Presentation on theme: "An Introductory Course in Waterloo Collegiate Institute"— Presentation transcript:
1 An Introductory Course in Waterloo Collegiate Institute SPH3UWThermal PhysicsAn Introductory Course inThermodynamicsWaterloo Collegiate Institute
2 Welcome to Thermodynamics Heat, Temperature, and PowerThermal ExpansionIdeal Gas LawKinetic Theory of GasesFirst Law of ThermodynamicsWork and PV DiagramsIsothermal ProcessesAdiabatic ProcessesIsobaric ProcessesIsochoric ProcessHeat Engines and the Second Law of Thermodynamics
3 Thermal Physics electrons and holes in semiconductors converting energy into workmagnetismthin films and surface chemistrythermal radiation (global warming)and much more…
4 Some DefinitionsAbsolute Zero: the lowest possible temperature, at which all molecular motion would cease and a gas would have zero volume.Calorie: the amount of heat required to raise the temperature of one gram of water by one Celsius degree.Calorimeter: device which isolates objects to measure temperature changes due to heat flow.Celsius (C): temperature scale in which the freezing point of water is 0 and the boiling point of water is 100Convection: heat transfer by the movement of a heated substance, due to the differences in density.Conduction: heat transfer from molecule to molecule in substances due to differences in temperature.
5 HeatHeat, represented by the variable Q, is a type of energy that can be transferred from one body to anotherHeat is measured in JoulesEnergy must be transferred in order to be called heat. (So heat may be gained or lost, but not possessed. It is incorrect to say, “a gas has 4000 J of heat”Internal energy:The sum of the energies of all of the molecules in a substance. Represented by the variable U (for example: the total of the Kinetic and Potential energy at the molecular level is called the Internal Energy of the system. U=KE+PE)Temperature:Related to the average kinetic energy per molecule of a substance.
6 What is Heat?Up to mid-1800’s heat was considered a substance -- a “caloric fluid” that could be stored in an object and transferred between objects. After 1850, kinetic theory.A more recent and still common misconception is that heat is the quantity of thermal energy in an object.The term Heat (Q) is properly used to describe energy in transit, thermal energy transferred into or out of a system from a thermal reservoir …(like cash transfers into and out of your bank account)QUSign of Q : Q > 0 system gains thermal energyQ < 0 system loses thermal energyW > 0 work done on systemW < 0 work done by the systemsSo we give Q + W a name: The Internal Energy
7 Still More HeatWe will be discussing three states of matter (solid, liquid, and gas). The molecules of a solid are fixed in a rigid structure. The molecules of a liquid are loosely bound and may mix with one another freely. (While a liquid has a definite volume, it still takes the shape of its container. The molecules of a gas interact with each other slightly, but usually move at higher speeds than that of solid of liquid.In all three states of matter the molecule are moving and therefore have Kinetic Energy. But, they also have Potential Energy due to the bonds between them. The sum of the potential and kinetic energies of the molecule of the substance is also known as its Internal Energy.When a warmer substance comes in contact with a cooler substance, some of the kinetic energy of the molecules in the warmer substance is transferred to the cooler substance. The energy representing this kinetic energy of the molecule that is transferred from the warmer to the cooler substance is called heat energy.
8 What is temperature? A mercury thermometer The mercury rises up the tube as it expands.This is movement.An object (say, a cup ofhot chocolate)The mercury isgaining (internal)energy fromthe hot chocolate.This transfer of energy is what we call heat.When the transfer stops, the objectsare in thermal equilibrium.
9 Temperature Scales Celsius (0C) Zero defined by an ice-water bath at 1 atm.Unit defined by water-steam (100ºC) at 1 atm.Kelvin (absolute K)Zero defined by absolute zero, but we cannotreach that temperature experimentallyK defined by the triple-point of water(0.01ºC at 4.58 mm of mercury, water can exist in all three states of matter)Unit is the same as the Celsius scaleFahrenheit (0F)Zero and unit based on salt-water, water freezes at 32 0F and boils at 212 0F )
10 Converting Between the Scales From Celsius to Kelvin:From Fahrenheit to Celsius:
11 ExampleYou place a small piece of melting ice in your mouth. Eventually, the water all converts from the ice at T1= F to body temperature, T2 = F. Express these temperatures as 0C and K.Plan: We convert Fahrenheit to Celsius temperature, then from Celsius to Kelvin
12 Mechanical Equivalent of Heat James Joule showed that mechanical energy could be converted to heat and arrived at the conclusion that heat was another form of energy.He showed that 1 calorie (c) of heat was equivalent to J of work. (that is 1 calorie is defined as the heat needed to raise the temperature of 1 gram of water 1 0C)1 cal = JKilocalorie(C) – the amount of heat needed to raise the temperature of 1,000 grams of water by 1 °C. (Used with food, Food calories (C) are determined by burning the food and measuring the amount of energy that is released.)British Thermal Units (BTUs) are the amount of heat to raise one pound of water by 1 °F.
