Physics 4311 ANSWERS: Sample Problems for Exam #2 (1)Short answer questions: (a) Consider an isolated system that consists of several subsystems interacting thermally and mechanically with each other. What does the Second Law of Thermodynamics say about the entropies of the subsystems when the system is undergoing a change? ANS: The Second Law says that the sum of the subsystem entropy changes can never decrease. It doesn’t say anything about the change for any individual subsystem. (b) Two thermally interacting subsystems form a thermally isolated composite system. The subsystems have identical total energies and pressures, but different values of $. Will net heat transfer occur between the subsystems, and if so, in which direction will it occur? ANS: Since the $ values are different, the subsystems are not in thermal equilibrium, and heat transfer will occur. Energy will be transferred from the subsystem with the lower value of $ to the subsystem with the higher value of $. (c) Briefly describe what happens to a macroscopic system’s microscopic quantum states and how the states are populated when the system’s energy changes by means of heat transfer or when work is done on (or by) the system. Use words not equations. ANS: Heat transfer changes which states are occupied, but doesn’t affect the energies of the states. Work changes the energies of the states. States may also be created or destroyed by varying external system parameters, i.e., by doing work. (d) When a system undergoes an adiabatic change of state, what must happen for its energy to change? ANS: For an adiabatic change, Q = 0, and the First Law states that )E = !W. For the energy to change, the system must do work. (e) When a system is put through a cyclic process, its final and initial states are identical. What is )E, the change in the system’s energy, for such a process? Briefly explain. ANS: )E = 0. The energy is a function of state, so it has the same value in the final and initial states in this case. (f) When work is performed quasi-statically (reversibly), what condition must be satisfied by the pressure of the system doing the work? ANS: It should differ only infinitesimally from the externally applied pressure on the system. (g) A system undergoes a change from an initial equilibrium macrostate to another. It is then returned to the initial state along a different path that is not the reverse of the initial path. What relation, if any, exists among the work done and heat absorbed by the system along the two different paths? ANS: The First Law is: )E = Q !W; )E is path independent. Let path 1 be the initial path and path 2 be the return path. )E has the same magnitude but opposite sign along path 1 and path 2. Thus, Q1 !W1 = !(Q2 !W2 ), or Q1 + Q2 = W1 + W2 . (h) A system consists of N1 molecules of type 1 and N2 molecules of type 2 confined within a box of volume V. The molecules interact very weakly and constitute an ideal gas mixture. Briefly explain why the number of states S(E) in the energy range between E and E+*E has the following dependence on the volume of the system:Ω ()EVV= NN12χ () E , where χ()E does not depend on V. ANS: Since the two gases interact weakly, the overallΩ ()E is the product of the separateΩ ii()E functions for each gas. Each individual Ni Ω i depends on Ei and V as Ω ii()EV= χii () E. X ⎛ ∂ ln Ω ⎞ (i) Use the expression for S(E) in (h) and the fundamental relation = ⎜ ⎟ , where ⎝ ⎠ kT∂ x E X is the generalized force corresponding to the external parameter x , to find the equation of state for this system, i.e., to find its mean pressure p as a function of V and T. ANS: p = (N1 + N2)kT/V (j) Prove that the energy E of an ideal gas depends on N and T, but not V. ANS: From the definition of $ and the form of S(E) for an ideal gas, S = BVNE"N, where " and B are constants that depend, in part, on the nature of the gas, but not on V, and N is the number of molecules. we find 1 αN β == Thus and there is no V dependence. kT E ENkT= α (k) The internal energy of an ideal rubber band is only a function of temperature T and does not depend on its length L. Determine whether positive heat is absorbed by an ideal rubber band or transferred to the surroundings when the rubber band is isothermally and quasi-statically stretched from a length L1 to a