DOCSLIB.ORG

Second Year Thermodynamics M. Coppins 1.1 Basic Concepts

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
Second Year Thermodynamics M. Coppins 1.1 Basic Concepts Second Year Thermodynamics M. Coppins 1.1 Basic Concepts 1.1.1 Jargon Microscopic: on the atomic scale; involving very few or single particles (atoms, molecules, electrons, etc). Macroscopic: on the everyday scale or larger; involving very large numbers of particles. Thermodynamics is a macrocscopic theory. System: thing whose properties we are interested in, e.g., gas in container. Surroundings: things outside the system; can interact with system. System + surroundings often called “the universe” in thermodynamics text books. Isolated system: amount of matter and total energy of system is fixed, e.g., gas in closed rigid, thermally insulating container. Closed system: amount of matter in system is fixed, but energy can enter or leave system, e.g., gas in closed rigid container with walls which conduct heat. Energy can enter as heat ⇒ P (= pressure) increases. Open system: matter and energy can enter and leave it, e.g., gas in open container. If heated P stays constant (e.g., 1 atm) but number of molecules in container decreases. 1.1.2 Equilibrium = state which things settle into if left alone. In equilibrium the macroscopic properties don’t change with time. Even when macroscopic properties have settled into their equilibrium values the microscopic properties (e.g., velocity of a given molecule) are not constant. • Could have mixture with several chemically different components. Ni = number of molecules of component i. Chemical reactions ⇒ Ni’s change. Eventually chemical reac- tions stop, and the system settles into chemical equilibrium: Ni’s are constant. • Could have unbalanced forces, e.g., push piston into a container of gas → V (= volume) gets smaller. Eventually settles into mechanical equilibrium: V is constant. • Could put system into thermal contact with something at different T (= temperature). Eventually settles into thermal equilibrium: T is constant. Chemical + mechanical + thermal = thermodynamic equilibrium. 1.1.3 State variables In equilibrium macroscopic properties are • constant in time (if left alone), • uniform in space (if no external force field present1). If an external force field is acting the macroscopic properties can vary spatially, e.g., in an isothermal atmosphere (→ 1st Year SoM course, Lecture 5) the density falls with height due to gravity. State variable = a macroscopic property of a system in equilibrium (e.g., for a gas the state variables include: P , V , T , N = number of molecules). Extensive state variables are proportional to the amount of material. Intensive state variables are independent of amount of material. Combine 2 identical volumes of gas. Produces a volume of gas with 2N, 2V , P , T , i.e., N and V double (extensive), but P and T are unchanged (intensive). Specific value: extensive variable per unit mass. Denoted with lower case letters. Chemists use “moles” instead of kilograms. 1 mole is amount of stuff that contains as many particles as there are atoms in 12 g of 12C. Chemists’ specific values are therefore “molar” values. We won’t use moles in the Thermodynamics course. The most familiar state variables are P , V , N, and T . There are many others, e.g., U = internal energy, β = thermal expansion coefficient. 1.1.4 Equation of state ...is a relationship between state variables (usually P , V , N, and T ). It might be derived theoretically or found experimentally. • Ideal gas: PV = NkBT (1.1.4.1) −23 −1 (where kB = Boltzmann’s constant = 1.38 × 10 JK ). Chemists write the ideal gas equation of state in terms of moles: PV = NmRT , where Nm = the number of moles and R is the “universal gas constant”. • Van der Waals gas: N 2 P + a (V − Nb) = NkBT (1.1.4.2) V 2 (where a and b are constants; → 1st Year SoM course, Lecture 9) 1This will assumed throughout the Thermodynamics course Thermodynamics applies to other macroscopic systems than gases, in which case the state variables P and V can be replaced with other more appropriate variables, e.g., • Stretched metal wire: In this case P is replaced with J = tension, and V is replaced with L = length. Under certain conditions the equation of state has the form: J L = L0 1 + + α(T − T0) (1.1.4.3) YA where Y = Young’s modulus, A = cross-sectional area, α = coefficient of linear expansion, T0 = reference temperature, and L0 = length when T = T0 and J = 0. 