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PHYS 352 Homework 1 Solutions PHYS 352 Homework 1 Solutions Aaron Mowitz (1 and 2) and Nachi Stern (3, 4, and 5) Problem 1 We will solve this problem using the microcanonical ensemble. The temperature of a thermody- namic system is defined by 1 @S = T @E N Each link in the polymer either points left or right, i.e. has two possible states. If n links are pointing left and n! are pointing right, the total number of possible configurations of the polymer is the number of ways one can arrange the left and right facing links. This is given by the binomial coefficient: N! Ω = n !n!! We then can write the entropy as S = kB ln Ω = kB (ln N! − ln n ! − ln n!!) ≈ kB (N ln N − n ln n − n! ln n!) where Stirling's approximation is used in the last line. We also have the following constraints on the total number of links and the total energy: N = n + n! E = −qEL = −qE (n! − n ) a Since we are working in the microcanonical ensemble, both of these quantities are fixed. We can use these constraints to rewrite the entropy in terms of E and N: 1 E E 1 E E S = k N ln N − N − ln N − − N + ln N + B 2 qEa qEa 2 qEa qEa Now we can find the temperature using the relation between temperature and entropy we wrote before: E 1 k 1 E 1 E k N − = B ln N − − ln N + = B ln qEa T 2qEa 2 qEa 2 qEa 2qEa E N + qEa The length of the polymer is related to the energy by E = −qEL, so we can invert the above relation to express the length of the polymer in terms of the temperature: ! 2 qEa L = Na 1 − 2qEa = Na tanh kBT 1 + e kB T Since tanh x is an increasing function of x, the length of the polymer will indeed decrease as temperature increases. 1 Problem 2 We will solve this problem using the canonical ensemble. Assuming the N systems in our collection are distinguishable, we can write the partition function of the entire system as a product of the partition functions of N three-level systems: N N −β −2β Z = Z1 = 1 + e + e We can then find the average energy of the system using this partition function: @ ln Z e−β + 2e−2β E = − = N @β 1 + e−β + e−2β This can be inverted to find T in terms of the energy: p !!−1 −3E2 + 6EN + N 22 + N − E T = − ln kB 2(2N − E) (Note: when inverting, there are two possible solutions. However, the other yields an imaginary temperature, which is unphysical) To find the entropy, we use the relation between the Helmholtz free energy and canonical partition function: F = E − TS = −kBT ln Z This gives us −β −2β −β S = kB ln Z + E=T = kB N ln 1 + e + e − E/ ln e N = k N ln e−β + 2e−2β − E/ ln e−β B E N = k N ln x + 2x2 − E/ ln x B E p −β −3E2+6EN+N 22+N−E where x = e = 2(2N−E) . One can also solve this problem via the microcanonical ensemble, similar to problem 1. However, since there are 2 constraints (total energy and total number of systems) but 3 unknowns (number of systems in each of the three states), there will be one free parameter (e.g. the number of systems with energy ). To find this, one must maximize the entropy with respect to this parameter, then one can proceed as in the first problem. 2 3) Quantum-Classical Correspondence in a Harmonic Oscillator 풑ퟐ ퟏ i) For the harmonic oscillator 푯 = + 풎흎ퟐ풙ퟐ, find the number of energy levels with energy less ퟐ풎 ퟐ than 푬. First consider the classical harmonic oscillator: Fix the energy level 퐻 = 퐸, and we may rewrite the energy relation as 푝2 1 1 푚휔2 퐸 = + 푚휔2푥2 → 1 = ( ) 푝2 + ( ) 푥2 2푚 2 2푚퐸 2퐸 In phase space, this equation corresponds to an ellipse, with semi-axes 2퐸 1 푎 = √2푚퐸 푏 = √ 푚 휔 The volume of phase space contained within the ellipse is just its surface. As lower energy trajectories are also ellipses contained within this ellipse, we have 2휋퐸 푑푥푑푝 퐸 푆 = ∫ 푑푥푑푝 = 휋푎푏 = → ∫ = 퐻<퐸 휔 퐻<퐸 2휋ℏ ℏ휔 Now for the quantum harmonic oscillator, whose energy level are given by 1 퐸 = ℎ휔 (푛 + ) ≈ ℎ휔푛 푛 2 Where the approximation is justified for high energy levels. The energy levels are equally spaced, so the number of energy levels below 퐸 is just 퐸 푑푥푑푝 푁 ≈ = ∫ ℏ휔 퐻<퐸 2휋ℏ 3 풑ퟐ ii) For the a more general potential 푯 = + 푽(풙), find the number of energy levels with energy less ퟐ풎 than 푬. Setting once more an energy level 퐻 = 퐸, one might express the momentum as 푝 = √2푚(퐸 − 푉(푥)) 푉(푥) is said to be monotonically decreasing and increasing on either side of 푥 = 0, in addition to diverging at infinity. For a fixed energy 퐸 there are only two solutions for 푝 = 0. The trajectory in phase space will be given by a closed curve, with momentum switching directions when 퐸 = 푉. Moreover, for lower energies, the trajectories again must be contained within the curve defined by energy level 퐸. We can thus integrate over the phase space surface, and the result must be proportional to the number of energy states inside it: 푆 = ∫ 푑푥푑푝 퐻<퐸 Recall Green’s theorem 휕퐿 휕푀 ∮(퐿푑푥 + 푀푑푦) = ∫ ( − ) 푑푥푑푦 휕푦 휕푥 퐶 푆 Set 퐿 = 푦 = 푝, 푀 = 0 and find 휕퐿 휕푀 휕푝 ∫ ( − ) 푑푥푑푦 = ∫ 푑푥푑푝 = ∫ 푑푥푑푝 = ∮ 푝푑푥 휕푦 휕푥 휕푝 퐻<퐸 푆 푆 퐻=퐸 Bohr-Sommerfeld quantization gives us a relation between the integral over the trajectory and the quantum energy level 푁. This condition is written as ∮ 푝푑푥 ≈ ℎ푁 = 2휋ℏ푁 퐻=퐸 Dividing by the prefactor, we retrieve the result of the first part 푑푥푑푝 푁 ≈ ∫ 퐻<퐸 2휋ℏ 4 4) Thermodynamics of Oxygen Production Air is well approximated by an ideal gas, so we may use the Sackur-Tetrode relation for the entropy: 푉 2휋푚푘푇 5 푆(푁, 푉) = 푘푁 [ln [ ( )3⁄2] + ] 푁 ℎ2 2 Defining 푁푂 and 푁푁 as the number of oxygen and nitrogen particles respectively, we make use of the fact that 80% of the mixture is nitrogen to write 푁 4푁 푁 = ; 푁 = 푂 5 푁 5 In the separated state, the volume is proportional to the number of particles 푉 4푉 푉 = = 1퐿 ; 푉 = = 4퐿 푂 5 푁 5 The entropy of the mixed system is 푉 2휋푚푘푇 3⁄2 5 푉 2휋푚푘푇 3⁄2 5 푆푖 = 푆(푁푂, 푉) + 푆(푁푁, 푉) = 푘푁푂 [ln [ ( 2 ) ] + ] + 푘푁푁 [ln [ ( 2 ) ] + ] 푁푂 ℎ 2 푁푁 ℎ 2 The entropy of the separated state is 푉푂 2휋푚푘푇 3⁄2 5 푉푁 2휋푚푘푇 3⁄2 5 푆푓 = 푆(푁푂, 푉푂) + 푆(푁푁, 푉푁) = 푘푁푂 [ln [ ( 2 ) ] + ] + 푘푁푁 [ln [ ( 2 ) ] + ] 푁푂 ℎ 2 푁푁 ℎ 2 The change of entropy between the states is 푉 푉 푘푁 5 1 ∆푆 = 푆푖 − 푆푓 = 푘푁푂 ln ( ) + 푘푁푁 ln ( ) = [ln(5) + 4 ln ( )] ≈ 푘푁 푉푂 푉푁 5 4 2 Using the ideal gas equation of state 푝푉 105[푃푎] × 5 ⋅ 10−3[푚3] 5 퐽 푘푁 = ≈ = 푇 300[퐾] 3 퐾 No work is done in the process and all energy goes to separating the gasses. By the 2nd law of thermodynamics 1 1 5 퐽 ∆퐸 ≥ 푇∆푆 ≈ 푘푇푁 ≈ × [ ] × 300[퐾] = 250퐽 2 2 3 퐾 In practice some energy will usually transform into heat, so the required energy to separate the gases is typically substantially higher. 5 5) A Thermodynamic Identity Show that 퐶푝 (휕푉/휕푝)푇 = 퐶푉 (휕푉/휕푝)푆 Nearly all of you solved this problem correctly, but I suspect many spent a lot of time trying guessing which relations should be used in order to get the right answer. Well, there is a rather easy way to do these kind of problems, that require little to no guesswork. It also works also for harder problems that contain more than 4 dynamic variables (in this case 푉, 푇, 푆, 푝). This general method is adapted from Gene Mazenko’s textbook “Equilibrium Statistical Mechanics”, and is based on the properties of Jacobians. The Jacobian for a 3-variable transformation (푢, 푣, 푤) → (푥, 푦, 푧) is defined as: 휕푢 휕푢 휕푢 휕푥 휕푦 휕푧 휕(푢, 푣, 푤) 휕푣 휕푣 휕푣 ≡ det 휕(푥, 푦, 푧) 휕푥 휕푦 휕푧 휕푤 휕푤 휕푤 (휕푥 휕푦 휕푧 ) And it may be easily generalized to more variables. The property that makes Jacobians useful is 휕푢 휕(푢, 푦, 푧) ( ) = 휕푥 푦,푧 휕(푥, 푦, 푧) By the properties of determinants, one can show that (replacing rows) 휕(푢, 푣, 푤) 휕(푣, 푢, 푤) = − 휕(푥, 푦, 푧) 휕(푥, 푦, 푧) And the generalized chain rule and reciprocal relations 휕(푢, 푣, 푤) 휕(푢, 푣, 푤) 휕(푟, 푠, 푡) 휕(푢, 푣, 푤) 휕(푥, 푦, 푧) −1 = ; = [ ] 휕(푥, 푦, 푧) 휕(푟, 푠, 푡) 휕(푥, 푦, 푧) 휕(푥, 푦, 푧) 휕(푢, 푣, 푤) Let us use these rules to show the identity in question: 휕푄 휕푆 휕(푆, 푝) 휕(푆, 푝) 휕(푇, 푉) 휕(푆, 푝) 휕푉 퐶푝 = ( ) = 푇 ( ) = 푇 = 푇 = 푇 ( ) 휕푇 푝 휕푇 푝 휕(푇, 푝) 휕(푇, 푉) 휕(푇, 푝) 휕(푇, 푉) 휕푝 푇 휕푉 Note: I chose the chain rule above because I wanted to find a relation containing ( ) .
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