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AS3105 Proses Astrofisika 1 Bab1 Pendahuluan Dhani Herdiwijaya v1.50 Aturan • Absensi >= 75% − < 75% tidak boleh ikut UAS • Prosentase nilai − NA = 40% UAS + 30% UTS + 20% tugas+kuis + 10% absensi • Tidak ada ujian susulan • Jadwal − Selasa jam 10.00-11.00, Ruang Seminar 2 − Kamis jam 09.00-11.00, Ruang Seminar 2 Referensi • Reif, F., Fundamentals of Statistical and Thermal Physics, McGraw Hill, 1987 • Reif, F., Statistical Physics, Berkeley Physics Course Vol. 5, McGraw Hill, 1987 • Landau, Lifshitz, Statistical Physics 3 rd ed. Pergamon Press (bacaan lanjut) • Padmanabhan, T., Theoretical Astrophysics Vol. 1. Astrophysical Processes, Cambridge University Press, 2000 • Mihalas, D. Stellar Atmosphere, W.H. Freeman, 1978 • Rybicky, G.B. and Ligthman, A.P., Radiative Processes in Astrophysics, John Wiley & Sons, 1979 (bacaan lanjut) Materi • Review termodinamika • Pendahuluan • Probabilitas • Distribusi Maxwell-Boltzmann • Distribusi Bose-Einstein • Distribusi Fermi-Dirac • Aplikasi − − − − Radiasi Benda Hitam Kondensat Bose Gas Fermi terdegenerasi Transfer Radiasi Bab 1 • Review Hukum Termodinamika • Termodinamika dan Fisika Statistik • Konsep makroskopik dan mikroskopik • Gas ideal Contoh Pertanyaan Termodinamika • Bagaimana kerja lemari es? Berapa efisiensi maksimumnya? • Berapa banyak energi diperlukan untuk memanaskan air dalam panci sehingga menjadi uap? Hukum Termodinamika • Hukum ke-0 • Hukum ke-1 • Hukum ke-2 • Hukum ke-3 Sistem • Sebuah sistem terpisah dari lingkungannya oleh dinding. Sistem tertutup, jika tidak ada perpindahan materi antara sistem dan lingkungan. • Dinding dibagi dua, yaitu − Adiabatic: tidak ada pertukaran panas − Diathermal: panas dapat mengalir lewat dinding Sistem yang umum ditinjau adalah • •sistem terisolasi dari lingkungan dengan dinding adiabatik (sistem adiabatik) • •sistem bersentuhan dengan sumber panas melalui dinding diatermal (sistem isotermal) Sebuah sistem terdiri dari variabel termodinamika, yang terbagi menjadi variabel extensive dan intensive. Variabel • Variabel Intensive: tidak bergantung ukuran sistem dan dapat berubah-ubah. Misalkan tekanan, temperatur, densitas • Variabel Extensive: bergantung kuran sistem. Misalkan, volume, jumlah partikel, massa, energi internal • Variabel Extensive and intensive cenderung berpasangan (conjugate variables). Variabel Conjugate: • • • • system generalised fluid Wire film intensive variable generalised force X pressure p tension F surface tension g extensive variable generalised displacement x volume V length l area A Sistem Termodinamika Open systems can exchange both matter and energy with the environment. Closed systems exchange energy but not matter with the environment. Isolated systems can exchange neither energy nor matter with the environment. Internal and external macroscopic parameters: temperature, volume, pressure, energy, electromagnetic fields, etc. (average values, fluctuations are ignored). No matter what is the initial state of an isolated system, eventually it will reach the state of thermodynamic equilibrium (no macroscopic processes, only microscopic motion of molecules). Hukum ke-0 Jika sistem A setimbang dengan sistem B dan sistem C, maka sistem B juga setimbang dengan sistem C. Berlaku kebalikannya eksistensi keadaan setimbang A very important macro-parameter: Temperature Temperature is a property associated with chaotic motion of many particles (it would be absurd to refer to the temperature of a single molecule, or to the temperature of many molecules moving with the same speed in the same direction). Introduction of the concept of temperature in thermodynamics is based on the the zeroth law of thermodynamics: A well-defined quantity called temperature exists such that two systems will be in thermal equilibrium if and only if both have the same temperature. The zeroth law implies