ProbSet7_solutions(2)

ProbSet7_solutions(2) - Chem C130/MCB C100A. Fall 2010....

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Unformatted text preview: Chem C130/MCB C100A. Fall 2010. Problem Set 7 UC Berkeley Page 1 Hand in the five problems marked with a * on Thursday, October 21. ANSWER KEY 1. What is the difference between an energy distribution and a microstate? Explain using simple diagrams. 2.* Consider a system of three identical and independent molecules, with energy levels that have values 0,1,2,3, …(arbitrary units of energy). Consider two states of the system, A and B, shown below. States A and B have equal energy (6 units). State A has one molecule in level 4 and two in level 1. State B has 1 molecule each in levels 1, 2 and 3. The multiplicity, W, of each state is given by N! / (n1! x n2! x n3! …x nj! x….), where N is the total number of molecules and nj is the number of molecules in state j. (i) Calculate the relative probability of observing the molecules in configuration A over configuration B. What is the change in entropy in going from state A to state B (assume kb=1). Assume that the 3 molecules in the system have access to only the 5 energy levels shown above. When the total energy of all 3 molecules is increased to (ii) (iii) Chem C130/MCB C100A. Fall 2010. Problem Set 7 UC Berkeley Page 2 the maximum possible value, explain whether the entropy increases or decreases with respect to state B. (iv) Next, assume that the system has an infinite number of possible energy levels of increasing energy. In this case, when the total energy of the 3 molecules is increased does the entropy always increase, or does it sometimes increase and sometimes decrease? Explain your answer. 3. A more accurate version of Stirling’s approximation is given below: ln N!= N ln N − N + ln( 2πN ) Estimate, by trial and error, how big N has to be for the error introduced by neglecting the last term to be less than 1% of the value of ln N! Chem C130/MCB C100A. Fall 2010. Problem Set 7 UC Berkeley Page 3 4. * A reversible process takes place within a system containing 2 moles of a molecule, with the absorption of 10 kJ mol-1 of heat from the surroundings at 300K. What is the ratio of the multiplicities of the initial and final states of the system? 5. Consider a chamber that is divided into two equal right and left halves by a semipermeable partition that allows A-type molecules to pass through, but blocks the passage of B-type molecules. There are 106 type B molecules on the left side of the partition. Consider the two situations described below: (i) There are 330,000 type A molecules on the left side of the partition, and 670,000 type A molecules on the right side of the partition. (ii) There are 320,000 type A molecules on the left side of the partition and 680,000 type A molecules on the right side of the partition. (A) Which situation is more likely, case (i) or case (ii)? What are their relative multiplicities? Use the probabilistic definition of the entropy to calculate the relative multiplicities. Answer: This is an insightful problem to work through because it addresses some important issues in calculating entropy quickly and with the least amount of numerical tedium. The situation described in the question is illustrated below: Chem C130/MCB C100A. Fall 2010. Problem Set 7 UC Berkeley Page 4 The first thing to note is that the question does not specify how many grid boxes the chamber is divided into (i.e., the value of M is not specified). This is because the value of M is not important in the calculation of differences in entropy, as long as M is sufficiently large (i.e., M >>N, where N is the total number of atoms). We explain why this is so below. The question also asks us to use the probabilistic definition