thegoodfellas/blogset-br huggingface dataset:
expected of a reversible process. 10. We concentrate on the first term of Eq. 20-4 (the second term is zero because the final and initial temperatures are the same, and because ln(1) = 0). Thus, the entropy change is DS = nR ln(Vf /Vi) . Noting that DS = 0 at Vf = 0.40 m3, we are able to deduce that Vi = 0.40 m3. We now examine the point in the graph where DS = 32 J/K and Vf = 1.2 m3; the above expression can now be used to solve for the number of moles. We obtain n = 3.5 mol. 11. (a) We refer to the copper block as block 1 and the lead block as block 2. The equilibrium temperature Tf satisfies m1c1(Tf - Ti,1) + m2c2(Tf - Ti2) = 0, which we solve for Tf : (b) Since the two-block system in thermally insulated from the environment, the change in internal energy of the system is zero. (c) The change in entropy is 12. We use Eq. 20-1: 13. The connection between molar heat capacity and the degrees of freedom of a diatomic gas is given by setting f = 5 in Eq. 19-51. Thus, , and . In addition to various equations from Chapter 19, we also make use of Eq. 20-4 of this chapter. We note that we are asked to use the ideal gas constant as R and not plug in its numerical value. We also recall that isothermal means constant-temperature, so T2 = T1 for the 1 2 process. The statement (at the end of the problem) regarding per mole may be taken to mean that n may be set identically equal to 1 wherever it appears. (a) The gas law in ratio form (see Sample Problem 19-1) is used to obtain . (b) The adiabatic relations Eq. 19-54 and Eq. 19-56 lead to . (c) Similarly, we find process 1 2 (d) The work is given by Eq. 19-14: W = nRT1 ln (V2/V1) = RT1 ln3 =1.10RT1. Thus, W/ nRT1= ln3 = 1.10. (e) The internal energy change is DEint = 0 since this is an ideal gas process without a temperature change (see Eq. 19-45). Thus, the energy absorbed as heat is given by the first law of thermodynamics: Q = DEint + W 1.10RT1, or Q/ nRT1= ln3 = 1.10. (f) DEint = 0 or DEint / nRT1=0 (g) The entropy change is DS = Q/T1 = 1.10R, or DS/R = 1.10. process 2 3 (h) The work is zero since there is no volume change. Therefore, W/nRT1 = 0 (i) The internal energy change is This ratio (-0.889) is also the value for Q/nRT1 (by either the first law of thermodynamics or by the definition of CV). (j) DEint /nRT1= -0.889. (k) For the entropy change, we obtain process 3 1 (l) By definition, Q = 0 in an adiabatic process, which also implies an absence of entropy change (taking this to be a reversible process). The internal change must be the negative of the value obtained for
it in the previous process (since all the internal energy changes must add up to zero, for an entire cycle, and its change is zero for process 1 2), so DEint = +0.889RT1. By the first law of thermodynamics, then, W = Q - DEint = -0.889RT1, or W /nRT1= -0.889. (m) Q = 0 in an adiabatic process. (n) DEint /nRT1= +0.889. (o) DS/nR=0. 14. (a) It is possible to motivate, starting from Eq. 20-3, the notion that heat may be found from the integral (or area under the curve) of a curve in a TS diagram, such as this one. Either from calculus, or from geometry (area of a trapezoid), it is straightforward to find the result for a straight-line path in the TS diagram: which could, in fact, be directly motivated from Eq. 20-3 (but it is important to bear in mind that this is rigorously true only for a process which forms a straight line in a graph that plots T versus S). This leads to Q = (300 K) (15 J/K) = 4.5103 J for the energy absorbed as heat by the gas. (b) Using Table 19-3 and Eq. 19-45, we find (c) By the first law of thermodynamics, 15. The ice warms to 0C, then melts, and the resulting water warms to the temperature of the lake water, which is 15C. As the ice warms, the energy it receives as heat when the temperature changes by dT