DOCSLIB.ORG

Contents 1 Entropy of Ideal

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
Contents 1 Entropy of Ideal PHY304 - Statistical Mechanics Spring Semester 2021 Dr. Anosh Joseph, IISER Mohali LECTURE 11 Wednesday, January 27, 2021 (Note: This is an online lecture due to COVID-19 interruption.) Contents 1 Entropy of Ideal Gas - Statistical Calculation 1 2 Gibbs’ Paradox 3 3 Counting Microstates using Quantum Mechanics 6 1 Entropy of Ideal Gas - Statistical Calculation Let us begin with the Hamiltonian of the ideal gas. We have N 3N X ~p2 X p2 H(q ; p ) = ν = ν : (1) ν ν 2m 2m ν=1 ν=1 We have numbered the coordinates and momenta from 1 to 3N. Let us evaluate !(E; V; N) according to the equation we encountered in the previous lecture Z !(E; V; N) = d3N qd3N p: H(qν ;pν )≤E The integral over the coordinates of the particles can be performed trivially, since H does not depend on them. We have Z !(E; V; N) = V N d3N p: (2) H(pν )≤E p The above integral is just the volume of a 3N-dimensional sphere of radius 2mE, since we PHY304 - Statistical Mechanics Spring Semester 2021 have H(pν) ≤ E implying 3N p 2 X 2 pν ≤ 2mE : (3) ν=1 It is given by N=2 π N VN (R) = N N R : (4) 2 Γ( 2 ) For !(E; V; N) this gives Z !(E; V; N) = V N d3N qd3N p H(pν ;qν ) 3N=2 π 3N=2 N = 3N 3N (2mE) V : (5) 2 Γ( 2 ) We have 1 @! Ω(E; V; N) = σ0 @E 1 π3N=2 = V N (2m)3N=2E(3N=2)−1: (6) σ 3N 0 Γ( 2 ) Once we have the number of microstates, we can find the entropy. We have " # 1 π3N=2 S(E; V; N) = k ln V N (2m)3N=2E(3N=2)−1 (7) B σ 3N 0 Γ( 2 ) Note that the argument of logarithm is indeed dimensionless. We can considerably simplify the above equation when N 1. In that limit we have E3N=2−1 ≈ E3N=2: (8) The logarithm of a Gamma function can be approximated according to Stirling’s formula ln Γ(n) ≈ (n − 1) ln(n − 1) − (n − 1) ≈ n ln n − n: (9) Let us also define a new constant 1=N σ ≡ σ0 : (10) Then the formula for the entropy of the ideal gas becomes " ( )# 3 V 4πmE 3=2 S(E; V; N) = Nk + ln : (11) B 2 σ 3N We can confirm from above that we get the right equations of state for the ideal gas. 2 / 10 PHY304 - Statistical Mechanics Spring Semester 2021 1 @S 3 1 3 = = Nk or E = Nk T: (12) T @E 2 B E 2 B V;N p @S Nk = = B or pV = Nk T: (13) T @V V B E;N The above expressions are independent of σ0. To our surprise, we can see that the above expression for entropy cannot be the correct entropy of an ideal gas. According to Eq. (11), S is not a purely extensive quantity. In the case of enlargement of the system by a factor α, each extensive factor in the logarithm causes the total S to be magnified by a factor α. We see that in the case at hand we will get an additional ln α factor. Our calculation above looks formally correct. There must be a principal error in the calculation of entropy. Let us clarify this problem by means of the well-known Gibbs paradox. 2 Gibbs’ Paradox Let us write down the expression for S we derived above as a function of T , V , and N 3 V S(T; V; N) = Nk + ln (2πmk T )3=2 : (14) B 2 σ B Consider a closed system consisting of two containers, which are separated by a wall and which contain two different ideal gases A and B. They are under the same pressure and temperature. See Fig. 1. gas A gas B T, p T, p VA, NA VB, NB Figure 1: Gibbs’ paradox. 3 / 10 PHY304 - Statistical Mechanics Spring Semester 2021 If we remove the wall, then both gases will spread out over the whole of the container until a new equilibrium situation is reached. We know that the internal energy of the ideal gas does not depend on the volume, but on temperature. Since the internal energy remains constant during the whole process, the temperature and pres- sure do not change either. The entropy however, increases. This is called the mixing entropy. We can easily calculate this using Eq. (14). Before we remove the separating wall, we have (0) (0) (0) Stotal = SA (T;VA;NA) + SB (T;VB;NB): (15) After removing the wall we have (1) (1) (1) Stotal = SA (T;VA + VB;NA) + SB (T;VA + VB;NB): (16) Now let us insert Eq. (14) for the entropy in the above equation. We get (1) (0) ∆S = Stotal − Stotal VA + VB VA + VB = NAkB ln + NBkB ln : (17) VA VB This looks promising. ∆S > 0, as it should be for this irreversible process. Now, let us perform the same analysis for two identical gasses in the two sub-systems. The initial entropy equation (0) (0) (0) Stotal = SA (T;VA;NA) + SB (T;VB;NB) is still right. However, in the final situation, we have (1) Stotal = S(T;VA + VB;NA + NB); (18) since NA + NB particles of the same gas are now distributed over the total volume VA + VB. So in this case also, we get exactly same entropy difference ∆S > 0 (Eq. (17)). However, this cannot be correct, since after the removal of the separating wall no macroscopically observable processes are happening at all. We can simply re-insert the separating wall and recover the initial situation. In the case of identical gases, the removal of the separating wall is a reversible process and we must have ∆S = 0. 4 / 10 PHY304 - Statistical Mechanics Spring Semester 2021 Let us try to understand what might be going on. In classical mechanics, the particles are distinguishable. We can label them as 1; 2; ··· ;NA and NA + 1;NA + 2; ··· ;NA + NB. Once we remove the wall, the gas particles (treat them as microscopic billiard balls) freely move around in the volume VA + VB. This is an irreversible process. We can mix the particles again and again, but we will never (almost never, in the case of finite number of particles) recover the initial situation. However, when we apply quantum mechanics, this argument does not work. We cannot attach a label to identical particles. They are all indistinguishable. But classical particles are distinguishable and we can identify them. This paradigm leads to the Gibbs’ paradox. We must note that two microstates should be considered different, if they differ in more than the enumeration of the particles. (Having other quantum numbers, for example.) For N particles, we can enumerate them in N! ways. Thus instead of σ(E; V; N) Ω(E; V; N) = ; (19) σ0 we now try a new definition of Ω 1 σ(E; V; N) Ω(E; V; N) = : (20) N! σ0 @! The calculation of σ(E) by σ(E) = @E is still unchanged. The factor 1=N! in Eq. (20) is called the Gibbs’ correction factor. After incorporating the Gibbs’ correction factor we have the expression for the entropy of an ideal gas " ( )# 3 V 4πmE 3=2 S(E; V; N) = Nk + ln − k ln N!: (21) B 2 σ 3N B For the case of very large N, we can apply the Stirling’s formula (ln N! ≈ N ln N − N), and we get " ( )# 5 V 4πmE 3=2 S(E; V; N) = Nk + ln : (22) B 2 Nσ 3N Thus we get the anticipated result that entropy is indeed an extensive quantity. 5 A factor 2 also appears. Let us now go back to our mixing experiment. We have 5 V S(E; V; N) = Nk + ln (2πmk T )3=2 : (23) B 2 Nσ B 5 / 10 PHY304 - Statistical Mechanics Spring Semester 2021 In the above we used 3 E = Nk T: 2 B (0) (1) Upon using the expression we found for Stotal and Stotal for different gases we obtain VA + VB VA + VB ∆S = NAkB ln + NBkB ln : (24) VA VB This is nothing but the expression we found for ∆S in Eq. (17). For identical gases, we get 5 VA + VB 3=2 δS = (NA + NB)kB + ln (2πmkBT ) 2 (NA + NB)σ 5 VA 3=2 −NAkB + ln (2πmkBT ) 2 NAσ 5 VB 3=2 −NBkB + ln (2πmkBT ) : (25) 2 NBσ As mentioned before we have thermal and mechanical equilibrium in the mixing process. Thus we have V V V + V A = B = A B : (26) NA NB NA + NB This implies, ∆S = 0: (27) Thus for two identical gases we get the entropy change ∆S = 0, as it should be. 3 Counting Microstates using Quantum Mechanics Let us determine Ω by counting the quantum mechanical states of distinguishable particles. This will give us an absolute number for Ω and thus an absolute entropy. We will also get a value for the unit surface (unit volume) σ0 in phase space. Let us revisit the quantum mechanical problem of a particle in a box of length L. See Fig. 2. The one-particle states have the wave function Ψnx;ny;nz = A sin(kxx) sin(kyy) sin(kzz) n πx n πy n πz = A sin x sin y sin z ; (28) L L L where nx; ny; nz = 1; 2; ··· : (29) 6 / 10 PHY304 - Statistical Mechanics Spring Semester 2021 E ⋮ E1 E0 0 L Figure 2: Quantum mechanical particle in a cubic box.
