Isothermal Heating: Purist and Utilitarian Views
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Recitation: 3 9/18/03
Recitation: 3 9/18/03 Second Law ² There exists a function (S) of the extensive parameters of any composite system, defined for all equilibrium states and having the following property: The values assumed by the extensive parameters in the absence of an internal constraint are those that maximize the entropy over the manifold of constrained equilibrium states. ¡¢ ¡ ² Another way of looking at it: ¡¢ ¡ U1 U2 U=U1+U2 S1 S2 S>S1+S2 ¡¢ ¡ Equilibrium State Figure 1: Second Law ² Entropy is additive. And... � ¶ @S > 0 @U V;N::: ² S is an extensive property: It is a homogeneous, first order function of extensive parameters: S (¸U; ¸V; ¸N) =¸S (U; V; N) ² S is not conserved. ±Q dSSys ¸ TSys For an Isolated System, (i.e. Universe) 4S > 0 Locally, the entropy of the system can decrease. However, this must be compensated by a total increase in the entropy of the universe. (See 2) Q Q ΔS1=¡ ΔS2= + T1³ ´ T2 1 1 ΔST otal =Q ¡ > 0 T2 T1 QuasiStatic Processes A Quasistatic thermodynamic process is defined as the trajectory, in thermodynamic space, along an infinite number of contiguous equilibrium states that connect to equilibrium states, A and B. Universe System T2 T1 Q T1>T2 Figure 2: Local vs. Total Change in Energy Irreversible Process Consider a closed system that can go from state A to state B. The system is induced to go from A to B through the removal of some internal constraint (e.g. removal of adiabatic wall). The systems moves to state B only if B has a maximum entropy with respect to all the other accessible states. -
Chapter 7 BOUNDARY WORK
Chapter 7 BOUNDARY WORK Frank is struck - as if for the first time - by how much civilization depends on not seeing certain things and pretending others never occurred. − Francine Prose (Amateur Voodoo) In this chapter we will learn to evaluate Win, which is the work term in the first law of thermodynamics. Work can be of different types, such as boundary work, stirring work, electrical work, magnetic work, work of changing the surface area, and work to overcome friction. In this chapter, we will concentrate on the evaluation of boundary work. 110 Chapter 7 7.1 Boundary Work in Real Life Boundary work is done when the boundary of the system moves, causing either compression or expansion of the system. A real-life application of boundary work, for example, is found in the diesel engine which consists of pistons and cylinders as the prime component of the engine. In a diesel engine, air is fed to the cylinder and is compressed by the upward movement of the piston. The boundary work done by the piston in compressing the air is responsible for the increase in air temperature. Onto the hot air, diesel fuel is sprayed, which leads to spontaneous self ignition of the fuel-air mixture. The chemical energy released during this combustion process would heat up the gases produced during combustion. As the gases expand due to heating, they would push the piston downwards with great force. A major portion of the boundary work done by the gases on the piston gets converted into the energy required to rotate the shaft of the engine. -
Muddiest Point – Entropy and Reversible I Am Confused About Entropy and How It Is Different in a Reversible Versus Irreversible Case
