Maxwell-Boltzmann Distribution

12 Key excerpts on "Maxwell-Boltzmann Distribution"

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  eBook - ePub

  Molecular Driving Forces

  Statistical Thermodynamics in Biology, Chemistry, Physics, and Nanoscience

  • Ken Dill, Sarina Bromberg(Authors)
  • 2010(Publication Date)
  • Garland Science

    (Publisher)

  10 The Boltzmann 10 Distribution Law Statistical Mechanics Gives Probability Distributions for Atoms and Molecules

  Now we begin statistical mechanics, the modeling and prediction of the properties of materials from the structures of the atoms and molecules of which they are composed. The core of statistical mechanics is modeling the probability distributions of the energies of atoms and molecules. The various averages over those distributions are what experiments measure. For example, to compute the properties of gases, you need the distributions of their energies and velocities (Chapter 11 ). You can predict chemical reaction equilibria if you know the distributions of the energies of the reactants and products (Chapter 13 ). And you can predict the average number of ligands bound to a DNA molecule if you know the distribution of energies of all the ligation states (Chapter 28 ).

  The central result of this chapter is the Boltzmann distribution law, which gives probability distributions from the underlying energy levels. We derive this result by bringing together the two main threads from earlier chapters. First, the principles of thermodynamics in Chapter 6 , Chapter 7 , Chapter 8 , Chapter 9 describe how to predict the state of equilibrium. Second, Chapter 5 relates a macroscopic property of the equilibrium (the entropy) to a microscopic property, the probability distribution.

  Here’s the kind of problem we want to solve. Example 8.2 describes a twodimensional model of a four-bead polymer chain that has four open conformations and one compact conformation. In that example, we computed the free energies of the open and compact states. Now we want to compute the probability distribution, the fraction of molecules that are in each conformation.

  To begin, we need to define the system and its energy levels. The system is one four-bead chain that has two energy levels (see Figure 10.1 ). Each energy level represents the number of bead–bead contacts that the chain can make. Let’s use the convention that zero is the lowest energy, the lowest rung on the ladder. Conformations with the maximum number of bead–bead contacts have zero energy. Breaking a contact increases the energy by an amount ε = ε0 > 0. We use ε to indicate the energy in general and ε0 to represent some particular constant value. We seek the distribution of probabilities p1 , p2 ,…, p5

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  Essential Statistical Physics

  • Malcolm P. Kennett(Author)
  • 2020(Publication Date)
  • Cambridge University Press

    (Publisher)

  We finish with a discussion of Brownian motion and diffusion. 5.1 Maxwell–Boltzmann Velocity Distribution We know from the canonical ensemble that for an ideal gas the probability that the system is in any particular microstate with energy between E and E + dE and partition function Z is given by e −βE Z . As we discussed in Section 4.5.3, we can understand the properties of an ideal gas by considering a single particle and then obtain the partition function of a many-particle system from the single-particle partition function Eq. (4.85): Z 1 = g s n Q V = g s V  mk B T 2π 2  3 2 . (5.1) 86 87 5.1 Maxwell–Boltzmann Velocity Distribution In order to find the probability that the system has energy E, we need to know how many microstates there are that have energy within the energy range [ E, E + dE], which we calculate below. 5.1.1 Density of States To obtain the density of states, we first pick an energy E ∗ and then calculate the number of states with energy E < E ∗ . For particle in a box states in three dimensions, for a cubic box with side length L and volume V = L 3 , we can write the energy eigenvalues as E =  2 2m π 2 V 2 3  n 2 x + n 2 y + n 2 z  , (5.2) where n x , n y and n z are positive integers, so we require n 2 x + n 2 y + n 2 z <  2mE ∗  2  V 2 3 π 2 , (5.3) which represents 1/8 of a sphere (in energy space) with radius r =  2mE ∗  2 V 1 3 π . Now, for a sphere of radius r , the number of points ( n x , n y , n z ) in the octant of the sphere, where n x , n y , n z > 0, is 1 8 × 4 3 πr 3 = 1 6 πr 3 , so the number of states in the relevant portion of the sphere with energy ≤ E is (including a factor g s to take into account spin degeneracy) is N ( E ) = g s π 6  2mE  2  3 2 V π 3 = g s 6π 2  2m  2  3 2 V E 3 2 . (5.4) If we look in the energy range [ E, E + dE], then the number of states is N ( E + dE ) − N ( E )  dN dE dE = g s 4π 2  2m  2  3 2 V E 1 2 dE.

