(a) Why cannot the Carnot cycle be applied in the real world? Because it would have to run infinitely slowly, which is not useful.

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1 PHSX 446 FINAL EXAM Spring 25 First, soe basic knowledge questions You need not show work here; just give the answer More than one answer ight apply Don t waste tie transcribing answers; just write on this exa sheet (a) Why cannot the Carnot cycle be applied in the real world? Because it would have to run infinitely slowly, which is not useful (b) Two systes in diffusive equilibriu have equal: a) teperatures, b) pressures, c) cheical potentials, d) entropies (c) What effect does a solute have on a liquid? ) It raises the boiling point 2) It lowers the boiling point 3) It raises the freezing point 4) It lowers the freezing point (d) How to theral fluctuations of the current in an electrical circuit scale with teperature? T (e) Ososis is driven by: a) a teperature gradient, b) a concentration gradient, c) a pressure gradient, d) an entropy gradient (f) The cheical potential of a Feri gas at low teperature is equal to: a) kt, b) pressure, c) the Feri energy, d) density of states, e) potential energy per particle (g) True or false? As the teperature of a Feri gas is lowered, the energy per particle goes to zero False (h) An increase in teperature of a photon gas leads to: a) an increase in pressure, b) an increase in the nuber of photons, c) negative cheical potential, d) decreased pressure (i) Which of the following are bosons? a) photons, b) electrons, c) neutrons, d) phonons (j) Which of the following are ferions? a) photons, b) electrons, c) neutrons, d) phonons (k) In solids at low teperatures, the doinant contribution to the heat capacity is: a) angular oentu states, b) phonons, c) electrons, d) translational otion (l) Bose-Einstein condensation is: a) a large increase in density at low T, b) occupation of states up to the Feri level, c) acroscopic occupation of the ground state, d) a transition fro a gas to a liquid

2 PHSX 446 FINAL EXAM Spring 25 2 To assist your grader, who wants you to do well on this exa, systeatically explain your logic and show your work in the following probles Without such assistance your grader, well, cannot grade Suggestion: if you run into technical difficulty with a proble, describe in words how you would proceed 2 A solid of ass s and initial teperature T s is iersed in a liquid of ass l and initial teperature T l The specific heat (heat capacity per unit ass) for the solid is c s =constant The specific heat of the liquid depends on teperature T as c l = c (T/T ), where c and T are constants As equilibriu is reached, no phase transitions occur (a) Find an algebraic expression for the final teperature of the syste Since no heat enters or leaves the syste: giving Tf dq = = dq l + dq s = l dt c l + s (b) Solve the expression fro part (a) Solve the quadratic in T f to obtain: T l Tf l c 2T (T 2 f T 2 l ) + s c s (T f T s ) = T s dt c s T f = [ (s ) sc s c 2 ( s T + T + Tl ) ] sc 2 /2 s T T s, l c l c l c where we ust choose the plus sign option against the square root to get a positive teperature 3 The vibrational energy levels of a diatoic olecule are well described as a onediensional quantu haronic oscillator with energy levels ɛ n = hω(n + /2) n =,, 2, In this proble, consider only the vibrational degrees of freedo of the olecule (a) Show that the partition function for vibration of a single olecule is Z = e β hω/2 e β hω

3 PHSX 446 FINAL EXAM Spring 25 3 As always, the parition function is: which can be written as Z = e β hω/2 Z = e βɛn, n= n= (b) Find the vibrational energy per olecule ( e β hω ) n = e β hω/2 e β hω E = ln Z [ ] β = hω 2 + e β hω (c) Evaluate the energy in the low-teperature liit kt << hω Include teperature dependence in this liit to leading order in e β hω The low-t liit corresponds to β hω >> The exponential in the expression for E becoes large, giving [ ] E = hω 2 + e β hω (d) Find the heat capacity C V per olecule in the low-teperature liit C V = kβ 2 E β = kβ2 ( hω) 2 e β hω (e) Describe and explain the low-teperature behavior of C V As T (β ), C V The reason is that the olecule is ost likely in its ground state, and is very likely to reain there If we add heat dq = C V dt, a large teperature change, of order hω >> kt, is needed for the syste to accept the energy dq, so C V is sall 4 Consider a long and narrow tube of length L and cross-sectional area A that extends fro x = to x = L The tube is filled with a photon gas in equilibriu at teperature T For this long and narrow tube, we can think of the photons as oving in one diension, along the x axis, that is, this is a one-diensional photon gas For this situation, the eleent of phase space is siply x h < p x < Integrals over this eleent of phase space give quantities per unit length In the following when you encounter an integral put it in diensionless for that is left unevaluated

4 PHSX 446 FINAL EXAM Spring 25 4 (a) Calculate the nuber of photons per unit length of tube, N/L Caution: photons are oving parallel and anti-parallel to the x axis Be careful how you write the energy of a particle Dropping the subscript x on the oentu, the energy of a photon, which is always positive, is ɛ = p c Integrating over all possible oenta, and recalling that a photon has two independent polarization states, gives for the nuber of photons in the tube: N = 2 h L e β p c = 4L h e βpc, where in the last step we used the fact that we have an integrand that is syetric in p over syetric liits If you don t use ɛ = p c, you get a divergent integral Introducing x βpc gives N L = 4 hβc e x (b) Calculate the total oentu of photons oving in the x > direction We should integrate over positive oenta only, so: P = 2L h p e βpc = 2L h(βc) 2 x e x (c) Note that the nuber density n of photons is given by n = N/(LA) Find the oentu flux J to x > in ters of physical constants, diensionless integrals, and A The oentu flux in the +x direction is the total oentu ties the nuber density of photons ties the speed at which the oentu is transported (the speed of light): J = P nc = 8(kT )3 L (hc) 2 A ( ) ( e x ) x e x (d) We ake the tube into a photon rocket by reoving the end of the tube at x = L Write the initial acceleration of the tube of ass M in ters of M, J, and A The instant the end of the tube is reoved, oentu leaves at a rate dp/dt = JA By Newton s Second Law: Ma = JA We see that a good photon rocket should be at high teperature, and that the thrust increases as T 3

5 PHSX 446 FINAL EXAM Spring Recall that a Feri gas at high teperature has a negative cheical potential As the teperature is lowered, the cheical potential passes through zero, and then approaches the Feri energy Zero cheical potential is interediate between the classical and quantu regies, and is the subject of this proble Assue the gas is non-relativistic The following integrals and nuerical values should be useful: x3/2 e x + = 5 x/2 e x + = = 72 (a) Calculate the nuber density n of particles for this gas with µ = Accounting for the two spin states of the ferions, the nuber of particles is: N = 2V Introducing x βp 2 /2 gives n = N V = 6π x/2 2β e x + (b) Calculate the energy density E/V E V = 2 4πp 2 (p 2 /2) e βp2 /2 + = 6π 4πp 2 e βp2 /2 + = (67) 6π x3/2 2β β e x + 2β = (5) 6π (c) Use the results of parts (a) and (b) to express the energy density in the for: E V = AnkT, where A is a constant and n is the nuber density Obtain A Cobining the results fro parts (a) and (b) gives: ( ) E 5 V = nkt = 72 nkt 67 (d) Is the energy density larger or saller than that of a classical onatoic gas? The energy density for a classical onatoic gas is 3nKT/2, so the energy density for this Feri gas is larger The reason is that at zero cheical potential there is still a bit of quantu degeneracy contributing to the energy 2β β