the existence of some universal property of systems in thermal equilibrium. This law, in conjunction with the concept of entropy, will help us to mathematically define temperature in terms of statistical ideas. Also, if system C is in thermal equilibrium with systems A and B, the latter two systems, when brought in contact, also will be in thermal equilibrium. - this allows us to obtain the temperature of a system without a direct comparison to some standard. Temperature Measurement Properties of a thermoscope (any device that quantifies temperature): 1. It should be based on an easily measured macroscopic quantity a (volume, resistance, etc.) of a common macroscopic system. 2. The function that relates the chosen parameter with temperature, T = f(a), should be monotonic. 3. The quantity should be measurable over as wide a range of T as possible. The simplest case – linear dependence T = Aa (e.g., for the ideal gas thermometer, T = PV/NkB). the ideal gas thermometer, T = PV/NkB T the resistance thermometer with a semi- conductor sensor, R exp T a Thermometer a thermoscope calibrated to a standard temp. scale The Absolute (Kelvin) Temp. Scale The absolute (Kelvin) temperature scale is based on fixing T of the triple point for water (a specific T = 273.16 K and P = 611.73 Pa where water can coexist in the solid, liquid, and gas phases in equilibrium). T,K 273.16 0 absolute zero P - for an ideal gas constant-volume T 273.16 K PTP thermoscope PTP P PTP – the pressure of the gas in a constant-volume gas thermoscope at T = 273.16 K Hukum ke-1 Energi internal sistem terisolasi, E dapat berubah jika terjadi aliran panas, Q atau melakukan usaha, W (tanda bisa pos. atau neg. bergantung konvensi) Dirumuskan sebagai ΔE=ΔQ+ΔW Banyak panas masuk ke sistem Usaha yg dilakukan dalam sistem Dq =CP dT atau dq = CV dT Dw = - pdV Temperatur dan Energi Kinetik The temperature T of a system of N particles is a quantity related to the averaged kinetic energy of the disordered motion of the particle in the frame of reference. T Ek ,ave 1 1 2 mi vi N i 2 Mass of particle i Velocity of particle i Thermal equilibrium: Average kinetic energy is the same in all regions of the system Hukum ke-2 • Clausius: Panas mengalir dari panas ke dingin • Kelvin: Tidaklah mungkin mengubah semua energi panas menjadi usaha • Carathéodory: lingkungan suatu keadaan stabil dari sistem yang terisolasi secara termal, terdapat suatu keadaan yang tidak teramati • Callen: Entropi tidak pernah turun dalam sistem terisolasi dan proses spontan Hukum ke-2 The entropy S of an isolated system increases during any spontaneous change. Spontaneous Low entropy gas S 0 Not observed High entropy Hukum ke-3 Nernst, Simon, dkk. • Entropi dalam suatu sistem akan mendekati nol, jika temperatur mendekati nol absolut Nernst’s Theorem (1906) The Third Law of Thermodynamics CV (T ) dT S (T ) S (0) T 0 T A non-trivial issue – what’s the value of S(0)? The answer is provided by Q.M. (discreteness of quantum states), it cannot be deduced from the other laws of thermodynamics – thus, the third law: Nernst’s Theorem: The entropy of a system at T = 0 is a well-defined constant. For any processes that bring a system at T = 0 from one equilibrium state to another, S = 0 . This is because a system at T = 0 exists in its ground state, so that its entropy is determined only by the degeneracy of the ground state. In other words, S(T) approaches a finite limit at T = 0 , which does not depend on specifics of processes that brought the system to the T = 0 state. A special case – systems with unique ground state, such as elementary crystals : S(0) = 0 ( = 1). However, the claim that S(0) is "usually" zero or negligible is wrong. Residual Entropy If S(0) 0, it’s called residual entropy. 