is dQ = mcI dT, where m is the mass of the ice and cI is the specific heat of ice. If Ti (= 263 K) is the initial temperature and Tf (= 273 K) is the final temperature, then the change in its entropy is Melting is an isothermal process. The energy leaving the ice as heat is mLF, where LF is the heat of fusion for ice. Thus, DS = Q/T = mLF/T = (0.010 kg)(333 103 J/kg)/(273 K) = 12.20 J/K. For the warming of the water from the melted ice, the change in entropy is where cw is the specific heat of water (4190 J/kg K). Thus, The total change in entropy for the ice and the water it becomes is Since the temperature of the lake does not change significantly when the ice melts, the change in its entropy is DS = Q/T, where Q is the energy it receives as heat (the negative of the energy it supplies the ice) and T is its temperature. When the ice warms to 0C, When the ice melts, When the water from the ice warms, The total energy leaving the lake water is Q = -222 J - 3.33 103 J - 6.29 102 J = -4.18 103 J. The change in entropy is The change in the entropy of the ice-lake system is DS = (15.27 - 14.51) J/K = 0.76 J/K. 16. (a) Work is done only for the ab portion of the process. This portion is at constant pressure, so the work done by the gas is (b) We use the first law:
DEint = Q - W. Since the process is at constant volume, the work done by the gas is zero and Eint = Q. The energy Q absorbed by the gas as heat is Q = nCV DT, where CV is the molar specific heat at constant volume and DT is the change in temperature. Since the gas is a monatomic ideal gas, . Use the ideal gas law to find that the initial temperature is and that the final temperature is . Thus, The change in the internal energy is DEint = 6p0V0 or DEint/p0V0=6.00. Since n = 1 mol, this can also be written Q = 6.00RT0. (c) For a complete cycle, DEint = 0 (d) Since the process is at constant volume, use dQ = nCV dT to obtain Substituting and using the ideal gas law, we write Thus, . Since n = 1, this is . (e) For a complete cycle, DEint = 0 and DS = 0. 17. (a) The final mass of ice is (1773 g + 227 g)/2 = 1000 g. This means 773 g of water froze. Energy in the form of heat left the system in the amount mLF, where m is the mass of the water that froze and LF is the heat of fusion of water. The process is isothermal, so the change in entropy is DS = Q/T = mLF/T = (0.773 kg)(333 103 J/kg)/(273 K) = -943 J/K. (b) Now, 773 g of ice is melted. The change in entropy is (c) Yes, they are consistent with the second law of thermodynamics. Over the entire cycle, the change in entropy of the water-ice system is zero even though part of the cycle is irreversible. However, the system is not closed. To consider a closed system, we must include whatever exchanges energy with the ice and water. Suppose it is a constant-temperature heat reservoir during the freezing portion of the cycle and a Bunsen burner during the melting portion. During freezing the entropy of the reservoir increases by 943 J/K. As far as the reservoir-water-ice system is concerned, the process is adiabatic and reversible, so its total entropy does not change. The melting process is irreversible, so the total entropy of the burner-water-ice system increases. The entropy of the burner either increases or else decreases by less than 943 J/K. 18. In coming to equilibrium, the heat lost by the 100 cm3 of liquid water (of mass mw = 100 g and specific heat capacity cw = 4190 J/kgK) is absorbed by the ice (of mass mi which melts and reaches Tf > 0 C). We begin by finding the equilibrium temperature: which yields, after using LF = 333000 J/kg and values cited in the problem, Tf = 12.24 which is equivalent to Tf = 285.39 K. Sample Problem 19-2 shows that for processes where DT = T2 T1, and Eq. 20-2 gives for the phase change experienced by the ice (with To = 273.15 K). The total entropy change is (with T in Kelvins) 19. We consider a three-step
