 1023
3= number of molecules x ave KE/molecule = N (3/2) kBT Summary of Kinetic Theory: The relationship between energy and temperature (for monatomic ideal gas)Root-mean-square speed :Careful with units: R = 8.31 J/mol/K m = molar mass in kg/mol vrms = speed in m/sInternal Energy U= number of molecules x ave KE/molecule= N (3/2) kBT= (3/2) n RT = (3/2) P V (ideal gas)
4Thermodynamicsa system (gas) interacts with surroundings via thermal processes (heat and work transferred into gas) :system surroundingsP,V,TExample: gas piston (car engine, bike pump, syringe)system = gas at pressure P and temperature T and volume Vsurroundings = piston and walls of containerstate of system: values for P,V, and T describe gas
5The 4 Laws of Thermodynamics 0. Law: Two objects at thermal equilibrium will have no heat flow between them and the same temperature T1 = T21st law: Total internal energy = heat put in + work done on it U = Q + W2nd law: Heat will spontaneously flow from hot to coldCarnot engine: ideal efficiency= (Hot- cold temp) /hot temp (otherwise: efficiency… use heat’s Q, not temp’s T)3rd law: You can’t get to absolute zero! (T= OK) (can’t remove all the energy w/o still adding some)
6First Law of Thermodynamics “The internal energy of a system tend to increase when HEAT is added and work is done ON the system.”Suggests a CHANGE or subtractionQ > 0 : heat is added to systemQ < 0 : heat is subtracted from systemW > 0 : work done on system by surroundingsW < 0 : work done on surroundings by system
7Internal Energy (DU) and Heat Energy (Q) All of the energy inside a system is called INTERNAL ENERGY, DU.When you add HEAT(Q), you are adding energy and the internal energy INCREASES.Both are measured in joules. But when you add heat, there is usually an increase in temperature associated with the change.
8Thermodynamic Systems and P-V Diagrams Ideal gas law: P V = n R TFor n fixed, P and V determine the “state” of the systemT = P V/ (n R)U = (3/2) n RT = (3/2) P VExamples:which point has highest T ?2which point has lowest U ?3to change the system from 3 to 2,energy must be added to system.Work done in a thermal process is the area underneath the P,V graph.PP1P3123V V2V
9Work done by a gasSuppose you had a piston filled with a specific amount of gas. As you add heat, the temperature rises and thus the volume of the gas expands. The gas then applies a force on the piston wall pushing it a specific displacement. Thus the gas does NEGATIVE work(uses its own internal energy to do work).
10Work is the AREA of a P vs. V graph W= -PDV = -P* (Vf- Vi) the work done on the gasCompress gas: gas expands itself smaller volume (DV=-) larger volume (DV=+) positive work more (W=+) negative work (W=-) gas has more energy gas loses energymove left on P-V diagram move right on P-V diagram-Work: clockwise +Work: counterclockwise
11Negative or positive Work Done ON gas? (P=constant) System: GasSurroundings:Piston, wallsIsobaric processyGasP,V2,T2GasP,V1,T1W = -F s = -P A D y = -P VV > 0expanding gas system does work on surroundings(isobaric process: pressure kept constant even though increase volume by adding heat – which offsets some but not all work done by gas)answer= negative work on gas
12What area is the total work done from 1 … to 4? VP1234Wtot = ??What area is the total work done from 1 … to 4?Isobaric Process: P=constant (horizontal line)Isochoric Process: V=constant (vertical line)VPW =- PDV (<0)1234DV > 0VPW = -PDV = 01234DV = 0VP1234Wtot < 0VPW =- PDV (>0)1234DV < 0VW = -PDV = 01234PDV = 0VP1234If we go the other way then Wtot > 0
13Now try this: 1 to 2 to 3 to 1 (one cycle) a Now try this: 1 to 2 to 3 to 1 (one cycle) a. What’s the total change in internal energy U? b. Is positive or negative work done? C. Is heat absorbed or released? D. Write an equation for work using P’s and V’sPP1P3123VV V2DU = 0 since no change overall (DT=0)overall net work is negative (clockwise)Heat is absorbed (Q>0) since work <0 and U =0Area = (V2-V1)x(P1-P3)/2
14Concept QuestionShown in the picture below are the pressure versus volume graphs for two thermal processes, in each case moving a system from state A to state B along the straight line shown. In which case is the work done by the system the biggest?1. Case 12. Case 23. SameAB4239V(m3)Case 1P(atm)Case 2correctNet Work = area under P-V curveArea the same in both cases!