1.1.5 Independent state variables Of the four state variables P , V , N, and T , we can specify any three; the equation of state then gives the value of the fourth one, e.g., ideal gas with P = 1.01 × 105 Pa (1 atm), N = 1025 molecules, T = 298 K → V = 0.41 m3 In thermodynamics we are usually interested in what happens when things change, e.g., change volume by ∆V , what is ∆T ? How many state variables can we vary independently? (1) Closed system: N = given constant. Can vary 2 of P , V , T , → equation of state gives 3rd one. Equation of State defines PVT surface, e.g. ideal gas Any point on this surface represents an equilibrium. Real substances have more complicated PVT surface, e.g., [Picture from University Physics (9th edition) by Young and Freedman.] (2) Open system: N = can change → 3 independent state variables. (3) N fixed, but two different phases (e.g., solid and liquid). Phase diagram is the projection of the PVT surface on the PT plane (→ 1st Year SoM course, Lecture 4). Two phases can only coexist in equilibrium along the lines on this diagram → extra constraint → only 1 independent state variable. [We consider the situation of two coexisting phases in detail in Lecture 3.5.] (4) Several chemically distinct components present ⇒ more independent state variables → chemistry. Most of the Thermodynamics course (Lectures 1.2 – 3.3) deals with a single component, single phase, closed system → 2 independent state variables. We will mostly concentrate on a fixed mass of gas in a container, partly because this par- ticular system is intrinsically important, and partly because it is easy to visualize what is happening. Remember, however, that thermodynamics doesn’t just apply to gases, but also to other macroscopic systems. Second Year Thermodynamics M. Coppins 1.2 T , U, S 1.2.1 Definitions The most important state variables in thermodynamics are: • T = temperature, • U = interal energy, • S = entropy. Like other state variables (e.g., P ) these are measurable macroscopic properties of things in equilibrium. T is in the equation of state (→ Lecture 1.1). U and S can be found from the other state variables. In thermodynamics we are usually concerned with changes in things, e.g., ∆T = Tfinal −Tinit. First law deals with ∆U (→ Lecture 1.4). Second law deals with ∆S (→ Lecture 2.3). Although thermodynamics is a macroscopic theory, this lecture will include a brief and very superficial look at T , U and S from a microscopic viewpoint. The microscopic theory will be covered in detail next term in the Statistical Physics course. 1.2.2 Temperature Microscopic view: T describes how particles of a system are distributed with respect to energy. Boltzmann law (→ 1st Year SoM course, Lecture 5): probability of particle having energy close to E, in −βE equilibrium, is proportional to e , where β = 1/kBT . From a theoretical point of view it would be better to replace the state variable T with β, but we are too used to temperature in everyday life. Example: gas molecules in Maxwell-Boltzmann velocity distribution function 1/2 2 m mvx f(vx) = exp − . 2πkBT 2kBT The probability that the x component of any given molecule’s velocity is between vx and vx +dvx is f(vx)dvx. Increase T ⇒ particles spread out more with respect to energy. We always use absolute temperatures. If β was used instead of T then absolute zero would correspond to β = ∞. Looked at this way the inaccessibility of absolute zero seems very natural. Macroscopic view: Two systems are in thermal equilibrium if they have the same T . T is sometimes approximately constant during some process. The theoretical idealization of this is an isothermal process, in which T is kept constant by placing the system in contact with a heat reservoir (i.e., an object so large that heat flow in or out of it does not change its temperature). Example: isothermal compression. Tgas stays equal to Tres during the compression because heat flows from the the gas into the reservoir. 1.2.3 Internal energy Microscopic view: U = total (kinetic + potential) energy of all the particles in the system. Ideal gas: particles do not interact (i.e., no potential energy of interaction) n ⇒ can use theorem of equipartition of energy: U = N d k T 2 B (nd = number of degrees of freedom: → 1st Year SoM course, Lecture 3) For a fixed mass of ideal gas it can be proved thermodynamically that the internal energy depends only on its temperature (→ Lecture 3.2). Microscopic theory required to obtain the functional form. Equipartition predicts n U = d Nk T (1.2.3.1) 2 B (nd = constant for give material, e.g., nd = 3 for a monatomic gas). Ideal gas: U = constant ⇔ T = constant. 3 N 2 Van der Waals gas: U = Nk T − a (depends on both T and V ). 2 B V Real gas: U depends on both T and V . 1.2.4 Entropy Microscopic view: S = kB ln W , where W = number of microstates corresponding to given macrostate. Increasing W ⇒ increasing disorder at microscopic level. Consider a system consting of N gas molecules in a box of volume V . Divide the box into a number of equal sized cells. Microscopic view: we know which cell each individual molecule is in.