Load more

Recommended publications
  • On Entropy, Information, and Conservation of Information
    entropy Article On Entropy, Information, and Conservation of Information Yunus A. Çengel Department of Mechanical Engineering, University of Nevada, Reno, NV 89557, USA; [email protected] Abstract: The term entropy is used in different meanings in different contexts, sometimes in contradic- tory ways, resulting in misunderstandings and confusion. The root cause of the problem is the close resemblance of the defining mathematical expressions of entropy in statistical thermodynamics and information in the communications field, also called entropy, differing only by a constant factor with the unit ‘J/K’ in thermodynamics and ‘bits’ in the information theory. The thermodynamic property entropy is closely associated with the physical quantities of thermal energy and temperature, while the entropy used in the communications field is a mathematical abstraction based on probabilities of messages. The terms information and entropy are often used interchangeably in several branches of sciences. This practice gives rise to the phrase conservation of entropy in the sense of conservation of information, which is in contradiction to the fundamental increase of entropy principle in thermody- namics as an expression of the second law. The aim of this paper is to clarify matters and eliminate confusion by putting things into their rightful places within their domains. The notion of conservation of information is also put into a proper perspective. Keywords: entropy; information; conservation of information; creation of information; destruction of information; Boltzmann relation Citation: Çengel, Y.A. On Entropy, 1. Introduction Information, and Conservation of One needs to be cautious when dealing with information since it is defined differently Information. Entropy 2021, 23, 779.
    [Show full text]
  • Practice Problems from Chapter 1-3 Problem 1 One Mole of a Monatomic Ideal Gas Goes Through a Quasistatic Three-Stage Cycle (1-2, 2-3, 3-1) Shown in V 3 the Figure
    Practice Problems from Chapter 1-3 Problem 1 One mole of a monatomic ideal gas goes through a quasistatic three-stage cycle (1-2, 2-3, 3-1) shown in V 3 the Figure. T1 and T2 are given. V 2 2 (a) (10) Calculate the work done by the gas. Is it positive or negative? V 1 1 (b) (20) Using two methods (Sackur-Tetrode eq. and dQ/T), calculate the entropy change for each stage and ∆ for the whole cycle, Stotal. Did you get the expected ∆ result for Stotal? Explain. T1 T2 T (c) (5) What is the heat capacity (in units R) for each stage? Problem 1 (cont.) ∝ → (a) 1 – 2 V T P = const (isobaric process) δW 12=P ( V 2−V 1 )=R (T 2−T 1)>0 V = const (isochoric process) 2 – 3 δW 23 =0 V 1 V 1 dV V 1 T1 3 – 1 T = const (isothermal process) δW 31=∫ PdV =R T1 ∫ =R T 1 ln =R T1 ln ¿ 0 V V V 2 T 2 V 2 2 T1 T 2 T 2 δW total=δW 12+δW 31=R (T 2−T 1)+R T 1 ln =R T 1 −1−ln >0 T 2 [ T 1 T 1 ] Problem 1 (cont.) Sackur-Tetrode equation: V (b) 3 V 3 U S ( U ,V ,N )=R ln + R ln +kB ln f ( N ,m ) V2 2 N 2 N V f 3 T f V f 3 T f ΔS =R ln + R ln =R ln + ln V V 2 T V 2 T 1 1 i i ( i i ) 1 – 2 V ∝ T → P = const (isobaric process) T1 T2 T 5 T 2 ΔS12 = R ln 2 T 1 T T V = const (isochoric process) 3 1 3 2 2 – 3 ΔS 23 = R ln =− R ln 2 T 2 2 T 1 V 1 T 2 3 – 1 T = const (isothermal process) ΔS 31 =R ln =−R ln V 2 T 1 5 T 2 T 2 3 T 2 as it should be for a quasistatic cyclic process ΔS = R ln −R ln − R ln =0 cycle 2 T T 2 T (quasistatic – reversible), 1 1 1 because S is a state function.