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  Modern Physics

  • Kenneth S. Krane(Author)
  • 2020(Publication Date)
  • Wiley

    (Publisher)

  The Distribution of Molecular Speeds From an experimental standpoint, it is often easier to measure the dis- tribution of speeds than the distribution of energies. So let’s use the Maxwell–Boltzmann distribution for the distribution of kinetic energies in a gas to obtain the distribution of molecular speeds, which then can be tested in the laboratory. That is, we wish to obtain an expression for the number of molecules with speeds in the interval dv at v (between v and v + dv), represented by dN = N(v) dv. The number of molecules dN in the interval is the same, whether we count them in terms of energy or speed, and so dN = N(E) dE = N(v) dv, or N(v) = N(E) dE∕dv. With E = 1 2 mv 2 and dE∕dv = mv, we have N(v) = N(E) dE dv = 2N √ (kT ) 3∕2 √ mv 2 2 e −mv 2 ∕2kT mv = N √ 2  ( m kT ) 3∕2 v 2 e −mv 2 ∕2kT (10.20) 318 Chapter 10 Statistical Physics Oven Molecular beam Detector FIGURE 10.10 Apparatus to measure the distribution of molecular speeds. dv v dN v p 0 N(v) FIGURE 10.9 The Maxwell speed distribution for gas molecules. The shaded strip represents the number of molecules with speeds between v and v + dv. This equation, which is known as the Maxwell speed distribution, is graphed in Figure 10.9. The shaded strip shows the number dN in the interval dv at v. The most probable speed occurs at the value v p = (2kT ∕m) 1∕2 . An example of an experiment to measure the distribution of molecular speeds is shown in Figure 10.10. A small hole in the side of an oven allows a stream of molecules to escape; we assume the hole to be small enough so that the distribution of speeds inside the oven is not changed. The beam of molecules is made to pass through a slot in a disk attached to an axle rotating at angular velocity . At the other end of the axle is a second slotted disk, but the slot is displaced from the first by an angle .

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  Spectroscopic Measurement

  An Introduction to the Fundamentals

  • Mark A. Linne(Author)
  • 2002(Publication Date)
  • Academic Press

    (Publisher)

  Moreover, the Maxwellian distribution introduces the statistical mechanics formalism in a simple and intuitive way. Next, the Boltzmann energy distribution is devel- oped. The presentation is simplified by taking advantage of the fact that many flowfields of interest can be considered ideal gases. Other short cuts are taken. Finally, equations that are useful for spectroscopic flowfield diagnostics are presented, using energy level expressions that will be developed in later chapters. 2.2. THE MAXWELLIAN VELOCITY DISTRIBUTION 11 v'l dV dv I v, /~Vx dvz v z Vx Figure 2.1 Single velocity value ~i in velocity space vx,Vy,Vz with differential volume in velocity space dVv- dvx dvy dvz. 2.2 The Maxwellian Velocity Distribution If the gas under investigation does not have appreciable bulk velocity (e.g. the Mach number is small), then the flowfield can be thought of as a collection of rapidly moving molecules with a relatively low (or zero) average velocity. The density of these molecules is sufficiently high for them routinely to collide with each other. Energy is transferred throughout the gas via these collisions, establishing an equilibrium dis- tribution of energy among the constituents of the gas (LTE). With respect to velocity, there are various values of speed and di- rection for the molecules that form the gas mixture. It is therefore necessary to develop a statistical distribution function that will pro- duce values of interest, such as the average velocity. We begin by assuming a uniform mass distribution in physical (Euclidean) space, but a nonuniform velocity distribution in velocity space (see Figure 2.1). The gas would then be represented in Figure 2.1 by a cloud of points (one for each molecule) with a centroid at the average value for V. Velocity distributions Next, we define a local density in velocity space (number of points per unit volume in velocity space) as F(~).

