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81 Pages

### Chapter 17 Problems

Course: PEP 111, Fall 2007
School: Stevens
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Word Count: 8916

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Model: 17.1. For a gas, the thermal energy is the total kinetic energy of the moving molecules. That is, Eth = Kmicro. Solve: The number of atoms is N= M 0.0020 kg = = 3.01 10 23 m 6.64 10 -27 kg Because helium atoms have an atomic mass number A = 4, the mass of each helium atom is m = 4 u = 4(1.661 10 -27 kg) = 6.64 10 -27 kg The average kinetic energy of each atom is 2 K avg = 1 mvavg = 2 1 2 (6.64 10...

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Model: 17.1. For a gas, the thermal energy is the total kinetic energy of the moving molecules. That is, Eth = Kmicro. Solve: The number of atoms is N= M 0.0020 kg = = 3.01 10 23 m 6.64 10 -27 kg Because helium atoms have an atomic mass number A = 4, the mass of each helium atom is m = 4 u = 4(1.661 10 -27 kg) = 6.64 10 -27 kg The average kinetic energy of each atom is 2 K avg = 1 mvavg = 2 1 2 (6.64 10 -27 kg)(700 m s) = 1.63 10-21 J 2 Thus the thermal energy of the gas is Eth = K micro = NK avg = (3.01 10 23 )(1.63 10 -21 J ) = 490 J 17.2. Model: For a gas, the thermal energy is the total kinetic energy of the moving molecules. Solve: Oxygen atoms have an atomic mass number A = 16, so the mass of each molecule is m = 32 u = 32(1.661 10-27 kg) = 5.32 10-26 kg The number of molecules in the gas is N= The thermal energy is M 0.0080 kg = 1.505 1023 = m 5.32 10 -26 kg 2(1700 J ) = 650 m/s (1.505 10 23 )(5.32 10 -26 kg) Eth = NK avg = N ( 1 2 2 mvavg vavg = ) 2 Eth = Nm 17.3. Model: The work done on a gas is the negative of the area under the pV curve. Visualize: Please refer to Figure Ex17.3. The gas is compressing, so we expect the work to be positive. Solve: The work done on the gas is W = - p dV = -(area under the pV curve) = - -(200 cm 3 )(200 kPa ) = (200 10 -6 m 3 ) 2.0 10 5 Pa = 40 J ( ) ( ) Assess: The area under the curve is negative because the integration direction is to the left. Thus, the environment does positive work on the gas to compress it. 17.4. Model: The work done on a gas is the negative of the area under the pV curve. Visualize: Please refer to Figure Ex17.4. The gas is expanding, so we expect the work to be negative. Solve: The area under the pV curve is the area of the rectangle and triangle. We have (200 10 -6 m 3 )(200 10 3 Pa ) + 1 2 (200 10 -6 m 3 )(200 10 3 Pa ) = 60 J Thus, the work done on the gas is W = -60 J. Assess: The environment does negative work on the gas as it expands. 17.5. Visualize: Please refer to Figure Ex17.5. Solve: The work done on gas in an isobaric process is W = - pV = - p(Vf - Vi ) Substituting into this equation, 80 J = -(200 10 3 Pa )(V1 - 3V1 ) Vi = 2.0 10 -4 m 3 = 200 cm 3 Assess: The work done to compress a gas is positive. 17.6. Model: Helium is an ideal gas that undergoes isobaric and isothermal processes. Solve: (a) Since the pressure (p i = pf = p) is constant the work done is Won gas = - pV = - p(Vf - Vi ) = - nRTi (Vf - V i ) Vi = - (0.10 mol)(8.31 J mol K )(573 K ) (b) For compression at a constant temperature, (1000 cm 3 - 2000 cm 3 ) 2000 cm 3 = 238 J Won gas = - nRT ln(Vf Vi ) 1000 10 -6 m 3 = -(0.10 mol)(8.31 J mol K )(573 K ) ln = 330 J 2000 10 -6 m 3 (c) For the isobaric case, p= nRTi = 2.38 10 5 Pa Vi For the isothermal case, pi = 2.38 10 5 Pa and the final pressure is pf = nRTf = 4.76 10 5 Pa Vf 17.7. Visualize: Solve: than the initial point, so Tf > Ti. Heat energy is thus transferred into the gas (Q > 0) and the thermal energy of the gas increases (Eth f > Eth i) as the temperature increases. Because W = - p dV and this is an isochoric process, W = 0 J. The final point is on a higher isotherm 17.8. Visualize: Solve: That is, the gas is compressed. Since the final point is on a lower isotherm than the initial point, Tf < Ti. In other words, the thermal energy decreases. For this to happen, the heat energy transferred out of the gas must be larger than the work done. Because this is an isobaric process W = - pdV = - p(Vf - Vi ) . Since Vf is smaller than Vi, W is positive. 17.9. Visualize: Solve: Because the process is isothermal, Eth = Eth f Eth i = 0 J. According to the first law of thermodynamics, Eth = W + Q. This can only be satisfied if W = Q. W is positive because the gas is compressing, hence Q is negative. That is, heat energy is removed from the gas. 17.10. Visualize: Solve: This is an adiabatic process of gas compression so no heat energy is transferred between the gas and the environment. That is, Q = 0 J. According to the first law of thermodynamics, the work done on a gas in an adiabatic process goes entirely to changing the thermal energy of the gas. The work W is positive because the gas is compressed. 