larger value L2 . ANS: Since the rubber band is being stretched isothermally, dE = 0. From the First Law, we then have dQ = dW = !FdL, where we consider F>0. Since FdL > 0 for the case at hand, it follows that dQ < 0, which means that positive heat is transferred to the surroundings from the rubber band. If you consider F<0, as is found in HW #8, Problem 1, then we write dW = FdL , because dW must always be negative for a rubber band being stretched. (l) What would happen to the temperature of an ideal rubber band that was stretched adiabatically? ANS:If the rubber band were stretched adiabatically, dQ = 0, and from the First Law dE = ! dW > 0 because dW is negative for a rubber band being stretched. The rubber band’s internal energy increases, which directly implies that its temperature increases. (m) A box is separated by a partition that divides its volume in the ratio 2:1. The larger portion contains 600 atoms of Ne and the smaller has 300 atoms of He. When a small hole is made in the partition, atoms can move between the two portions. At equilibrium, what are the mean numbers of atoms of each type expected on each side of the partition? ANS: The atoms should be evenly distributed throughout the box, so the larger portion should have 2/3 of each type and the smaller portion should have 1/3. Thus, 200 He and 400 Ne in the large side, and 100 He and 200 Ne in the small side. (n) Refer to (m). What is the probability of finding simultaneously, 300 He atoms and 0 Ne atoms in the smaller portion of the box? ANS: Treat He and Ne atoms as steps in two independent random walks with p=2/3 (being p in the large side) and q=1/3 (being in the small side). Let P be the overall probability. Then P 300 600 300 600 = W300(0)W600(600)= q p = (1/3) (2/3) .1.6×10!249. (2) In the pressure-volume diagram, the single step expansion and compression of an ideal gas at constant pressure are illustrated. The gas may exchange heat with a reservoir that is at temperature T. The system pressure is p , and the system volume is V , and T is the absolute temperature. (a) For the single step expansion (path 1) against a constant opposing pressure the gas goes from the initial equilibrium macrostate A (p1 , V1 , T) to the final equilibrium macrostate B (p2 , V2 , T). What is the largest mass M that can be lifted through a height h against gravity in this expansion? Let g be the gravitational acceleration constant. What is the heat Q absorbed by the system in this expansion? ANS: W = Ipop dV and here the constant opposing pressure must be p2, otherwise the gas would not reach the correct final state when it was done expanding. Thus, we have st W = p2(V2 !V1) = Mgh, and we find easily that M = p2(V2 !V1)/(gh). To find Q, use the 1 Law, )E = Q !W, and realize that )E = 0 because the initial and final states of this ideal gas have the same T . Thus, Q = W = Mgh > 0. The gas absorbed positive heat from the surroundings while doing work. (That’s what enabled its temperature and internal energy to remain constant.) (b)(10 pts.) Now the gas is restored to its initial state by a single step compression (path 2). What is the smallest mass m that must fall through the height h to restore the system? What is the heat Q absorbed by the system in this compression? ANS: W = Ipop dV and here the constant opposing pressure must be p1 , otherwise the gas would not reach the correct final state when it was done being compressed. Thus, we have st W = p1(V1 !V2) = !mgh, and we find easily that m = p1(V2 !V1)/(gh). To find Q, use the 1 Law, )E = Q !W, and realize that, again, )E = 0 because the initial and final states of this ideal gas have the same T . Thus, Q = W = !mgh < 0. The gas transfers positive heat to the surroundings while doing work. (That’s what enabled its temperature and internal energy to remain constant.) (c)(2 pts.) Is M greater or smaller than m? ANS: M < m Take the ratio to see this M/m = p2/p1 <1 (d)(2pts.) Along which path is net positive heat removed from the surroundings? ANS: Path 1; in part (a), the gas absorbs positive heat from the surroundings. (e)(2pts.) Along which path is net positive heat removed from the gas? ANS: Path 2; in part (b) positive heat is transferred to the surroundings. (3) Consider two Einstein solids A and B with the same vibrational frequency <.