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  Kinetic Theory of Gases

  • Walter Kauzmann(Author)
  • 2013(Publication Date)
  • Dover Publications

    (Publisher)

  u ) can be regarded as the proportionality factor relating the range to the probability:

  We shall now show that for a gas at temperature T consisting of molecules of mass m

  where A is a quantity independent of u , but dependent on m and T , and where k is Boltzmann’s constant (Chapter 2 , Section 1d ). This is the Maxwell-Boltzmann Distribution function . Evidently it consists of two factors: (1) a factor Au 2 , which increases monotonically as u increases, and (2) a Boltzmann factor exp (−mu 2 / 2kT ), which decreases monotonically from 1 at u = 0 to zero as u

  FIG. 4-2 General shape of the distribution function for molecular velocities , P (u ) = df/du , where f (u )

  is the function plotted in Fig. 4-1

  .

  FIG. 4-3 Dissection of the Maxwell-Boltzmann Distribution function into the factors Au 2 and exp ( − mu 2 /2kT ).

   

  increases. The behavior of these two factors and of their product, P (u ) is shown in Fig. 4-3 .

  Although the Maxwell-Boltzmann Distribution was originally derived using classical mechanical concepts, it is somewhat easier to derive it if one approaches the problem from a quantum theoretical point of view. Furthermore, some of the mathematical procedures employed in this derivation will prove useful in discussing other subjects. Before proceeding with the derivation, however, it will be helpful to consider some of the quantum principles governing the motions of free particles in closed regions of space.

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  Modern Physics for Scientists and Engineers

  • Stephen Thornton, Andrew Rex, Carol Hood, , Stephen Thornton, Stephen Thornton, Andrew Rex, Carol Hood(Authors)
  • 2020(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  9.2 Maxwell Velocity Distribution As Laplace pointed out, we could, in principle, know everything about an ideal gas by knowing the position and instantaneous velocity of every molecule. This entails knowing six parameters per molecule, three for position (x, y, z) and three for velocity (v x , v y , v z ). Many relevant physical quantities must depend on one or more of these six parameters. It is useful to think of them as the compo- nents of a six-dimensional phase space. Maxwell focused on the three velocity components because he was most in- terested in the thermal properties of ideal gases. The velocity components of the molecules of an ideal gas are more important than the (random) instantaneous positions, because the energy of a gas depends only on the velocities and not on the instantaneous positions. The crucial question for Maxwell was: what is the distribution of velocities for an ideal gas at a given temperature? Let us define a velocity distribution function f ( v S ) (see Appendix 3B) such that f ( v ) S d 3 v S 5 the probability of finding a particle with velocity between v S and v S 1 d 3 v S where d 3 v S 5 dv x dv y dv z . Note that because v S is a vector quantity, the preceding statement implies three separate conditions. The vector v S has components v x , v y , and v z . Therefore f ( v S ) d 3 v S is the probability of finding a particle with v x between v x and v x 1 dv x , with v y between v y and v y 1 dv y , and with v z between v z and v z 1 dv z . We can think of the distribution function f ( v S ) as playing a role analogous to the probability density C*C in quantum theory. Maxwell was able to prove* that the probability distribution function is pro- portional to exp(2 1 2 mv 2 ykT ), where m is the molecular mass, v is the molecular Phase space *There are numerous ways of demonstrating that the distribution is proportional to exp(2 1 2 mv 2 ykT ). See, for example, Daniel Schroeder, Thermal Physics, Addison-Wesley (1999), pp.