13 Heat Transfer and Temperature Change The change in temperature that a substance experiences depends upon two things: its identity (specific heat) and the amount of material (mass). The equation that connects the amount of heat, Q, and the resulting temperature change , T in 0C, is:Where Q is the quantity of heat (calories) m is the mass of the sample in grams and c is the intrinsic property called specific heat capacity in 1 cal/g°C.Note: that positive Q is interpreted as heat coming in (T is positive, so T increases), while negative Q corresponds to heat going out (T is negative, so T decreases).
14 Heat Transfer ExampleDuring a bout with the flu an 80. kg man ran a fever of C instead of the normal body temperature of C. Assuming that the human body is mostly water (c=1cal/g 0C), how much heat, in calories and Joules, is required to raise his temperature by that that amount?We could also use:
15 Heat Transfer and Phase Changes Consider an ice cube. Since water freezes at 0 0C, the temperature of the ice cube is 0 0C. If we add heat to the ice, its temperature does not rise. Instead the thermal energy absorbed by the ice goes into loosening the intermolecular bonds of the ice, thereby transforming it into a liquid. The temperature remains at 0 0C.In each phase change (solid to liquid, liquid to gas), absorbed heat causes no temperature change so Q=mcT does not apply. The equation we use is:Where L is the latent heat of transformation (solid to liquid, or vice versa L is latent heat of fusion. From liquid to gas, L is called latent heat of vaporization). This equation tells us how much heat must be transferred in order to cause a sample of mass m to undergo a phase change.
16 Example on Temperature and Phase You want to cool 0.25kg of water, initially at 25 0C, by adding ice, initially at -20 0C. How much ice should you add so that the final temperature will be 0 0C with all the ice melted [cwater = 4190 J/kg K, cice=2100 J/kg K, L=334000J/kg]?The ice and water are the objects that exchange heat. The water undergoes a temperature change only, while the ice undergoes both a temperature and phase change. We require the mass of the ice.Let’s first determine the negative heat added to the water.For Ice, first we determine the heat needed to warm the ice.
17 Example on Temperature and Phase You want to cool 0.25kg of water, initially at 25 0C, by adding ice, initially at -20 0C. How much ice should you add so that the final temperature will be 0 0C with all the ice melted [cwater = 4190 J/kg K, L=334000J/kg]?For Ice, now we need the heat to phase shift it from solid to liquid.The sum of these quantities must be zero
18 Heat Transfer and Thermal Expansion When a substance undergoes a temperature change, it changes in size. Changes in length due to temperature is governed by:Where is the coefficient of linear expansion of the materialSimilarly, there are formulas for changes in Area and volumesWater contracts as it is cooled until it reaches 4 °C. From then on, cooling causes expansion. This is because freezing water has a crystalline structure.
19 UnderstandingYou need to slide an aluminum ring onto a rod. At room temperature (20oC), the internal diameter of the ring is 50.0 mm and the diameter of the rod is 50.1 mm.Should you heat or cool the ring to make it fit onto the rod?If the coefficient of linear expansion for aluminum is 2.5 x 10-5 oC-1, at what temperature will the ring just barely fit onto the rod?a) Heat, even the hole will expandb)Therefore the final temperature is 20oC+80oC or 100oC
20 The Kinetic Theory of Gases Unlike the condensed phases of matter (solid or liquid), molecules of a gas move freely and rapidly. A confined gas exerts a force on the walls of its container. The moving molecules strike the walls and rebound. The magnitude of the force per unit area is called pressure, and is denoted by P:The SI unit for pressure is the N/m2, the pascal (Pa).We also need a way to talk about the vast number of molecules in a given sample of gas. One mole of atoms or molecules (or teachers)contains of these elementary quantities. NA is known as Avogadro’s Number.
21 Molecular Picture of Gas Gas is made up of many individual moleculesNumber density is Number of molecules/Volume:N/V = r/mr is the mass densitym is the mass for one moleculeNumber of moles: n = N / NANA = Avogadro’s Number = 6.022x1023 mole-1Mass of 1 mole of “stuff” in grams = molecular mass in ue.g., 1 mole of N2 has mass of 2x14=28 gramsAtomic mass unit
22 The Ideal Gas LawThree physical properties – Pressure (P), volume (V), and temperature (T) describe a gas. At low densities, all gasses approach ideal behaviour; This means that these three variables are related by the equation:Where n is the number of moles of gas and R is the constant (8.31 J/mol K), called the universal gas constant. This equation is known as the Ideal Gas Law.