17.11. Visualize: Solve: lower isotherm than the initial point, so Tf < Ti. Heat energy is transferred out of the gas (Q < 0 J) and the thermal energy of the gas decreases as the temperature falls. That is, Eth f < Eth i. To bring about this process: (1) The locking pin is removed so the piston can slide up and down. (2) The masses on the top of the piston are not changed. This keeps the pressure constant. (3) The ice block is brought into contact with the bottom of the cylinder. This is a case of gas compression and therefore W = - pdV is a positive quantity. The final point is on a 17.12. Visualize: Solve: For the isothermal process T = 0 K. This means the first law of thermodynamics can only be satisfied if W = -Q. When the gas compresses, W > 0 J implying Q < 0 J. Heat energy is transferred out of the gas, but the temperature of the gas does not change. For the isochoric process, W = 0 J because the volume does not change. Since the final point is on a lower isotherm, the final termperature is lower. That is, heat energy is transferred out of the gas (Q < 0) and hence the thermal energy of the gas decreases. To bring about this process: (1) Place the cylinder on the ice. The gas will begin to contract as heat energy is transferred to the ice. (2) Add masses on the top of the piston to increase the pressure and keep pV constant as the volume decreases. (3) At the desired volume and pressure, insert a locking pin into the piston to fix the volume. (4) Keep the ice in contact with the bottom of the cylinder until the original pressure is obtained in the gas. 17.13. Solve: The first law of thermodynamics is Eth = W + Q -200 J = 500 J + Q Q = -700 J The negative sign means a transfer of energy from the system to the environment. Assess: Because W > 0 means a transfer of energy into the system, Q must be less than zero and larger in magnitude than W so that Eth f < Eth i. 17.14. Solve: This is an isobaric process. W > 0 because the gas is compressed. This transfers energy into the system. Also, 100 J of heat energy is transferred out of the gas. The first law of thermodynamics is Eth = W + Q = - pV + Q = (4.0 105 Pa)(200 600) 10-6 m3 100 J = 60 J Thermal energy increases by 60 J. 17.15. Model: The removal of heat from the ice reduces its thermal energy and its temperature. Solve: M = iceV = (920 kg/m3)(0.06 0.06 0.06) m3 = 0.199 kg The specific heat of ice from Table 17.2 is cice = 2090 J/kg K, so Q = (0.199 kg)(2090 J/kg K)(243 K 273 K) = 12,500 J Thus, the energy removed from the ice block is 12,500 J. Assess: The negative sign with Q means loss of energy. The heat needed to change an object's temperature is Q = McT. The mass of the ice cube is 17.16. Model: The spinning paddle wheel does work and changes the water's thermal energy and its temperature. Solve: (a) The temperature change is T = Tf Ti = 25C - 21C = 4 K. The mass of the water is M = (200 10-6 m3)(1000 kg/m3) = 0.20 kg The work done is W = Eth = McwaterT = (0.20 kg)(4190 J/kg K)(4 K) = 3350 J (b) Q = 0. No energy is transferred between the system and the environment because of a difference in temperature. 17.17. Model: Heating the mercury at its boiling point changes its thermal energy without a change in temperature. Solve: The mass of the mercury is M = 20 g = 2.0 10-2 kg, the specific heat cmercury = 140 J/kg K, the boiling point Tb = 357C, and the heat of vaporization LV = 2.96 105 J/kg. The heat required for the mercury to change to the vapor phase is the sum of two steps. The first step is Q1 = McmercuryT = (2.0 10-2 kg)(140 J/kg K)(357C - 20C) = 944 J The second step is Q2 = MLV = (2.0 10-2 kg)(2.96 105 J/kg) = 5920 J The total heat needed is 6864 J. 17.18. Model: Heating the mercury changes its thermal energy and its temperature. Solve: (a) The heat needed to change the mercury's temperature is Q = McHgT T = Q 100 J = = 35.7 K = 35.7C McHg (0.020 kg)(140 J kg K ) (b) The amount of heat required to raise the temperature of the same amount of water by the same number of degrees is Q = McwaterT = (0.020 kg)(4190 J/kg K)(35.7 K) = 2990 J Assess: Q is directly proportional to cwater and the specific heat for water is much higher than the specific heat for mercury. This explains why Qwater > Qmercury. 17.19. Model: Changing ethyl alcohol at 20C to solid ethyl alcohol at its melting point requires two steps: lowering its temperature from 20C to -114C, then changing the ethyl alcohol to its solid phase at -114C. Solve: The change in temperature is -114C - 20C = -134C = -134 K. The mass is M = V = (789 kg / m 3 )(200 10 -6 m 3 ) = 0.1578 kg The heat needed for the two steps is Q1 = McalcoholT = (0.1578 kg)(2400 J/kg K)(-134 K)= -5.075 104 J Q2 = -MLf = - (0.1578 kg)(1.09 105 J/kg) = -1.720 104 J The total heat required is Q = Q1 + Q2 = -6.79 104 J Thus, the minimum amount of energy that must be removed is 6.79 104 J. Assess: The negative sign with Q indicates that 6.79 104 J will be removed from the system. 17.20. Model: Changing solid lead at 20C to liquid lead at its melting point (Tm = 328C) requires two steps: raising the temperature to Tm and then melting the solid at Tm to a liquid at Tm. Solve: The equation for the total heat is Q = Q1 + Q2 1000 J = Mclead(Tf Ti) + MLf 1000 J = M(128 J/kg K)(328 20) K + M(0.25 105 J/kg) M = The maximum mass of lead you can melt with 1000 J of heat is 15.5 g. (64,424 J kg) 1000 J = 15.5 g 17.21. Model: We have a thermal interaction between the copper pellets and the water. Solve: The conservation of energy equation Qc + Ww = 0 is Mc cc (Tf - 300C) + Mw cw (Tf - 20C) = 0 J Solving this equation for the final temperature Tf gives M c (300C) + Mw cw (20C) Tf = c c M c cc + M w c w = (0.030 kg)(385 J / kg K) (300C) + (0.10 kg)(4190 J / kg K) (20C) = 27.5C (0.030 kg)(385 J / kg K) + (0.10 kg)(4190 J / kg K) The final temperature of the water and the copper is 27.5C. 17.22. Model: We have a thermal interaction between the copper block and water. Solve: The conservation of energy equation Qcopper + Qwater = 0 J is Mcopperccopper (Tf Ti copper) + Mwatercwater(Tf Ti water) = 0 J Both the copper and the water reach the common final temperature Tf = 25.5C. Thus Mcopper(385 J/kg K)(25.5C 300C) + (1.0 10-3 m3)(1000 kg/m3)(4190 J/kg K)(25.5C 20C) = 0 J M copper = 0.218 kg 17.23. Model: We have a thermal interaction between the