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  Modern Physics

  • Gary N. Felder, Kenny M. Felder(Authors)
  • 2022(Publication Date)
  • Cambridge University Press

    (Publisher)

  • The equation for the Boltzmann distribution is P = (1/Z)e −E/(k B T) , where 1/Z is a normalization constant. Z is called the “partition function” and e −E/(k B T) is called the “Boltzmann factor.” • Remember that this formula is not the probability that your system will be in energy E; it is the probability that your system will be in any given microstate with energy E. Multiply that by the total number of microstates with energy E (the “degeneracy”) to find the total probability of that energy level. For this reason, it is possible for a high energy level to be more probable than a low level. 498 10 Statistical Mechanics • The Boltzmann distribution predicts that, for very low temperatures, all the particles crowd into the lowest-energy microstates possible. For very high temperatures, the probabilities even out somewhat; all states with energy much less than k B T are roughly equally likely. Section 10.5 Some Applications of the Boltzmann Distribution The Boltzmann distribution can be used to predict (among other things) the average thermal energy of a system of particles, the heat capacity of a gas, and the distribution function of molecular speeds. • As a general guideline, the average thermal energy of a particle is typically of order k B T above its ground state. • That guideline applies when the “density of states” – the number of available microstates per unit energy – is reasonably uniform, and reasonably high. A very non-uniform density of states can push the average energy much higher or much lower than k B T. (This should make sense if you think about it.) A very low density of states can push the average energy much lower, because the first excited state may be beyond reach. • A more accurate estimate of thermal energy is provided for some systems by the “equipartition theorem”: the average value of any energy term that depends quadratically on one of the system’s degrees of freedom is (1/2)k B T.

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  Principles of Thermodynamics

  • Myron Kaufman(Author)
  • 2002(Publication Date)
  • CRC Press

    (Publisher)

  5.6 Velocity Distributions The kinetic theory of gases deals with the translational motions of gaseous molecules. Translational motion of molecules can be considered separately from rotational and internal motions and has energy e tr ¼ 1 2 mc 2 ¼ 1 2 mðv 2 x þ v 2 y þ v 2 z Þ ð50Þ where c is the molecular speed. Because classical mechanics is appropriate for treating translational energy in macroscopic containers, all values of velocity are possible and we should use Eq. (19), the distribution function for continuous- energy levels. f ðeÞd e ¼ gðeÞ q exp  e kT   d e ¼ gðeÞ q exp  mc 2 2kT   d e ¼ gðeÞ q exp  mv 2 x 2kT   exp  mv 2 y 2kT   exp  mv 2 z 2kT   d e ð51Þ The exponential terms tell us how the molecules are distributed in velocity space, and the degeneracy, gðeÞ d e, is proportional to the volume in velocity space that gives rise to energies between e and e þ d e: gðeÞ d e ¼ C4pc 2 dc ð52Þ C is a normalization factor determined by integrating over all of velocity space: C q 4p ð 1 0 c 2 exp  mc 2 2kT   dc ¼ 1 ð53Þ Two formulas that may be found in most tables of integrals and are very useful in kinetic theory are ð 1 0 x 2n expðbx 2 Þ dx ¼ 1 2 ffiffiffi p p ð2nÞ!b ðnþ1=2Þ 2 2n n! ð54Þ and ð 1 0 x 2nþ1 expðbx 2 Þ dx ¼ 1 2 n!b ðnþ1Þ ð55Þ Copyright © 2002 by Taylor & Francis Group LLC where n is an integer. From Eq. (53), this gives C q ¼ m 2pkT   3=2 ð56Þ and f ðeÞ d e ¼ 4p m 2pkT   3=2 c 2 exp  mc 2 2kT   dc ¼ F ðcÞ dc ð57Þ F ðcÞ, the distribution over molecular speeds, is known as the Maxwell–Boltzmann velocity distribution. In Fig. 1a, the distribution are shown for N 2 and H 2 , both at 300 K. Note that because the distributions are normalized, the areas under both curves are the same. The most probable speed, average speed, and root-mean- Figure 1 (a) Speed distribution of N 2 and H 2 at 300 K. Copyright © 2002 by Taylor & Francis Group LLC square speed are marked on the H 2 distribution curve.

