CEM 05 Solutions for assignment #4 5. A heat engine based on a Carnot cycle does.50 kj of work per cycle and has an efficiency of 45.0%. What are q and q C for one cycle? Since the engine does work on the surroundings it loses that amount of energy so w = -.50 kj. he efficiency is given as w, q so we can solve for q : w q.50kj 0.45 5.6kJ Now we can find q C from the expression for the efficiency: q C q Solve for q C : q q C 5.6kJ0.55.kJ 5. A heat engine running in reverse is called a heat pump (e.g., a refrigerator is a type of heat pump). A heat pump uses an input of work, w cyc, to cause the system to absorb heat, q C, from a cold reservoir at C and emit heat, q, into a hot reservoir at. he efficiency of a heat pump, P, is called the coefficient of performance and is defined by q /w cyc. he efficiency of a refrigerator, R, is called the coefficient of performance and is defined by q C /w cyc. For a reversible Carnot cycle heat pump and refrigerator, express (a) P and (b) R in terms of C and. (c) Show that ε P is always greater than. (d) Suppose a reversible heat pump transfers heat from the outdoors at 7 K to a room at 9 K. For each joule of work input to the heat pump, how much heat will be deposited in the room? (e) What happens to R as C goes to 0 K? (Note that w cyc is the algebraic sum of all the work steps around the cycle. Whereas w cyc < 0 for heat engines, w cyc > 0 for heat pumps and refrigerators.) a. For the Carnot cycle, U q q w cyc C cyc internal energy is a state function 0 so we can write the work in terms of the heat: w q q cyc C Substituting this into the expression for P : q P wcyc q q q C qc q q C C in class we showed that q C C b. Now do the same for R : Copyright 08 by Eric B. Brauns
CEM 05 Solutions for assignment #4 qc R wcyc qc q q C q qc q q substitute from C C q q C C C C C C C C c. Since, by definition, > C, the denominator in ε P from (a) is always positive and less than so that ε P will always be greater than. d. For this part, we want to relate q (the heat deposited into the room) to w cyc (the net work input). his is exactly what the coefficient of performance describes. Equating the expression derived in (a) with the definition of ε P gives q P w cyc C using = 9 K and C = 7 K q w cyc C 4.7w cyc According to this result, if w cyc = + J, then q = -4.7 J. In other words, for each joule of work input, 4.7 J of heat are deposited into the room. e. With R = C /( C ), it s clear that R 0 as C 0. 5. he enthalpy of vaporization of argon at 87. K and atm (its normal boiling point) is 6.5 x 0 J mol -. At this temperature and pressure, what is (a) the entropy of vaporization, Δ vap S and (b) the entropy of condensation, Δ cond S? a. Δ vap S is related to Δ vap according to: vap S vap vap 6.50 Jmol 87.K 74.8JK mol b. Because entropy is a state function and condensation is the opposite of vaporization, S S 74.8JK mol cond vap Copyright 08 by Eric B. Brauns
CEM 05 Solutions for assignment #4 5.4 At atm and between 00 K and 400 K, the constant pressure molar heat capacity of O gas is given by the empirical relationship C P,m = a + b where a = 5.75 J K - mol - and b = 0.0 J K - mol -. What is ΔS when mol of O is heated from 00 K to 400 K at a constant pressure of atm? he entropy change associated with a change in temperature from to is CP,m S N d substitute the given expression for C P,m N a d bd N aln b 400K mol5.75jk mol ln 0.0JK mol 400K00K 00K 7.4 JK 5.5 Determine whether the following are true or false. If a statement is false, explain why or cite a counter example. (a) For a closed system, ΔS sys can never be negative. (b) For a reversible process in a closed system, ΔS sys must be zero. (c) For a reversible process in a closed system, ΔS tot must be zero. (d) For an adiabatic process in a closed system, ΔS sys cannot be negative. (e) For a process in an isolated system, ΔS tot cannot be negative. (f) For an adiabatic process in a closed system, ΔS sys must be zero. (g) An adiabatic process cannot decrease the entropy of a closed system. (h) For a closed system, equilibrium has been reached when S sys has been maximized. a. F. Cooling a gas that is held in an impermeable container would make ΔS sys < 0. b. F. Reversible isothermal expansion of an ideal gas corresponds to ΔS sys > 0. c. d. e. f. his is true if the process is reversible and false if it is irreversible. g. h. F. Entropy is maximized at equilibrium for an isolated system (not a closed system). In a closed system, equilibrium is a balance between maximizing the entropy and minimizing the energy. 