PHYS-4420 THERMODYNAMICS & STATISTICAL MECHANICS FINAL EXAMINATION SPRING 2003 Friday, May 9, 2003 Your grade will be sent to you by e-mail by 5:00 PM, Monday, May 12, 2003 NAME: ______ANSWERS_________________________ There are five pages to this examination. Check to see that you have them all. To receive credit for a problem, you must show your work, or explain how you arrived at your answer. 1. (10%) State the First, Second, and Third Laws of Thermodynamics in some reasonable form. 1st. In any process, the change in internal energy of a system is equal to the heat that goes into the system minus the work that comes out of the system plus the change due to the increase in number of particles in the systetm. (Conservation of energy.) dE = dQ – dW + dN 2nd. For any two interacting systems, the entropy of the combined systems cannot decrease. S 0 3rd. The entropy of a system goes to zero as the temperature of the system goes to zero. 2. (15%) When a single die (one half of a pair of dice) is tossed, the probability that the face with only one dot on it will turn up is one in six. A naïve person might believe that tossing the die a second time would double the probability of the “one” face turning up to two chances in six. You can quickly see the fallacy of that reasoning if you extend it to six tosses, since it predicts six chances in six; a sure thing. Calculate the actual probability of at least one “one” turning up in six tosses of a die. (That means: the probability of a “one” showing up once or more than once in six tosses.) 6 5 Easy way: Prob = 1 – Prob (0 “ones”) = 1 = 1 – 0.335 6 n 6n N! 6! 1 5 p nq N n Harder way: Prob of n “ones” is P(n) n!( N n)! n!(6 n)! 6 6 From this: P(1) = 0.402, P(2) = 0.201, P(3) = 0.054, P(4) = 0.008, P(5) = 0.00064, and P(6) = 2.1 ×10-5. Then, Probability = P(1) + P(2) + P(3) + P(4) + P(5) + P(6) Either way, the result is the same. Probability = 0.665 = 66.5% 1 NAME: ______ ANSWERS _________________________ 3. (25%) Consider a collection of N identical, indistinguishable harmonic oscillators, all of frequency . The energies that one these oscillators can take on (measured relative to the ground state) are n = nh. 1 a) (5%) Find the partition function for one of these oscillators. (Hint: x n for x < 1) n 0 1 x n0 n0 e nh e h n 1 1 e h 1 1 e h b) (5%) Find the partition function for the collection of N oscillators. Z 1 N 1 1 N! N ! 1 e h N Z 1 1 h N! 1 e N c) (5%) Find the average energy of the collection of oscillators. E h e h N h ln Z ln N ! N ln(1 e h ) N h h 1 e e 1 E N h e 1 h For this system suppose the number of oscillators is N = 1000, and the temperature is such h that kT (i.e. h = 2). 2 d) (10%) How many oscillators are likely to be found in the n = 0 state, the n = 1 state and the n = 2 state? The probability of finding an oscillator in a particular state is given by: e nh e nh p(n) e nh (1 e h ) e 2 n (1 e 2 ) nh e Then, N(n) = Ne-2n(1 – e-2) = 1000 e-2n(1 – e-2) = 864.7 e-2n From this, we get: N(0) = ______865___________ N(1) = ______117___________ N(2) = _______16___________ 2 NAME: ______ ANSWERS _________________________ 4. (40%) One mole of an ideal monatomic gas traverses the cycle shown in the figure. Process 12 is isothermal, process 23 takes place at constant pressure, and process 31 takes place at constant volume. (Hint: for one mole of an ideal monatomic gas, the internal energy is E 32 RT .) The temperatures of the gas at points 1, 2, and 3 are: T1 = 300 K, T2 = 300 K, T3 = 1500 K a) (5%) Find V1/V2, the ratio of the volume of the gas at point 1 to the volume at point 2. (Hint: V1 = V3) V V T 1500 K 23 takes place at constant pressure, so V T. Then, 1 3 3 V2 V2 T2 300 K V1/V2 = _____5______ b) (10%) Find the magnitude of the heat added to the gas in the process 23. Express the answer in terms of the gas constant R. Constant pressure process, so: Q23 = Cp(T3 – T2) = 52 R(1500 K 300 K) Q23 = __(3000 K)R___ 3 NAME: ______ ANSWERS _________________________ c) (5%) Find the magnitude of the heat removed from gas in the process 31. Express the answer in terms of the gas constant R. Constant volume process, so: Q31 = CV(T1 – T3) = 32 R(300 K 1500 K) = – (1800 K)R (Magnitude) Q31 = (1800 K)R d) (10%) Find the magnitude of the heat removed from gas in the process 12. Express the answer in terms of the gas constant R. Process is isothermal, so E = 0, and Q12 = W12 V2 V2 V dV 1 W12 pdV RT RT ln 2 RT ln RT ln 5 R(300 K) ln 5 V1 V1 V 5 V1 (Magnitude) Q12 = R(300 K) ln 5 = (483 K)R e) (5%) How much work is done by the gas in one complete cycle? Express the answer in terms of the gas constant R. W = Qin – Qout = (3000 K)R – [(1800 K)R + (483 K)R] W = (717 K)R f) (5%) Find the efficiency of a heat engine operating with this cycle. W (717 K) R Qin (3000 K) R = 0.239 = 23.9% 4 NAME: ______ ANSWERS _________________________ 5. (10%) A Carnot refrigerator is to be used to lower the temperature of a container that has a constant heat capacity of C. When an amount of heat dQ2 is removed from the container, its dQ2 temperature decreases by an amount dT2 . An amount of work dW is done to remove C dQ2 from the container, at T2, and deposit dQ1 into the high temperature reservoir at temperature T1. The temperature of the high temperature reservoir does not change when heat is added to it. Initially, the container is at the same temperature as the reservoir, T1. Show that to lower the temperature of the container from T1 to some lower temperature Tf, the amount of work required is: T W C T1 T f T1 ln 1 T f This is easy if you remember the coefficient of performance for a Carnot refrigerator. T Q T2 T T c 2 . Then, W Q2 1 2 Q2 1 1 . For an infinitessimal amount of work, W T1 T2 T2 T2 T T dW dQ2 1 1 CdT2 1 1 . Then, just integrate this from T1 to Tf. T2 T2 T T T dT2 dT2 , and with a little work, W C T1 ln f (T f T1 ) . dW C T1 T1 T T2 T f f 1 1 T A few more manipulations yields: W C T1 T f T1 ln 1 . T f If you did not recall the coefficient of performance, you could proceed: dW dQ1 dQ2 dQ1 CdT2 . To integrate this, dQ1 must be expressed in terms of dT2. Q T dQ1 T1 T T , and dQ1 1 dQ2 1 CdT2 . Then, For a Carnot cycle, 1 1 , so Q2 T2 dQ2 T2 T2 T2 dW dQ1 dQ2 T T1 CdT2 CdT2 C 1 dT2 dT2 . Then, proceed as shown above. T2 T2 5