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Cornell University, Physics Department PHYS-3341 Statistical Physics Fall 2014 Prof. Itai Cohen Solutions to Problem Set 7 David C. Tsang, Woosong Choi 7.1 Dilute Gas in an Enclosure (a) By symmetry we see v̄z 0 . By the maxwell-boltzmann distribution we have R∞ r R∞ 2 2 |v |e−mvz /2kT dvz 2 0 vz e−mvz /2kT dvz 2kT /m 2kT −∞ z = R ∞ −mv2 /2kT =p |vz | = R ∞ −mv2 /2kT = z z πm e dvz dvz 2 0 e 2πkT /m −∞ R ∞ 2 −mv2 /2kT 2kT 3/2 1√ z π e dv 2 v kT z z 2 m 0 v 2 = R ∞ −mv2 /2kT = √ = 1/2 z m 2 0 e dvz π 2kT m (b) DefineR φ(vz ) as the # of molecules escaping per unit time per unit z-velocity. That is ∞ Φ = 0 φ(vz )dvz . Now φ(vz ) = Avz n(vz ) where n(vz ) is the number of molecules with z-velocity vz per unit volume per unit z-velocity. 2 N e−mvz /2kT n(vz ) = R ∞ −mv2 /2kT dv z z V −∞ e r R∞ 2 AN 0 e−mvz /2kT dvz AN kT AN R∞ ⇒Φ= = |v | = z 2 2V 0 e−mvz /2kT dvz 2V V 2πm (c) Now assume we cover the hole with a plate and assume elastic collisions on that plate. Then the froce on that plate per unit time is F = Φ∆p = 2mΦvz,esc where ∆p is the average change in momentum due to an elastic collision. F 2m ⇒P = = Φvz,esc A A (d) vz,esc AP PV = = 2Φm Nm r πm kT = 2kT m r πm = 2kT r πkT π = |vz | 2m 2 For an ideal gas with maxwell-boltzmann distribution we have vz,esc > |vz |. Another way of seeing this is to recognize that the escaping molecules are “pushed” preferentially on one side, giving them a greater average speed. 2 1 7.2 Two-Dimensional Gas Reif §7.7: Monoatomic molecules adsorbed on a surface are free to move on this surface and can be traeted as a classical ideal two-dimensional gas. At absolute temperature T , what is the heat capacity per mole of molecules thus adsorbed on a surface of fixed size? In an ideal gas the particles are non-interacting , and there are no external potentials, therefore V = 0. Hence we have the energy E = K the kinetic energy. The equipartition theorem gives us 1 kT = N kT Ē = K̄ = 2N 2 where 2 comes from the number of dimensions and N is the number of particles. Thus we have ∂ Ē C = N k = νR ⇒ =R C= ∂T n 2 7.3 More Oscillators Reif §7.10: A system consists of N very weakly interacting particles at a temperature T sufficiently high so that classical statistical mechanics is applicable. Each particle has mass m and is free to perform one-dimensional oscillations about its equilibrium position. Calculate the heat capacity of this system of particles at this temperature in each of the following cases: (a) The force effective in resotring each particle to its equilibrium position is proportional to its displacement x from this position. (b) The restoring force is proportional to x3 . (a) We have an energy of the form E= N X p2 1 + ko x2i m 2 i i Thus there are two seperate quadratic degrees of freedom per particle, hence we have by equipartition ∂E Ē = N kT ⇒ C= = Nk ∂T 2 (b) For this system we have the energy E= N X p2 1 + ko x4i m 4 i i The classical partition function is then Z ∞ N Z 1 Y ∞ −βp2i /2m 4 e−βko xi /4 dxi e dpi Z = N h0 i −∞ −∞ N N/2 Z ∞ 2πm −ko y 4 /4 −1/4 = e dy (where y = β 1/4 x) β 2 ho β −∞ N/2 Z N 2πm −ko y 4 /4 = e dy β −3N/4 2 ho Since the first two terms in the product do not depend on β we must have Ē = − ∂ ln Z 3 = N kT ∂β 4 ⇒ 3 C = Nk 4 2 7.4 Specific heat of graphite Reif §7.11: Assume the following highly simplified model for calculating the specific heat of graphite, which has a highly anisotropic crystalline layer structure. Each carbon atom in this structure can be regarded as performing simple harmonic oscillations in three dimensions. The restoring forces in directions parallel to a layer are very large; hence the natural frequencies of oscillations in the x and y directions lying within the plane of a layer are both equal to a value ωk which is so large that ~ωk 300k. On the other hand, the restoring force perpendicular to a layer is quite small; hence the frequency of oscillation ω⊥ of an atom in the z direction perpendicular to a layer is so small that ~ω⊥ 300k. On the basis of this model, what is the molar specific heat (at constant volume) of graphite at 300◦ K? For harmonic oscillations we have energy E = ~ω(n + 12 ). Adding the terms for all directions we get 1 E = ~ω⊥ (n1 + n2 + 1) + ~ωk (nk + ) 2 But since the energy terms are indepedent in each direction, we can separate direction dependence of the partition function Z = Z⊥ Zk 3 where both Z⊥ and Zk are partition functions for harmonic oscillator(s). Now using the partition function to find the specific heat is simple cV ∂ 2 ln Z ∂ Ē =− ∂T ∂T ∂β 2 ∂ ln Z⊥ ∂ 2 ln Zk = − − ∂T ∂β ∂T ∂β ¯ ∂ E⊥ ∂ Ēk = + ∂T ∂T = At 300◦ K, since ~ωk 300k and ~ω⊥ 300k following the argument at p.253 of Reif we get E¯⊥ = Ē1 + Ē2 ' ~ω⊥ (1 + 2e−β~ω⊥ ) Ēk ' kT Taking the derivative of these with respect to T , dβ ∂ E¯⊥ 1 ∂ E¯⊥ = = − 2 ~ω⊥ (−2~ω⊥ e−β~ω⊥ ) ∂T dT ∂β kT 2 −β~ω⊥ = 2k(β~ω⊥ ) e ) cV = k + ~ω⊥ (−2~ω⊥ ) ∂ Ēk = k ∂T Combining this we get cV = k 2(β~ω⊥ )2 e−β~ω⊥ + 1 Introducing the definition of Einstein temperature θE = in a mole(since above result was only for one atom), ~ω⊥ k and summing over all the atoms " # 2 θE −θE /T cV = R 2 e +1 T 2 4 7.5 Ferro Fluids Reif §7.14: Consider an assembly of N0 weakly interacting magnetic atoms per unit volume at a temperature T and describe the situation classically. Then each magnetic moment µ can make any arbitrary angle θ with respect to a given direction (call it the z direction). In the absence of a magnetic field, the probability that this angle lies between θ and θ + dθ is simply proportional to the solid angle 2πsinθdθ enclosed in this range. In the presence of a magnetic field H in the z direction, this probability must further be proportional to the Boltzmann factor e−βE , where E is the magnetic energy of the moment µ making this angle θ with the z axis. Use this result to calculate the classical expression for the mean magnetic moment M̄z of these N0 atoms. Following the argument given in the problem we have P (θ)dθ ∝ 2πsinθdθeβµHcosθ for one magnetic moment. Since they are only weakly interacting, we can consider the atoms individually and sum up in the end. Since we have the proportionality of P (θ) we can find the partition function as follows: Z π dθ2πsinθdθe Z= βµHcosθ Z 1 βµHcosθ d(cosθ)e = 2π −1 0 eβµHcosθ = 2π βµH 1 =− −1 4π sinh(βµH) βµH To find the average magnetization of the magnetic moment in z-direction, µ¯z = ∂ ln Z 1 = µcoth(βµH) − ∂(βH) βH thus for N0 atoms M̄z = N0 µ coth(βµH) − 1 βµH 2 5 7.6 Mean Values Reif §7.19: A gas of molecules, each of mass m, is in thermal equilibrium at the aboslute temperature T . Denote the velocity of a molecule by ~v , its three cartesian components by vx , vy and vz , and its speed by v. What are the following mean values: (a) vx (b) vx2 (c) v 2 vx (d) vx3 vy (e) (vx + bvy )2 where b is a constant (f ) vx2 vy2 (a) vx = 0 since vx is an odd function. (b) vx2 = kT m by the equipartition thereom. (c) v 2 vx = 0 since v 2 vx is an odd function (d) vx3 vy = 0 since vx3 vy is odd in both vx and vy . (e) (vx + bvy )2 = vx2 + 2bvx vy + bvy2 = (f ) vx2 vy2 = vx2 vy2 = kT 2 m kT m +0+ bkT m = (1 + b) kT . m since the distributions of vx and vy can be considered independent. 2 7.7 Isotope Separation Reif §7.26: A vessel is closed off by a porous partition through which gases can pass by effusion and then be pumped off to some collecting chamber. The vessel itself is filled with a dilute gas consisting of two types of molecules which differ because they contain two different atomic isotopes and have correspondingly masses m1 and m2 . The concentrations of these molecules are c1 and c2 , respectively, and are matintained constant inside the vessel by constantly replenishing the cupply of gas in it by a steady slow flow of fresh gas through the vessel. (a) Let c01 and c02 denote the concentrations of the two types of molecules in the collecting chamber. What is the ratio c02 /c01 ? 6 (b) By using the gas U F6 , one can attempt to separate U 235 from U 238 , the first of these isotopes being the one useful in initiation of nuclear-fission reactions. The molecules in the vessel are then U 238 F619 and U 235 F619 . (The concentrations of these molecules, corresponding to the natural abundance of the two uranium isotopes, are c238 = 99.3 percent and c235 = 0.7 percent.) Calculate the corresponding ratio c0235 /c0238 of the molecules collected after effusion in terms of their original concentration ratio c235 /c238 . (a) The ratio of concentration of the molecules in the collecting chamber will be the ratio of the molecules effused through the porous partition. From (7.11.13) average number of particles that will hit, i.e. pass through, the holes per unit area is nkT p̄ =√ Φ0 = √ 2πmkT 2πmkT where p̄ is the mean pressure of the molecules, thus can be substituted to nkT using the equation of state. T is the same for the two types of molecules thus Φ0 ∝ √nm is the relation relevant to the ratio. r c1 m2 c01 = c02 c2 m1 (b) For U F6 , using the result of (a) we get c0235 c235 = 0 c238 c238 r m238 c235 = m235 c238 r 352 c235 = 1.004 × 349 c238 Although the change in concentration ratio is small, by repeatedly using effusion we can increase the concentration of U 235 in the mixture. 2 7