thermometer and the water. Solve: The conservation of energy equation Qthermo + Qwater = 0 J is Mthermocthermo(Tf Ti thermo) + Mwatercwater(Tf Ti water) = 0 J The thermometer slightly cools the water until both have the same final temperature Tf = 71.2C. Thus (0.050 kg)(750 J / kg K)(71.2C - 20.0C) + (200 10 -6 m 3 )(1000 kg / m 3 )( 4190 J / kg K)(71.2C - Ti water ) = 1920 J + 838 (J / K)(71.2C - Ti water ) = 0 J Ti water = 73.5C Assess: The thermometer reads 71.2C for a real temperature of 73.5C. This is reasonable. The conservation of energy equation QAl + Qwater = 0 J is MAl cAl(Tf Ti Al) + Mwater cwater(Tf Ti water) The pan and water reach a common final temperature Tf = 24.0C Solve: 17.24. Model: We have a thermal interaction between the aluminum pan and the water. (0.750 kg)(900 J / kg K)(24.0C - Ti Al ) + (10 10 -3 m 3 )(1000 kg / m 3 )(4190 J / kg K)(24.0C - 20.0C ) = (675.0 J / K)(24.0C - Ti Al ) + 167,600 J = 0 J Ti Al = 272C = [(272) (9/5) + 32]F = 522F 17.25. Model: We have a thermal interaction between the metal sphere and the mercury. Solve: The conservation of energy equation Qmetal + QHg = 0 J is Mmetalcmetal(Tf Ti metal) + MHgcHg(Tf Ti Hg) = 0 J The metal and mercury reach a common final temperature Tf = 99.0C. Thus (0.500 kg)cmetal(99C 300C) + (300 10-6 m3)(13,600 kg/m3)(140 J/kg K)(99C 20C) = 0 J We find that cmetal = 449 J kg K . The metal is iron. 17.26. Model: Use the models of isochoric and isobaric heating. Note that the change in temperature on the Kelvin scale is the same as the change in temperature on the Celsius scale. Solve: (a) The atomic mass number of argon is 40. That is, Mmol = 40 g/mol. The number of moles of argon gas in the container is n= The amount of heat is M 1.0 g = = 0.025 mol Mmol 40 g mol Q = nCV T = (0.025 mol)(12.5 J/mol K)(100C) = 31.25 J (b) For the isobaric process Q = nCP T becomes 31.25 J = (0.025 mol)(20.8 J/mol K)T T = 60C 17.27. Model: The heating processes are isobaric and isochoric. O2 is a diatomic ideal gas. Solve: (a) The number of moles of oxygen is n= For the isobaric process, M 1.0 g = = 0.03125 mol Mmol 32 g mol Q = nCP T = (0.03125 mol)(29.2 J/mol K)(100C) = 91.2 J (b) For the isochoric process, Q = nCV T = 91.2 J = (0.03125 mol)(20.9 J/mol K)T T = 140C 17.28. Model: The heating is an isochoric process. Solve: The number of moles of helium is n= For the isochoric processes, M 2.0 g = = 0.50 mol Mmol 4 g mol QHe = nCV T = (0.50 mol)(12.5 J mol K )T Because QHe = QO2 , M QO2 = nCV T = (20.9 J mol K )T 32 g mol (0.50 mol)(12.5 J mol K) = M (20.9 J mol K ) M = 9.57 g 32 g mol 17.29. Model: The O2 gas has = 1.40 and is an ideal gas. Solve: (a) For an adiabatic process, pV remains a constant. That is, V V pi Vi = pf Vf pf = pi i = (3.0 atm) i Vf 2Vi 1.40 1 = (3.0 atm) 2 1.40 = 1.14 atm (b) Using the ideal-gas law, the final temperature of the gas is calculated as follows: p V pi Vi pV 1.14 atm 2Vi = 321.5 K = 48.5C = f f Tf = Ti f f = ( 423 K ) 3.0 atm Vi pi Vi Ti Tf 17.30. Model: We assume the gas is an ideal gas and = 1.40 for a diatomic gas. Solve: Using the ideal-gas law, Vi = For an adiabatic process, nRTi (0.10 mol)(8.31 J mol K )( 423 K ) = = 1.157 10 -3 m 3 pi 3 1.013 10 5 Pa ( ) pi Vi = pf Vf p Vf = Vi i pf 1 p = (1.157 10 -3 m 3 ) i 0.5 pi 1 1.40 = 1.90 10 -3 m 3 To find the final temperature, we use the ideal-gas law once again as follows: Tf = Ti 0.5 pi 1.90 10 -3 m 3 pf Vf = ( 423 K ) = 346.9 K = 73.9C -3 3 pi Vi pi 1.157 10 m 17.31. Model: is 1.40 for a diatomic gas and 1.67 for a monoatomic gas. Solve: (a) We will assume that air is a diatomic gas. For an adiabatic process, Tf Vf -1 = Ti Vi -1 Thus 1 1 Vi Tf -1 1123 K 1.40 -1 = = 26.4 = 303 K Vf Ti (b) For argon, a monatomic gas, 1 Vi 1123 K 1.67-1 = 7.07 = 303 K Vf 17.32. Model: Changing steam to ice requires four steps: removing heat to convert steam into water at 100C, removing heat to lower the water temperature from 100C to 0C, converting water at 0C to ice at 0C, and lowering the temperature of ice to -20C. Solve: Number of moles of steam is n= The mass of steam is 2 1.013 10 5 Pa (1530 10 -6 m 3 ) pV = = 0.10 mol RT (8.31 J mol K)(373 K) ( ) M = nMmol = (0.10 mol)(18 g mol) = 1.8 g = 0.0018 kg The heat needed for each step is Q1 = - MLV = -(0.0018 kg) 22.6 10 5 J kg = -4068 J Q2 = Mcwater Twater = (0.0018 kg)( 4190 J kg K )(0C - 100C) = -754.2 J Q3 = - MLf = -(0.0018 kg) 3.33 10 5 J kg = -599.4 J Q4 = Mcice Tice = (0.0018 kg)(2090 J kg K )( -20C - 0C) = -75.24 J The total heat required in this process is ( ) ( ) Q = Q1 + Q2 + Q3 + Q4 = -5497 J Assess: Approximately 75% of the heat is removed from steam at 100C to make water at 100C. This is consistent with our experience that it takes much longer for a pan of water to boil away than it does to reach boiling. 17.33. Solve: The area of the garden pond is A = (2.5 m )2 = 19.635 m 2 and its volume is V = A(0.30 m ) = 5.891 m 3 . The mass of water in the pond is M = V = (1000 kg m 3 )(19.635 m 3 ) = 5891 kg The water absorbs all the solar power which is (400 W m )(19.635 m ) = 7854 W 2 2 This power is used to raise the temperature of the water. That is, Q = (7854 W )t = Mcwater T = (5891 kg)( 4190 J kg K )(10 K ) t = 31,425 s = 8.73 hr 17.34. Model: The potential energy of the bowling ball is transferred into the thermal energy of the mixture. We assume the starting temperature of the bowling ball to be 0C. Solve: The potential energy of the bowling ball is U g = M ball gh = (11 kg)(9.8 m s 2 )h = (107.8 kg m s 2 )h This energy is transferred into the mixture of ice and water and melts 5 g of ice. That is, (107.8 kg m s2 )h = Eth = Mw Lf h = (0.005 kg)(3.33 10 5 2 J kg (107.8 kg m s ) ) = 15.45 m 17.35. Model: Heating the water and the kettle raises the temperature of the water to the boiling point and raises the temperature of the kettle to 100C. Solve: The amount of heat energy from the electric stove's output in 3 minutes is Q = (2000 J s)(3 60 s) = 3.6 10 5 J This heat energy heats the kettle and brings the water to a boil. Thus, Q = Mwater cwater T + M kettle ckettle T Substituting the given values into this equation, 3.6 10 5 J = Mwater ( 4190 J kg K )(100C - 20C) + (0.750 kg)( 449 J kg K )(100C - 20C) Mwater = 0.994 kg The volume of water in the kettle is V= Mwater 0.994 kg = = 0.994 10 -3 m 3 = 994 cm 3 water 1000 kg m 3 Assess: 1 L = 103 cm3, so V 1 L. This is a reasonable volume of water. 