of entropy. To do this, we assume that each side of the chamber is divided into M grid boxes. Each grid box can be in one of 3 states: (i) occupied by A, (ii) occupied by B, or (iii) empty. Notice that the number of B molecules in each chamber does not change in State i or State ii. In this case, the change in entropy depends only on the entropy of the A molecules. Let us first understand why this is so. We can ignore the B molecules because: Let the number of molecules be denoted A1,left, B1,left, etc. For the left side, probability of a grid box being in each of these three states is given by: B1, left M − ( A1, left + B1, left ) , pB1, left = and pempty, left = ≈1 M M M The last term is essentially equal to 1 because the number of A and B molecules is very small compared to M . The entropy on the left in case 1 is therefore given by: pA1, left = A1, left S1, left = − Mk B ⎡ pA1, left ln pA1, left + pB1, left ln pB1, left + pempty, left ln( pempty, left ) ⎤ ⎣ ⎦ Because pempty, left ≈ 1, the last term is zero (ln1=0). Hence: S1, left = − Mk B ⎡ pA1, left ln pA1, left + pB1, left ln pB1, left ⎤ ⎣ ⎦ Likewise, the entropy of the right side in state 1 is given by: S1, right = − Mk B ⎡ pA1, right ln pA1, right ⎤ (note that there are no B molecules on the right). ⎣ ⎦ So, the total entropy in case 1 is: S1 = S1, left + S1, right = − Mk B ⎡ pA1, left ln pA1, left + pB1, left ln pB1, left + pA1, right ln pA1, right ⎤ ⎣ ⎦ By similar reasoning, the entropy for case 2 is: S2 = − Mk B ⎡ pA 2, left ln pA 2, left + pB 2, left ln pB 2, left + pA 2, right ln pA 2, right ⎤ ⎣ ⎦ But the number of B molecules is the same in state 1 and state 2. Therefore, when we calculate the difference in entropy between the two states the term involving B molecules will cancel out. So we only need to consider the entropy of the A molecules. The next thing to realize is that when calculating differences in entropy, the value of M does not matter when the probabilistic definition of entropy is used (i.e., M is very large). Value of M does not matter: Chem C130/MCB C100A. Fall 2010. Problem Set 7 UC Berkeley Page 5 S1 = − Mk B ⎡ pA1, left ln pA1, left + pA1, right ln pA1, right ⎤ ⎣ ⎦ ⎡ A1, left ⎛ A1, left ⎞ A1, right ⎛ A1, right ⎞ ⎤ × ln ⎜ = − Mk B ⎢ ⎟ + M × ln ⎜ M ⎟ ⎥ ⎝M⎠ ⎝ ⎠⎦ ⎣M Cancelling the M terms we get: ⎡ ⎛ A1, left ⎞ ⎛ A1, right ⎞ ⎤ S1 = − k B ⎢ A1, left × ln ⎜ ⎟ + A1, right × ln ⎜ M ⎟ ⎥ ⎝M⎠ ⎝ ⎠⎦ ⎣ = − kB ⎡ A1, left × (ln A1, left − ln M ) + A1, right × (ln A1, right − ln M ) ⎤ ⎣ ⎦ = − kB ⎡ A1, left × (ln A1, left ) + A1, right × (ln A1, right ) − ( A1, left + A1, right )ln M ⎤ ⎣ ⎦ = − kB ⎡ A1, left × (ln A1, left ) + A1, right × (ln A1, right ) ⎤ + kB ( Atotal )ln M ⎣ ⎦ In the last step Atotal is the total number of A molecules. Likewise, the entropy of state 2 is given by: S2 = − k B ⎡ A2, left × (ln A2, left ) + A2, right × (ln A2, right ) ⎤ + kB ( Atotal )ln M ⎣ ⎦ Comparing these expressions for S1 and S2 we can see that the last term, which is the only term involving M, is the same in both cases. So when we take the difference between the entropies (i.e., S1-S2) the term involving M disappears. Bottom line: to compare the entropies we only need to calculate the terms involving the numbers of A molecules. Now we can work out which state has higher entropy. S1 = − k B ⎡ A1, left × (ln A1, left ) + A1, right × (ln A1, right ) ⎤ + kB ( Atotal )ln M ⎣ ⎦ = − kB ⎡ 0.33 × 10 6 × ln(0.33 × 10 6 ) + 0.67 × 10 6 × ln(0.67 × 10 6 ) ⎤ + k B ( Atotal )ln M ⎣ ⎦ = − kB ⎡ 4.193 × 10 6 + 8.988 × 10 6 ⎤ + kB ( Atotal )ln M ⎣ ⎦ = − kB ⎡13.181 × 10 6 ⎤ + kB ( Atotal )ln M ⎣ ⎦ And: S2 = − k B ⎡ A2, left × (ln A2, left ) + A2, right × (ln A2, right ) ⎤ + kB ( Atotal )ln M ⎣ ⎦ = − kB ⎡ 0.32 × 10 6 × ln(0.32 × 10 6 ) + 0.68 × 10 6 × ln(0.68 × 10 6 ) ⎤ + k B ( Atotal )ln M ⎣ ⎦ = − kB [13.188 ] + kB ( Atotal )ln M The difference in entropy is given by: ΔS = S2 − S1 = − k B [(13.188 − 13.181) × 10 6 ] = − k B [0.007 × 10 6 ] Since the answer is negative, it means that State ii is less likely than State i (State ii has lower entropy). Now we can calculate the relative value of their multiplicities: Chem C130/MCB C100A. Fall 2010. Problem Set 7 UC Berkeley Page 6 ⎛W ⎞ ΔS = k B (ln W2 − ln W1 ) = kB ln ⎜ 2 ⎟ ⎝ W1 ⎠ ⇒ ⎛W ⎞ ΔS = ln ⎜ 2 ⎟ = −0.007 × 10 6 = −7000 kB ⎝ W1 ⎠ ⎛W ⎞ ⇒ ⎜ 2 ⎟ = e−7000 ⎝ W1 ⎠ (B) What is the entropy change, in units of J K-1, on going from case (i) to case (ii)? ΔS = − k B × 7000 = 1.38 × 10 −23 × 7000 = 9660 × 10 −23 J K -1 6. Consider a system with just two energy levels, 1 and 2. The system has 1 mole (6.0 x 1023) molecules distributed between these two energy levels, E1 and E2. kB = 1.38 x 10-23 JK-1 E1 = 0.0 J, E2 = 3.0 x 10-21 J, , where p2 is the probability of finding a molecule in level 2. (i) (ii) Calculate the total energy of the system at equilibrium at a temperature of T=300 K (state A) The energy of the system is doubled with respect to the energy in part (i), and the system is allowed to reach equilibrium. Calculate the new temperature of the system at equilibrium. Answer: The answer is that the temperature is negative, which appears to be absurd. This is a consequence of the fact that there are only two energy levels that the energy can go into. As the energy increases the entropy decreases, which corresponds to a “negative” temperature. Although normal systems do not exhibit this behavior, there are some situations, such as magnetic spins, where there are only two energy levels and a “negative” apparent temperature can manifest itself as the energy increases. 7. *Consider the following distributions of molecular energy. (i) Which has higher multiplicity? Chem C130/MCB C100A. Fall 2010. Problem Set 7 UC Berkeley Page 7 (ii) What is the ratio of the multiplicity for the two distributions? (iii) Calculate the difference in the entropy of the two distributions in units of J mol-1 K-1. Answer: Because the number of molecules is large enough, this question is best answered using the probabilistic definition of entropy, rather than trying to calculate factorials. For the first distribution: N total = 10000 + 8000 + 6000 + 1000 = 25000 10000 8000 = 0.4 p1 = = 0.32 25000 25000 6000 1000 p2 = = 0.24 p3 = = 0.04 25000 25000 S = − N total × [ 0.4 ln 0.4 + 0.32 ln 0.32 + 0.24 ln 0.24 + 0.04 ln 0.04 ] kB p0 = = − N total × [ −0.367 − 0.365 − 0.343 − 0.129 ] = N total × 1.204 For the second distribution: N total = 10000 + 7000 + 5000 + 3000 = 25000 10000 7000 = 0.4 p1 = = 0.28 25000 25000 5000 3000 p2 = = 0.20 p3 = = 0.12 25000 25000 S = − N total × [ 0.4 ln 0.4 + 0.28 ln 0.28 + 0.20 ln 0.20 + 0.12 ln 0.12 ] kB p0 = = − N total × [ −0.367 − 0.356 − 0.322 − 0.254 ] = N total × 1.299 Now calculate the ratio of multiplicities: ΔS W = ln( 2 ) = N total (1.299 − 1.204) = 25000 × 0.095 = 2375 kB W1 ⇒ W2 = e2375 W1 Now calculate the entropy difference: Chem C130/MCB C100A. Fall 2010. Problem Set 7 UC Berkeley Page 8 ΔS = 2375 kB ⇒ ΔS = k B × 2375 = 1.38 × 10 −23 × 2375 = 3.28 × 10 −20 J K -1 8.* Consider the following distributions of molecular energy. Which has higher entropy? Answer: Both systems have the same entropy because the numbers of molecules in each energy level are the same (therefore the probabilities and the multiplicity will be the same). 