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  Physics of Matter

  • George C. King(Author)
  • 2023(Publication Date)
  • Wiley

    (Publisher)

  The number density of the molecules is n and they all have the same velocity v along the direction of flow. A volume vtA of gas will pass a given point on the tube, say z = z 0 , in time t. This volume contains vtAn molecules, and hence the number of molecules that pass a given point per unit time per unit area is vn Thermal energy of atoms and molecules 95 We have from Equation (3.51) dn dz mg kT n 0 e mgz kT Substituting for dn/dz from Equation (3.51) into Equation (3.59), we find that the number of molecules that cross the plane at height z but do not cross the plane at height z + dz is vmg kT n 0 e mgz kT dz (3.60) This is just the number of molecules given by Equation (3.58). Hence, v mg kT n 0 e mgz kT dz n 0 vP v d v (3.61) Since mgz mv 2 2 and gdz vdv, we obtain P v dv m v kT e mv 2 2kT dv , (3.62) where the factor m v kT is a constant. Equation (3.62) is the probability distribution for the molecules having velocity in the range v to v + dv, in accord with Boltzmann’s law. 3.6 Derivation of the Maxwell–Boltzmann speed distribution In Section 3.4, we obtained the probability distribution for a single component of molecular velocity. As noted there, it is usually more useful to know the speed distribution of the molecules. The difference is, of course, that speed is the magnitude of velocity and does not depend on the direction of travel. In this section, we derive the Maxwell–Boltzmann speed distribution. However, instead of going directly to the speed distribution of a three-dimensional gas, we first deal with the case of a two-dimensional gas, for which the analysis is easier to visualise. The two-dimensional case may seem rather artificial, but, in fact, there are physical situations that can be described in this way. For example, it describes the behaviour of gases on the surface of a catalyst. We can imagine the molecules skating on and confined to the flat surface of the catalyst.

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  Dynamics Of Complex Systems

  • Yaneer Bar-yam(Author)
  • 2019(Publication Date)
  • CRC Press

    (Publisher)

  Using the Boltzmann probability distribution for the velocity, we could calculate the average velocity of the particle as: Thermodynamics and statistical mechanics 85 p p~ 2 2 2 Jv~e-2mkTd3pd3x!h3 Jvie-2mkTdpl. 3kT =3=3 = =-(1.3.83) _ _i_ p~' m Je 2mkTd3pix!h3 Je-2mkTdpl. which is the same result as we obtained for the ideal gas in the last part of Question 1.3.2. We could even consider one coordinate of one particle as a separate system and arrive at the same conclusion. Our description of systems is actually a de-scription of coordinates. There are differences when we consider the particle to be a member of an en-semble and as one particle of a gas. In the ensemble, we do not need to consider the distinguishability of particles. This does not affect any of the properties of a single particle. This discussion shows that the ideal gas model may be viewed as quite close to the basic concept of an ensemble. Generalize the physical particle in three dimensions to a point with coordinates that describe a complete system. These coordinates change in time as the system evolves according to the rules of its dynamics. The ensemble rep-resents this system in the same way as the ideal gas is the ensemble of the particle. The lack of interaction between the different members of the ensemble, and the existence of a transfer of energy to and from each of the systems to generate the Boltzmann probability, is the same in each of the cases. This analogy is helpful when thinking about the nature of the ensemble. 1.3.4 Phase transitions-first and second order In the previous section we constructed some of the underpinnings of thermody-namics and their connection with microscopic descriptions of materials using statis-tical mechanics. One of the central conclusions was that by minimizing the free en-ergy we can find the equilibrium state of a material that has a fixed number of particles, volume and temperature.