5.6 For each of the following processes deduce whether each of the quantities ΔS sys and ΔS tot is positive, zero, or negative. (a) Reversible melting of solid benzene at atm and the normal melting temperature. (b) Reversible melting of ice at atm and 7 K. (c) Reversible adiabatic expansion of an ideal gas. (d) Reversible isothermal expansion of an ideal gas. (e) Adiabatic expansion of an ideal gas into a vacuum. (f) Joule-homson expansion of an ideal gas. (g) Reversible heating of an ideal gas at constant pressure. (h) Reversible cooling of an ideal gas at constant volume. (i) Combustion of benzene in a sealed container with rigid, adiabatic walls. (j) Adiabatic expansion of an nonideal gas into vacuum. he answers are on the next page. Keep the following in mind: q q rev irr 0if reversible ds, ds, S S S rev irr tot sys sur, 0 if irreversible 0 if reversible and for adiabatic processes Ssys 0 if irreversible Copyright 08 by Eric B. Brauns
CEM 05 Solutions for assignment #4 a. ΔS sys > 0 and ΔS tot = 0 b. ΔS sys > 0 and ΔS tot = 0 c. ΔS sys = 0 and ΔS tot = 0 d. ΔS sys > 0 and ΔS tot = 0 e. ΔS sys > 0 and ΔS tot > 0 f. ΔS sys > 0 and ΔS tot > 0 (Comment: Joule-homson expansion is adiabatic and irreversible) g. ΔS sys > 0 and ΔS tot = 0 h. ΔS sys < 0 and ΔS tot = 0 i. ΔS sys > 0 and ΔS tot > 0 (Comment: combustion is irreversible) j. ΔS sys > 0 and ΔS tot > 0 5.7 (a) What is ΔS for each step of a Carnot cycle? (b) What is ΔS tot for each step of a Carnot cycle? a. For both adiabatic steps, ΔS = 0. For the isothermal compression at : q V f Pi S NRln NRln V P i f For the isothermal expansion at C : q V C f Pi S NRln NRln V P C i f b. All the steps of a Carnot cycle are reversible, so ΔS tot = 0 for each. 5.8 One mol of an ideal gas is taken through the cycle shown (this is a diesel cycle). Step is an adiabatic expansion to a volume of 0 L from a pressure of 5 atm and an initial volume of 0 L. During step, the gas cools by releasing kj of heat. he gas then undergoes an adiabatic compression back to the original pressure of 5 atm during step 4. he final step, 4, is an expansion to the original volume of 0 L against a constant external pressure of 5 atm. For each step and the overall cycle, calculate ΔS for the system, surroundings, and total. Sketch what this cycle would look like if vs. S were plotted instead of P vs. V. For this problem, assume that C P,m = 5R/ and that all the steps are reversible. All the steps are reversible so ΔS tot = 0 and ΔS sur = -ΔS sys. All we have to do is calculate ΔS sys for each step. : For a reversible adiabatic expansion ΔS sys = 0 and ΔS sur = 0. : For a constant volume decrease in temperature from to we use S NC ln sys V,m We need C V,m,, and. We are given C P,m, so C V,m is just C P,m R = R/. o find and, we will need. he gas is ideal, so this is just: PV NR 5atm0L mol8.06 0 L atmk mol 609.K Copyright 08 by Eric B. Brauns 4
CEM 05 Solutions for assignment #4 Since the process from to is adiabatic, we can now find using V V 5 0L 609.K 0L 8.85K o find, we need to be a little more creative. We re told that going from to, the system releases 000 J of heat. Since this is a constant volume process, the heat is equal to the change in internal energy which can be calculated from C V,m,, and : q U NC 000J note that q is negative because this is the heat released by the system V,m Solve for : NC q V,m NCV,m mol 8.4JK mol 8.84K000J mol 8.4JK mol 0.67K Now we have what we need to calculate the entropy change for the system: 0.67K S mol 8.4JK mol sys ln 8.85K.9JK 4: his is a reversible adiabatic process (compression in this case) so ΔS sys = 0. 