23 The Ideal Gas Law Therefore Boltzmann’s constant An important consequence of this equation is that, for a fixed volume of gas, an increase in P gives a proportional increase in T. By applying Newton’s Second Law we can find that the pressure – exerted by N molecules of gas in a container of volume V is related to the average kinetic energy of the molecules by the equation:Boltzmann’s constantThereforem=molecular mass (kg)M=molar mass (kg/mole)root-mean-square
24 The Ideal Gas LawWe recall that the sum of the Kinetic and Potential energy (at the molecular level) is called the Internal energy of the system.With the average kinetic energy of a molecule related to temperature by:Atoms of a single elementThen for an ideal monatomic gas (No potential energy) containing N molecules, the Internal energy will just be N time this equationn is the number of Moles and R the Ideal gas constant.
25 The Ideal Gas LawCalculate the internal energy of the air in a typical room with a volume of 50 m3. treat the air as if it were a monatomic ideal gas at 1 atm = 1.01 x 105 PAHint:
26 P V = N kB T P V = n R T The Ideal Gas Law P = pressure in N/m2 (or Pascals)V = volume in m3N = number of moleculesT = absolute temperature in Kk B = Boltzmann’s constant = 1.38 x J/KNote: P V has units of N-m or J (energy!)P V = n R Tn = number of molesR = ideal gas constant = NAkB = 8.31 J/(mol*K)20
27 Ideal Gas Law ACT II PV = nRT You inflate the tires of your car so the pressure is 30 psi, when the air inside the tires is at 20 degrees C. After driving on the highway for a while, the air inside the tires heats up to 38 C. Which number is closest to the new air pressure?1) 16 psi 2) 32 psi 3) 57 psiCareful, you need to use the temperature in KP = P0 (38+273)/(20+273)23
28 Boyle’s LawThe quantities P, V, n, and T aren’t independent but are related by an equation of state. You can perform various experiments where two of these quantities are held fixed and a relation between the other two is determined. These mini-laws have names like Boyle’s Law and Charles’s Law.PV=constantn and T held fixed
29 Charles’s LawThe quantities P, V, n, and T aren’t independent but are related by an equation of state. You can perform various experiments where two of these quantities are held fixed and a relation between the other two is determined. These mini-laws have names like Boyle’s Law and Charles’s Law.V/T=constantn and P held fixed
30 Ideal Gas Law: ACT 1 PV = nRT A piston has volume 20 ml, and pressure of 30 psi. If the volume is decreased to 10 ml, what is the new pressure? (Assume T is constant.)1) ) ) 15When n and T are constant, we have PV = constant (Boyle’s Law)Change this to the pressure change with temperature, but give the temperature in C.V=20P=30V=10P=??
31 Balloon is still open to atmospheric pressure, so it stays at 1 atm Balloon ACT 1What happens to the pressure of the air inside a hot-air balloon when the air is heated? (Assume V is constant)1) Increases 2) Same 3) DecreasesBalloon is still open to atmospheric pressure, so it stays at 1 atm
32 FB = r V g, r is density of outside air! Balloon ACT 2What happens to the buoyant force on the balloon when the air is heated? (Assume V remains constant)1) Increases 2) Same 3) DecreasesFB = r V g, r is density of outside air!32
33 P and V are constant. If T increases N decreases. Balloon ACT 3What happens to the number of air molecules inside the balloon when the air is heated? (Assume V remains constant)1) Increases 2) Same 3) DecreasesPV = NkTP and V are constant. If T increases N decreases.34
34 Balloon SummaryIn terms of the ideal gas law, explain briefly how a hot air balloon works.Hot air has less mole density than cool air. So less hot air is required in order to achieve the same pressure as cool air. This makes the density of hot air less allowing it to float.When temperature increases the volume of the gas increases, thus reducing the density of the gas making it lighter that then surrounding air, which causes the balloon to rise.Note! this is not a pressure effect, it is a density effect. As T increases, the density decreases the balloon then floats due to Archimedes principle. The pressure remains constant!
35 The Laws of Thermodynamics We have learned about two ways in which energy may be transferred between a system and its environment. One is work (force act over a distance), the other is heat (energy transferred due to a difference in temperature). The study of energy transfers involving work and heat, and the resulting changes in internal energy, temperature, volume, and pressure is called thermodynamics.The order of discovery: (A hint at the logic of Thermodynamics)The 2nd Law was discovered FirstThe 1st Law was discovered SecondThe 0th Law was the third to be discoveredThe 4th Law is called the 3rd Law
36 The Zeroth Law of Thermodynamics When two objects are brought into contact, heat will flow from the warmer object to the cooler one until they reach thermal equilibrium.If Objects 1 and 2 are each in thermal equilibrium with Object 3, then Objects 1 and 2 are in thermal equilibrium with each other.