reversible process as follows: the supercooled water drop (of mass m) starts at state 1 (T1 = 268 K), moves on to state 2 (still in liquid form but at T2 = 273 K), freezes to state 3 (T3 = T2), and then cools down to state 4 (in solid form, with T4 = T1). The change in entropy for each of the stages is given as follows: DS12 = mcw ln (T2/T1), DS23 = -mLF/T2, DS34 = mcI ln (T4/T3) = mcI ln (T1/T2) = -mcI ln (T2/T1). Thus the net entropy change for the water drop is 20. (a) We denote the mass of the ice (which turns to water and warms to Tf) as m and the mass of original-water (which cools from 80o down to Tf) as m. From SQ = 0 we have LF m + cm (Tf 0o) + cm (Tf 80o) = 0 . Since LF = 333 103 J/kg, c = 4190 J/(kgCo), m = 0.13 kg and m = 0.012 kg, we find Tf = 66.5oC, which is equivalent to 339.67 K. (b) Using Eq. 20-2, the process of ice at 0o C turning to water at 0o C involves an entropy change of = = 14.6 J/K . (c) Using Eq. 20-1, the process of m = 0.012 kg of water warming from 0o C to 66.5o C involves an entropy change of . (d) Similarly, the cooling of the original-water involves an entropy change of (e) The net entropy change in this calorimetry experiment is found by summing the previous results; we find (by using more precise values than those shown above) Snet = 4.39 J/K. 21. We note that the connection between molar heat capacity and the degrees of freedom of a monatomic gas is given by setting f = 3 in Eq. 19-51. Thus, , and . (a) Since this is an ideal gas, Eq. 19-45 holds, which implies DEint = 0 for this process. Eq. 19-14 also applies, so that by the first law of thermodynamics, Q = 0 + W = nRT1 ln V2/V1 = p1V1 ln 2 Q/p1V1= ln2 = 0.693. (b) The gas law in ratio form (see Sample Problem 19-1) implies that the pressure decreased by a factor of 2 during the isothermal expansion process to V2=2.00V1, so that it needs to increase by a factor of 4 in this step in order to reach a final pressure of p2=2.00p1. That same ratio form now applied to this constant-volume process, yielding 4.00 = T2T1 which is used in the following: or . (c) The work done during the isothermal expansion process may be obtained by using Eq. 19-14: W = nRT1 ln V2/V1= p1V1 ln 2.00 W/p1V1= ln2 = 0.693. (d) In step 2 where the volume is kept constant, W = 0. (e) The change in internal energy can be calculated by combining the above results and applying the first law of thermodynamics: or DEint/p1V1 = 9/2 = 4.50. (f) The change in entropy may be computed by using Eq. 20-4: The second
approach consists of an isothermal (constant T) process in which the volume halves, followed by an isobaric (constant p) process. (g) Here the gas law applied to the first (isothermal) step leads to a volume half as big as the original. Since , the reasoning used above leads to Q = p1V1 ln 2.00 (h) To obtain a final volume twice as big as the original, in this step we need to increase the volume by a factor of 4.00. Now, the gas law applied to this isobaric portion leads to a temperature ratio T2/T1 = 4.00. Thus, or Q/p1V1 = 15/2 = 7.50. (i) During the isothermal compression process, Eq. 19-14 gives W = nRT1 ln V2/V1= p1V1 ln (-1/2.00) = -p1V1 ln 2.00 W/p1V1= -ln2 = -0.693. (j) The initial value of the volume, for this part of the process, is , and the final volume is Vf = 2V1. The pressure maintained during this process is p = 2.00p1. The work is given by Eq. 19-16: (k) Using the first law of thermodynamics, the change in internal energy is or DEint/p1V1 = 9/2 = 4.50. The result is the same as that obtained in part (e). (l) Similarly, the same as that obtained in part (f). 