Physics 4311 ANSWERS: Sample Problems for Exam #2. (1)Short answer questions:

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1 (1)Short answer questions: Physics 4311 ANSWERS: Sample Problems for Exam #2 (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 =, 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 =. 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, Q 1!W 1 =!(Q 2!W 2 ), or Q 1 + Q 2 = W 1 + W 2. (h) A system consists of N 1 molecules of type 1 and N 2 molecules of type 2 confined within a box of volume V. The molecules interact very weakly and constitute an ideal gas mixture.

2 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 ( E) V N 1 N2 = V χ ( E), where χ( E) does not depend on V. ANS: Since the two gases interact weakly, the Ω ( E) Ω i ( E i ) N Ω i depends on E i and V as Ω i ( E i ) = V i χ ( E ). overall is the product of the separate functions for each gas. Each individual i X (i) Use the expression for S(E) in (h) and the fundamental relation =, where kt ln Ω 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 = (N 1 + N 2 )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 = BV N E "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 E = αnkt and there is no V dependence. kt E (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 L 1 to a larger value L 2. ANS: Since the rubber band is being stretched isothermally, de =. From the First Law, we then have dq = =!FdL, where we i dw dq consider F>. Since FdL > for the case at hand, it follows that <, which means that positive heat is transferred to the surroundings from the rubber band. If you consider F<, 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 =, and from the First Law de =! dw > 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 6 atoms of Ne and the smaller has 3 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, 2 He and 4 Ne in the large side, and 1 He and 2 Ne in the small side.

3 (n) Refer to (m). What is the probability of finding simultaneously, 3 He atoms and 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 = W 3 ()W 6 (6)= q 3 p 6 = (1/3) 3 (2/3) !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 (p 1, V 1, T) to the final equilibrium macrostate B (p 2, V 2, 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 = Ip op dv and here the constant opposing pressure must be p 2, otherwise the gas would not reach the correct final state when it was done expanding. Thus, we have W = p 2 (V 2!V 1 ) = Mgh, and we find easily that M = p 2 (V 2!V 1 )/(gh). To find Q, use the 1 st Law, )E = Q!W, and realize that )E = because the initial and final states of this ideal gas have the same T. Thus, Q = W = Mgh >. The gas absorbed positive heat from the surroundings while doing work. (That s what enabled its temperature and internal energy to remain constant.) (b)(1 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 = Ip op dv and here the constant opposing pressure must be p 1, otherwise the gas would not reach the correct final state when it was done being compressed. Thus, we have W = p 1 (V 1!V 2 ) =!mgh, and we find easily that m = p 1 (V 2!V 1 )/(gh). To find Q, use the 1 st Law, )E = Q!W, and realize that, again, )E = because the initial and final states of this ideal gas have the same T. Thus, Q = W =!mgh <. 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 = p 2 /p 1 <1 (d)(2pts.) Along which path is net positive heat removed from the surroundings? ANS: Path 1;

4 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.

5 (3) Consider two Einstein solids A and B with the same vibrational frequency <. Solid A has N A atoms and energy E A, while solid B has N B atoms and energy E B. After these two solids are placed in thermal contact and allowed to come to thermal equilibrium, their final energies are E A and E B, respectively. To sufficient accuracy, the form of S for an Einstein solid is ln S = (M+3N)ln (M+3N)!M lnm!3n ln3n, where N is the number of atoms in the solid and M is related to the energy E by the equation E = h< (M+3N /2). (a) Find expressions for E A and E B in terms of N A, N B, and the total initial energy E = E A + E B. ANS: At equilibrium $ A = $ B, where we readily find that h<$ = ln(1 + 3N/M). It thus follows that M A /N A = M B /N B and then that E A /N A = E B /N B. Since E A + E B = E, we find E A = X A E and E B = X B E, where X A and X B are the respective number fractions of atoms in solids A and B, X A = N A /(N A +N B ), X B = 1!X A. (b) What is the heat Q absorbed by solid A in going from its initial state to the final equilibrium state? Write your answer in terms of E A and E B. ANS: Q = E A!E A = X A E B!X B E A after a bit of algebra. (c)what condition has to be satisfied to ensure that Q is positive? Give a simple physical interpretation of this condition. ANS: Clearly, the condition X A E B >X B E A must be satisfied. Rewrite this as E B /N B > E A /N A, from which it follows that M B /N B > M A /N A, and finally that $ A > $ B, or equivalently that T B > T A. (d) Is the entropy change for this process positive, negative, or zero. Briefly explain.ans: The entropy change must be positive. An isolated system has undergone internal heat transfer to reach a new state of equilibrium. By the 2 nd Law, such a change is always accompanied by an increase in the total entropy of the system. (4) Consider a thermally isolated system consisting of two chambers of volumes V and 2V separated by a thermally conductive partition that initially is not free to move. Each chamber contains an ideal gas at absolute temperature T. The pressure in the small chamber is p; the pressure in the large chamber is 3p. Now the partition is allowed to move without the gases mixing. (a)when equilibrium is established what are the final pressures, temperatures and volumes of each chamber? Express your answers in terms of p, V, and T. ANS: T is unchanged. V goes to 3/7 V and 2V goes to 18/7 V. Final pressure is 7/3 p in each chamber. (b) What are the changes in the total internal energy and total entropy of the system? Assume the small chamber contains N molecules. ANS: E is unchanged. Δ S = Nk [6ln(9/7) + ln(3/7)] where N is the number of molecules in the smaller chamber. (c) Is this process reversible? Briefly explain. ANS: No. The surroundings are unaffected and the total entropy change is positive. It will not be possible to restore the system to its original state without changing the surroundings.

6 (5) An ideal gas is compressed quasi-statically and isothermally from its original pressure p and volume V to a final volume V /4. (a) What is the final pressure of the gas? ANS: p f = 4p (b) Calculate the work performed by the surroundings on the gas to achieve this change of state. What is the work W performed by the gas during this process? Is it positive or negative? ANS: V /4 V /4 dv 1 Wsurr = p V oppdv = NkT = NkT = NkT V W =!W surr =!NkT ln4; W is negative. V ln ln 4 4 (c) What is the change in the internal energy of the gas? ANS: )E = (isothermal ideal gas) (d) Find the heat Q absorbed by the gas during this process. Is it positive or negative? ANS: )E = Q! W =, so Q = W =!NkT ln4; Q is negative. (Positive heat is transferred to the surroundings to keep the gas at temperature T.) (e) What is the change in the entropy of the gas? What is the change in entropy of the surroundings? ANS: Because the process is isothermal, )S gas = Q/T =!Nk ln4 <. Because the process is quasi-static, )S gas + )S surr =, so )S surr = Nk ln4 >. The gas entropy decreased, while the entropy of the surroundings increased.