Load more

Recommended publications
  • Bose-Einstein Condensation of Photons and Grand-Canonical Condensate fluctuations
    Bose-Einstein condensation of photons and grand-canonical condensate fluctuations Jan Klaers Institute for Applied Physics, University of Bonn, Germany Present address: Institute for Quantum Electronics, ETH Zürich, Switzerland Martin Weitz Institute for Applied Physics, University of Bonn, Germany Abstract We review recent experiments on the Bose-Einstein condensation of photons in a dye-filled optical microresonator. The most well-known example of a photon gas, pho- tons in blackbody radiation, does not show Bose-Einstein condensation. Instead of massively populating the cavity ground mode, photons vanish in the cavity walls when they are cooled down. The situation is different in an ultrashort optical cavity im- printing a low-frequency cutoff on the photon energy spectrum that is well above the thermal energy. The latter allows for a thermalization process in which both tempera- ture and photon number can be tuned independently of each other or, correspondingly, for a non-vanishing photon chemical potential. We here describe experiments demon- strating the fluorescence-induced thermalization and Bose-Einstein condensation of a two-dimensional photon gas in the dye microcavity. Moreover, recent measurements on the photon statistics of the condensate, showing Bose-Einstein condensation in the grandcanonical ensemble limit, will be reviewed. 1 Introduction Quantum statistical effects become relevant when a gas of particles is cooled, or its den- sity is increased, to the point where the associated de Broglie wavepackets spatially over- arXiv:1611.10286v1 [cond-mat.quant-gas] 30 Nov 2016 lap. For particles with integer spin (bosons), the phenomenon of Bose-Einstein condensation (BEC) then leads to macroscopic occupation of a single quantum state at finite tempera- tures [1].
    [Show full text]
  • Thermodynamics of Radiation Pressure and Photon Momentum
    Thermodynamics of Radiation Pressure and Photon Momentum Masud Mansuripur† and Pin Han‡ † College of Optical Sciences, the University of Arizona, Tucson, Arizona, USA ‡ Graduate Institute of Precision Engineering, National Chung Hsing University, Taichung, Taiwan [Published in Optical Trapping and Optical Micromanipulation XIV, edited by K. Dholakia and G.C. Spalding, Proceedings of SPIE Vol. 10347, 103471Y, pp1-20 (2017). doi: 10.1117/12.2274589] Abstract. Theoretical analyses of radiation pressure and photon momentum in the past 150 years have focused almost exclusively on classical and/or quantum theories of electrodynamics. In these analyses, Maxwell’s equations, the properties of polarizable and/or magnetizable material media, and the stress tensors of Maxwell, Abraham, Minkowski, Chu, and Einstein-Laub have typically played prominent roles [1-9]. Each stress tensor has subsequently been manipulated to yield its own expressions for the electromagnetic (EM) force, torque, energy, and linear as well as angular momentum densities of the EM field. This paper presents an alternative view of radiation pressure from the perspective of thermal physics, invoking the properties of blackbody radiation in conjunction with empty as well as gas-filled cavities that contain EM energy in thermal equilibrium with the container’s walls. In this type of analysis, Planck’s quantum hypothesis, the spectral distribution of the trapped radiation, the entropy of the photon gas, and Einstein’s and coefficients play central roles. 1. Introduction. At a fundamental