1. For compounds, for example, there is frequently a significant amount of entropy that comes from a multiplicity of possible molecular orientations in the crystal (degeneracy of the ground state), even at absolute zero. Structure of the hexagonal ice: - H atoms, --- - covalent O-H bonds, - - - weak H bonds. Six H atoms can “tunnel” simultaneously through a potential barrier. 2. Nernst’s Theorem applies to the equilibrium states only! Glasses aren’t really in equilibrium, the relaxation time – huge. They do not have a welldefined T or CV. Glasses have a particularly large entropy at T = 0. S supercooled liquid liquid glass residual entropy crystal T Example (Pr. 3.14) For a mole of aluminum, CV = aT + bT3 at T < 50 K (a = 1.35·10-3 J/K2, b = 2.48·10–5 J/K4). The linear term – due to mobile electrons, the cubic term – due to the crystal lattice vibrations. Find S(T) and evaluate the entropy at T = 1K,10 K. CV (T ) dT aT bT 3 dT b S (T ) aT T 3 T T 3 0 0 T T 1 S (1K ) 1.35 10 3 J/K 2 1K 2.48 10 5 J/K 4 1K 3 1.36 10 3 J/K 3 - at low T, nearly all the entropy comes from the mobile electrons T = 1K T = 10K 1 S (10 K ) 1.35 10 3 J/K 2 10K 2.48 10 5 J/K 4 103 K 3 2.18 10 2 J/K 3 - most of the entropy comes from lattice vibrations S (1K ) 1.35 103 J/K 20 10 kB 1.38 1023 J/K S (10 K ) 2.18 10 2 J/K 21 1 . 6 10 kB 1.38 1023 J/K - much less than the # of particles, most degrees of freedom are still frozen out. C 0 as T 0 CV (T ) dT S (T ) S (0) T 0 T C (T ) 0 as T 0 For example, let’s fix P : Q CP (T )dT T 0 CP (T ) dT T 0 T Q T - finite, CP(0) = 0 Similar, considering V = const - CV(0) = 0 . Thus, the specific heat must be a function of T. We know that at high T, CV approaches a universal temperature-independent limit that depends on N and # of degrees of freedom. This high - T behavior cannot persist down to T = 0 – quantum effects will come into play. Entropi dS surroundings dSuniverse dS dS surroundings dSuniverse 0 dS System Heat: dqsurroundings T Surroundings dSuniverse 0 For an equilibrium process System + Surroundings = Universe The universe seeks higher entropy Overall entropy production Entropi dq dS 0 T dq : heat supplied to the system dS : Change of entropy of system due to internal processes. Termodinamika Termodinamika merupakan kerangka umum dengan hanya melihat sifat-sifat makroskopik atau variabel termodinamika dan mengabaikan sifat mikroskopik (setiap molekul). Makroskopik • Kondisi makroskopik memperhatikan sistem secara keseluruhan. Sistem yang setimbang dinyatakan dalam beberapa variabel, misalkan: tekanan, volume, temperatur, energi internal dan entropi. • Kondisi makro dapat berubah dengan 2 cara, yaitu usaha pada sistem dan/atau panas yang mengalir dalam sistem. • Variabel makroskopik termodinamika • Misalkan, tabung gas He, maka dapat diketahui volume, jumlah molekul, energi internalnya Macroscopic Description is Qualitatively Different! Why do we need to consider macroscopic bodies as a special class of physical objects? Because the behavior of macroscopic bodies differs from that of a single particle in a very fundamental respect: For a single particle: all equations of classical mechanics, electromagnetism, and quantum mechanics are time-reversal invariant (Newton’s second law, F = dp/dt, looks the same if the time t is replaced by –t and the momentum p by –p). For macroscopic objects: the processes are often irreversible (a timereversed version of such a process never seems to occur). Examples: (a) living things grow old and die, but never get younger, (b) if we drop a basketball onto a floor, it will bounce several times and eventually come to rest - the arrow of time does exist. Conservation of energy does not explain why the time-reversed process does not occur - such a process also would conserve the total energy. Though the total energy is conserved, it has been transferred from one degree of freedom (motion of ball’s center of mass) to many degrees of freedom (associated with the individual molecules of the ball and the floor) – the distribution of energy