37 The First Law of Thermodynamics Simply put, the first Law of Thermodynamics is a statement of the Conservation of energy that includes heat.So what does the First Law say? In words, the internal energy of a body (such as a gas) can be increased by heating it or by doing mechanical work on itJames Joule showed that mechanical energy could be converted to heat and arrived at the conclusion that heat was another form of energy. He showed that 1 calorie of heat was equivalent to J of work.1 cal = J
38 Changing the Internal Energy U is a “state” function --- depends uniquely on the state of the system in terms of p, V, T etc.(e.g. For a classical ideal gas, U = NkT )There are two ways to change the internal energy of a system:WORK done by the system on the environmentWby = -WonHEAT is the transfer of thermal energy into the system from the surroundingsQThermal reservoirWork and Heat are process energies, not state functions.
39 The First Law of Thermodynamics The first law deals with changes in the internal energy of the system. It is important to note that changes in the internal energy of a system can only occur if the system is not isolated. This tells us that the system is imbedded in its surroundings in a way that there can be energy transfer. There are two types of energy transfer (Work and Heat), and the difference between these two is determined only by what occurs in the surroundings of the system.Work: is an energy transfer between a system and its surroundings that is the result of an organized motion in the surroundings.For example:You can increase the internal energy of a piece of plastic by vigorously rubbing it.You can decrease the internal energy of a gas by letting it expand against some external pressure (such as a piston).Piston examples are commonly used to illustrate many concepts of Thermodynamics, so Let’s look into this more deeply.
40 The First Law of Thermodynamics Consider an example system of a piston and cylinder with an enclosed dilute gas characterized by P,V,T & n.
41 The First Law of Thermodynamics What happens to the gas if the piston is moved inwards?
42 The First Law of Thermodynamics If the container is insulated the temperature will rise, the atoms move faster and the pressure rises.Is there more internal energy in the gas?Yes!
43 The First Law of Thermodynamics External agent did work in pushing the piston inward.W = Fd=(PA)DxW =PDVDx
44 The First Law of Thermodynamics Work done on the gas equals the change in the gases internal energy,W = DUDx
45 The First Law of Thermodynamics Direction of arrow (left to right is very importantWe can represent the force the gas exerts on the piston or the external force that compresses the piston by a PV diagram.VP12V V2(+) area(+) work done by gas(-) work done on gasThis tells us that the work done by the gas during the expansion (from 1 to 2) is just the area under the curve. By convention, this area is positive for positive work done by the gas. This also corresponds to negative work done on the gas. We must be careful to distinguish the work done on the gas and the work done by the gas.VP12V V2(-) area(-) work done by gas(+) work done on gas
46 The First Law of Thermodynamics Example:One mole of monatomic ideal gas is enclosed under a frictionless piston. A series of processes occur, and finally the state of this ideal gas returns to its initial state (see PV diagram). Answer the following questions in terms of P0, V0, and R.Determine the temperature at each vertex.Determine the change in internal energy (U) for each process.Determine the work done by the gas for each process.(a) UseVP123V V V0P01/2P0
47 The First Law of Thermodynamics Example:One mole of monatomic ideal gas is enclosed under a frictionless piston. A series of processes occur, and finally the state of this ideal gas returns to its initial state (see PV diagram). Answer the following questions in terms of P0, V0, and R.Determine the temperature at each vertex.Determine the change in internal energy (U) for each process.Determine the work done by the gas for each process.(a) UseVP123V V V0P01/2P0
48 The First Law of Thermodynamics Example:One mole of monatomic ideal gas is enclosed under a frictionless piston. A series of processes occur, and finally the state of this ideal gas returns to its initial state (see PV diagram). Answer the following questions in terms of P0, V0, and R.Determine the temperature at each vertex.Determine the change in internal energy (U) for each process.Determine the work done by the gas for each process.(a) Use area under graph for each segmentVP123V V V0P01/2P0Note:No work for constant volume.Total work 3/4P0V0 is same as area enclosed by triangle.
49 The First Law of Thermodynamics Summary:TUWbyWon123(PoVo)/R4.5PoVo3PoVo-3PoVo23-2(PoVo)/R31-1(PoVo)/R-1.5PoVo-2.25PoVo+2.25PoVoVP123V V V0P01/2P0
50 First Law of Thermodynamics Let’s change the situation:Keep the piston fixed at its original location.Place the cylinder on a hot plate.What happens to gas?