    [Show full text]
  • Statistical Mechanics I Fall 2005 Mid-Term Quiz Review Problems
    8.333: Statistical Mechanics I Fall 2005 Mid-term Quiz� Review Problems The Mid-term quiz will take place on Monday 10/24/05 in room 1-190 from 2:30 to 4:00 pm. There will be a recitation with quiz review on Friday 10/21/05. All topics up to (but not including) the micro-canonical ensemble will be covered. The exam is ‘closed book,’ but if you wish you may bring a two-sided sheet of formulas. The enclosed exams (and solutions) from the previous years are intended to help you review the material. Solutions to the midterm are available in the exam s section . ******** Answer all three problems, but note that the first parts of each problem are easier than its last parts. Therefore, make sure to proceed to the next problem when you get stuck. You may find the following information helpful: Physical Constants 31 27 Electron mass me 9:1 10− Kg Proton mass mp 1:7 10− Kg ∝ × 19 ∝ × 34 1 Electron Charge e 1:6 10− C Planck’s const./2α h¯ 1:1 10− J s− ∝ × 8 1 ∝ × 8 2 4 Speed of light c 3:0 10 ms− Stefan’s const. δ 5:7 10− W m− K− ∝ × 23 1 ∝ × 23 1 Boltzmann’s const. k 1:4 10− J K− Avogadro’s number N 6:0 10 mol− B ∝ × 0 ∝ × Conversion Factors 5 2 10 4 1atm 1:0 10 N m− 1A˚ 10− m 1eV 1:1 10 K ∞ × ∞ ∞ × Thermodynamics dE = T dS +dW¯ For a gas: dW¯ = P dV For a wire: dW¯ = J dx − Mathematical Formulas pλ 1 n �x n! 1 0 dx x e− = �n+1 2! = 2 R 2 2 2 � � 1 x p 2 β k dx exp ikx 2β2 = 2αδ exp 2 limN ln N ! = N ln N N −� − − − !1 − h i h i R n n ikx ( ik) n ikx ( ik) n e− = 1 − x ln e− = 1 − x n=0 n! h i n=1 n! h ic � � P 2 4 � � P 3 5 cosh(x) = 1 + x + x + sinh(x) = x + x + x + 2! 4! ··· 3! 5! ··· 2λd=2 Surface area of a unit sphere in d dimensions Sd = (d=2 1)! − 1 8.333: Statistical Mechanics I� Fall 1999 Mid-term Quiz� 1.
    [Show full text]
  • Solutions – PHY293F – Fall 2009 – Page 1 of 4 PHY 293F
    PHY 293F – WAVES AND PARTICLES DEPARTMENT OF PHYSICS, UNIVERSITY OF TORONTO PROBLEM SET #8 – SOLUTIONS Marked Q2 (parts (a) and (b); out of 7 points) and Q4 (out of 3 points) for a total of 10. 1. Calculate the Helmholtz free energy of a photon gas, such as the one we used to investigate blackbody radiation by doing the following: a. Calculate the Helmholtz free energy directly from F = U – TS. Simplify your expression to get the most compact expression possible. We know U and S in terms of T, V and α (related to a used in class by a=αV): 4 8 5 kT V " ( ) 4 U = 3 = #VT 15(hc) 5 3 32" $ kT ' 4 3 S = V& ) k = #VT , 45 % hc ( 3 8" 5k 4 where # = 3 . 15 hc ( ) Then we calculate the Helmholtz free energy and get: F =U " TS ! 4 = #VT 4 " T $ #VT 3 3 4% 4 ( = T '# V " #V * & 3 ) 1 1 = " #VT 4 = " U. 3 3 b. Verify your answer in part (a) by calculating the entropy, S and comparing it to the result we got ! 5 4 3 in class. To get a compact answer in terms of V and T, you can use " = 8# k 15(hc) . The entropy is equal to: $ #F ' S = "& ) . ! T % # (V So, using the result from part (a), we calculate the entropy as: $ #F ' # $ 1 4 ' ! "& ) = " & " *VT ) #T #T 3 % (V % ( 4 = *VT 3 = S. 3 Problem Set #8 – Solutions – PHY293F – Fall 2009 – Page 1 of 4 ! The expression is the same as the one we derived in class, except this one is in terms of α, V and T instead of a (α⋅V) and T.