changes in an irreversible manner. Contoh Pertanyaan lain • Mengapa sifat air berbeda dengan uap, meskipun molekul keduanya sama? • Mengapa besi kehilangan sifat kemagnetannya pada temperatur tertentu? Mempertanyakan variabel mikroskopik Sifat dari molekul dengan jumlah sangat besar diperlukan statistika atau fisika statistik Mikroskopik • Kondisi mikroskopik dinyatakan oleh banyak sekali variabel, karena skala molekul. Misalkan kondisi gas dinyatakan dari posisi dan momentum setiap saat. Untuk N partikel mempunyai N variabel. Untuk 1 cm3 gas at STP, N ~ 1019. • Posisi dan momentum setiap partikel memenuhi hukum mekanika. Praktis, tidaklah mungkin melakukan komputasi, sehingga diperlukan distribusi statistik mekanika statistik • Dalam tabung Helium, perlu diketahui posisi dan kecepatan setiap (dan fungsi gelombang) molekul. Energi total merupakan jumlah energi semua molekul • Gaya setiap molekul, F=ma - terdapat 6N koordinat saat t0: ri and vi - dengan N ≡ moles (1023) ! Ruang fase {p} momenta N particles 6N dimensional space {q} Coordinates Trajectory Up One-dimensional harmonic oscillator 2N dimensional phase space Alam Mikroskopik dan Makroskopik Mekanika klasik Mekanika Kuantum Interaksi Microscopic Ruang Phase Fungsi gelombang Statistical Averaging Fisika Statistik Makroskopik Observables: Dynamic and Thermodynamic quantities Fisika Statistika • Fisika Statistik merupakan jembatan antara alam mikroskopik dan makroskopik ada hubungan hukum termodinamika dan sifat statistik molekul. • Fisika statistik menjelaskan sifat makroskopik dalam perhitungan detil mikroskopik. • Fisika statistik mempunyai kemampuan prediksi, karena adanya model mikroskopik yang dapat menghitung sesuatu yang terukur. Fisika Statistika sbg Jembatan • Dalam mikroskopik, energi setiap atom bernilai dikrit ata kontinu • Dalam kesetimbangan, atom terdistribusi dalam tingkatan energi. Dalam distribusi tersebut dapat diketahui berapakah atom yang mempunyai energi dalam rentang E dan E + dE. Probabilitas distribusi atom ini cukup untuk menentukan variabel makroskopik termodinamika • Jadi Fisika Statistik menghubungkan probabilitas mikroskopik menjadi variabel makroskopik dalam suatu sistem. • Orang yang banyak berjasa menyatukan fisika termal (termodinamika) dan mekanika adalah Boltzmann dan Gibbs Fisika Statistik dan Termodinamika Statistical description of a large system of identical (mostly, non-interacting) particles Equation of state for macrosystems (how macroparameters of the system and the temperature are interrelated) all microstates of an isolated system occur with the same probability, the concepts of multiplicity (configuration space), entropy Irreversibility of macro processes, the 2nd Law of Thermodynamics Concepts of Statistical Physics 1. The macrostate is specified by a sufficient number of macroscopically measurable parameters. 2. The microstate is specified by the quantum state of each particle in a system. 3. The multiplicity is the number of microstates that are consistent with a given macrostate. For each macrostate, there is an extremely large number of possible microstates that are macroscopically indistinguishable. 4. The Fundamental Assumption: for an isolated system, all accessible microstate are equally likely. 5. The probability of a macrostate is proportional to its multiplicity. This will sufficient to explain irreversibility. Hukum ke-1 dalam bentuk Probabilitas Fisika Statistik Boltzmann hypothesis: the entropy of a system is related to the probability of its state; the basis of entropy is statistical. Fisika Kuantum • Dalam fisika kuantum, keadaan mikroskopik dinyatakan dalam fungsi gelombang • Mikroskopik dinyatakan dalam bilanganbilangan kuantum Quantum numbers • The state of the system is fully described by the wave function. • The state of the system is fully described by a set of quantum numbers (n, m, l, …..). • The total energy