51 The First Law of Thermodynamics Heat flows into the gas.Atoms move faster, internal energy increases.Q = heat in JoulesDU = change in internal energy in Joules.Q = DU
52 The First Law of Thermodynamics Heat:Heat is an energy transfer between a system and its surroundings that is the result of random motion in the surroundings. (Note: there is a difference between a work process [organized motion in surroundings] and a heat process [random motion in surroundings]).Heat will always flow spontaneously from the system at higher temperature to the system of lower temperature, but heat can be made to flow in the opposite direction as well if work is done in the process (a refrigerator).
53 The First Law of Thermodynamics What if we added heat and pushed the piston in at the same time?
54 The First Law of Thermodynamics Work is done on the gas, heat is added to the gas and the internal energy of the gas increases!DU=Q+WF
55 First Law of Thermodynamics Energy Conservation :The change in internal energy of a system (U) is equal to the heat flow into the system (Q) minus the work done by the system (W)DU = Q - WWork done by systemIncrease in internal energy of systemVP123V V2P1P3ideal gasHeat flow into systemEquivalent way of writing 1st Law:DU = Q + WWork doneon systemIncrease in internal energy of systemHeat flow into system07
56 First Law of Thermodynamics If we re-examine our PV diagram and now calculate the heat transfer (Q) for each process.Since the work done by the gas and the change in internal energy has already been calculated, it is very easy to calculate the heat transfer from the First Law:DU = Qinto + WonVP123V V V0P01/2P0
57 First Law of Thermodynamics If we re-examine our PV diagram and now calculate the heat transfer (Q) for each process.Since the work done by the gas and the change in internal energy has already been calculated, it is very easy to calculate the heat transfer from the First Law:DU = Qinto + WonVP123V V V0P01/2P0
58 First Law of Thermodynamics If we re-examine our PV diagram and now calculate the heat transfer (Q) for each process.Since the work done by the gas and the change in internal energy has already been calculated, it is very easy to calculate the heat transfer from the First Law:DU = Qinto + WonVP123V V V0P01/2P0
59 First Law of Thermodynamics UWbyQin123(PoVo)/R4.5PoVo3PoVo7.5PoVo23-2(PoVo)/R-3PoVo31-1(PoVo)/R-1.5PoVo-2.25PoVo-3.75PoVoVP123V V V0P01/2P0
60 (1) Positive, heat flows into soup (2) Zero (is close enough) Signs ExampleYou are heating some soup in a pan on the stove. To keep it from burning, you also stir the soup. Apply the 1st law of thermodynamics to the soup. What is the sign of :1) Q2) W3) DU(1) Positive, heat flows into soup(2) Zero (is close enough)(3) Positive, Soup gets warmer
61 SignsEnergy is also used to mean the potential to do work; the greater the ability of something to change the things around it, the more energy it has. Mathematically, the first law is expressed: U= Q+W.Where Q is the change in the quantity of heat added to the system, U is the change in the internal energy and W is the work done on or by the system.The sign convention here is that if U is positive the amount of internal energy increases. This means that Q stands for the heat energy put into the system and W for the work done on the system. This is known as the ‘physicists’ convention’.WorkW>0: Work is done on the system by the surroundingsW<0: Work is done by the system on the surroundingsHeatQ>0: Heat is added to the system from the surroundingsQ<0: Heat is released by the system to the surroundings
62 First Law of Thermodynamics For each process in this cycle, indicate in the table below whether the quantities W, Q, and U are positive (+), negative (-), or zero (0). W is the work done BY the GAS.ProcessWQU1 -> 22 -> 33 -> 1+++-----VP123V V V0P01/2P0
63 W = P V :For constant Pressure Work Done by a Gas ACTMMyThe work done by a gas as it contracts isA) Positive B) Zero C) NegativeW = F d cosq < 0= P A d = P A Dy = P DVNote that equation only works for constant pressure!W = P V :For constant PressureW > 0 if V > 0 expanding system does positive workW < 0 if V < 0 contracting system does negative workW = 0 if V = 0 system with constant volume does no work
64 Work Done by a Gas ACTA gas is kept in a cylinder that can be compressed by pushing down on a piston. You add 2500 J of heat into the system, and then you push the piston 1.0 m down with a constant force of 1800 N. What is the change in the gas’s internal energy.We define the variables in the First Law, being careful of the signs:Heat added : Q=+2500 JWe did work on gas, so W is positive:Therefore:
65 Pressure as a Function of Volume Work is the area under the curve of a PV-diagram.Work depends on the pathtaken in “PV space.”When a process is depicted on a PV diagram, directional arrows are used on the graph. In a process that moves to the right, the gas will do positive work on the surroundings; in a process that moves to the left, the gas will do negative work on the surroundings (ie the surroundings do work on the gas)The precise path serves to describe the kind of process that took place.