    [Show full text]
  • 21 Apr 2021 Photon Gas . L18–1 the Photon Gas General
    21 apr 2021 photon gas . L18{1 The Photon Gas General Considerations • Goal: Consider the quantized electromagnetic field in a box of volume V , in thermal equilibrium at temperature T . As a thermodynamical system we have a gas of photons, but the number N is not fixed; photons can be absorbed and emitted by the walls. Notice that in a real system this equilibrium description is often not a very good approximation because there are no photon-photon interactions (for our purposes), and the photon interactions with matter (the walls) play a central role in establishing equilibrium, but they are strongly dependent on the material photons interact with. We would like to calculate E¯ and the energy density u = E=V¯ , the p and S equations of state, and the black-body spectrum (Planck's radiation law). • States: Photons are spin-1 massless particles whose 1-particle quantum states are specified by the pair α = (k; γ). In a box of volume V = L1L2L3 with periodic boundary conditions, the allowed values for the wave vector components are ki = 2πni=Li, ni 2 Z for i = 1, 2, 3, while the polarization has two values we can label γ = ±1. We will use the Fock representation and label general states in the total Hilbert space by the occupation numbers for each (k; γ), j N ;N ; :::; N ; :::i, where each N = 0, 1, 2, ... k1,γ1 k2,γ2 kj ,γj kj ,γj • Hamiltonian: Photons are, to an excellent approximation, non-interacting particles. The single-particle mode (k; γ) has energy α = k =h! ¯ with ! = ck, so the Hamiltonian operator can be written as ^ X ^ X y H = ¯h! Nk,γ = ¯h! a^k,γ a^k,γ ; k,γ k,γ ^ y y where Nα =a ^α a^α is the number operator for the mode α, anda ^α anda ^α the creation and annihilation operators for a photon in that mode, respectively.
    [Show full text]
  • Physics 831: Statistical Mechanics
    Physics 831: Statistical Mechanics Russell Bloomer1 University of Virginia Note: There is no guarantee that these are correct, and they should not be copied 1email: [email protected] Contents 1 Problem Set 1 1 1.1 Kittel 8.1: Heat pump ........................................ 1 1.2 Kittel 8.6: Room air conditioner .................................. 1 1.3 Kittel 8.9: Cooling of nonmetallic solid to T = 0 ......................... 2 1.4 Sterling Heat Engine ......................................... 2 1.5 Unavailability for work ........................................ 3 1.6 Gibbs Free Energy .......................................... 3 2 Problem Set 2 5 2.1 Spin model .............................................. 5 2.2 Paramagnetism of a system of N localized spin-1/2 particles ................... 6 2.3 Kittel 2.3: Quantum harmonic oscillator .............................. 7 2.4 Review Thermal mechanics ..................................... 7 2.5 Review Thermal mechanics ..................................... 8 3 Problem Set 3 9 3.1 Kittel 3.2: Magnetic susceptibility ................................. 9 3.2 Kittel 3.3: Free energy of a harmonic oscillator .......................... 10 3.3 Kittel 3.4: Energy fluctuations ................................... 10 3.4 Kittel 3.10: Elasticity of polymers ................................. 11 3.5 Ising spin chain ............................................ 12 4 Problem Set 4 13 4.1 Application of equal partition theorem ............................... 13 4.2 Thermal Equilibrium of the Sun and Earth ............................ 14 4.3 Kittel 4.3: Average temperature of the interior of the Sun .................... 14 4.4 Kittel 4.6: Pressure of thermal radiation .............................. 15 4.5 Kittel 4.7: Free energy of a photon gas ............................... 15 4.6 Kittel 4.18: Isentropic expansion of photon gas .......................... 16 5 Problem Set 5 17 5.1 5.1: Kittel 4.14: Heat capacity of liquid 4He at low temperature ..................
    [Show full text]
  • Complete Quantum Thermodynamics of the Black Body Photon
    COMPLETE QUANTUM THERMODYNAMICS OF THE BLACK BODY PHOTON GAS Vladan Pankovi´c, Darko V. Kapor Department of Physics, Faculty of Sciences, 21000 Novi Sad, Trg Dositeja Obradovi´ca 4, Serbia, [email protected] PACS number: 03.65.Ta, 05.70.-a keywords: photon gas thermodynamics, black body, black hole entropy Abstract Kelly and Leff demonstrated and discussed formal and conceptual similarities between basic thermodynamic formulas for the classical ideal gas and black body photon gas. Leff pointed out that thermodynamic formulas for the photon gas cannot be deduced completely by thermodynamic methods since these formulas hold two characteristic pa- rameters, r and b, whose accurate values can be obtained exclusively by accurate methods of the quantum statistics (by explicit use of the Planck’s or Bose-Einstein distribution). In this work we prove that the complete quantum thermodynamics of the black body pho- ton gas can be done by simple, thermodynamic (non-statistical) methods. We prove that both mentioned parameters and corresponding variables (photons number and pressure) can be obtained very simply and practically exactly (with relative error about few per- cent), by non-statistical (without any use of the Planck’s or Bose-Einstein distribution), quantum thermodynamic methods. Corner-stone of these methods represents a quantum thermodynamic stability condition that is, in some degree, very similar to quantum sta- bility condition in the Bohr quantum atomic theory (de Broglie’s interpretation of the Bohr quantization postulate). Finally, we discuss conceptual similarities between black arXiv:1103.3161v1 [physics.gen-ph] 16 Mar 2011 body photon gas entropy and Bekenstein-Hawking black hole entropy.