22. (a) The final pressure is (b) We use the ratio form of the gas law (see Sample Problem 19-1) to find the final temperature of the gas: For later purposes, we note that this result can be written exactly as Tf = Ti (2e1). In our solution, we are avoiding using the one mole datum since it is not clear how precise it is. (c) The work done by the gas is (d) Consideration of a two-stage process, as suggested in the hint, brings us simply to Eq. 20-4. Consequently, with (see Eq. 19-43), we find 23. We solve (b) first. (b) For a Carnot engine, the efficiency is related to the reservoir temperatures by Eq. 20-13. Therefore, which is equivalent to 68C. (a) The temperature of the cold reservoir is TL = TH 75 = 341 K 75 K = 266 K. 24. Eq. 20-13 leads to quoting more figures than are significant. As a percentage, this is e = 99.99995%. 25. (a) The efficiency is We note that a temperature difference has the same value on the Kelvin and Celsius scales. Since the temperatures in the equation must be in Kelvins, the temperature in the denominator is converted to the Kelvin scale. (b) Since the efficiency is given by e = |W|/|QH|, the work done is given by 26. The answers to this exercise do not depend on the engine being of the Carnot design. Any heat engine that intakes energy as heat (from, say, consuming fuel) equal to |QH| = 52 kJ and exhausts (or discards) energy as heat equal to |QL| = 36 kJ will have these values of efficiency e and net work W. (a) Eq. 20-12 gives (b) Eq. 20-8 gives 27. With TL = 290 k, we find which yields the (initial) temperature of the
high-temperature reservoir: TH = 483 K. If we replace e = 0.40 in the above calculation with e = 0.50, we obtain a (final) high temperature equal to . The difference is 28. (a) Eq. 20-13 leads to We recall that a Watt is Joule-per-second. Thus, the (net) work done by the cycle per unit time is the given value 500 J/s. Therefore, by Eq. 20-11, we obtain the heat input per unit time: (b) Considering Eq. 20-8 on a per unit time basis, we find (4.67 0.500) kJ/s = 4.17 kJ/s for the rate of heat exhaust. 29. (a) Energy is added as heat during the portion of the process from a to b. This portion occurs at constant volume (Vb), so Qin = nCV DT. The gas is a monatomic ideal gas, so and the ideal gas law gives DT = (1/nR)(pb Vb pa Va) = (1/nR)(pb pa) Vb. Thus, . Vb and pb are given. We need to find pa. Now pa is the same as pc and points c and b are connected by an adiabatic process. Thus, and The energy added as heat is (b) Energy leaves the gas as heat during the portion of the process from c to a. This is a constant pressure process, so or . The substitutions Va Vc = Va 8.00 Va = 7.00 Va and were made. (c) For a complete cycle, the change in the internal energy is zero and W = Q = 1.47 103 J 5.54 102 J = 9.18 102 J. (d) The efficiency is e = W/Qin = (9.18 102 J)/(1.47 103 J) = 0.624 = 62.4%. 30. From Fig. 20-28, we see QH = 4000 J at TH = 325 K. Combining Eq. 20-11 with Eq. 20-13, we have = 1 W = 923 J . Now, for = 550 K, we have 31. (a) The net work done is the rectangular area enclosed in the pV diagram: Inserting the values stated in the problem, we obtain W = 2.27 kJ. (b) We compute the energy added as heat during the heat-intake portions of the cycle using Eq. 19-39, Eq. 19-43, and Eq. 19-46: where, to obtain the last line, the gas law in ratio form has been used (see Sample Problem 19-1). Therefore, since W = p0V0, we have Qabc = 13W/2 = 14.8 kJ. (c) The efficiency is given by Eq. 20-11: (d) A Carnot engine operating between Tc and Ta has efficiency equal to where the gas law in ratio form has been used. (e) This is greater than our result in part (c), as expected from the second law of thermodynamics. 32. (a) Using Eq. 19-54 for process D A gives which leads to . The result (see 19-9 and 19-11) implies the gas is monatomic. (b) The input heat is that absorbed during process A B: and the exhaust heat is that liberated during process C D: where in the last step we have used the fact that (from the gas law in ratio form see Sample Problem 19-1).
