1 Exam Thermodynamics 12 April 2018 Please, hand in your answers to problems 1, 2, 3 and 4 on separate sheets. Put your name and student number on each sheet. The examination time is 12:30 until 15:30. There are 4 problems, with each 4 subproblems, a list of constants and a formulae sheet, 5 pages in total. All 16 subproblems have equal weight for the final grade. Problem 1 a) Give a definition and/or short description of the following concepts perfect gas isobaric process activity thermodynamic equilibrium constant electromotive force b) Indicate for each of the following quantities whether we are dealing with a state function; answer using only yes or no: the pressure P, the difference of internal energy and work U W, the product of temperature and entropy T S, the ratio of heat and temperature Q T, the product of volume, pressure and temperature V P T. c) Give the meaning of all symbols in the following formula as well as a description of its use in thermodynamic problems. q = i exp ɛ i d) Consider an isobaric process for which 2 mol of a perfect gas is heated from t 1 = 50 C to t 2 = 100 C. Use the Boltzmann definition of entropy to determine the ratio, W (t2) W (t 1), for the number of micro states of the gas. Do not forget to fill in the online survey to the course.
2 Problem 2 Consider a system consisting of 1 mol of lead, Pb, at standard pressure. In the table below a number of thermodynamic parameters are given. T fus fus H T vap vap H S (298 K) Pb 600.6 K 4.774 kj/mol 2013 K 177.82 kj/mol 64.78 J/mol K The heat capacity for all phases is given by (T is the temperature in K) C P (J/mol K) = A + BT + CT 2, in which the parameters A, B and C, for temperatures 10 K T 6000 K, in the three phases s(olid), l(iquid) and g(as) are given in the table below Pb A(J/molK) B(J/molK 2 ) C(J/molK 3 ) s (T > 10 K) 25.01 5.442 10 3 4.061 10 6 l 38.00 14.62 10 3 7.255 10 6 g (T < 6000 K) 85.66 69.282 10 3 13.318 10 6 a) At very high temperature the system behaves like a perfect gas. What are the values of A, B and C in that case? b) Which of the following quantities have the same value for both phases (melt and solid) at constant pressure, at T = T fus, where the melt and the solid are in equilibrium? Answer with only yes or no. the entropy, the enthalpy, the Helmholtz free energy, the Gibbs free energy, the chemical potential. c) Calculate the entropy change vap S at the evaporation temperature T vap. d) Calculate the entropy of this system at 1000 K; you can neglect the contributions for T < 10 K. Problem 3 A Daniell cell is an electrochemical cell consisting of a half cell with a copper electrode in a copper(ii) solution and a half cell with a zinc electrode in a zinc solution. We consider a Daniell cell at P = P, with initially a 0.200 M Cu(NO 3 ) 2 aqueous solution and a 0.010 M Zn(NO 3 ) 2 aqueous solution. We connect an external (load) resistance, R L = 100 Ω, to this cell and keep the cell at constant temperature, T = 298 K. The cell voltage will decrease until the cell is in equilibrium. During this discharging process of the cell, the NO 3 -ions will migrate through the semipermeable membrane separating the two half cells to keep the net charge in the half cells equal to zero. The electrodes can be considered as large enough to never become depleted. The internal resistance of the cell is R int = 20.0 Ω and can be considered to represent all internal losses in the cell during discharge; R int can be considered as constant. The following data can be used E Cu 2+ /Cu = +0.34 V at T = 298 K, = 0.76 V at T = 298 K. E Zn 2+ /Zn fus H(H 2 O) = 6.008 kj/mol at T = 273.15 K. Assume that all activities can be approximated by the molarities.