    [Show full text]
  • Lecture 4: 09.16.05 Temperature, Heat, and Entropy
    3.012 Fundamentals of Materials Science Fall 2005 Lecture 4: 09.16.05 Temperature, heat, and entropy Today: LAST TIME .........................................................................................................................................................................................2� State functions ..............................................................................................................................................................................2� Path dependent variables: heat and work..................................................................................................................................2� DEFINING TEMPERATURE ...................................................................................................................................................................4� The zeroth law of thermodynamics .............................................................................................................................................4� The absolute temperature scale ..................................................................................................................................................5� CONSEQUENCES OF THE RELATION BETWEEN TEMPERATURE, HEAT, AND ENTROPY: HEAT CAPACITY .......................................6� The difference between heat and temperature ...........................................................................................................................6� Defining heat capacity.................................................................................................................................................................6�
    [Show full text]
  • Entropy: Ideal Gas Processes
    Chapter 19: The Kinec Theory of Gases Thermodynamics = macroscopic picture Gases micro -> macro picture One mole is the number of atoms in 12 g sample Avogadro’s Number of carbon-12 23 -1 C(12)—6 protrons, 6 neutrons and 6 electrons NA=6.02 x 10 mol 12 atomic units of mass assuming mP=mn Another way to do this is to know the mass of one molecule: then So the number of moles n is given by M n=N/N sample A N = N A mmole−mass € Ideal Gas Law Ideal Gases, Ideal Gas Law It was found experimentally that if 1 mole of any gas is placed in containers that have the same volume V and are kept at the same temperature T, approximately all have the same pressure p. The small differences in pressure disappear if lower gas densities are used. Further experiments showed that all low-density gases obey the equation pV = nRT. Here R = 8.31 K/mol ⋅ K and is known as the "gas constant." The equation itself is known as the "ideal gas law." The constant R can be expressed -23 as R = kNA . Here k is called the Boltzmann constant and is equal to 1.38 × 10 J/K. N If we substitute R as well as n = in the ideal gas law we get the equivalent form: NA pV = NkT. Here N is the number of molecules in the gas. The behavior of all real gases approaches that of an ideal gas at low enough densities. Low densitiens m= enumberans tha oft t hemoles gas molecul es are fa Nr e=nough number apa ofr tparticles that the y do not interact with one another, but only with the walls of the gas container.