of the system is given from Schrödinger equation. An observable such as the energy E corresponds to an operator such as H. r1 , r2 ,..., t H E Model Gas Ideal Our first model of a many-particle system: the Ideal Gas Models of matter: gas models (random motion of particles) lattice models (positions of particles are fixed) Air at normal conditions: ~ 2.71019 molecules in 1 cm3 of air (Pr. 1.10) Size of the molecules ~ (2-3)10-10 m, distance between the molecules ~ 310-9 m The average speed - 500 m/s The mean free path - 10-7 m (0.1 micron) The number of collisions in 1 second - 5 109 The ideal gas model - works well at low densities (diluted gases) • all the molecules are identical, N is huge; • the molecules are tiny compared to their average separation (point masses); • the molecules do not interact with each other; • the molecules obey Newton’s laws of motion, their motion is random; • collisions between the molecules and the container walls are elastic. The quantum version of the ideal gas model helps to understand the blackbody radiation, electrons in metals, the low-temperature behavior of crystalline solids, etc. The Equation of State of Ideal Gases An equation of state - an equation that relates macroscopic variables (e.g., P, V, and T) for a given substance in thermodynamic equilibrium. In equilibrium ( no macroscopic motion), just a few macroscopic parameters are required to describe the state of a system. Geometrical representation of the equation of state: P an equilibrium state the equationof-state surface f (P,V,T) = 0 V T The ideal gas equation of state: PV nRT P V n T – – – – pressure volume number of moles of gas the temperature in Kelvins R – a universal constant [Newtons/m2] [m3] [K] R 8.31 J mol K The Ideal Gas Law In terms of the total number of molecules, N = nNA Avogadro’s number NA 6.0220451023 PV NkBT the Boltzmann constant kB = R/NA 1.3810-23 J/K (introduced by Planck in 1899) The equations of state cannot be derived within the frame of thermodynamics: they can be either considered as experimental observations, or “borrowed” from statistical mechanics. Avogadro’s Law: equal volumes of different gases at the same P and T contain the same amount of molecules. The P-V diagram – the projection of the surface of the equation of state onto the P-V plane. isotherms The van der Waals model of real gases For real gases – both quantitative and qualitative deviations from the ideal gas model 4 U(r) 3 Electric interactions between electro-neutral molecules : Energy 2 1 0 -1 -2 -3 Veff V Nb repulsion attraction aN 2 Peff P 2 V 1.5 2.0 2.5 r 3.0 3.5 4.0 distance van der Waals attraction (~ r -7) aN 2 P 2 V Nb NkBT V The van der Waals equation of state for real gases (the only model with interactions that we’ll consider) Connection between Ktr and T for Ideal Gases ? T of an ideal gas the kinetic energy of molecules Pressure – the result of collisions between the molecules and walls of the container. Momentum Strategy: Pressure = Force/Area = [p / t ]/Area px = 2 m vx Intervals between collisions: t = 2 L/vx For each (elastic) collision: Piston area A vx Volume = LA L N For N molecules - 2mvx 1 1 2 1 Pi p x v x mvx 2 L / vx A V V PV mvx2 N m v x2 i no-relativistic motion Connection between Ktr and T for Ideal Gases (cont.) N PV mvx2 N m v x2 i PV NkBT Average kinetic energy of the translational motion of molecules: 3 K tr k BT 2 3 U K tr NkBT 2 2 PV U 3 K tr m v x2 k BT 1 1 3 2 2 2 2 m v m vx v y vz m vx2 2 2 2 - the temperature of a gas is a direct measure of the average translational kinetic energy of its molecules! The internal energy U of a monatomic ideal gas is independent of its volume, and depends only on T (U =0 for an isothermal process, T=const). - for an ideal gas of non-relativistic particles, kin. energy (velocity)2 . Units for Energy, Temperature 3 K tr k BT 2 - the kinetic energy is proportional to the temperature, and