4 : he gas expands reversibly against a constant external pressure of 5 atm and its temperature drops from 4 to. he entropy change for the system is S NC ln sys P,m 4 owever, we first need to determine 4 using V 4 4 V which requires V 4. We can relate the pressures and volumes of states and 4 using: P V V 4 P 4 Substituting this into the formula for 4 gives: P 4 P 4 where P is easily calculated using the ideal gas equation: NR P V mol8.06 0 L atmk mol 0.67K 0L.5atm Now we can calculate 4 :.5 atm 0.66K 4 5atm 59.4K 5 Copyright 08 by Eric B. Brauns 5
CEM 05 Solutions for assignment #4 Finally, we can calculate the entropy change for the system: 5 609.K S mol 8.4JK mol sys ln 59.4K.9JK As a final check, we can note that the sum of the entropy changes for the system around the cycle is zero which it should be since the entropy is a state function. he second part of the question asks us to sketch the Diesel cycle on a vs. S plot: 5.9 he constant pressure heat capacity for some solid substance at 0 K is 0.6 J K - mol -. (a) Calculate its third law entropy, S, at this temperature, (b) then find C P,m and S at = 6 K. a. A 0 K, the third law entropy is given by 0K C P,m S S d 0 0 0 0K CP,m d since S 0 0 0 o integrate, we need to know how the heat capacity varies with temperature between 0 K and 0 K. Because the temperature is so low, we can use the Debye approximation, C P,m = a, and use the given heat capacity at 0 K to determine the constant, a: C a P,m solve for a C P,m 0.6JK mol 4-4 - a 6.0 JK mol 0K he entropy at 0 K can now be calculated: 0K S a 0 d 0 0K a a 0K 0K 4 4 6. 0 JK mol 0K 0.07 JK mol b. At = 6 K, we use the same expression in (a), but with 6 K as the upper integration limit: S a6k 6 4 4 6. 0 JK mol 6K 0.045JK mol Copyright 08 by Eric B. Brauns 6
CEM 05 Solutions for assignment #4 5.0 Consider the reaction CO(g) + O (g) CO (g). For temperatures between 7 K and 000 K, the heat capacity for each species is described by the empirical formula C P,m () = a + b + c + d where a, b, c, and d are empirically determined constants summarized in the following table: 4 a JK mol b JK mol c JK mol d JK mol O g 5.67.0.76 0 7. 0 CO g 8.74.79 0.05 0 4.9 0 CO g.64 6.6 0 4.06 0 9.70 0 6 5 9 5 9 (a) Evaluate Δ r S 98 for this reaction. he third law entropy, S 98, for O (g), CO(g), and CO (g) are 05.4 J K - mol -, 97.67 J K - mol -, and.74 J K - mol -, respectively. (b) Using the given expression for C P,m (), evaluate Δ r S 000. (c) Calculate ΔC P,m (98 K) using the given data, then use that to reevaluate Δ r S 000 assuming that the heat capacities are temperature independent. a. he reaction entropy at 98 K is S S CO,g S CO,g S O,g r 98 98 98 98.74JK mol 97.67 JK mol 05.4JK mol 7.00JK mol b. he reaction entropy at = 000 K is related to the reaction entropy at = 98 K by C P,m S S d where C C CO C CO C O P,m P,m P,m P,m use the given formula for each C P,m a b c d CO a b c d CO C a b c d P O a b c d a b c d a b c d CO CO CO CO CO CO CO CO O O O O group the terms with like powers of a a a b b b CO CO O CO CO O c c c d d d CO CO O CO CO O a b c d for convenience, use Δa, Δb, Δc, and Δd for the quantities in parentheses a b c d Substitute this into the expression for the reaction entropy at 000 K: S S a d b d c d d d S aln b c d With = 98 K, = 000 K, Δ r S 98K from (a), and Δa, Δb, Δc, and Δd calculated from the data in the table: 000K S 7.00 JK mol 9.87 JK mol ln 0.75JK mol 000K 98K 000 98K 5 9.844 0 JK mol 000K 98K 8 4.805 0 JK mol 000K 98K 74.54JK mol c. Use the expression from (b) for ΔC P,m () with = 98 K: Copyright 08 by Eric B. Brauns 7
CEM 05 Solutions for assignment #4 C a b c d P,m 5 9.87 JK mol 0.75JK mol 98K 9.844 0 JK mol 98K 8 4.86JK mol If the heat capacities are independent of temperature, the reaction entropy at is related to the entropy at through: S S C d P,m S C ln P,m Using ΔC P,m () we just calculated, with = 000 K and = 98 K: 000K S 7.00JK mol.86jk mol ln 000 98K 88.57 JK mol.805 0 JK mol 98K Copyright 08 by Eric B. Brauns 8