17.36. Model: Each car's kinetic energy is transformed into thermal energy. Solve: For each car, K = 1 Mv 2 = Eth = Mccar T T = 2 Assume ccar = ciron. The speed of the car is v2 2ccar 2 v = 80 km hr = (22.22 m s) = 0.55C 80 1000 m = 22.22 m s T = 2( 449 J kg K ) 3600 s Solve: The aluminum, copper, and alcohol form a closed system, so Q = QAl + QCu + Qeth = 0 J. The mass of the alcohol is 17.37. Model: There are three interacting systems: aluminum, copper, and ethyl alcohol. Meth = V = (790 kg m 3 )(50 10 -6 m 3 ) = 0.0395 kg Expressed in terms of specific heats and using the fact that T = Tf Ti, the Q = 0 J condition is MAl cAl TAl + MCu cCu TCu + Meth ceth Teth = 0 J Substituting into this expression, (0.010 kg)(900 J kg K)(298 K - 473 K) + (0.020 kg)(385 J kg K)(298 K - T ) + (0.0395 kg)(2400 J kg K )(298 K - 288 K ) = -1575 J + (7.7 J K )(298 - T ) + 948 J = 0 J T = 216.6 K = -56.4C 17.38. Model: There are two interacting systems: aluminum and ice. The system comes to thermal equilibrium in four steps: (1) the ice temperature increases from -10C to 0C, (2) the ice becomes water at 0C, (3) the water temperature increases from 0C to 20C, and (4) the cup temperature decreases from 70C to 20C. Solve: The aluminum and ice form a closed system, so Q = Q1 + Q2 + Q3 + Q4 = 0 J. These quantities are Q1 = Mice cice T = (0.100 kg)(2090 J kg K )(10 K ) = 2090 J Q2 = Mice Lf = (0.100 kg) 3.33 10 5 J kg = 33,300 J Q3 = Mice cwater T = (0.100 kg)( 4190 J kg K )(20 K ) = 8380 J Q4 = MAl cAl T = MAl (900 J kg K )( -50 K ) = -( 45,000 J kg) M Al The Q = 0 J equation now becomes 43,770 J (45,000 J/kg)MAl = 0 J The solution to this is MAl = 0.973 kg. ( ) Solve: The metal, aluminum container, and water form a closed system, so Qm + QAl + Qw = 0 J, where Qm is the heat transferred to the metal sample. This equation can be written: MmcmTm + MAlcAlTAl + MwcwT = 0 J Substituting in the given values, 17.39. Model: There are three interacting systems: metal, aluminum, and water. (0.512 kg)cm (351 K - 288 K) + (0.100 kg)(900 J kg K)(351 K - 371 K) + (0.325 kg)( 4190 J kg K )(351 K - 371 K ) = 0 J cm = 900 J kg K From Table 17.2, we see that this is the specific heat of aluminum. 17.40. Solve: For a monatomic gas, the molar specific heat at constant volume is CV = 12.5 J mol K . From Equation 17.22, 12.5 J mol K = The gas is therefore neon. Mmol 625 J kg K Mmol = 20 g mol 1000 g kg 17.41. Model: Heating the water raises its thermal energy and its temperature. Solve: A 5.0 kW heater has power P = 5000 W. That is, it supplies heat energy at the rate 5000 J/s. The heat supplied in time t is Q = 5000t J. The temperature increase is TC = (5/9)TF = (5/9)(75) = 41.67C. Thus Q = 5000 t J = Mcw T = (150 kg)(4190 J / kg K) (41.67C) t = 5283 s = 87.3 min Assess: A time of 1.5 hours to heat 40 gallons of water is reasonable. 17.42. Model: Heating the material increases its thermal energy. Visualize: Please refer to Figure P17.42. Heat raises the temperature of the substance from 40C to 20C, at which temperature a solid to liquid phase change occurs. From 20C, heat raises the liquid's temperature up to 40C. Boiling occurs at 40C where all of the liquid is converted into the vapor phase. Solve: (a) In the solid phase, Q = McT c = (b) In the liquid phase, Q 1 20 kJ 1 = = 2000 J kg K T M 20 K 0.50 kg 1 Q 1 80 kJ = = 2667 J kg K M T 0.50 kg 60 K (c) The melting point Tm = -20C and the boiling point Tb = +40C . (d) The heat of fusion is c= Lf = The heat of vaporization is Q 20,000 J = = 4.0 10 4 J kg M 0.50 kg Q 60,000 J = = 1.2 10 5 J kg M 0.50 kg Lv = 17.43. Model: Heating the material increases its thermal energy. Visualize: Please refer to Figure P17.43. The material melts at 300C and undergoes a solid-liquid phase change. The material's temperature increases from 300C to 1500C. Boiling occurs at 1500C and the material undergoes a liquid-gas phase change. Solve: (a) In the liquid phase, the specific heat of the liquid can be obtained as follows: Q = McT c = (b) The latent heat of vaporization is 20 kJ 1 1 Q = = 83.3 J kg K M T 0.200 kg 1200 K Lv = Q 40 kJ = 2.0 10 5 J kg = M (0.200 kg) to -196C, and then liquefying it at -196C. Assume the cooling occurs at a constant pressure of 1 atm. Solve: The mass of 1.0 L of liquid nitrogen is M = V = 810 kg m 3 10 -3 m 3 = 0.810 kg . This mass corresponds to 17.44. Model: The liquefaction of the nitrogen occurs in two steps: lowering nitrogen's temperature from 20C ( )( ) n= M 810 g = = 28.9 mols Mmol 28 g mol At constant atmospheric pressure, the heat to be removed from 28.93 mols of nitrogen is Q = MLv + nCP T = -(0.810 kg) 1.99 10 5 J kg + (28.9 mols)(29.1 J mol K )(77 K - 293 K ) = -3.43 10 5 J ( ) 17.45. Model: There are two interacting systems: coffee (i.e., water) and ice. Changing the coffee temperature from 90C to 60C requires four