    [Show full text]
  • Photon Gas in a Finite Box: Thermodynamics and Finite Size Effects
    Photon gas in a finite box: thermodynamics and finite size effects A. A. Sokolsky and M. A. Gorlach Physics Department of the Belarusian State University, 4 Nezalejnosty av., Minsk, 220030, Belarus. Thermodynamic properties of a photon gas in a small box are explored taking into account finite size effects. General thermodynamic relations are derived for this finite system. Photon gas thermodynamic functions are calculated for the case of cuboid cavity on the basis of first principles. New finite size effects are discussed. Introduction As it is known, the phenomenon of blackbody radiation plays an important role in physics. In 1884 L. Boltzmann derived that internal energy of radiation in a cavity is proportional to the fourth power of the temperature; this statement is known as Stefan- Boltzmann law ([1]). In 1900, basing of quantum hypothesis, M. Planck obtained a formula describing blackbody radiation spectrum ([2]); this formula was thought to be correct for arbitrary temperatures and cavity volumes for a long period of time. But it turns out that Planck’s formula, Stefan-Boltzmann law, Wien’s displacement law are not accurate in the case of sufficiently small cavities and low temperatures. This remark was apparently first done by Bijl ([3]). He also found a criterion of the Planck’s formula validity: ~c TV 1/3 ≫ B ≡ ≈ 0.2290 cm · K (1) kB The following step was done in [4] where first correction terms to the Stefan- Boltzmann law were calculated for the case of cubic cavity with ideally conducting walls. Since that time numerous efforts were made in order to take into account finite size effects in thermal radiation theory.
    [Show full text]
  • Bose-Einstein Condensation of Photons in a Microscopic Optical Resonator: Towards Photonic Lattices and Coupled Cavities
    Bose-Einstein Condensation of Photons in a Microscopic Optical Resonator: Towards Photonic Lattices and Coupled Cavities Jan Klaers, Julian Schmitt, Tobias Damm, David Dung, Frank Vewinger, and Martin Weitz Institute for Applied Physics, University of Bonn, Wegelerstr. 8, 53115 Bonn, Germany Bose-Einstein condensation has in the last two decades been observed in cold atomic gases and in solid-state physics quasiparticles, exciton-polaritons and magnons, respectively. The perhaps most widely known example of a bosonic gas, photons in blackbody radiation, however exhibits no Bose-Einstein condensation, because the particle number is not conserved and at low temperatures the photons disappear in the system’s walls instead of massively occupying the cavity ground mode. This is not the case in a small optical cavity, with a low-frequency cutoff imprinting a spectrum of photon energies restricted to values well above the thermal energy. The here reported experiments are based on a microscopic optical cavity filled with dye solution at room temperature. Recent experiments of our group observing Bose-Einstein condensation of photons in such a setup are described. Moreover, we discuss some possible applications of photon condensates to realize quantum manybody states in periodic photonic lattices and photonic Josephson devices. 1. INTRODUCTION When a gas of particles is cooled, or its density is increased up to the point that the associated de Broglie wavepackets spatially overlap, quantum statistical effects come into play. Specifically, for material particles with integer spin (bosons), the phenomenon of Bose-Einstein condensation into a single quantum state then minimizes the free energy. Bose-Einstein condensation was first experimentally achieved in 1995 by laser and subsequent evaporative cooling of a dilute cloud of alkali atoms [1,2].