66 Work Done by a GasWork done by gas equals area inside graph
67 P-V DiagramsWhen you are describing a process that involves a gas expanding or compressing, you will need to use specific nomenclature. A process is called isobaric if the pressure remains constant throughout. It is called isothermal if the temperature remains constant. If no heat energy enters or leaves the gas, it is called adiabatic.
68 Different Thermodynamic Paths The work done depends on the initial and finalstates and the path taken between these states.
69 Thermodynamic Systems and P-V Diagrams ideal gas law: PV = nRT (nR = NkB)for n fixed, P and V determine “state” of monatomic ideal gas systemT = PV/nRU = (3/2)nRT = (3/2)PVExamples:which point has highest T?Bwhich point has lowest U?Cto change the system from C to B,energy must be added to systemVPABCV V2P1P3
70 First Law of Thermodynamics Isobaric Example 2 moles of monatomic ideal gas is takenfrom state 1 to state 2 at constant pressureP=1000 Pa, where V1 =2m3 and V2 =3m3(R=8.31 J/k mole) Find:1) T12) T23)DU4) W5) QVP12V V21. PV1 = nRT1 T1 = PV1/nR = 120K2. PV2 = nRT2 T2 = PV2/nR = 180K3. DU = (3/2) nR DT = 1500 JDU = (3/2) P DV = 1500 J (has to be the same)4. W = P DV = J5. Q = DU + W = = 2500 J21
71 First Law of Thermodynamics Isochoric 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.PP2P12Q = DU + WDU = (3/2) nR DT = 1500 JW = P DV = 0 JQ = DU + W = = 1500 J1VVRequires less heat to raise T at const. volume than at const. pressure (no energy used for work)
72 Heat Transfer and Temperature Change for Gases The value of the heat transfer in a system, Q, is also path-dependent. We recall for a solid or liquid Q=mcT. For gases, however, the system is a little more complicated, because the value of the proportionality constant between Q and T depends on whether the volume or pressure is kept constant.If the volume remains constant (isochoric) during heat transfer, thenCV is the molar heat capacity at constant volumeIf the pressure remains constant (isobaric) during the heat transferCP is the molar heat capacity at constant pressure
73 Total Work Done V P Wtot = ?? V P W = PDV (>0) DV > 0 V P 1234Wtot = ??VPW = PDV (>0)1234DV > 0VPW = PDV = 01234DV = 0VP1234Wtot > 0VPW = PDV (<0)1234DV < 0VW = PDV = 01234PDV = 0General rule: work done is area under P-V curve (even if not horizontal).
74 Question25 L of gas is enclosed in a cylinder/piston apparatus at 2 atm of pressure and 300 K. If 100 kg of mass is placed on the piston causing the gas to compress to 20 L at constant pressure. This is done by allowing heat to flow out of the gas. What is the work done on the gas? What is the change in internal energy of the gas? How much heat flowed out of the gas?Po = 202,600 Pa, Vo = m3, To = 300 K, Pf = 202,600 Pa, Vf=0.020 m3,W =-PDV = -202,600 Pa (0.020 – 0.025)m3=1013 J energy added to the gas.
75 Question25 L of monatomic gas is enclosed in a cylinder/piston apparatus at 2 atm of pressure and 300 K. If 100 kg of mass is placed on the piston causing the gas to compress to 20 L at constant pressure. This is done by allowing heat to flow out of the gas. What is the work done on the gas? What is the change in internal energy of the gas? How much heat flowed out of the gas?
76 Question25 L of gas is enclosed in a cylinder/piston apparatus at 2 atm of pressure and 300 K. If 100 kg of mass is placed on the piston causing the gas to compress to 20 L at constant pressure. This is done by allowing heat to flow out of the gas. What is the work done on the gas? What is the change in internal energy of the gas? How much heat flowed out of the gas?DU = Won - QoutQout = Won - DU = 1013J – (-1518J) = 2531 J heat out
77 Second Law of Thermodynamics Entropy, S, is a measure of the disorder, or randomness of a system. The greater the disorder of a system, the greater the entropy. If a system is highly ordered (like particles is a solid) we say that the entropy is low.The second law of thermodynamics states that all spontaneous processes proceeding in an isolated system lead to an increase in entropy.The increase or decrease in entropy can be found byWhere Q is heat flow into or out of a system and T is the average Kelvin temperature
78 Thermal Efficiency of a Heat Engine The thermal efficiency, e, of the heat engine is equal to the ratio of the heat we get out to the heat we put in.Where Qin is heat absorbed, and Qout is heat dischargedUnless Qout=0, the engines efficiency is always less than 1.