steps: (1) raise the temperature of ice from -20C to 0C, (2) change ice at 0C to water at 0C, (c) raise the water temperature from 0C to 60C, and (4) lower the coffee temperature from 90C to 60C. Solve: For the closed coffee-ice system, Q = Qice + Qcoffee = (Q1 + Q2 + Q3 ) + (Q4 ) = 0 J Q1 = Mice cice T = Mice (2090 J kg K )(20 K ) = Mice ( 41,800 J kg) Q2 = Mice Lf = Mice (330,000 J kg) Q3 = Mice cwater T = Mice ( 4190 J kg K )(60 K ) = Mice (251,400 J kg) Q4 = Mcoffee ccoffee T = (300 10 -6 m 3 )(1000 kg m 3 )( 4190 J kg K )( -30 K ) = -37, 000 J The Q = 0 J equation thus becomes Mice ( 41, 800 + 330, 000 + 251, 400) J kg - 37, 710 J = 0 J Mice = 0.0605 kg = 60.5 g Visualize: 60.5 g is the mass of approximately 1 ice cube. 17.46. Model: There are two interacting systems: the nuclear reactor and the water. The heat generated by the nuclear reactor is used to raise the water temperature. Solve: For the closed reactor-water system, energy conservation per second requires Q = Qreactor + Qwater = 0 J The heat from the reactor in t = 1 s is Qreactor = -2000 MJ = -2.0 10 9 J and the heat absorbed by the water is -2.0 10 J + mwater ( 4190 J kg K )(12 K ) = 0 J mwater = 3.98 10 4 kg 9 Qwater = mwater cwater T = mwater ( 4190 J kg K )(12 K ) Each second, 3.98 104 kg of water is needed to remove heat from the nuclear reactor. Thus, the water flow per minute is kg 60 s 1m 3 1L 3.98 10 4 -3 3 = 2.39 106 L/min s min 1000 kg 10 m 17.47. Model: We have three interacting systems: the aluminum, the air, and the firecracker. The energy released by the firecracker raises the temperature of the aluminum and the air. We will assume that air is basically N2. Solve: For the closed firecracker + air + aluminum system, energy conservation requires that Q = Qfirecracker + QAl + Qair = 0 J QAl = mAlcAlT = (2.0 kg)(900 J/kg K)(3 K) = 5400 J pV Qair = nCV T = C T RT V (8.31 J / mol K)(298 K) = 51 J Thus, Qfirecracker = -QAl - Qair = -5450 J = (1.01 10 5 Pa (20 10 -3 m 3 ) ) (20.8 J / mol K)(3 K) That is, 5450 J of energy are released on explosion of the firecracker. Assess: The negative sign with Qfirecracker means that the firecracker has lost energy. 17.48. Model: We have two interacting systems: the water and the gas. For the closed system comprised of water and gas to come to equilibrium, heat is transferred from one interacting system to the other. Solve: Energy conservation requires that Qair + Qwater = 0 J ngasCV (Tf Ti gas) + mwaterc(Tf Ti water) = 0 J Using the ideal-gas law, Ti gas = pgas Vgas ngas R = (10 1.013 10 Pa)(4000 10 5 -6 (0.40 mol)(8.31 J mol K) m3 ) = 1219 K The energy conservation equation with Ti water = 293 K becomes (0.40 mol)(12.5 J mol K)(T - 1219 K) + (20 10 -3 kg)(4190 J kg K)(T - 293 K) = 0 J Tf = 345 K We can now use the ideal-gas equation to find the final gas pressure. pi Vi pV T 345 K = f f pf = pi f = (10 atm) = 2.83 atm Ti Tf Ti 1219 K 17.49. Model: For a gas, the thermal energy is the total kinetic energy of the moving molecules. That is, Eth = Kmicro. Also, the work W done on an expanding gas is negative. Solve: (a) The thermal energy of N molecules is Eth = NKavg, where K avg = 1 m vavg 2 ( ) 2 is the average kinetic energy per molecule. The mass of a hydrogen molecule is The number of molecules in a 1 g = 0.001 kg sample is M 0.001 kg N= = = 3.01 10 23 m 3.32 10 -27 kg The average kinetic energy per molecule is m = 2 u = 2 (1.661 10 -27 kg) = 3.32 10 -27 kg K avg = 1 m vavg 2 ( ) 2 = 1 2 (3.32 10 -27 kg)(700 m s) = 8.13 10 -22 J 2 Thus, the thermal energy in the 1-gram sample of the gas is Eth = NK avg = (3.01 10 23 )(8.13 10 -22 J ) = 245 J (b) The first law of thermodynamics tells us the change of thermal energy when work is done and heat is added: Eth = W + Q = -300 J + 500 J = 200 J Here W is negative because energy is transferred from the system to the environment. The work and heat raise the thermal energy of the gas by 200 J to Eth = 445 J. Now the average kinetic energy is K avg = Eth N = 1.487 10 -21 J = 1 m vavg 2 Solving for the new average speed gives ( ) 2 vavg = 2 K avg m = 2(1.478 10 -21 J ) 3.32 10 -27 kg = 944 m s 17.50. These Model: are isothermal and isobaric ideal-gas processes. Solve: (a) The work done at constant temperature is Vf Vf nRT W = - pdV = - dV = - nRT (ln Vf - ln Vi ) = - nRT ln(Vf Vi ) Vi Vi V = -(2.0 mol)(8.31 J mol K )(303 K ) ln( 1 ) = 5530 J 3 (b) The work done at constant pressure is W = - p dV = - p(Vf - Vi ) = - p Vf Vi Vi 2 - Vi = pVi 3 3 = 2 2 nRT = (2.0 mol)(8.31 J mol K )(303 K ) = 3360 J 3 3 (c) For an isothermal process in which Vf = 1 Vi , the pressure changes to pf = 3 pi = 4.5 atm. 3 17.51. Model: This is an isothermal process. The work done is positive for a compression. Solve: For an isothermal process, W = - nRT ln(Vf Vi ) For the first process, W = 500 J = - nRT ln( 1 ) nRT = 721.35 J 2 For the second process, 1 1 W = - nRT ln( 10 ) W = -(721.35 J ) ln( 10 ) = 1660 J 17.52. Visualize: Solve: (a) The gas exerts a force on the piston of magnitude Fgas on piston = pgas A = (3 atm 101,300 Pa atm) (0.080 m ) [ 2 ] = 6110 N This force is directed toward the right. (b) The piston is in static equilibrium, so the environment must exert a force on the piston of equal magnitude Fenviron on piston = 6110 N but in the opposite direction, toward the left. (c) The work done by the environment is r r r Wenviron = Fenviron on piston r = - Fenviron on piston x = -(6110 N )(0.10 m ) = -611 J The work is negative because the force and the displacement are in opposite directions. (d) The work done by the gas is r r Wgas = Fgas on piston r = + Fgas on piston x = (6110 N )(0.10 m ) = 611 J This work is positive because the force and the displacement are in the same direction. (e) The first law of thermodynamics is W + Q = Eth where W is Wenviron. So here W = -611 J and we find Q = Eth - W = (196 J ) - ( -611 J ) = 807 J Thus, 807 J of heat is added to the gas. 17.53. Model: This is an isobaric process. Visualize: Solve: (a) The initial conditions are p1 = 10 atm = 1.013 106 Pa, T1 = 50C = 323 K, V1 = r2L1 = (0.050 m)2 (0.20 m) = 1.57 10-3 m3. The gas is heated at a constant pressure, so heat and temperature change are related by Q = nCPT. From the ideal gas law, the number of moles of gas is n= 6 -3 3 p1V1 (1.013 10 Pa )(1.57 10 m ) = 0.593 mol = RT1 (8.31 J mol K)(323 K) The temperature change due to the addition of Q = 2500 J of heat is thus Q 2500 J T = = = 203 K nCP (0.593 mol)(20.8 J mol K ) The final temperature is T2 = T1 + T = 526 K = 253C. (b) Noting that the volume of a cylinder is V = r2L and that r doesn't change, the ideal gas relationship for an isobaric process is V2 V1 L L T 526 K = 2 = 1 L2 = 2 L1 = (20 cm) = 32.6 cm 323 K T2 T1 T2 T1 T1 17.54. Model: The process in part (a) is isochoric and the process in part (b) is isobaric. Solve: (a) Initially V1 = (0.20 m)3 = 0.0080 m3 = 8.0 L and T1 = 293 K. Helium has an atomic mass number A = 4, so 3 g of helium is n = M/Mmol = 0.75 mole of helium. We can find the initial pressure from the ideal-gas law: nRT1 (0.75 mol)(8.31 J mol K )(293 K ) p1 = = = 228 kPa = 2.25 atm V1 0.0080 m 3 Heating the gas will raise its temperature. A constant volume process has Q = nCVT, so Q 1000 J T = = = 107 K nCV (0.75 mol)(12.5 J mol K ) This raises the final temperature to T2 = T1 + T = 400 K. Because the process is isochoric, p2 p T 400 K = 1 p2 = 2 p1 = (2.25 atm) = 3.07 atm 293 K T2 T1 T1 (b) The initial conditions are the same as part a, but now Q = nCPT. Thus, Q 1000 J T = = = 64.1 K nCP (0.75 mol)(20.8 J mol K ) Now the final temperature is T2 = T1 + T = 357 K. Because the process is isobaric, V2 V1 T 357 K = V2 = 2 V1 = 0.0080 m 3 = 0.00975 m 3 = 9.75 L 293 K T2 T1 T1 (c) ( ) Solve: At T = 0 K, atoms of the gas have no thermal energy ((Eth)i = 0) and no velocity. Consequently, the pressure due to the atoms of the gas is zero. If we start the gas in a volume of nRTf (2.0 mol)(8.31 J mol K )(310 K ) = Vi = Vf = = 0.05086 m 3 pf 1.013 10 5 Pa 17.55. Model: For a gas, the thermal energy is the total kinetic energy of the moving molecules. ( ) and increase the temperature from Ti = 0 K to Tf = 310 K at constant volume, then the pressure will rise with T as heat Q is added. The amount of heat added at constant volume is Q = nCV T = (2.0 mol)(12.5 J mol K )(310 K ) = 7750 J No work is done, so the first law is Eth = Q + W = Q = 7750 J. The change in thermal energy is Eth = ( Eth )f - ( Eth )i . But (Eth)i = 0, so the thermal energy of the gas is (Eth)f = 7750 J. Solve: (a) The final temperature is T2 = T1 because the process is isothermal. (b) The work done on the gas is 17.56. Model: This is an isothermal process. V2 W = - pdV = - V1 V2 nRT1 V dV = - nRT1 ln 2 = -nRT1ln 2 V V1 V1 (c) From the first law of thermodynamics Eth = W + Q = 0 J because T = 0 K. Thus, the heat energy transferred to the gas is Q = - W = nRT1 ln 2 . Solve: (a) The initial conditions are p1 = 3.0 atm = 304,000 Pa and T1 = 293 K. Nitrogen has a molar mass Mmol = 28 g/mol, so 5 g of nitrogen gas has n = M/Mmol = 0.1786 mol. From this, we can find the initial volume: 17.57. Model: The gas is an ideal gas and it goes through an isobaric and an isochoric process. nRT1 (0.1786 mol)(8.31 J mol K )(293 K ) = = 1.430 10 -3 m 3 = 1430 cm 3 304,000 Pa p1 V1 = The volume triples, so V2 = 3V1 = 4290 cm3. The expansion is isobaric (p 2 = p1 = 3.0 atm), so V2 V1 V = T2 = 2 T1 = (3)293 K = 879 K = 606C T2 T1 V1 (b) The process is isobaric, so Q = nCP T = (0.1786 mol)(29.1 J mol K )(879 K - 293 K ) = 3050 J (c) The pressure is decreased at constant volume (V3 = V2 = 4290 cm3 ) until the original temperature is reached (T3 = T1 = 293 K). For an isochoric process, p3 T p 293 K = 2 p3 = 3 p2 = (3.0 atm) = 1.0 atm 879 K T3 T2 T2 (d) The process is isochoric, so Q = nCV T = (0.1786 mol)(20.8 J mol K )(293 K - 879 K ) = -2180 J So, 2180 J of heat was removed to decrease the pressure. (e) 17.58. Model: The gas is an ideal gas. Visualize: Please refer to Figure P17.58. Call the upper right corner (on process A) 2 and the lower left corner (on process B) 3. Solve: The change in thermal energy is the same for any gas process that has the same T. Processes A and B have the same T, since they start and end at the same points, so (Eth)A = (Eth)B. The first law is then (Eth)A = QA + WA = (Eth)B = QB + WB QA QB = WB WA In process B, work W = pV = pi(2Vi Vi) = piVi is done during the isobaric process i 3. No work is done during the isochoric process 3 f. Thus WB = piVi. Similarly, no work is done during the isochoric process i 2 of process A, but W = pV = 2pi(2Vi Vi) = 2piVi is done during the isobaric process 2 f. Thus WA = 2piVi. Combining these, QA QB = WB WA = piVi (2piVi) = piVi 17.59. Model: The two processes are isochoric and isobaric. Visualize: Please refer to Figure P17.59. Solve: Process A is isochoric which means From the ideal-gas equation, Tf Ti = pf pi Tf = Ti ( pf pi ) = Ti (1 atm 3 atm) = 1 Ti 3 Ti = 5 -6 3 pi Vi (3 1.013 10 Pa )(2000 10 m ) = = 731.4 K Tf = 1 Ti = 243.8 K 3 nR (0.10 mol)(8.31 J mol K) Tf - Ti = -487.6 K Thus, the heat required for process A is QA = nCV T = (0.10 mol)(20.8 J mol K )( -487.6 K ) = -1010 J Process B is isobaric which means From the