15First Law of Thermodynamics Example VP12V V22 moles of monatomic ideal gas is takenfrom state 1 to state 2 at constant pressureP=1000 Pa, where V1 =2m3 and V2 =3m3. FindT1, T2, DU, W, Q.1. P V1 = n R T1 T1 = P V1/(nR) = 120K2. P V2 = n RT2 T2 = P V2/(nR) = 180K3. DU = (3/2) n R DT = 1500 J or DU = (3/2) P DV = 1500 J4. W = -P DV = J (neg work done on gas)5. Q = DU - W = 1500 J J = 2500 J > 0 heat gained by gas
16First Law of Thermodynamics Example 2 moles of monatomic ideal gas is takenfrom state 1 to state 2 at constant volumeV=2m3, where T1=120K and T2 =180K. Find Q.PP2P121. P V1 = n R T1 P1 = n R T1/V = 1000 Pa2. P V2 = n R T1 P2 = n R T2/V = 1500 Pa3. DU = (3/2) n R DT = 1500 J4. W = -P DV = 0 J (no change in volume)5. Q = DU - W = 1500 – 0 = 1500 J=> It requires less heat to raise T at const. volume than at const. pressure.1VV
17ExampleSketch a PV diagram and find the work done on the gas during the following stages.A gas is expanded from a volume of 1.0 L to 3.0 L at a constant pressure of 3.0 atm.The gas is then cooled at a constant volume until the pressure falls to 2.0 atm-600 J
18Example continuedThe gas is then compressed at a constant pressure of 2.0 atm from a volume of 3.0 L to 1.0 L.The gas is then heated until its pressure increases from 2.0 atm to 3.0 atm at a constant volume.+400 J
19Example continued What is the net work ON gas? -600 J + 400 J = -200 J NET work isthe area inside the shape.-600 J J = -200 JRule of thumb: If the system rotates CW, the NET work is negative.If the system rotates CCW, the NET work is positive.Side note: work by gas = +200J
20ExampleA series of thermodynamic processes is shown in the pV-diagram. In process ab 150 J of heat is added to the system, and in process bd , 600J of heat is added. Fill in the chart. On gas.150150 J600-240360 J750-240510 J-90600600-90510 J
21Classification of Thermal Processes (using work on gas definition) Isobaric : P = constant W = -P DVIsochoric : V = constant: W = 0Isothermal : T = constant( DU = 0): W = -QAdiabatic : no heat flow (Q = 0) W = - DU* Remember: work is area under P-V curve (positive work if compress gas to less volume)
22Thermodynamic Processes - Isothermal To keep the temperature constant both the pressure and volume change to compensate. (Volume goes up, pressure goes down)“BOYLES’ LAW”Same temp so DU=0DU= Q + W= 0 soQ= -WDV = + since expandsW = -PV = neg workQ= -W = positive heat added to maintain temp while gas uses own energy to expand to larger volume
23Thermodynamic Processes - Isobaric More heat Q is added to the gas than work energy used by the gas internal energy (U) decreases since PV is smaller. W= + (compress) Q= - release heat DU = - (colder at end)U= Q + W Q < W
24Thermodynamic Processes - Isovolumetric No work since constant volumeW= 0DU = QQ = + since higher tempDU = + (higher int. energy)
25Thermodynamic Processes - Adiabatic ADIABATIC- (GREEK- adiabatos- "impassable")In other words, NO HEAT can leave or enter the system.Q= 0 by definition U= Q + W U = W onlyW = - since gas expands hence U < 0 (tem drops) gas loses temp as expands since don’t provide heat to offset work done by gas
27Second Law of Thermodynamics “Heat will not flow spontaneously from a colder body to a warmer body AND heat energy cannot be transformed completely into mechanical work.”The bottom line:Heat always flows from a hot body to a cold bodyNothing is 100% efficient
28Engines Heat flows from a HOT reservoir to a COLD reservoir QH = remove from, absorbs = hotQC= exhausts to, expels = cold
29Engine EfficiencyIn order to determine the thermal efficiency of an engine you have to look at how much ENERGY you get OUT based on how much you energy you take IN. In other words:
30Rates of Energy UsageSometimes it is useful to express the energy usage of an engine as a RATE.For example:The RATE at which heat is absorbed!The RATE at which heat is expelled.The RATE at which WORK is DONE
32Is there an IDEAL engine model? Our goal is to figure out just how efficient such a heat engine can be: what’s the most work we can possibly get for a given amount of fuel?The efficiency question was first posed—and solved—by Sadi Carnot in 1820, not long after steam engines had become efficient enough to begin replacing water wheels, at that time the main power sources for industry. Not surprisingly, perhaps, Carnot visualized the heat engine as a kind of water wheel in which heat (the “fluid”) dropped from a high temperature to a low temperature, losing “potential energy” which the engine turned into work done, just like a water wheel.
33Carnot EfficiencyCarnot a believed that there was an absolute zero of temperature, from which he figured out that on being cooled to absolute zero, the fluid would give up all its heat energy. Therefore, if it falls only half way to absolute zero from its beginning temperature, it will give up half its heat, and an engine taking in heat at T and shedding it at ½T will be utilizing half the possible heat, and be 50% efficient. Picture a water wheel that takes in water at the top of a waterfall, but lets it out halfway down. So, the efficiency of an ideal engine operating between two temperatures will be equal to the fraction of the temperature drop towards absolute zero that the heat undergoes.
34Carnot EfficiencyCarnot temperatures must be expressed in KELVIN!!!!!!The Carnot model has 4 partsAn Isothermal ExpansionAn Adiabatic ExpansionAn Isothermal CompressionAn Adiabatic CompressionThe PV diagram in a way shows us that the ratio of the heats are symbolic to the ratio of the 2 temperatures
35ExampleA particular engine has a power output of 5000 W and an efficiency of 25%. If the engine expels 8000 J of heat in each cycle, find (a) the heat absorbed in each cycle and (b) the time for each cycle10,667 J2667 J0.53 s
36ExampleThe efficiency of a Carnot engine is 30%. The engine absorbs 800 J of heat per cycle from a hot temperature reservoir at 500 K. Determine (a) the heat expelled per cycle and (b) the temperature of the cold reservoir240 J560 J350 K