    [Show full text]
  • Gibbs' Paradox and the Definition of Entropy
    Entropy 2008, 10, 15-18 entropy ISSN 1099-4300 °c 2008 by MDPI www.mdpi.org/entropy/ Full Paper Gibbs’ Paradox and the Definition of Entropy Robert H. Swendsen Physics Department, Carnegie Mellon University, Pittsburgh, PA 15213, USA E-Mail: [email protected] Received: 10 December 2007 / Accepted: 14 March 2008 / Published: 20 March 2008 Abstract: Gibbs’ Paradox is shown to arise from an incorrect traditional definition of the entropy that has unfortunately become entrenched in physics textbooks. Among its flaws, the traditional definition predicts a violation of the second law of thermodynamics when applied to colloids. By adopting Boltzmann’s definition of the entropy, the violation of the second law is eliminated, the properties of colloids are correctly predicted, and Gibbs’ Paradox vanishes. Keywords: Gibbs’ Paradox, entropy, extensivity, Boltzmann. 1. Introduction Gibbs’ Paradox [1–3] is based on a traditional definition of the entropy in statistical mechanics found in most textbooks [4–6]. According to this definition, the entropy of a classical system is given by the product of Boltzmann’s constant, k, with the logarithm of a volume in phase space. This is shown in many textbooks to lead to the following equation for the entropy of a classical ideal gas of distinguishable particles, 3 E S (E; V; N) = kN[ ln V + ln + X]; (1) trad 2 N where X is a constant. Most versions of Gibbs’ Paradox, including the one I give in this section, rest on the fact that Eq. 1 is not extensive. There is another version of Gibbs’ Paradox that involves the mixing of two gases, which I will discuss in the third section of this paper.
    [Show full text]
  • IB Questionbank
    Topic 3 Past Paper [94 marks] This question is about thermal energy transfer. A hot piece of iron is placed into a container of cold water. After a time the iron and water reach thermal equilibrium. The heat capacity of the container is negligible. specific heat capacity. [2 marks] 1a. Define Markscheme the energy required to change the temperature (of a substance) by 1K/°C/unit degree; of mass 1 kg / per unit mass; [5 marks] 1b. The following data are available. Mass of water = 0.35 kg Mass of iron = 0.58 kg Specific heat capacity of water = 4200 J kg–1K–1 Initial temperature of water = 20°C Final temperature of water = 44°C Initial temperature of iron = 180°C (i) Determine the specific heat capacity of iron. (ii) Explain why the value calculated in (b)(i) is likely to be different from the accepted value. Markscheme (i) use of mcΔT; 0.58×c×[180-44]=0.35×4200×[44-20]; c=447Jkg-1K-1≈450Jkg-1K-1; (ii) energy would be given off to surroundings/environment / energy would be absorbed by container / energy would be given off through vaporization of water; hence final temperature would be less; hence measured value of (specific) heat capacity (of iron) would be higher; This question is in two parts. Part 1 is about ideal gases and specific heat capacity. Part 2 is about simple harmonic motion and waves. Part 1 Ideal gases and specific heat capacity State assumptions of the kinetic model of an ideal gas. [2 marks] 2a. two Markscheme point molecules / negligible volume; no forces between molecules except during contact; motion/distribution is random; elastic collisions / no energy lost; obey Newton’s laws of motion; collision in zero time; gravity is ignored; [4 marks] 2b.
    [Show full text]
  • Gibbs Paradox: Mixing and Non Mixing Potentials T
    Physics Education 1 Jul - Sep 2016 Gibbs paradox: Mixing and non mixing potentials T. P. Suresh 1∗, Lisha Damodaran 1 and K. M. Udayanandan 2 1 School of Pure and Applied Physics Kannur University, Kerala- 673 635, INDIA *[email protected] 2 Nehru Arts and Science College, Kanhangad Kerala- 671 314, INDIA (Submitted 12-03-2016, Revised 13-05-2016) Abstract Entropy of mixing leading to Gibbs paradox is studied for different physical systems with and without potentials. The effect of potentials on the mixing and non mixing character of physical systems is discussed. We hope this article will encourage students to search for new problems which will help them understand statistical mechanics better. 1 Introduction namics of any system (which is the main aim of statistical mechanics) we need to calculate Statistical mechanics is the study of macro- the canonical partition function, and then ob- scopic properties of a system from its mi- tain macro properties like internal energy, en- croscopic description. In the ensemble for- tropy, chemical potential, specific heat, etc. malism introduced by Gibbs[1] there are of the system from the partition function. In three ensembles-micro canonical, canonical this article we make use of the canonical en- and grand canonical ensemble. Since we are semble formalism to study the extensive char- not interested in discussing quantum statis- acter of entropy and then to calculate the tics we will use the canonical ensemble for in- entropy of mixing. It is then used to ex- troducing our ideas. To study the thermody- plain the Gibbs paradox. Most text books Volume 32, Number 3, Article Number: 4.