Therefore, Eq. 20-12 leads to 33. (a) We use . The heat absorbed is (b) The heat exhausted is then (c) Now we have (d) Similarly, . 34. All terms are assumed to be positive. The total work done by the two-stage system is W1 + W2. The heat-intake (from, say, consuming fuel) of the system is Q1 so we have (by Eq. 20-11 and Eq. 20-8) Now, Eq. 20-10 leads to where we assume Q2 is absorbed by the second stage at temperature T2. This implies the efficiency can be written 35. (a) The pressure at 2 is p2 = 3.00p1, as given in the problem statement. The volume is V2 = V1 = nRT1/p1. The temperature is (b) The process 2 3 is adiabatic, so . Using the result from part (a), V3 = 4.00V1, V2 = V1 and g =1.30, we obtain . (c) The process 4 1 is adiabatic, so . Since V4 = 4.00V1, we have (d) The process 2 3 is adiabatic, so or . Substituting V3 = 4.00V1, V2 = V1, p2 = 3.00p1 and g =1.30, we obtain (e) The process 4 1 is adiabatic, so and where we have used V4 = 4.00V1. (f) The efficiency of the cycle is e = W/Q12, where W is the total work done by the gas during the cycle and Q12 is the energy added as heat during the 1 2 portion of the cycle, the only portion in which energy is added as heat. The work done during the portion of the cycle from 2 to 3 is W23 = o p dV. Substitute to obtain Substitute V2 = V1, V3 = 4.00V1, and p3 = 3.00p1 to obtain Similarly, the work done during the portion of the cycle from 4 to 1 is No work is done during the 1 2 and 3 4 portions, so the total work done by the gas during the cycle is The energy added as heat is Q12 = nCV (T2 T1) = nCV (3T1 T1) = 2nCVT1, where CV is the molar specific heat at constant volume. Now g = Cp/CV = (CV + R)/CV = 1 + (R/CV), so CV = R/(g 1). Here Cp is the molar specific heat at constant pressure, which for an ideal gas is Cp = CV + R. Thus, Q12 = 2nRT1/(g 1). The efficiency is With g = 1.30, the efficiency is e = 0.340 or 34.0%. 36. Eq. 20-10 still holds (particularly due to its use of absolute values), and energy conservation implies |W| + QL = QH. Therefore, with TL = 268.15 K and TH = 290.15 K, we find which (with |W| = 1.0 J) leads to 37. A Carnot refrigerator working between a hot reservoir at temperature TH and a cold reservoir at temperature TL has a coefficient of performance K that is given by . For the refrigerator of this problem, TH = 96 F = 309 K and TL = 70 F = 294 K, so K = (294 K)/(309 K 294 K)
= 19.6. The coefficient of performance is the energy QL drawn from the cold reservoir as heat divided by the work done: K = |QL|/|W|. Thus, |QL| = K|W| = (19.6)(1.0 J) = 20 J. 38. (a) Eq. 20-15 provides which yields |QH| = 49 kJ when KC = 5.7 and |QL| = 42 kJ. (b) From 20-5 we obtain if we take the initial 42 kJ datum to be accurate to three figures. The given temperatures are not used in the calculation; in fact, it is possible that the given room temperature value is not meant to be the high temperature for the (reversed) Carnot cycle since it does not lead to the given KC using Eq. 20-16. 39. The coefficient of performance for a refrigerator is given by K = |QL|/|W|, where QL is the energy absorbed from the cold reservoir as heat and W is the work done during the refrigeration cycle, a negative value. The first law of thermodynamics yields QH + QL W = 0 for an integer number of cycles. Here QH is the energy ejected to the hot reservoir as heat. Thus, QL = W QH. QH is negative and greater in magnitude than W, so |QL| = |QH| |W|. Thus, The solution for |W| is |W| = |QH|/(K + 1). In one hour, The rate at which work is done is (1.57 106 J)/(3600 s) = 440 W. 40. (a) Using Eq. 20-14 and Eq. 20-16, we obtain (b) A similar calculation (being sure to use absolute temperature) leads to 0.50 J in this case. (c) With TL = 100 K, we obtain |W| = 2.0 J. (d) Finally, with the low temperature reservoir at 50 K, an amount of work equal to |W| = 5.0 J is required. 