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Chem 452B – HW 6 (A) Fall 2012 Homework 6A KEY Q1) For the following processes state whether the listed quantities are greater, less than or equal to zero. Indicate your reasoning in a brief sentence and state any assumptions you made. Unless stated otherwise, assume all gases to be ideal. a) A gas sample goes through a Carnot cycle: T, q, w, U, H, S, G and A. By going through a cycle the system is back to its original state, so all state functions must be zero, so this leaves only q and w to change, and q=-w is required. If the system did work then q is positive. Heat flowed through and was used to produce work. b) A sample of hot water is mixed with a sample of cold water in an insulated container of fixed volume: q, w, U, H and S. This is adiabatic but irreversible, so q=0, DU=0=DH, as the internal energy is conserved, and DS>0 because it goes to equilibrium spontaneously by having all terms have the same final temperature. Presumably no work, w=0. c) An ideal gas expands adiabatically and reversibly: VT, q, w, U, H and S Adiabatic reversible change is DS=0 (by definition) and q=q_rev=0. Expansion implies the system does work so w<0, and DV>0 (expands), DT<0, DU<0, DH<0, internal energy is consumed. Q2) We have examined cases where the entropy is always positive for irreversible adiabatic transformations. Now we want to show there are several ways to compute the entropy because it is path independent. Consider the adiabatic, irreversible transformation of one mole of an ideal monatomic gas starting at P1 ,V1 , T1 ends up compressed in an irreversible one step process against a constant external pressure, which is Pext P2 2P1 The problem begins by equating the two parts of the first laws: CV dT dU q w w P2 V2 V1 T2 P2 3 T2 1 2 T1 T1 P1 T2 7 V2 7 and T1 5 V1 10 1A) Find the entropy change of the system for this process. There are at least two approaches to solving this problem, which is a challenging one. The first approach is to generate a path that helps one solve for the entropy change, which is defined for the reversible path. So we have the starting and ending points of the process. From the Carnot cycle we learned that we can connect any two states for an I.G. by an isothermal reversible process followed by an adiabatic reversible process. We must find the point, state 3, that connects 1 and 2 by these two steps. 5 For the isothermal step: PV . 3 3 PV 1 1 , and for the adiabatic process P3 V3 P2 V2 3 V These give 2 equations and 2 unknowns from which you can get 3 and all other terms. Find V2 the ratio’s for state 3 relative to either 1 or 2. 1 Chem 452B – HW 6 (A) Fall 2012 P3 V3 P2 V2 P3 V3 P2 V2 P1 V1 P1 V1 PV 3 3 1 PV 1 1 V3 V1 1 5 3 53 P2 V2 2 0.7 P1 V1 3 53 2 V3 2 0.7 1.16 V1 1B) The entropy change is the sum of the two terms. The entropy change over the reversible adiabatic part is zero, so the total entropy is found from the reversible isothermal change from T1 to T3. (I got S nR 0.148 ; see if you get something similar). It is rather striking that very little about the initial and final states are specified and yet we get a numerical value for the entropy. V3 V w PdV S nR ln 3 nR 0.148 T V1 T V1 1C) There is an alternative path (among many) which is also possible. In this path the temperature is raised from T1 to T2 at constant volume (held at V1). Then in the second step, with the temperature held at T2 the volume is changed (i.e. decreased) from V1 to V2. This is just one of several paths one might imagine and does not require finding an intermediate point on the path. This works because the entropy is path independent. The answer lies in knowing the derivatives and integrating to find the change. For example for the first part of this change: V2 T2 S S . And for the second part of the path: SV dT S V1 T dV T2 . The total T V T V T1 V1 entropy is the sum of these two and should be the same as part 1B). Using the definition of these partial derivatives (as applied to the I.G.) solve for this entropy change and compare with 1B. I find this method much easier because I don’t have to think hard about the path, and I checked and got the same answer as above. Pretty slick. T2 7 T1 5 and V2 7 V1 10 S SV ST S S dT T V T1 S dV V T V1 V2 T2 V1 S nR 0.148 2 T2 3 T V nR ln 2 ln 2 V1 2 T1 Chem 452B – HW 6 (A) Fall 2012 2) Find the entropy change of the environment and the total entropy change for this process. Because