    [Show full text]
  • Thermodynamics of Ideal Gases
    D Thermodynamics of ideal gases An ideal gas is a nice “laboratory” for understanding the thermodynamics of a fluid with a non-trivial equation of state. In this section we shall recapitulate the conventional thermodynamics of an ideal gas with constant heat capacity. For more extensive treatments, see for example [67, 66]. D.1 Internal energy In section 4.1 we analyzed Bernoulli’s model of a gas consisting of essentially 1 2 non-interacting point-like molecules, and found the pressure p = 3 ½ v where v is the root-mean-square average molecular speed. Using the ideal gas law (4-26) the total molecular kinetic energy contained in an amount M = ½V of the gas becomes, 1 3 3 Mv2 = pV = nRT ; (D-1) 2 2 2 where n = M=Mmol is the number of moles in the gas. The derivation in section 4.1 shows that the factor 3 stems from the three independent translational degrees of freedom available to point-like molecules. The above formula thus expresses 1 that in a mole of a gas there is an internal kinetic energy 2 RT associated with each translational degree of freedom of the point-like molecules. Whereas monatomic gases like argon have spherical molecules and thus only the three translational degrees of freedom, diatomic gases like nitrogen and oxy- Copyright °c 1998{2004, Benny Lautrup Revision 7.7, January 22, 2004 662 D. THERMODYNAMICS OF IDEAL GASES gen have stick-like molecules with two extra rotational degrees of freedom or- thogonally to the bridge connecting the atoms, and multiatomic gases like carbon dioxide and methane have the three extra rotational degrees of freedom.
    [Show full text]
  • 5.5 ENTROPY CHANGES of an IDEAL GAS for One Mole Or a Unit Mass of Fluid Undergoing a Mechanically Reversible Process in a Closed System, the First Law, Eq
    Thermodynamic Third class Dr. Arkan J. Hadi 5.5 ENTROPY CHANGES OF AN IDEAL GAS For one mole or a unit mass of fluid undergoing a mechanically reversible process in a closed system, the first law, Eq. (2.8), becomes: Differentiation of the defining equation for enthalpy, H = U + PV, yields: Eliminating dU gives: For an ideal gas, dH = and V = RT/ P. With these substitutions and then division by T, As a result of Eq. (5.11), this becomes: where S is the molar entropy of an ideal gas. Integration from an initial state at conditions To and Po to a final state at conditions T and P gives: Although derived for a mechanically reversible process, this equation relates properties only, and is independent of the process causing the change of state. It is therefore a general equation for the calculation of entropy changes of an ideal gas. 1 Thermodynamic Third class Dr. Arkan J. Hadi 5.6 MATHEMATICAL STATEMENT OF THE SECOND LAW Consider two heat reservoirs, one at temperature TH and a second at the lower temperature Tc. Let a quantity of heat | | be transferred from the hotter to the cooler reservoir. The entropy changes of the reservoirs at TH and at Tc are: These two entropy changes are added to give: Since TH > Tc, the total entropy change as a result of this irreversible process is positive. Also, ΔStotal becomes smaller as the difference TH - TC gets smaller. When TH is only infinitesimally higher than Tc, the heat transfer is reversible, and ΔStotal approaches zero. Thus for the process of irreversible heat transfer, ΔStotal is always positive, approaching zero as the process becomes reversible.