the Boltzmann constant kB is the coefficient of proportionality that provides one-to-one correspondence between the units of energy and temperature. Theorists usually assume that kB = 1 - the same units for energy and temperature – which makes a lot of sense. If the temperature is measured in Kelvins, and the energy – in Joules: kB = 1.3810-23 J/K In many sub-fields of physics that deal with microscopic particles (including condensed matter physics, physics of high energies, astrophysics, etc.), a convenient unit for energy is an electron-Volt (the kinetic energy acquired by an electron accelerated by the electrostatic potential difference in 1 V). K = eV 1 eV = 1.6 10-19 J At T = 300K 1.38 1023 J / K 300 K k BT 26 meV 19 1.6 10 J / eV Pressure of a photon gas 1 2 2 PV N m v N K tr - valid for non-relativistic particles 3 3 1 PV N vp - valid also for relativistic particles (m = f(v)) 3 In particular, it applies to the gas of photons that move randomly within some cavity with mirror walls. For photons, p E ph c E ph cp h PV 1 E ph 3 pre-factor 1/3 (instead 2/3) - due to a different relation between E and p for relativistic particles We will use this equation later in the course when we consider the thermal radiation. Degrees of Freedom Indeed, the polyatomic molecules have more than just three degrees of freedom – they rotate and vibrate. The degrees of freedom of a system are a collection of independent variables required to characterize the system. Thus, for instance, the degrees of freedom of an ideal gas, in a thermodynamic description, are P and V (or a pair of other variables). The independent coordinates that describe the position of a mechanical system in space - e.g., for a rigid body, it is sufficient to specify the coordinates of its center of mass (x, y, z) and three angles (x, y, z). K Ix – the moment of inertia for rotations around the x-axis, etc. Polyatomic molecules: 6 transl.+rotat. degrees of freedom 1 1 M x 2 y 2 z 2 I x x2 I y y2 I z z2 2 2 Diatomic molecules: 3 + 2 = 5 transl.+rotat. degrees of freedom (QM: degrees of freedom corresponding to rotations that leave the molecule completely unchanged do not count ) Degrees of Freedom (cont.) Plus all vibrational degrees of freedom. The one-dimensional vibrational motion counts as two degrees of freedom (kin. + pot. energies): K U x 1 1 m x 2 k x 2 2 2 For a diatomic molecule (e.g., H2), 5 transl.+rotat. degrees of freedom plus 2 vibrational degrees of freedom = total 7 degrees of freedom Among 7 degrees of freedom, only 3 (translational) degrees correspond to a continuous energy spectrum, the other 5 – to a discrete energy spectrum. U(x) U(x) 4 E4 3 Energy 2 1 E3 0 E2 -1 -2 -3 1.5 2.0 2.5 3.0 x distance 3.5 4.0 E1 x Equipartition of Energy “Quadratic” degree of freedom – the corresponding energy = f(x2, vx2) [ translational motion, (classical) rotational and vibrational motion, etc. ] Equipartition Theorem: At temperature T, the average energy of any “quadratic” degree of freedom is 1/2kBT. - holds only for a system of particles whose kinetic energy is a quadratic form of x2, vx2 (e.g., the equipartition theorem does not work for photons, E = cp) Piston – a mechanical system with one degree of freedom. Thus, vx m v 2x 2 M u2 2 1 k BT 2 M – the mass of a piston, u2 the average u2, where u is the piston’s speed along the x-axis. Thus, the energy that corresponds to the one-dimensional translational motion of a macroscopic system is the same as for a molecule (in this respect, a macrosystem behaves as a giant “molecule”). “Frozen” degrees of freedom U /kBT one mole of H2 7/2N Vibration 5/2N Rotation For an ideal gas 3/2N PV = NkBT U = f/2 NkBT Example of H2: Translation 10 100 1000 T, K An energy available to a H2 molecule colliding U(x) E4 with a wall at T=300 K: 3/2kBT ~ 40meV. If the E3 difference between energy levels is >> kBT, E2 then a typical collision cannot