9. Regarding the two distributions in Question 8, we have not specified whether they represent the same system at the same temperature or not. Are these two distributions possible alternate distributions for the same isolated system at the same temperature? Choose the best response from the choices below, and justify your answer. (i) The two distributions are both possible instantaneous snapshots of the system because all energy distributions are equally likely. (ii) Neither distribution is possible, because neither corresponds to a Boltzmann distribution. (iii) If one distribution is a possible distribution for the system at a particular temperature, then the other one cannot be because this would violate the law of conservation of energy. (iv) It is unlikely that upper energy levels would be populated without populating lower energy levels, so the distribution on the right is not possible. Chem C130/MCB C100A. Fall 2010. Problem Set 7 UC Berkeley Page 9 10*. A system consists of 25 molecules distributed among energy levels in the following way. Would you expect to get the same value of the entropy if you used the statistical definition or the probabilistic definition? Explain your answer. 11. For a system of molecules at constant volume and number of molecules we assume that the total entropy is given by the following equation: Chem C130/MCB C100A. Fall 2010. Problem Set 7 UC Berkeley Page 10 Wtotal = W positional × Wenergy Suppose that the system consists of molecules that can form hydrogen bonds with each other. Would this equation be true for such a system? Explain your answer. For the following questions, assume that the value of the Boltzmann constant, kB is 1.0. 12. A system starts with a multiplicity of 2000. 2kJ of heat are transferred into the system reversibly at 298K. What is the multiplicity now? Answer: Initial entropy = kB ln W = 1*ln2000 = 7.6 J/K Change in entropy = 2000J/298K = 6.7 J/K Final entropy = 14.3 J/K Final multiplicity = eS/kB = 1 623 346 13. A system starts with a multiplicity of 1099. Heat is transferred out of the system reversibly at 298K such that the final multiplicity is 105. How much heat was lost? Answer: Change in entropy = kBlnW2- kBlnW1 = 1*ln(105) – 1*ln(1099) = -216.4 J/K qrev = ∆SxT = -216.4J/K*298K = -64 487.2 J = 64.5 kJ was lost from the system 14. A system of non-interacting atoms has a positional multiplicity of 120 and an energetic multiplicity of 5. What is the total multiplicity? What are the positional, energetic and total entropy? Answer: Total multiplicity = positional multiplicity x energetic multiplicity = 120 x 5 = 600 positional entropy = kb ln 5 = 1.6 J/K; energetic = kb ln 120 = 4.8 J/K total entropy = positional + energetic entropy = 1.6+4.8 = 6.4 J/K 15. A system with one mole of an ideal gas changes its volume from 10L to 1L by compression. What is the change in entropy? Answer: ∆S = nR ln (V2/V1) = 1mol x 8.31 J/Kmol x ln (1L/10L) = -19.13 J/K Chem C130/MCB C100A. Fall 2010. Problem Set 7 UC Berkeley Page 11 16. Consider this system with five molecules and three energy levels. The energy levels are such that a molecule at energy level 1 contributes 1J of energy to Utotal, a molecule at energy level 2 contributes 2J of energy to Utotal. Which of state A or state B has more internal energy? State A State B Answer: For State A: Utotal = ∑NiUi = 3xN3+2xN2+1xN1 =3x1+2x3+1x1 = 10 For State B Utotal = ∑NiUi = 3xN3+2xN2+1xN1 =3x0+2x2+1x3 = 7 Therefore, system A has greater energy 17. Consider the system and the two states from question 16. Which state has greater multiplicity? Answer: For State A: W = 5!/1!3!1! = 20 For State B: W = 5!