(i) T, p, N Gibbs free energy G (ii) T, p, µ no thermodynamic potential, since T, p, µ are not independent of each other (iii) S, p, N Enthalpy H

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1 Solutions exam 2 roblem 1 a Which of those quantities defines a thermodynamic potential Why? 2 points i T, p, N Gibbs free energy G ii T, p, µ no thermodynamic potential, since T, p, µ are not independent of each other iii S, p, N Enthalpy H b Suppose we have a box with two chambers In each of the two chambers we have the same ideal gas Which conditions need to be fulfilled by i Thermal equilibrium? Heat exchange or T 1 T 2 ii Chemical equilibrium? article exchange or µ 1 µ 2 iii Mechanical equilibrium? olume exchange or points c What is extremized in the microcanonical, the canonical, and the grand canonical ensemble? Is it a maximum or a minimum? 3 points microcanonical: entropy is maximal or partition function maximal canonical: Helmholtz free energy is minimal, or entropy is maximal st boundary conditions, or partition function maximal grand-canonical: Grand-potential is minimal, or entropy is maximal st boundary conditions, or partition function maximal d Define the Bose-Einstein condensation What are the volume and chemical potential in the condensed phase? Explain why 4 points Below a critical temperature T c a phase transition sets in, ie particles go from the gas phase in the condensate At low temperature most particles occupy the lowest energy level at T all particles, see problem f with N N 3/2 T 1 1 T c The volume of the condensate is cond and the chemical potential is µ e lot qualitatively the Carnot-cycle in a T -S and a p- diagram Show where the heat is going in and out of the system 4 points shaded areas not to scale Heat goes into the system at step 1 2 isothermal expansion and out at step 3 4 isothermal compression Steps 2 3 and 4 1 are adiabatic dq expansion and adiabatic dq compression, respectively f lot qualitatively the Fermi-Dirac and Bose-Einstein as a function of βe µ for T K and T K 4 points 1

2 left: Fermi distribution at differen temperatures, middle Fermi, Bose and Boltzmann distribution at T K, right: Fermi, Bose and Boltzmann distribution at higher temperaure g What is the Maxwell construction for the van der Waals gas? Why is it used? 5 points Below a critical temperature T c a gas can condensate to a liquid The equation for the van der Waals gas that inclludes interaction such that in principle the phase transition to liquid phase is possible is cubic in the voume and hence has three solutions, possible volumes for a given pressure One of those can be dismissed because of the positive slope of the vdw curve, corresponding to unphysical behaviour The other two solutions correspond to liquid phase lower volume and gas phase higher volume The Maxwell construction finds the corresponding pressure form the horizontal line that equates the areas between the line and the vdw isotherme below and above This is the pressure at wchich, fo the given temperature liquid and gas phase coexist roblem 2 - Gas with two energy states A classical gas in a volume is composed of N non-interacting and indistinguishable particles The single particle Hamiltonian ish p 2 /2m + ɛ, with m the mass of the particle and p the absolute value of the momentum Moreover, for each particle, we find two internal energy levels: a ground state with energy ɛ and degeneracy g 1, and an excited state with energy ɛ ɛ 1 and degeneracy g 2 a Determine the canonical and grand canonical partition function for N particles Show that the Helmholtz free energy has the form [ N A k B T ln h 3N + 3N 2πm N! 2 ln + Nlng 1 Nβɛ lng 2 β 1 points When computing the canonical partition function, we have to consider the continuous part of the Hamiltonian p 2 /2m, plus the contribution coming from the internal degrees of freedom The partition function for the N particles is then the product of N single particle partition functions Z N T,, N N 3N p2 β h 3N e 2m dp g 1 + g 2 e βɛ N N! 2πm N h 3N N! β 3N 2 g1 + g 2 e βɛ N Using eax2 dx π a The factorial N! accounts for the classical indistiguishability of the particles From the partition function we compute the free energy [ N A k B T lnz N k B T ln h 3N + 3N 2πm N! 2 ln + Nlng 1 Nβɛ lng 2 β 2