3 a) Give the chemical equation for the net cell reaction and determine the cell voltage at time zero, E(t = 0), the moment just before the load resistance is attached. b) Calculate the concentrations in the two half cells at the end of the discharge process. c) Calculate the efficiency of the process. d) Next, we want to use the cell during winter time. Assuming that all paramaters are independent of the temperature, estimate the lowest temperature, which allows the cell to be used without the solution of one of the half cells to solidify; Do this both for t = 0 and the final equilibrium state. Problem 4 The process of protein unfolding generally involves a strong temperature dependence of the thermodynamic parameters. We study the equilibrium between the folded, N, and unfolded, D, at P = P, of λ-repressor, a protein from λ phage that binds to DNA and regulates transcription. Some of the thermodynamic parameters for the unfolding of λ-repressor are G 17.7 kj/mol at T = 298 K, H 90.4 kj/mol at T = 298 K. In Figure 1 the fraction, f D, of D(enatured) (unfolded) protein is plotted as a function of temperature for λ-repressor. In this problem we approximate activitites by concentrations, unless mentioned otherwise and we suppress the label r by writing r G as G, etc. f D 1.0 0.8 0.6 0.4 0.2 0.0 0 20 40 60 80 T ( o C) Figuur 1: Fraction, f D, of denatured protein vs temperature for λ-repressor. a) Show that the fraction, f D, of denatured (unfolded) protein can be written in terms of the thermodynamic equilibrium constant, K, of the equilibrium, according to f D = [D] [D] + [N] = K 1 + K. b) Estimate the so-called melting temperature, T m, of the protein, for which half of the proteins is unfolded using the thermodynamic data at T = 298 K and compare your value with the graph in Figure 1. The reason for the deviation in temperature is due to the strong temperature dependence of the thermodynamic parameters, mentioned above. In Figure 2 this temperature dependence for λ- repressor is presented.
4 The following values were found for these plots G max = 4.8 kcal/mol at T = 288 K, H = -88 kcal/mol at T = 233 K, H = 100 kcal/mol at T = 353 K, T S = -80 kcal/mol at T = 233 K, T S = 100 kcal/mol at T = 353 K. G (kcal/mol) 5 2.5 0 2.5 5 40 20 80 ( C) T o (kcal/mol) 100 50 0 50 100 40 H T S 20 80 ( C) T o Figuur 2: Temperature dependence of G, and H and S for λ-repressor. c) Use the data to Figure 2 to determine the temperature dependence of the difference in heat capacity C P = C D,P C N,P between the unfolded and folded state. Assume that the relevant plot can be considered as a linear function. d) Calculate the ratio of activity coefficients γ D γn on the concentration scale in the most stable state of the folded protein. In this most stable state f D = 1.000 10 3. List of constants Elementary charge e 1.602 10 19 C Faraday s constant F 9.648 10 4 Cmol 1 Boltzmann s constant k 1.381 10 23 JK 1 Planck s constant h 6.626 10 34 Js Bohr Magneton µ B 9.274 10 24 JT 1 Atomic mass constant m u 1.661 10 27 kg Amadeo Avogadro di Quaregna e Ceretto s constant N A 6.022 10 23 mol 1 Gas constant R 8.314 JK 1 mol 1 Free fall acceleration g 9.807 ms 2 Unit of energy 1 cal = 4.184 J Standard pressure P 1 bar = 10 5 Nm 2 = 0.9869 atm = 750 Torr
5 Formulae P V = nrt = N U = 3 2 nrt = 3 2 N U = W + Q dw = P ext. dv + dw and dw max = (dg) P,T dq P = C P dt and dq V = C V dt Q 1 Q 2 = T 1 T 2 ds = dqrev dq T T ds tot = ds + ds omg 0 du = P dv + T ds + i H = U + P V dh = V dp + T ds + i A = U T S da = P dv SdT + i G = H T S dg = V dp SdT + i r G = ( ) G = r G + RT ln Q, where Q = ξ P,T i RT ln K = r G E = E RT νf ln Q, and dw = Edq and E = IR and P = EI µ i = µ i + RT ln a i = µ i + RT ln P i P a νi i G P,T = i µ i n i n j dµ j = 0 j ( RT 2 T = trs H ) x B S = nr (x A ln x A + x B ln x B ) Π = [B]RT = n B V RT S = k ln W n i N = exp ɛi q, where q = i exp ɛ i and < X >= N < x >= N i x i n i N