    [Show full text]
  • Physics 12C: Problem Set 6
    Physics 12c: Problem Set 6 Due: Thursday, May 23, 2019 1. Light bulb in a refrigerator A refrigerator that draws 50 W of power is contained in a room at temperature 300◦K. A 100 W lightbulb is left burning inside the refrigerator. Find the steady-state temperature inside the refrigerator assuming it operates reversibly and is perfectly insulated. 2. Photonic heat engine Consider a heat engine undergoing a Carnot cycle, where the working fluid is a photon gas rather than a classical ideal gas. In the first stage, the gas expands isothermally at temperature τh from the initial volume V1 to the final volume V2. In the second stage, it expands isentropically to volume V3, cooling to temperature τl. In the third stage it is compressed isothermally at temperature τl to volume V4, and in the fourth stage it is compressed isentropically back to volume V1, heating to temperature τh. (a) The energy per unit volume of a photon gas is U=V = Aτ 4, where A = 2 3 3 π =15~ c . Use the thermodynamic identity dU = τdσ − pdV (1) to find the entropy of the gas, expressed in terms of A; τ, and V . Assume that the entropy is zero at τ = 0. (b) Use the thermodynamic identity again to express the pressure p in terms of A; τ; and V . (c) Calculate the work done W12 and the heat added Q12 during the first stage of the cycle, expressed in terms of A; τh;V1, and V2. Verify that Q12 − W12 is the change in the internal energy of the gas.
    [Show full text]
  • Blackbody Radiation (Ch
    Lecture 26. Blackbody Radiation (Ch. 7) Two types of bosons: (a) Composite particles which contain an even number of fermions. These number of these particles is conserved if the energy does not exceed the dissociation energy (~ MeV in the case of the nucleus). (b) particles associated with a field, of which the most important example is the photon. These particles are not conserved: if the total energy of the field changes, particles appear and disappear. We’ll see that the chemical potential of such particles is zero in equilibrium, regardless of density. Radiation in Equilibrium with Matter Typically, radiation emitted by a hot body, or from a laser is not in equilibrium: energy is flowing outwards and must be replenished from some source. The first step towards understanding of radiation being in equilibrium with matter was made by Kirchhoff, who considered a cavity filled with radiation, the walls can be regarded as a heat bath for radiation. The walls emit and absorb e.-m. waves. In equilibrium, the walls and radiation must have the same temperature T. The energy of radiation is spread over a range of frequencies, and we define uS (ν,T) dν as the energy density (per unit volume) of the radiation with frequencies between ν and ν+dν. uS(ν,T) is the spectral energy density. The internal energy of the photon gas: ∞ = , dTuTu νν ()∫ S ( ) 0 In equilibrium, uS (ν,T) is the same everywhere in the cavity, and is a function of frequency and temperature only. If the cavity volume increases at T=const, the internal energy U=u(T) V also increases.
    [Show full text]
  • Bose Systems
    Chapter 7 Bose Systems In this chapter we apply the results of section 5.3 to systems of particles satisfying Bose-Einstein statistics. The examples are Black Body radiation (the photon gas), atomic vibration in solids (the phonon gas) and alkali atoms in traps and liquid 4He (a Bose gas and fluid). Bosons in traps and the superfluid state of liquid 4He is believed to be an example of Bose-Einstein condensation in which a large fraction of the Bosons occupy (condense into) the ground state. 7.1 Black Body Radiation We are all familiar with the idea that hot objects emit radiation {a light bulb, for example. In the hot wire ¯lament, an electron, originally in an excited state drops to a lower energy state and the energy di®erence is given o® as a photon, (² = hº = ²2 ¡ ²1). We are also familiar with the absorption of radiation by surfaces. For example, clothes in the summer absorb photons from the sun and heat up. Black clothes absorb more radiation than lighter ones. This means, of course that lighter colored clothes reflect a larger fraction of the light falling on them. A black body is de¯ned as one which absorbs all the radiation incident upon it {a perfect absorber. It also emitts the radiation subsequently. If radia- tion is falling on a black body, its temperature rises until it reaches equilibrium with the radiation. At equilibrium it re-emitts as much radiation as it absorbs so there is no net gain in energy and the temperature remains constant. In this case the surface is in equilibrium with the radiation and the temperature of the surface must be the same as the temperature of the radiation.
    [Show full text]