79 EfficiencyIf we re-examine our PV diagram and now calculate the efficiency of the system.The efficiency of a cycle is defined as the ratio of the work done by the gas to the heat Qin that flows into the system. Any heat that is expelled into the surroundings is not included in the calculation of Qin. From the point of view of efficiency, this expelled heat is lost and its energy is not used by the system:WbyQin123PoVo7.5PoVo23-3PoVo31-2.25PoVo-3.75PoVoTotal0.75 PoVoVP123V V V0P01/2P0
80 Heat Engine: Efficiency The objective: turn heat from hot reservoir into workThe cost: “waste heat”1st Law: Qin -Qout = Wefficiency e W/Qin=W/Qin= (Qin-Qout)/Qin= 1-Qout/QinTHTCQinQoutWHEAT ENGINE13
81 A hot (98 C) slab of metal is placed in a cool (5C) bucket of water. ACTA hot (98 C) slab of metal is placed in a cool (5C) bucket of water.What happens to the entropy of the metal?A) Increase B) Same C) DecreasesWhat happens to the entropy of the water?DS = Q/THeat leaves metal: Q<0Heat enters water: Q>0
82 Description of Disorder Isolated systems tend toward greater disorder, and entropy is a measure of that disorderS = kB ln (W)kB is Boltzmann’s constantW is a number proportional to the probability that the system has a particular configurationThis version is a statement of what is most probable rather than what must beThe Second Law also defines the direction of time of all events as the direction in which the entropy of the universe increases
83 Second Law of Thermodynamics The entropy change (Q/T) of the system+environment 0never < 0order to disorderConsequencesA “disordered” state cannot spontaneously transform into an “ordered” stateNo engine operating between two reservoirs can be more efficient than one that produces 0 change in entropy. This is called a “Carnot engine”31
85 CarnotNow suppose you want to make an engine that runs on energy from a source of high temperature heat (at temperature TH) such as a burner. And, furthermore, suppose you want your heat engine to work in the most efficient way possible.The first thing you have to do is to get energy into your working fluid. Obviously, the best process to do this will be a constant temperature (isothermal) heat transfer, which will result in an expansion of the working gas at TH (shown on the graph as A-->B). Any other expansion process that results in a reduction of the temperature of the working gas below TH, will also result in a reduction of efficiency. [An analogy can be made here between temperature and pressure. Imagine a water turbine running from a dam -- ideally you want to keep the water level at the maximum so that you have the maximum pressure to turn the turbine. If you draw off water too fast then the water level (and therefore pressure) will drop and the power output of the turbine will be reduced.]
86 CarnotThe isothermal expansion (A-->B) produces a work output -- yay! -- but at the end of the process (point B) you are faced with a dilemma. You want to get your piston back to it's original position, so that you can repeat the expansion/work output process again. But if you simply compress the working gas, then you have to do exactly the same amount of work to get back to the original state as you produced during the expansion process (and, of course, you end up rejecting your thermal energy back to the high temperature heat source). This would mean that you would get no net work output from your heat engine, which would be very unsatisfactory.
87 CarnotThe answer, of course, is to perform another expansion process with the high temperature heat source disconnected from the system (shown on the graph as B-->C). The temperature (and pressure) of the working gas will fall, but because the system is not exchanging heat then you are not losing any of the 'potential' of the high temperature heat from the heat source.This then allows you to compress the working gas at low temperature and pressure (shown on the graph as C-->D), which requires much less work to return the piston almost to its original position. Even though you have to put work back into the engine to do this, it is much less than the work output during the isothermal expansion process (A-->B) -- and so, overall, the engine produces a NET work output. This compression process must, of course, reject heat -- which it does to a low temperature heat sink at temperature TC (usually the ambient environment). Again, an isothermal process is the best way to do this, in order to keep the temperature difference between the heat source and sink always at the maximum value.
88 CarnotOf course, the engine is not quite back to its original state, as the temperature of the working gas is still at TC. To return the temperature to TH, we disconnect the system from the low temperature heat sink, and perform another compression process (shown on the graph as D-->A). This requires a work input, which will be of exactly the same value as the work output in B-->C. But, of course -- to reiterate -- the engine will still have an overall net work output, since the isothermal work output A-->B is much greater than the isothermal work input C-->D.Thus the engine has produced a work output, and returned to its original state, ready to do the whole thing all over again. This is how the Carnot cycle works.
89 Carnot CycleIdealized Heat EngineNo FrictionDS = Q/T = 0Reversible ProcessProcess A is expansionProcess B is compressionABTUQW1->2Isothermal expansionQH absorbedT=THQH>0QH2->3Adiabatic expansionNo heat exchange,Temp drops to TCTC - TH-W33->4Isothermal compressionHeat QC discardedT=TCQC<0-QC4->1Adiabatic compressionNo heat exchange, Temp rises to THTH - TCW4Note: work from (2->3) and (4->1) balance, yet more work by during (1->2) than on during (3->4).