3 b Compute the energy E of the total system as a function of the temperature T 5 points lnzn E β,n 3 2 Nk BT + N g 2ɛ e ɛ/k BT g1 + g2 e ɛ/k BT c Compute specific heat C as a function of T 5 points An the specific heat E C 3,N 2 Nk B + d Ng2 ɛ e ɛ/k BT 3 dt g 1 + g 2 ɛ e ɛ/k BT 2 Nk Ng 1 g 2 ɛ 2 e ɛ/k BT B + k B T 2 g 2 + g 1 e ɛ/k BT 2 d Analyze the limit of low temperatures of the specific heat and explain this result 5 points For low temperatures, we find Ng 1 g 2 ɛ 2 e ɛ/k BT k B T 2 g 2 + g 1 e ɛ/k BT g 1 g 2 ɛ 2 2 k B T 2 g1 2ɛ e2ɛ/k BT That is the expected result because in such limit only the ground state the one with energy is populated roblem 3 - Relativistic fermi gas In high energy physics it is convenient to take into account relativistic effects We now want to calculate the energy and pressure of the non-interacting relativistic fermi gas In this situation, the energy of one particle is given by ɛ 2 p p 2 c 2 +m 2 c 4 Assume that we are dealing with spin-1/2 particles, further let be ρɛ the density of states and Ωɛ the number of states with energy smaller than ɛ a The grand-canonical partition sum for fermions reads as Ξ 1 + e βɛpµ p Assuming we have a large system and small differences between the energy levels, we can transform the sum over all states, that appears in the corresponding grand canonical potential, into an integral: p dɛ ρɛ erform this transformation and show that the grand canonical potential for a general system of fermions can then be expressed as ΦT,, µ k B T dɛ Ωɛ 1 + e βɛµ Explain all vanishing terms that appear during your calculation 9 points 3

4 The expression for the grand-canonical potential reads as ΦT,, µ k B T ln Ξ k B T p ln 1 + e βɛpµ k B T dɛ ρɛ ln 1 + e βɛµ k B T Ωɛ ln 1 + e βɛµ + k B T dɛ Ωɛ 1 + e βɛµ β eβɛµ dɛ Ωɛ 1 + e βɛµ Here we performed a partial integration The boundary term vanish since for ɛ we get Ω from a b Calculate the number of states in 3 dimensions Ωɛ 2 h 3 d 3 p ɛ p<ɛ with the fermion energy ɛ p being smaller than ɛ 5 points As usual: ΩE 2 h 3 8π 3h 3 ɛ p<e E 2 d 3 p 8π h 3 c 2 m2 c 2 p 2 c 2 <E 2 m 2 c 4 p 2 dp 3/2 8π 3h 3 p3 p E 2 /c 2 m 2 c 2 c Go to the limit of highly relativistic particles ɛ mc 2 Show that the density of states is then given by ρɛ 3Ωɛ/ɛ 4 points In the highly relativistic limit we have E 2 /c 2 m 2 c 2 E 2 /c 2 We then just calculate the derivative: ρe dωe de 8π 3h 3 d E 3 de c 3 8π E2 h 3 c 3 3ΩE E d Use your result from a and the expression for the number of fermions per state nɛ dn/dω to show that Φ E/3, where E dnɛn is the total energy of the system the integral over all single fermion energies 5 points From the lecture we know that nɛ e βɛµ It thus follows with part b that ΦT,, µ 1 3 dɛ ɛρɛ nɛ 1 3 dɛ ɛ dωɛ dɛ dn dω 1 3 dn ɛn 1 3 E e Use the thermodynamic definition of the grand-canonical potential to obtain an expression for the pressure pe, 2 points We know that Φ E T S µn T S p + µn T S µn p, hence p E/3 4

5 roblem 4 - Gas with hard core interaction A gas obeys the equation of state Nk B T exp, where and b are the parameters of system and N is kept fixed in the following a Find the Maxwell relation involving S/ T 3 points Using Helmholtz free energy we have da de T S SdT d S/ T / b Use a differential for S to calculate det, Show that for the upper equation of state E is a function of T only 9 points We have S S de T ds d T dt + d d S T S T dt + [ T [ dt + T Using equation of state we have: d Nk B dt exp T Then for de we can write S de T dt + [T T d T T S T S dt d d Then we can see that E is just a function of T, ie E ET H c Using C and the definition of the enthalpy show that: γ C /C 1 + Nk B C exp 9 points Using enthalpy definition H E + we have C H E + E + Also because E ET for heat capacity C at constant volume we have E E C Then for C we can write C C + 5

6 For finding it is easiest to calculate rewrite equation of state for T T Nk B exp 1 / Then for this reason we Then we have exp Nk B [ 1 Nk B Nk B Nk B 2 exp exp With substituting as a state function in above relation we have Then 1 [ NkB T Nk B exp [ T 2 2 T exp Then for C and substituting as a state function we can write C C + C + C + 2 Nk B T T exp C + Nk B exp 2 T Then for γ we have γ C 1 + Nk B C C exp d In which limit does γ go to the value obtained from an ideal gas? 4 points In the limit of we have γ 1 + Nk B C That is the same as the ideal gas relation for γ 6