ideal-gas equation, Tf Vf = Ti Vi Tf = Ti (Vf Vi ) = Ti (3000 cm 3 1000 cm 3 ) = 3Ti Ti = 5 -6 3 pi Vi (2 1.013 10 Pa )(1000 10 m ) = = 243.8 K nR (0.10 mol)(8.31 J mol K) Tf = 3Ti = 731.4 K Tf - Ti = 487.6 K Thus, heat required for process B is QB = nCP T = (0.10 mol)(29.1 J mol K )( 487.6 K ) = 1419 J Assess: Heat is transferred out of the gas in process A, but transferred into the gas in process B. 17.60. Model: We have an adiabatic and an isothermal process. Visualize: Please refer to P17.60. Solve: For the adiabatic process, no heat is added or removed. That is Q = 0 J. Isothermal processes occur at a fixed temperature, so T = 0 K. Thus Eth = 0 J, and the first law of thermodynamics gives Q = -W = nRT ln(Vf Vi ) The temperature T can be obtained from the ideal gas equation as follows: pi Vi = nRT T = Substituting into the equation for Q we get 1.013 10 5 Pa (3000 10 -6 m 3 ) pi Vi = 366 K = nR (0.10 mol)(8.31 J mol K) ( ) 1000 10 -6 m 3 Q = (0.10 mol)(8.31 J mol K )(366 K ) ln = -334 J 3000 10 -6 m 3 That is, 334 J of heat energy is removed from the gas. 17.61. Model: The monatomic gas is an ideal gas which is subject to isobaric and isochoric processes. Visualize: Please refer to Figure P17.61. Solve: (a) For this isobaric process, p1 = 4.0 atm, V1 = 800 10-6 m3, p2 = 4.0 atm, and V2 = 1600 10-6 m3. The temperature T1 of the gas is obtained from the ideal-gas equation as: pV T1 = 1 1 = 390 K nR where n = 0.10 mol. Also, T2 = T1 1600 10 -6 m 3 V2 = T1 = 2T1 = 780 K V1 800 10 -6 m 3 Thus, the heat required for the process 1 2 is Q = nCP (T2 - T1 ) = (0.10 mol)(29.1 J mol K )(390 K ) = 1135 J This is heat transferred to the gas. (b) For the isochoric process, V2 = V3 = 1600 10-6 m3, p2 = 4.0 atm, p3 = 2.0 atm, and T2 = 780 K. T3 can be obtained from the ideal-gas equation as follows: pV p2 V2 2.0 atm = 390 K = 3 3 T3 = T2 ( p3 p2 ) = (780 K ) 4.0 atm T2 T3 The heat required for the process 2 3 is Q = nCV (T3 - T2 ) = (0.10 mol)(20.8 J / mol K )(390 K - 780 K ) = -811 J Because of the negative sign, this is the amount of heat removed from the gas. (c) The change in the thermal energy of the gas is Eth = (Q12 + Q23 ) + (W12 + W23 ) = 1135 J - 811 J + W12 + 0 J = 324 J - pV = 324 J (4.0 1.013 105 Pa)(1600 10-6 m3 800 10-6 m3) = 0 J Assess: This result was expected since T3 = T1. 17.62. Model: Assume that the gas is an ideal gas. Visualize: The volume of container A is a constant. On the other hand, heating container B causes the volume to change, but the pressure remains the same. Solve: (a) For the heating of the gas in container A, TA = QA / nCV . Similarly, for the gas in container B, TB = QB / nCP . Because QA = QB and CP > CV, we see that TA > TB . The gases started at the same temperature, so TA > TB. (b) (c) The pressure in container B exerted by the gas is equal to the pressure on the gas by the piston. That is, pB = patmos + w piston Apiston = 1.013 10 5 Pa + (10 kg)(9.8 m s 2 ) 1.0 10 -4 m 2 = 1.081 106 Pa Container A has the same volume, temperature, and number of moles of gas as container B, so PA = PB = 1.081 10 6 Pa. (d) The heating of container B is isobaric, so Vf Vi T = Vf = Vi f Tf Ti Ti We have Ti = 293 K, and Tf can be obtained from Q = Pt = nCP (Tf - Ti ) The number of moles of gas is n = PVi / RTi = 0.355 mol. Thus i (25 W )(15 s) = (0.355 mol)(20.8 J mol K)(Tf - 293 K) Tf = 344 K Vf = (8.0 10 -4 m 3 )(344 K 293 K ) = 9.39 10-4 m3 = 939 cm3 17.63. Model: Assume that the gas is an ideal gas. A diatomic gas has 1.40. Solve: (a) For container A, ViA = nRTiA (0.10 mol)(8.31 J mol K )(300 K ) = 8.20 10 -4 m 3 = 3.0 1.013 10 5 Pa piA For an isothermal process pfAVfA = piAViA. This means TfA = TiA = 300 K and VfA = ViA ( piA pfA ) = (8.20 10 -4 m 3 )(3.0 atm 1.0 atm) = 2.46 10 -3 m 3 The gas in container B starts with the same initial volume. For an adiabatic process, p pfBVfB = piBViB VfB = ViB iB pfB 1 3.0 atm = (8.20 10 -4 m 3 ) 1.0 atm 1 1.40 = 1.80 10 -3 m 3 The final temperature TfB can now be obtained by using the ideal-gas equation: TfB = TiB (b) pfB VfB 1.0 atm 1.80 10 -3 m 3 = (300 K ) = 220 K 3.0 atm 8.20 10 -4 m 3 piB ViB 17.64. Model: The gas is assumed to be an ideal gas that is subjected to isobaric and isochoric processes. Visualize: Please refer to Figure P17.64. Solve: (a) The initial conditions are p1 = 3.0 atm = 304,000 Pa, V1 = 100 cm3 = 1.0 10-4 m3, and T1 = 100C = 373 K. The number of moles of gas is n= -4 3 p1V1 (304,000 Pa )(1.0 10 m ) = 9.81 10 -3 mol = RT1 (8.31 J mol K)(373 K) At point 2 we have p2 = p1 = 3.0 atm and V2 = 300 cm3 = 3V1. This is an isobaric process, so V2 V1 V = T2 = 2 T1 = 3(373 K) = 1119 K T2 T1 V1 The gas is heated to raise the temperature from T1 to T2. The amount of heat required is Q = nCP T = (9.81 10 -3 mol)(20.8 J mol K )(1119 K - 373 K ) = 152 J This amount of heat is added during process 1 2. (b) Point 3 returns to T3 = 100C = 373 K. This is an isochoric process, so Q = nCV T = (9.81 10 -3 mol)(12.5 J mol K )(373 K - 1119 K ) = -91.5 J This amount of heat is removed during process 2 3. 