90 The Carnot Cycle in Slow Motion Stage 1In the first stage the piston moves upward while the chamber absorbs heat from a source and the gas begins to expand. The portion of the graphic from point 1 to point 2 represents this behaviour. Because the temperature of the gas does not change, all the heat drawn in from the source goes into work performed by the expansion of the gas.There is only heat QH flowing into the system, there is no change in Potential Energy, therefore sinceDU = Q + W, the work by the system is –Q (the gas does positive work).
91 The Carnot Cycle in Slow Motion Stage 2In the second stage the heat source is removed; and the piston continues to move upward and the gas is still expanding while cooling (lowering in temperature from TH to TC). The portion of the graphic from point 2 to point 3 represents this behaviour. This stage is adiabatic expansion the gas (no heat transfer).Because the system expands it does negative work (the gas does positive work), its internal energy and temperature decrease because it receives no influx of heat from the surroundings
92 The Carnot Cycle in Slow Motion Stage 3In the third stage the piston begins to move downward and the cool gas (TC) is placed in thermal contact with heat reservoir at temperature TC, the gas is isothermally compressed at this temperature TC , thus the gas expels heat QC to the reservoir (the engine gives energy to the environment).Because the system contracts (decrease in volume and increase in pressure) the system does positive work on the gas (gas does negative work) and, rather than increasing its internal energy, the gas discards heat to the low temperature reservoir.
93 The Carnot Cycle in Slow Motion Stage 4In the final stage the piston moves downward and the cool gas is compressed to its original state. Its temperature also rises to its orignal state (point 4 to point 1). No exchange of heat with the surroundings take place.Because the system is compressed, its internal energy and temperature increase (since no heat is discarded), and the work done on the gas by the environment is W42 (The gas does negative work)
94 Carnot Cycle ReviewDuring step one the gas does a positive amount of work. Step two is adiabatic, with the gas doing a positive amount of work. During step three there is a negative amount of work done by the gas. Finally in step four, which is adiabatic, the work done by the gas is negative. Notice that the total work done (the remaining area) is positive because positive work is done at high temperatures and negative work is done at lower temperatures. During step one heat is absorbed (Q > 0) and during step three heat is released (Q < 0). More heat is absorbed than is released for the entire cycle. This is the basis of how engines work: Heat (from the hot reservoir) is transferred into mechanical work (piston moving).
95 Engines and the 2nd LawTHTCQHQCWHEAT ENGINEThe objective: turn heat from hot reservoir into workThe cost: “waste heat”1st Law: QH -QC = Wefficiency e W/QH =W/QH = 1-QC/QHS = QC/TC - QH/TH 0S = 0 for CarnotTherefore, QC/QH TC/ THQC/QH = TC/ TH for CarnotTherefore e = 1 - QC/QH 1 - TC/ THe = 1 - TC/ TH for Carnote = 1 is forbidden!e largest if TC << TH36
96 ExampleConsider a hypothetical refrigerator that takes 1000 J of heat from a cold reservoir at 100K and ejects 1200 J of heat to a hot reservoir at 300K.1. How much work does the refrigerator do?2. What happens to the entropy of the universe?3. Does this violate the 2nd law of thermodynamics?THTCQHQCWQC = 1000 JQH = 1200 JSince QC + W = QH, W = 200 JDSH = QH/TH = (1200 J) / (300 K) = 4 J/KDSC = -QC/TC = (-1000 J) / (100 K) = -10 J/KDSTOTAL = DSH + DSC = -6 J/K decreases (violates 2nd law)
97 ReviewWhich of the following is forbidden by the second law of thermodynamics?1. Heat flows into a gas and the temperature falls2. The temperature of a gas rises without any heat flowing into it3. Heat flows spontaneously from a cold to a hot reservoir4. All of the aboveNot 1, because volume could increase and lower temperatureNot 2, because increase of pressure can increase temperature43
98 Summary First Law of Thermodynamics Ideal Gas Law Average KE of Gas MoleculeInternal energy of ideal GasExpansion WorkHeat Engine EfficiencyCarnot Efficiency
99 Summary II Isobaric Work: Isochoric Work: Isothermal Work: Adiabatic Work:Isothermal Internal EnergyThe internal energy of an ideal gas depends only on temperature
100 Summary III Cyclic process (originates and ends at same state) U=0 Wby=QintoIsovolumetric (constant volume)Qinto=UIsobaric (constant pressure)Isothermal (constant temperature)Adiabatic (no heat exchange)Q=0U=WonU=-Wby