Ex3009: Entropy and heat capacity of quantum ideal gases

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1 Ex009: Entropy and heat capacity of quantum ideal gases Submitted by: Yoav Zigdon he problem: Consider an N particle ideal gas confined in volume V at temperature. Find a the entropy S and b the heat capacity C, highlighting its dependence on the temperature. 1 Consider classical gas. Consider Fermi gas at low temperatures, using leading order Sommerfeld expansion. Consider Bose gas below the condensation temperature. Consider Bose gas above the condensation temperature. 5 What is C Bose /C classical at the condensation temperature? 6 For temperatures that are above but very close to the condensation temperature, find an approximation for C V in terms of elementary functions. Hints: In use the Grand-Canonical formalism to express N and E as a function of the temperature and the fugacity z. Use the equation for N in order to deduce an expression for z N Note that the derivative of the polylogarithmic function L α z is 1/zL α 1 z. Final results should be expressed in terms of N, V,, but it is allowed to define and use the notations λ and ɛ F and c. In item the final result can include ratios of polylogarithmic functions, with the fugacity z as an implicit variable. Note that such ratios are all of order unity throughout the whole temperature range provided α > 1, while functions with α < 1 are singular at z = 1. he solution: 1 Classical Gas: Defining the thermal wavelength: λ = 1 π m 1 he partition function: Z N = 1 N! V λ N he free energy is: 1 V N F = ln N! λ We have used the Stirling approximation: V N ln λ + N ln N N ln N! N ln N N a he entropy is given by: F V = N ln Nλ

2 b he heat capacity: C V = S = N 6 Fermi gas at low temperature ɛ F. In order to calculate the grand free energy, lets start by recalling the Sommerfeld expansion of the number of particles and the energy: N = V 1 6π m µ 1 + π 7 8 µ E = V 1 5 6π m 5 µ 1 + 5π 8 he pressure is related to the energy by: µ 8 P = E V By extensiveness, the grand free energy is: F G = P V = V 1 5 6π m 5 µ 1 + 5π 8 µ 9 10 a he entropy is: to the first order FG µ,v = π V 1 6π m µ 1 11 We would like to express the entropy using the number of particles because we are interested in a closed system, where the number of particles is fixed. It is convenient to express the result in terms of the Fermi energy: µ = ɛ F 1 π 1 1 ɛ F In leading order the O correction to µ can be ignored and we get the leading linear approximation S = π N ɛ F 1 b he heat capacity to the first order C V = S = π N ɛ F 1

3 Bose gas C > : In order to find the grand free energy, we need the pressure, which is: P = 5 λ ζ By extensiveness: F G = P V = V λ ζ a he entropy: FG µ,v = 5 V m π 5 ζ Formally one can regard the entropy S of a Bose gas, as the sum over entropies S r of harmonic oscillators that have frequencies ω r = ɛ r µ. Below the condensation temperature the zero-frequency mode ω 0 = 0 contributes an infinite offset S 0 to the total entropy. his offset should be excluded if the system is closed, because then the occupation of the lower orbital is not an independent variable but is dictated by the occupations of the excited orbitals. Differently phrased, one may say that in the regime < c the Bose gas is formally equivalent to a cavity with photons/phonons. b he heat capacity: 17 C V = S V = 15 V m π 5 ζ 18

4 Bose gas > C. he system is closed, so in order to find the grand free energy, we use the expressions for the number of particles and the energy of Bose gas > C. N = V m π z = constant 19 he energy is: E = V m π he grand free energy, as before: 5 L 5 z 0 F G = P V = E 1 herefore: F G = V m π 5 L 5 z a he entropy is: FG µ,v = m 5 V π L 5 z + 5 L 5 z In order to calculate L 5 z We use: and And get: L 5 z = 1 z z z = 1 z lnz 5 µ L 5 z = 1 z lnz 6 Substituting to the entropy expression: m 5 L 5 z S = V L z π z lnz his equation is true also for < C since µ = 0 and hence z = 1. Identifying the prefactor in the last equation as the number of particles, 5 L 5 z S = N z lnz 8 In the Boltzmann regime z 1 this agree with the classical gas result Eq5. 7

5 b Heat capacity: Our system is closed, so the number of particles should stay constant. We calculate the heat capacity using the energy. Alternatively we could have done it using the entropy. C V = E = 15 m V L 5 z + V m 5 π π L 5 z 9 Regarding in 19 as an implicit equation for z as a function of the free variables N and we get by differentiation: 1 z + L z z = 0 0 Now, we know that L z = 1 z z 1 Hence. 1 z = z N z z 1 Lets substitute at Eq9: C V = 15 m V L 5 z + m π V 5 z z π z C V = m V z π 15 L 5 z z 9 z Expressing the heat capacity using the number of particles, We finally get: 15 L 5 z C V = N z 9 z z z In the Boltzman approximation z 1 and L α z z, and by substitution one can see: C V = N - in agreement with the classical heat capacity. 5 1 Note that this derivative in constant number is different from the same derivative in constant chemical potential. 5

6 5 In the critical temperature: C V = N 15ζ 5 ζ 1.9N 6 his result can be deduced using Eq18 or alternatively Eq5. In the latter equation, the right term in the right-hand-side vanish since z diverges at z 1. hus, C Bose C Classical So it is easier to heat a classical gas by 8% than Bose quantum gas. 6 We basically need to approximate polylogarithmic raios: look at Eq. 5. For z 1 +, C, µ 0 we shall use the formula: F α<1 e βµ = Setting α = 1 we get that for C: 0 z/ z, L 5 z/ z near z = 1 dx xα 1 e x z 1 = ΓαL αz 1 1 α βµ 1 α 8 z 1 π βµ 9 o get an approximation for z, we remember Eq.1 and integrate: So: z ζ z z ζ On top of that we need: L 5 z µ 0 π C π µ C z ζ 5 ζ π µ C ζ 5 ζ 1 + πζ µ But the latter expressions depend on µ, which we would like to relate to the small parameter: C 1 t = C C So we approximate the number of particles N z near the condensation temperature: = C 1 + C C 1 + t C Putting this together with Eq. 0 in the expresion for N: N = V mc 1 + π t ζ µ π C 5 6

7 After a little algebra, µ πζ t C 8 6 Now we are ready to get the heat capacity: 15 ζ 5 C V = N ζ 5 ζ + ζ 7πζ t N t 7 6 Comparing with item - the heat capacity is continuous. 7

Physics 112 The Classical Ideal Gas

Physics 112 The Classical Ideal Gas Peter Young (Dated: February 6, 2012) We will obtain the equation of state and other properties, such as energy and entropy, of the classical ideal gas. We will start