/2!3! = 10 Therefore, system A has greater multiplicity 18. A system has 100 000 molecules at energy level 1, 10 000 molecules at energy level 2, and 1 000 molecules at energy level 3. What is the entropy of the system? (hint: use the probabilistic definition) Answer: Total molecules = Nt =111 000 Probability of level 1 = N1/Nt = 100000/111000 = 0.90 Chem C130/MCB C100A. Fall 2010. Problem Set 7 UC Berkeley Page 12 Probability of level 2 = N2/Nt = 10000/111000 = 0.09 Probability of level 3 = N3/Nt = 1000/111000 = 0.01 Entropy = -Ntkb∑pilnpi = -111 000 J/K x ( 0.90 ln 0.90 + 0.09 ln 0.09 + 0.01 ln 0.01) = -111 000 J/K x (-0.095-0.217-0.046) = -111 kJ/K x (-0.359) = 39.849 kJ/K 19. A system at 293K with 100 000 molecules has two energy levels (A and B). At first, the two energy levels are populated equally. After a reversible process, energy level A is populated by 65% of the molecules. i) What is the change in entropy? ii) iii) How much heat was added or removed from the system? What is the difference in energy between the two levels? Answer: i) Initial entropy = -Ntkb∑pilnpi = -100000J/K x (2x0.5 ln 0.5) = 693 kJ/K Final entropy = -100000J/K x (0.65 ln 0.65+0.35 ln 0.35) = 647 kJ/K The change in entropy (∆S) is -46 kJ/K ii) ∆S = qrev/T ; qrev = ∆SxT = -46kJ/K x 293K = -13478 kJ As a check, we can verify that heat is removed from the system because the entropy of the system decreases during this process. iii) From Boltzmann distribution: − (U 2 −U1 ) P2 0. 5 = = e RT P1 0. 5 ΔU = − RT ln( = +1.53 kJ 0. 5 ) = −2.5 kJ * ln(0.54 ) 0. 5 20. What is the entropy of the following energy distribution? What is the multiplicity? E3 = 200 molecules E2 = 1100 molecules E1 = 10500 molecules Answer: We first compute the entropy using the probabilistic definition Initial entropy = -Ntkb∑pilnpi = -11800 J/K x (0.02 ln 0.02 + 0.09 ln 0.09 + 0.89 ln 0.89) = -11800 J/K x (-0.40) = 1220 J/K Chem C130/MCB C100A. Fall 2010. Problem Set 7 UC Berkeley Page 13 S = -kblnW; W = eS/kb = e 1220 ~ 1060 . 21. A system has a partition function of 1.3 at 293K. At that temperature, an energy level is populated by 10% of the molecules. What is the energy of that level? Answer: e-u/kbT = p x Q; u = -kb x T x ln (p x Q) = -1J/K x 293K x ln (0.1x1.3) = -598J 22. Multiple Choice/TrueFalse 1) For a system converting from state 1 to state 2: (kBlnW2−kBlnW1)= qrev/T True/False 2) If the second lowest energy level is separated from the ground state by 0.5kbT, then the second energy level will not be occupied appreciably. True/False 3) There are many equivalent microstates corresponding to a particular energy distribution True/False 4) Which of the following deNinitions of entropy are equivalent for a large system: a) probabilistic deNinition b) thermodynamic deNinition c) statistical deNinition d) all of the above 5) The probabilistic deNinition of entropy is more accurate than the statistical deNinition for small systems True/False 6) After spontaneous heat transfer between systems, the overall multiplicity is: a) lower than it was before the transfer of heat b) zero c) maximized d) minimized Chem C130/MCB C100A. Fall 2010. Problem Set 7 UC Berkeley Page 14 7) The partition function (Q) is needed to describe: a) the volume of two gases on either side of a partition b) the extent to which many different energy levels are occupied c) how the statistical deNinition of entropy is equivalent to the probabilistic deNinition of entropy d) the relationship between work and pressure Fill in the blank (5) 1) The Boltzmann distribution describes the energy of molecules at __equilibrium_______________. 2) Boltzmann constant is the gas constant (R) divided by __Avogadro’s Number__________ . 3) Kinetic energy is due to the __motion________ of atoms. Potential energy is due to the _conNiguration or relative positions__________ of atoms. 4) The direction of spontaneous change is governed by an increase in ___entropy_________, thus explaining how energy can be transferred “uphill”. 5) If two systems of different temperatures are brought together, there will be a net transfer of heat from the system of __higher________ temperature to the system of __lower_______ temperature ...
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