41. The efficiency of the engine is defined by e = W/Q1 and is shown in the text to be . The coefficient of performance of the refrigerator is defined by K = Q4/W and is shown in the text to be . Now Q4 = Q3 W, so (Q3 W)/W = T4/(T3 T4). The work done by the engine is used to drive the refrigerator, so W is the same for the two. Solve the engine equation for W and substitute the resulting expression into the refrigerator equation. The engine equation yields W = (T1 T2)Q1/T1 and the substitution yields Solving for Q3/Q1, we obtain With T1 = 400 K, T2 = 150 K, T3 = 325 K, and T4 = 225 K, the ratio becomes Q3/Q1=2.03. 42. (a) Eq. 20-13 gives the Carnot efficiency as 1 TL /TH . This gives 0.222 in this case. Using this value with Eq. 20-11 leads to W = (0.222)(750 J) = 167 J. (b) Now, Eq. 20-16 gives KC = 3.5. Then, Eq. 20-14 yields |W| = 1200/3.5 = 343 J. 43. We are told K = 0.27KC where where the Fahrenheit temperatures have been converted to Kelvins. Expressed on a per unit time basis, Eq. 20-14 leads to Appendix D indicates 1 But/h = 0.0003929
hp, so our result may be expressed as |W|/t = 0.25 hp. 44. The work done by the motor in t = 10.0 min is |W| = Pt = (200 W)(10.0 min)(60 s/min) = 1.20 105 J. The heat extracted is then 45. We need nine labels: Label Number of molecules on side 1 Number of molecules on side 2 I 8 0 II 7 1 III 6 2 IV 5 3 V 4 4 VI 3 5 VII 2 6 VIII 1 7 IX 0 8 The multiplicity W is computing using Eq. 20-20. For example, the multiplicity for label IV is and the corresponding entropy is (using Eq. 20-21) In this way, we generate the following table: Label W S I 1 0 II 8 2.9 1023 J/K III 28 4.6 1023 J/K IV 56 5.6 1023 J/K V 70 5.9 1023 J/K VI 56 5.6 1023 J/K VII 28 4.6 1023 J/K VIII 8 2.9 1023 J/K IX 1 0 46. (a) We denote the configuration with n heads out of N trials as (n; N). We use Eq. 20-20: (b) There are 2 possible choices for each molecule: it can either be in side 1 or in side 2 of the box. If there are a total of N independent molecules, the total number of available states of the N-particle system is With N = 50, we obtain Ntotal = 250 =1.13 1015. (c) The percentage of time in question is equal to the probability for the system to be in the central configuration: With N = 100, we obtain (d) W(N/2, N) = N!/[(N/2)!]2 = 1.01 1029, (e) Ntotal = 2N =1.27 1030, (f) and p(N/2;N) = W(N/2, N)/ Ntotal = 8.0%. Similarly, for N = 200, we obtain (g) W(N/2, N) = 9.25 1058, (h) Ntotal =1.61 10 60, (i) and p(N/2; N) = 5.7%. (j) As N increases the number of available microscopic states increase as 2N, so there are more states to be occupied, leaving the probability less for the system to remain in its central configuration. Thus, the time spent in there decreases with an increase in N. 47. (a) Suppose there are nL molecules in the left third of the box, nC molecules in the center third, and nR molecules in the right third. There are N! arrangements of the N molecules, but nL! are simply rearrangements of the nL molecules in the right third, nC! are rearrangements of the nC molecules in the center third, and nR! are rearrangements of the nR molecules in the right third. These rearrangements do not produce a new configuration. Thus, the multiplicity is (b) If half the molecules are in the right half of the box and the other half are in the left half of the box, then the multiplicity is If one-third of the molecules are in each third of the box, then the multiplicity is The ratio is (c) For N = 100, 48. Using Hookes law, we find the spring constant to be . To find the rate of change of