1 Muddiest Point { Entropy and Reversible I am confused about entropy and how it is different in a reversible versus irreversible case. Note: Some of the discussion below follows from the previous muddiest points comment on the general idea of a reversible and an irreversible process. You may wish to have a look at that comment before reading this one. Let's talk about entropy first, and then we will consider how \reversible" gets involved. Generally we divide the universe into two parts, a system (what we are studying) and the surrounding (everything else). In the end the total change in the entropy will be the sum of the change in both, dStotal = dSsystem + dSsurrounding: This total change of entropy has only two possibilities: Either there is no spontaneous change (equilibrium) and dStotal = 0, or there is a spontaneous change because we are not at equilibrium, and dStotal > 0. Of course the entropy change of each piece, system or surroundings, can be positive or negative. However, the second law says the sum must be zero or positive. Let's start by thinking about the entropy change in the system and then we will add the entropy change in the surroundings. Entropy change in the system: When you consider the change in entropy for a process you should first consider whether or not you are looking at an isolated system. Start with an isolated system. An isolated system is not able to exchange energy with anything else (the surroundings) via heat or work. Think of surrounding the system with a perfect, rigid insulating blanket. -
Isothermal Process It Is the Process in Which Other Physical Quantities Might Change but the Temperature of the System Remains Or Is Forced to Remain Constant
Sajit Chandra Shakya Department of Physics Kathmandu Don Bosco College New Baneshwor, Kathmandu Isothermal process It is the process in which other physical quantities might change but the temperature of the system remains or is forced to remain constant. For example, the constant temperature of human body. Under constant temperature, the volume of a gas system is inversely proportional to the pressure applied, the phenomena being called Boyle's Law, written in symbols as 1 V ∝ P 1 Or, V = KB× where KB is a constant quantity. P Or, PV = KB …………….. (i) This means whatever be the values of volume and pressure, their product will be constant. So, P1V1 = KB, P2V2 = KB, P3V3 = KB, etc, Or, P1V1 = P2V2 = P3V3, etc. The requirements for an isothermal process are as follows: 1. The process should be carried very slowly so that there is an ample time for compensation of heat in case of any loss or addition. 2. The boundaries of the system should be highly conducting so that there is a path for heat to flow into or flow away from a closed space in case of any energy loss or oversupply. 3. The boundaries should be made very thin because the resistance of the substance for heat conduction will be less for thin boundaries. Since in an isothermal change, the temperature remains constant, the internal energy also does not change, i.e. dU = 0. So if dQ amount of heat is given to a system which undergoes isothermal change, the relation for the first law of thermodynamics would be dQ = dU + dW Or, dQ = 0 + PdV Or, dQ = PdV This means all the heat supplied will be utilized for performing external work and consequently its value will be very high compared to other processes. -
The First Law of Thermodynamics Continued Pre-Reading: §19.5 Where We Are
Lecture 7 The first law of thermodynamics continued Pre-reading: §19.5 Where we are The pressure p, volume V, and temperature T are related by an equation of state. For an ideal gas, pV = nRT = NkT For an ideal gas, the temperature T is is a direct measure of the average kinetic energy of its 3 3 molecules: KE = nRT = NkT tr 2 2 2 3kT 3RT and vrms = (v )av = = r m r M p Where we are We define the internal energy of a system: UKEPE=+∑∑ interaction Random chaotic between atoms motion & molecules For an ideal gas, f UNkT= 2 i.e. the internal energy depends only on its temperature Where we are By considering adding heat to a fixed volume of an ideal gas, we showed f f Q = Nk∆T = nR∆T 2 2 and so, from the definition of heat capacity Q = nC∆T f we have that C = R for any ideal gas. V 2 Change in internal energy: ∆U = nCV ∆T Heat capacity of an ideal gas Now consider adding heat to an ideal gas at constant pressure. By definition, Q = nCp∆T and W = p∆V = nR∆T So from ∆U = Q W − we get nCV ∆T = nCp∆T nR∆T − or Cp = CV + R It takes greater heat input to raise the temperature of a gas a given amount at constant pressure than constant volume YF §19.4 Ratio of heat capacities Look at the ratio of these heat capacities: we have f C = R V 2 and f + 2 C = C + R = R p V 2 so C p γ = > 1 CV 3 For a monatomic gas, CV = R 3 5 2 so Cp = R + R = R 2 2 C 5 R 5 and γ = p = 2 = =1.67 C 3 R 3 YF §19.4 V 2 Problem An ideal gas is enclosed in a cylinder which has a movable piston. -