cause transitions kBT E1 x to the higher (excited) states and thus cannot transfer energy to this degree of freedom: it is “frozen out”. The rotational energy levels are ~15 meV apart, the difference between vibrational energy levels ~270 meV. Thus, the rotational degrees start contributing to U at T > 200 K, the vibrational degrees of freedom - at T > 3000 K. Adiabatic Process in an Ideal Gas adiabatic (thermally isolated system) dU W12 Q12 0 The amount of work needed to change the state of a thermally isolated system depends only on the initial and final states and not on the intermediate states. V2 W12 P(V , T )dV P V1 2 V2 1 PV= NkBT2 PV= NkBT1 V1 V to calculate W1-2 , we need to know P (V,T) for an adiabatic process U f Nk BT 2 2 dP 1 0 f P f Nk B dT PdV 2 ( f – the # of “unfrozen” degrees of freedom ) PV NkBT PdV VdP NkB dT dV V dU , 2 1 f PdV VdP 2 PdV f V P dV dP 0 V P1 P V1 V P ln ln 1 PV P1V1 const P V1 PV Adiabatic Process in an Ideal Gas (cont.) PV P1V1 const P 2 V2 1 PV= NkBT2 PV= NkBT1 V1 V An adiabata is “steeper” than an isotherma: in an adiabatic process, the work flowing out of the gas comes at the expense of its thermal energy its temperature will decrease. P1V1 P(V , T )dV dV V V1 V1 V2 W12 V2 1 1 1 P1V1 1 1 1 V2 V1 1+2/31.67 (monatomic), 1+2/5 =1.4 (diatomic), 1+2/6 1.33 (polyatomic) (again, neglecting the vibrational degrees of freedom) The root-mean-square speed vrms v 2 3k BT m - not quite the average speed, but close... For H2 molecules (m ~21.710-27 kg ) at 300K: vrms~ 1.84 103 m/s For N2 – vrms= 493 m/s (Pr. 1.18), for O2 – vrms= 461 m/s This speed is close to the speed of sound in the gas – the sound wave propagates due to the thermal motion of molecules. D(v) hydrogen molecules in the “tail” escape the Earth gravitation field vrms v Problem Imagine that we rapidly compress a sample of air whose initial pressure is 105 Pa and temperature is 220C (= 295 K) to a volume that is a quarter of its original volume (e.g., pumping bike’s tire). What is its final temperature? Rapid compression – approx. adiabatic, no time for the energy exchange with the environment due to thermal conductivity P1V1 Nk BT1 P2V2 Nk BT2 P2 P1V1 P2V2 P1V1 V2 P1V1 P1V1 Nk T T2 B 2 V2 1 T1 For adiabatic processes: V1 T2 T1 V2 1 1 V1 V2 1 T1 V1 T2 V2 also P 1 / T const 1 const 295 K 40.4 295 K 1.74 514 K - poor approx. for a bike pump, works better for diesel engines T2 T1 Homework 1: Thermodynamics Let p be the pressure and V be the volume for a real gas of n moles. If the equation of state is given by pV = φ(T), where T is the temperature of the system. The internal energy of the system U is also a function of T only, i.e. (∂U/∂V )T = 0. Show that the function φ(T) must be linear function of the temperature, i.e. φ(T) = cT, where c is a constant. Solution HW1 Homework 2: The “exponential” atmosphere Consider a horizontal slab of air whose thickness is dz. If this slab is at rest, the pressure holding it up from below must balance both the pressure from above and the weight of the slab. Use this fact to find an expression for the variation of pressure with altitude, in terms of the density of air, . Assume that the temperature of atmosphere is independent of height, the average molecular mass m. P(z+dz)A area A z+dz z P(z)A Mg M=rAdz Solution HW 2: The “exponential” atmosphere Consider a horizontal slab of air whose thickness is dz. If this slab is at rest, the pressure holding it up from below must balance both the pressure from above and the weight of the slab. Use this fact to find an expression for the variation of pressure with altitude, in terms of the density of air, . Assume that the temperature of atmosphere is independent of height, the average molecular mass m. P(z+dz)A area A z+dz z P(z)A Mg P( z dz ) A Mg P( z ) A Mg P( z dz ) P( z ) A dP M Adz g dz M Nm PV Pm N the density of air: V V k BT k BT mgz P( z ) P(0) exp kT dP mg P dz kT