it is adiabatic (albeit and 1 step) q=0, so the environment DSenv=0, Therefore: STot SSys nR 0.148 0 3) Verify the sign of the change of the total entropy and comment on the relation to the fundamental inequality of the second law. The total entropy change is positive indicating that this is a spontaneous process as written. Q3) We discussed the difference between a rubber band (as an entropic spring) and a steel spring (as an enthalpic spring). a) Qualitatively describe why heat flows out of an enthalpic spring upon contraction but flows into an entropic spring upon contraction. (Check your rubber band again to be sure of this.) Enthalpic spring: As the spring is stretched the potential energy is stored into the spring as internal (potential) energy. So when the spring is allowed to spring back that energy must come out. If the spring is not doing work, then the energy comes out as heat, the contraction is exothermic; negative q, heat out. Entropic spring: When the rubber band is stretched the potential energy is stored in the rubber band in entropic order. It is a more ordered system, the internal energy does not change. Because the entropy goes down on stretching, the heat must be negative, so stretching is exothermic. Now when the rubber band can relax it can still do work but the order goes to disorder and so the entropy goes up, so the reversible heat must be positive, even done with no work then maximal heat is absorbed because the rubber band is more disordered. b) We rewrote dU in terms of the length of the rubber band and the temperature in the room: dU TdS Fd . In this form, the force (the negative of the tension) is F k . And we noted that the Thermodynamic equation of state (T.EoS) should be found by replacing P by –F: U F F T T T There are two possibilities for using this expression. A) Use the T.EoS to verify that if the spring constant is independent of the temperature one recovers the usual rule for a Hook’s Law spring. Explain and show how the internal energy of the spring, U, depends on the length in this case. Explain this case in terms of the potential energy. F k T T 0 T T U k F T F k T T This shows that the internal energy increases as the spring is stretched. 3 Chem 452B – HW 6 (A) Fall 2012 B) Assume that the spring constant depends on temperature: k T where is independent of temperature. Find the dependence of the internal energy, U, for this case. Compare your answer to that of an ideal gas for an isothermal process. F T T T T F T T U k F T FF 0 T T This shows that the internal energy of a rubber band does not depend on the extension of the rubber band. C) We did not talk about what happens when we heat the rubber band when it has a fixed mass attached. Will the rubber band contract or expand on heating? Use the cyclic rule to bolster your thinking. In this case you must relate this change to the change in the force with changing temperature (at fixed length) and how the length changes with changing force (at fixed temperature). F T 0 F T T T F T This shows that the length must get shorter as the rubber band is warmed. Again, counterintuitive, but just another way to represent the fact that the spring constant increases with temperature. Optional Problem Q4) Consider a system undergoing a phase transition from phase to phase . The phase transition occurs at temperature T0 and pressure P0 . The heat absorbed per mole of the substance undergoing the phase transition is q 0 and there is a molar volume change of V 0 . The molar heat capacities (independent of temperature) for the phases and are C p , and C p , respectively. I will be assigning this problem on hw6B, so I have started it here and will finish it on the next assignment. a) Find expressions for w, U , H , S and G for converting 1 mole of phase to phase at T0 and P0 . 4 Chem 452B – HW 6 (A) Fall 2012 w Po V 0 U q0 w, H q0 q S = H To G=0 The rest to be continued. See if you can make progress on it before I post my answers. b) Find expressions for H and S for converting 1 mole of phase to phase at temperature, T * different from T0 , but at the same pressure, P0 . c) The denaturation (unfolding) transition of a globular protein can be approximated as a phase change of to Calculate, H , S and G for q0 =638 kJ mole-1 , T0 =70 C, P0=1atm and C p, -C p, =8.37 kJ mole-1 K 1 . d) Calculate, H , S and G for the same transition at 37 °C and 1 atm. Assume that C p , -C p , is independent of temperature. 5