    [Show full text]
  • Gibb's Paradox for Distinguishable
    Gibb’s paradox for distinguishable K. M. Udayanandan1, R. K. Sathish1, A. Augustine2 1Department of Physics, Nehru Arts and Science College, Kerala-671 314, India. 2Department of Physics, University of Kannur, Kerala 673 635, India. E-mail: [email protected] (Received 22 March 2014, accepted 17 November 2014) Abstract Boltzmann Correction Factor (BCF) N! is used in micro canonical ensemble and canonical ensemble as a dividing term to avoid Gibbs paradox while finding the number of states and partition function for ideal gas. For harmonic oscillators this factor does not come since they are considered to be distinguishable. We here show that BCF comes twice for harmonic oscillators in grand canonical ensemble for entropy to be extensive in classical statistics. Then we extent this observation for all distinguishable systems. Keywords: Gibbs paradox, harmonic oscillator, distinguishable particles. Resumen El factor de corrección de Boltzmann (BCF) N! se utiliza en ensamble microcanónico y ensamble canónico como un término divisor para evitar la paradoja de Gibbs, mientras se busca el número de estados y la función de partición para un gas ideal. Para osciladores armónicos este factor no viene sin que éstos sean considerados para ser distinguibles. Mostramos aquí que BCF llega dos veces para osciladores armónicos en ensamble gran canónico para que la entropía sea ampliamente usada en la estadística clásica. Luego extendemos esta observación para todos los sistemas distinguibles. Palabras clave: paradoja de Gibbs, oscilador armónico, partículas distinguibles. PACS: 03.65.Ge, 05.20-y ISSN 1870-9095 I. INTRODUCTION Statistical mechanics is a mathematical formalism by which II. CANONICAL ENSEMBLE one can determine the macroscopic properties of a system from the information about its microscopic states.
    [Show full text]
  • Chapter 3 3.4-2 the Compressibility Factor Equation of State
    Chapter 3 3.4-2 The Compressibility Factor Equation of State The dimensionless compressibility factor, Z, for a gaseous species is defined as the ratio pv Z = (3.4-1) RT If the gas behaves ideally Z = 1. The extent to which Z differs from 1 is a measure of the extent to which the gas is behaving nonideally. The compressibility can be determined from experimental data where Z is plotted versus a dimensionless reduced pressure pR and reduced temperature TR, defined as pR = p/pc and TR = T/Tc In these expressions, pc and Tc denote the critical pressure and temperature, respectively. A generalized compressibility chart of the form Z = f(pR, TR) is shown in Figure 3.4-1 for 10 different gases. The solid lines represent the best curves fitted to the data. Figure 3.4-1 Generalized compressibility chart for various gases10. It can be seen from Figure 3.4-1 that the value of Z tends to unity for all temperatures as pressure approach zero and Z also approaches unity for all pressure at very high temperature. If the p, v, and T data are available in table format or computer software then you should not use the generalized compressibility chart to evaluate p, v, and T since using Z is just another approximation to the real data. 10 Moran, M. J. and Shapiro H. N., Fundamentals of Engineering Thermodynamics, Wiley, 2008, pg. 112 3-19 Example 3.4-2 ---------------------------------------------------------------------------------- A closed, rigid tank filled with water vapor, initially at 20 MPa, 520oC, is cooled until its temperature reaches 400oC.
    [Show full text]
  • Classical Particle Indistinguishability, Precisely
    Classical Particle Indistinguishability, Precisely James Wills∗1 1Philosophy, Logic and Scientific Method, London School of Economics Forthcoming in The British Journal for the Philosophy of Science Abstract I present a new perspective on the meaning of indistinguishability of classical particles. This leads to a solution to the problem in statisti- cal mechanics of justifying the inclusion of a factor N! in a probability distribution over the phase space of N indistinguishable classical par- ticles. 1 Introduction Considerations of the identity of objects have long been part of philosophical discussion in the natural sciences and logic. These considerations became particularly pertinent throughout the twentieth century with the develop- ment of quantum physics, widely recognized as having interesting and far- reaching implications concerning the identity, individuality, indiscernibility, indistinguishability1 of the elementary components of our ontology. This discussion continues in both the physics and philosophy literature2. ∗[email protected] 1. This menagerie of terms in the literature is apt to cause confusion, especially as there is no clear consensus on what exactly each of these terms mean. Throughout the rest of the paper, I will be concerned with giving a precise definition of a concept which I will label `indistinguishability', and which, I think, matches with how most other commentators use the word. But I believe it to be a distinct debate whether particles are `identical', `individual', or `indiscernible'. The literature I address in this paper therefore does not, for example, overlap with the debate over the status of Leibniz's PII. 2. See French (2000) for an introduction to this discussion and a comprehensive list of references.
    [Show full text]