Ch 19. the First Law of Thermodynamics
Ch 19. The First Law of Thermodynamics Liu UCD Phy9B 07 1 19-1. Thermodynamic Systems Thermodynamic system: A system that can interact (and exchange energy) with its surroundings Thermodynamic process: A process in which there are changes in the state of a thermodynamic system Heat Q added to the system Q>0 taken away from the system Q<0 (through conduction, convection, radiation) Work done by the system onto its surroundings W>0 done by the surrounding onto the system W<0 Energy change of the system is Q + (-W) or Q-W Gaining energy: +; Losing energy: - Liu UCD Phy9B 07 2 19-2. Work Done During Volume Changes Area: A Pressure: p Force exerted on the piston: F=pA Infinitesimal work done by system dW=Fdx=pAdx=pdV V Work done in a finite volume change W = final pdV ∫V initial Liu UCD Phy9B 07 3 Graphical View of Work Gas expands Gas compresses Constant p dV>0, W>0 dV<0, W<0 W=p(V2-V1) Liu UCD Phy9B 07 4 19-3. Paths Between Thermodynamic States Path: a series of intermediate states between initial state (p1, V1) and a final state (p2, V2) The path between two states is NOT unique. V2 W= p1(V2-V1) +0 W=0+ p2(V2-V1) W = pdV ∫V 1 Work done by the system is path-dependent. Liu UCD Phy9B 07 5 Path Dependence of Heat Transfer Isotherml: Keep temperature const. Insulation + Free expansion (uncontrolled expansion of a gas into vacuum) Heat transfer depends on the initial & final states, also on the path. -
Thermodynamic Propertiesproperties
ThermodynamicThermodynamic PropertiesProperties PropertyProperty TableTable -- from direct measurement EquationEquation ofof StateState -- any equation that relates P,v, and T of a substance jump ExerciseExercise 33--1212 A bucket containing 2 liters of R-12 is left outside in the atmosphere (0.1 MPa) a) What is the R-12 temperature assuming it is in the saturated state. b) the surrounding transfer heat at the rate of 1KW to the liquid. How long will take for all R-12 vaporize? See R-12 (diclorindifluormethane) on Table A-2 SolutionSolution -- pagepage 11 Part a) From table A-2, at the saturation pressure of 0.1 MPa one finds: • Tsaturation = - 30oC 3 • vliq = 0.000672 m /kg 3 • vvap = 0.159375 m /kg • hlv = 165KJ/kg (vaporization heat) SolutionSolution -- pagepage 22 Part b) The mass of R-12 is m = Volume/vL, m = 0.002/0.000672 = 2.98 kg The vaporization energy: Evap = vap energy * mass = 165*2.98 = 492 KJ Time = Heat/Power = 492 sec or 8.2 min GASGAS PROPERTIESPROPERTIES Ideal -Gas Equation of State M PV = nR T; n = u mol Universal gas constant is given on Ru = 8.31434 kJ/kmol-K = 8.31434 kPa-m3/kmol-k = 0.0831434 bar-m3/kmol-K = 82.05 L-atm/kmol-K = 1.9858 Btu/lbmol-R = 1545.35 ft-lbf/lbmol-R = 10.73 psia-ft3/lbmol-R EExamplexample Determine the particular gas constant for air (28.97 kg/kmol) and hydrogen (2.016 kg/kmol). kJ 8.1417 kJ = Ru = kmol − K = 0.287 Rair kg M 28.97 kg − K kmol kJ 8.1417 kJ = kmol − K = 4.124 Rhydrogen kg 2.016 kg − K kmol IdealIdeal GasGas “Law”“Law” isis aa simplesimple EquationEquation ofof StateState PV = MRT Pv = RT PV = NRuT P V P V 1 1 = 2 2 T1 T2 QuestionQuestion …...….. -
Chapter 11 ENTROPY