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  Discover Entropy And The Second Law Of Thermodynamics: A Playful Way Of Discovering A Law Of Nature

  A Playful Way of Discovering a Law of Nature

  • Arieh Ben-naim(Author)
  • 2010(Publication Date)
  • World Scientific

    (Publisher)

  CHAPTER 5 Discover the Maxwell–Boltzmann Distribution In this chapter, we shall carry out a new (and final) group of experiments. The struc-ture of this chapter is similar to that of Chapter 4, and the experiments are almost the same. As in Chapter 4, we have a system of marbles distributed in cells. The experiment involves the “shaking” of the system, with exactly the same protocol as when we carried out the experiments in Chapters 3 and 4. However, unlike the exper-iments in Chapter 3, where we had no “string attached” to the marbles, and unlike the experiments of Chapter 4, where we had strings attached to the marbles, in this chapter we impose “cloth attached” to the marbles. The reason for attaching cloth rather than string will become clear only in Chap-ter 7. For now, we will just describe the new condition imposed on the marble, do the experiments, and examine the evolution of the system. Or, if you prefer, think of the new game as being a little harder than the game in Chapter 4, which was harder than the game in Chapter 3. Imagine that we attach a piece of square cloth to each marble. The area of the cloth is simply the square of the cell’s level. Thus, the cloth attached to the marble in cell 1 (i.e. in level 0) has area 0. To the marbles in cell 2 (level 1), the attached cloth has a unit area. To cell 3, the area is 2 2 = 4 and so on. In general, to the marble in cell k we attach a cloth with area ( k − 1 ) 2 , as shown in Fig. 5.1. The experiment we now carry out is essentially the same as in the previous chap-ters. We choose a marble at random and move it to a new, randomly chosen cell. We require that the total areas of all the cloths attached to the marbles in the system be conserved. This means that if we move a marble— say, from cell 2 to cell 3 — the cloth area changes from 1 to 4. Therefore, to conserve the total area of the cloth we must make another move — say, of three marbles from cell 2 to cell 1.

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  Problems and Solutions on Thermodynamics and Statistical Mechanics

  • Yung-Kuo Lim(Author)
  • 1990(Publication Date)
  • WSPC

    (Publisher)

  Derive an expression for W . Solution: ( was co nsin) The Maxwell distribution of velocity is given by We take the z-axis normal to the solid surface. Then the distribution of the component v, of velocity is Statistical Physic8 371 Hence nvzP(v,)dv, = n ( K ) ’ ” e x p (--) MU; 27rM 2kT where n is the molecular number density. 2181 A gas in a container consists of molecules of mass m. The gas has a well defined temperature T. What is (a) the most probable speed of a molecule? (b) the average speed of the molecules? ( c ) the average velocity of the molecues? Solution: The Maxwell velocity distribution is given by dw = (-) m 3/2 exp[-mv2/2kT]dv,dv,dv, . 2 s k T (a) Let f ( v ) = v2 exp(-mv2/2kT). The most probable speed is given df ( v ) by = 0, as u = ( $ ) 112 . (b) The average speed is (c) The average velocity V = (is,, isy, is,) is given by 5, = (n1/27rkT)~/’ v, exp[-mv2/2kT]dv,dv,dv, = 0 , . 01 and isy = is, = 0. Thus V = 0. 372 Problems €4 Solutiom on Thermodynamics d Statistical Mechanics 2182 Find the rate of wall collisions (number of atoms hitting a unit area on the wall per second) for a classical gas in thermal equilibrium in terms of the number density and the mean speed of the atoms. Solution: rate of collision is W I T ) Take the z-axis perpendicular to the wall, pointing towards it. The I' = /m dv, Jrn dv, /urn u, fdv, , -m -rn m 312 m where f = n (a) exp [-m (us + v i + u:)] , and n is the number 1 density of the atoms. Integrating we obtain I' = ?nG, where V is the mean speed, v = ( 8 k T / ~ r n ) ~ / ~ . 4 - 2183 At time t = 0, a thin walled vessel of volume V , kept at constant tem- perature, contains NO ideal gas molecules which begin to leak out through a small hole of area A. Assuming negligible pressure outside the vessel, cal- culate the number of molecules leaving through the hole per unit time and the number remaining at time t. Express your answer in terms of No, A, V , and the average molecular velocity, 8.

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