Chapter 11 ENTROPY There is nothing like looking, if you want to find something. You certainly usually find something, if you look, but it is not always quite the something you were after. − J.R.R. Tolkien (The Hobbit) In the preceding chapters, we have learnt many aspects of the first law applications to various thermodynamic systems. We have also learnt to use the thermodynamic properties, such as pressure, temperature, internal energy and enthalpy, when analysing thermodynamic systems. In this chap- ter, we will learn yet another thermodynamic property known as entropy, and its use in the thermodynamic analyses of systems. 250 Chapter 11 11.1 Reversible Process The property entropy is defined for an ideal process known as the re- versible process. Let us therefore first see what a reversible process is all about. If we can execute a process which can be reversed without leaving any trace on the surroundings, then such a process is known as the reversible process. That is, if a reversible process is reversed then both the system and the surroundings are returned to their respective original states at the end of the reverse process. Processes that are not reversible are called irreversible processes. Student: Teacher, what exactly is the difference between a reversible process and a cyclic process? Teacher: In a cyclic process, the system returns to its original state. That’s all. We don’t bother about what happens to its surroundings when the system is returned to its original state. In a reversible process, on the other hand, the system need not return to its original state. -
12/8 and 12/10/2010
PY105 C1 1. Help for Final exam has been posted on WebAssign. 2. The Final exam will be on Wednesday December 15th from 6-8 pm. First Law of Thermodynamics 3. You will take the exam in multiple rooms, divided as follows: SCI 107: Abbasi to Fasullo, as well as Khajah PHO 203: Flynn to Okuda, except for Khajah SCI B58: Ordonez to Zhang 1 2 Heat and Work done by a Gas Thermodynamics Initial: Consider a cylinder of ideal Thermodynamics is the study of systems involving gas at room temperature. Suppose the piston on top of energy in the form of heat and work. the cylinder is free to move vertically without friction. When the cylinder is placed in a container of hot water, heat Equilibrium: is transferred into the cylinder. Where does the heat energy go? Why does the volume increase? 3 4 The First Law of Thermodynamics The First Law of Thermodynamics The First Law is often written as: Some of the heat energy goes into raising the temperature of the gas (which is equivalent to raising the internal energy of the gas). The rest of it does work by raising the piston. ΔEQWint =− Conservation of energy leads to: QEW=Δ + int (the first law of thermodynamics) This form of the First Law says that the change in internal energy of a system is Q is the heat added to a system (or removed if it is negative) equal to the heat supplied to the system Eint is the internal energy of the system (the energy minus the work done by the system (usually associated with the motion of the atoms and/or molecules). -
BASIC THERMODYNAMICS REFERENCES: ENGINEERING THERMODYNAMICS by P.K.NAG 3RD EDITION LAWS of THERMODYNAMICS
Module I BASIC THERMODYNAMICS REFERENCES: ENGINEERING THERMODYNAMICS by P.K.NAG 3RD EDITION LAWS OF THERMODYNAMICS • 0 th law – when a body A is in thermal equilibrium with a body B, and also separately with a body C, then B and C will be in thermal equilibrium with each other. • Significance- measurement of property called temperature. A B C Evacuated tube 100o C Steam point Thermometric property 50o C (physical characteristics of reference body that changes with temperature) – rise of mercury in the evacuated tube 0o C bulb Steam at P =1ice atm T= 30oC REASONS FOR NOT TAKING ICE POINT AND STEAM POINT AS REFERENCE TEMPERATURES • Ice melts fast so there is a difficulty in maintaining equilibrium between pure ice and air saturated water. Pure ice Air saturated water • Extreme sensitiveness of steam point with pressure TRIPLE POINT OF WATER AS NEW REFERENCE TEMPERATURE • State at which ice liquid water and water vapor co-exist in equilibrium and is an easily reproducible state. This point is arbitrarily assigned a value 273.16 K • i.e. T in K = 273.16 X / Xtriple point • X- is any thermomertic property like P,V,R,rise of mercury, thermo emf etc. OTHER TYPES OF THERMOMETERS AND THERMOMETRIC PROPERTIES • Constant volume gas thermometers- pressure of the gas • Constant pressure gas thermometers- volume of the gas • Electrical resistance thermometer- resistance of the wire • Thermocouple- thermo emf CELCIUS AND KELVIN(ABSOLUTE) SCALE H Thermometer 2 Ar T in oC N Pg 2 O2 gas -273 oC (0 K) Absolute pressure P This absolute 0K cannot be obtatined (Pg+Patm) since it violates third law. -
W 15 “Isotherms of Real Gases”
Fakultät für Physik und Geowissenschaften Physikalisches Grundpraktikum W 15 “Isotherms of Real Gases” Tasks 1. Measure the isotherms of a substance for eight temperatures. Describe the processes that occur close to the critical point. 2. Plot the isotherms and determine the saturation vapour pressure ps in the region of the Maxwell line (coexistence of fluid and vapour). 3. Plot ln(ps) as a function of (1/T). Fit the vapour-pressure equation to the data and determine the average molar latent heat of vaporization of the substance under study. 4. Determine the amount of substance of the substance under study. 5. Use the Clausius-Clapeyron equation to determine the molar heat of vaporization as a function of temperature. Plot the latent heat as a function of the reduced temperature T/TK and fit a power law to the data. Literature Physikalisches Praktikum, Hrsg. W. Schenk, F. Kremer, 13. Auflage, Wärmelehre 2.0.1, 2.0.3 Physics, M. Alonso and E. J. Finn, 15.6 Accessories Instrument for investigation of the critical point, thermostat Keywords for preparation: - Ideal and real gas, kinetic theory, molar volume - Isothermal and adiabatic state transformation - Isothermal, isobaric, isochoric diagrams in p,V,T-space - Isotherms of an ideal and real gas in the p-V diagram - Equation of state after van der Waals, description in the p-V diagram - Maxwell construction, critical point - Compressibility factor z (ideal gas, real gas) - Vapour-pressure curve, Clausius-Clapeyron equation Remarks The “instrument for the investigation of the critical point” (company PHYWE) is already prepared for the measurement of a p-V diagram. -
Increasing Temperature Increases Disorder, Because the Entropy Dominates the Free Energy at High Temperatures, Whereas Enthalpy Dominates at Low Temperatures
Increasing temperature increases disorder, because the entropy dominates the free energy at high temperatures, whereas enthalpy dominates at low temperatures. That is, in the expression ΔG = ΔH TΔS, − the entropy term (TΔS) becomes more important (either for ΔS positive or ΔS negative) as T increases; hence at high temperatures the system will become more disordered to increase its entropy, even at the cost of increasing enthalpy. 3. The difference between the energy and enthalpy changes in expanding an ideal gas. How much heat is required to cause the quasi-static isothermal expansion of one mole of an ideal gas at T = 500 K from PA =0.42 atm, VA = 100 liters to PB =0.15 atm? (a) What is VB? (b) What is ΔU for this process? (c) What is ΔH for this process? (a) For an ideal gas, in any isothermal process ΔU = 0. Therefore ΔU = q + w =0 = q = w. ⇒ − For such a process, when it is also quasi-static, V2 nRT V w = pdV = dV = nRT ln 2 , − − V1 V − V1 so V q = nRT ln 2 V1 1 1 V2 = (1 mol)(2 cal K− mol− )(500 K) ln 100 L Using the ideal gas law, V2 p1 p1V1 = nRT = p2V2 = = . ⇒ V1 p2 79 Therefore we can write p q = nRT ln 1 p2 0.42 atm = (1000 cal) ln 0.15 atm = 1030 cal 1000 cal. ≈ (b) From the ideal gas law, p1 0.42 atm V2 = V1 = (100 L) p2 0.15 atm = 280 L. (c) For any isothermal process of an ideal gas, ΔU =0 [ΔU = C (T T )].