17.65. Model: Assume the gas to be an ideal gas. Visualize: Please refer to Figure P17.65. Solve: (a) The work done on the gas is the negative of the area under the p-versus-V graph, that is W = -area under curve = -50.7 J (b) The change in thermal energy is Eth = nCV T = nCV (Tf - Ti ) Using the ideal-gas law to calculate the initial and final temperatures, 5 -6 3 pi Vi ( 4.0 1.013 10 Pa )(100 10 m ) = = 325 K nR (0.015 mol)(8.31 J mol K) Ti = Tf = 1.013 10 5 Pa (300 10 -6 m 3 ) pf Vf = = 244 K nR (0.015 mol)(8.31 J mol K) ( ) Eth = (0.015 mol)(12.5 J mol K )(244 K - 325 K ) = -15.2 J (c) From the first law of thermodynamics, Eth = Q + W Q = Eth - W = -15.2 J - ( -50.7 J ) = 35.5 J That is, 35.5 J of heat energy is transferred to the gas. 17.66. Model: Assume that the gas is an ideal gas and that the work, heat, and thermal energy are connected by the first law of thermodynamics. Visualize: Please refer to Figure P17.66. Solve: (a) For point 1, V1 = 1000 cm3 = 1.0 10-3 m3, T1 = 133C = 406 K, and the number of moles is n= Thus, the pressure p1 is 120 10 -3 g M = = 0.030 mol Mmol 4 g / mol nRT1 = 1.012 10 5 Pa = 1.0 atm V1 The process 1 2 is isochoric (V2 = V1) and p2 = 5p1 = 5.0 atm. Thus, p1 = T2 = T1 ( p2 p1 ) = ( 406 K )(5) = 2030 K = 1757C The process 2 3 is isothermal (T2 = T3), so V3 = V2(p2/p3) = V2(p2/p1) = 5V2 = 5000 cm3 p (atm) Point 1 Point 2 Point 3 1.0 5.0 1.0 T (C) 133 1757 1757 V (cm3) 1000 1000 5000 (b) The work W12 = 0 J because it is an isochoric process. The work in process 2 3 can be found using Equation 17.16 as follows: W23 = - nRT2 ln(V3 V2 ) = -(0.030 mol)(8.31 J mol K )(2030 K ) ln(5) = -815 J The work in the isobaric process 3 1 is W31 = - p(Vf - Vi ) = -(1.012 10 5 Pa )(1.0 10 -3 m 3 - 5.0 10 -3 m 3 ) = 405 J (c) The heat transferred in process 1 2 is Q12 = nCV T = (0.030 mol)(12.5 J mol K )(2030 K - 406 K ) = 609 J The heat transferred in the isothermal process 2 3 is Q23 = - W23 = 815 J . The heat transferred in the isobaric process 3 1 is Q31 = nCP T = (0.030 mol)(20.8 J mol K )( 406 K - 2030 K ) = -1013 J 17.67. Model: The air is assumed to be an ideal diatomic gas that is subjected to an adiabatic process. Solve: The air admitted into the cylinder at T0 = 30C = 303 K and p0 = 1 atm = 1.013 105 Pa has a volume V0 = 600 10-6 m3 and contains pV n = 0 0 = 0.024 mol RT0 Using Equation 17.36 and the fact that Q = 0 J for an adiabatic process, Eth = Q + W = nCV T W = nCVT 400 J = (0.024 mol)(20.8 J/mol K)(Tf 303 K) Tf = 1100 K For an adiabatic process Equation 17.40 is 1 1 Tf V -1 f =TV -1 0 0 T -1 303 K 1.4 -1 = (6.0 10 -4 m 3 ) Vf = V0 0 = 2.39 10 -5 m 3 = 23.9 cm 3 1100 K Tf Assess: Note that W is positive because the environment does work on the gas. 17.68. Model: The gas is an ideal gas that is subjected to an adiabatic process. Solve: (a) For an adiabatic process, V p ln(2.5) = 1.32 pf V = pi Vi f = i 2.5 = (2.0) = pi Vf ln(2.0) f (b) Equation 17.40 for an adiabatic process is Tf V -1 f = Ti Vi -1 T V f = i Ti Vf -1 = (2.0) 1.32 -1 = (2.0) 0.32 = 1.25 17.69. Model: Air is assumed to be an ideal diatomic gas that is subjected to an adiabatic process. Solve: (a) Equation 17.40 for an adiabatic process is Tf V -1 f = Ti Vi -1 V T -1 f = i Vi Tf 1 For the temperature to increase from Ti = 20C = 293 K to Tf = 1000C = 1273 K, the compression ratio will be Vf 293 K 1.4 -1 V 1 = = 0.02542 max = = 39.3 1273 K Vi Vmin 0.02542 (b) From the Equation 17.39, 1 V p 1.4 pf V = pi Vi f = i = (39.3) = 171 pi Vf f 17.70. Model: The helium gas is assumed to be an ideal gas that is subjected to an isobaric process. Solve: (a) The number of moles in 2.0 g of helium is n= M 2.0 g = = 0.50 mol Mmol 4.0 g mol nRTi = 0.0153 m 3 = 15.3 L pi At Ti = 100C = 373 K and pi = 1.0 atm = 1.013 105 Pa, the gas has a volume Vi = For an isobaric process (pf = pi) that doubles the volume Vf = 2Vi, (Tf Ti ) = (Vf Vi ) = 2 Tf = 2Ti = 2(373 K ) = 746 K = 473C (b) The work done by the environment on the gas is W = - pi (Vf - Vi ) = - pi Vi (2 - 1) = -(1.013 10 5 Pa )(0.0153 m 3 ) = -1550 J (c) The heat input to the gas is Q = nCP (Tf - Ti ) = (0.50 mol)(20.8 J mol K )(746 K - 373 K ) = 3880 J (d) The change in the thermal energy of the gas is Eth = Q + W = 3880 J - 1550 J = 2330 J (e) Assess: The internal energy can also be calculated as follows: Eth = nCV T = (0.5 mol)(12.5 J mol K )(746 K - 373 K ) = 2330 J This is the same result as we got in part (d). 17.71. Model: The helium gas is assumed to be an ideal gas that is subjected to an isothermal process. Solve: (a) The number of moles in 2.0 g of helium gas is n= M 2.0 g = = 0.50 mol Mmol 4.0 g mol At Ti = 100C = 373 K and pi = 1.0 atm = 1.013 105 Pa, the gas has a volume nRTi Vi = = 0.0153 m 3 = 15.3 L pi For an isothermal process (Tf = Ti) that doubles the volume Vf = 2Vi, pf Vf = pi Vi pf = pi (Vi Vf ) = (1.0 atm)( 1 ) = 0.50 atm 2 (b) The work done by the environment on the gas is W = - nRTi ln(Vf Vi ) = -(0.50 mol)(8.31 J mol K )(373 K ) ln(2) = -1074 J (c) Because Eth = Q + W = 0 J for an isothermal process, the heat input to the gas is Q = -W = 1074 J. (d) The change in internal energy Eth = 0 J. (e) 17.72. Model: The nitrogen gas is assumed to be an ideal gas that is subjected to an adiabatic process. Solve: (a) The number of moles in 14.0 g of N2 gas is n= M 14.0 g = = 0.50 mol Mmol 28 g mol At Ti = 273 K and pi = 1.0 atm = 1.013 105 Pa, the gas has a volume nRTi Vi = = 0.0112 m 3 = 11.2 L pi For an adiabatic process that compresses to a pressure pf = 20 atm, we can use Equation 17.39 and Equation 17.40 as follows: Tf V -1 f = Ti Vi -1 T V f = i Ti Vf pi ) -1 -1 p V pf V = pi Vi f = i Vf pi f 1.4 -1.0 1.4 1 Combining the above two equations yields (Tf Ti ) = ( pf (b) The work done on the gas is (c) The heat input to the gas is Q = 0 J. (d) From the above equation, Tf = Ti (20) = Ti (2.3535) = 643 K W = Eth = nCV (Tf - Ti ) = (0.50 mol)(20.8 J mol K )(643 K - 273 K ) = 3850 J 1 Vi pf V = = (20) 1.4 = 8.50 = max Vf pi Vmin 1 (e) 17.73. Model: The gas is assumed to be an ideal gas that is subjected to an isochoric process. Solve: (a) The number of moles in 14.0 g of N2 gas is n= M 14.0 g = = 0.50 mol Mmol 28 g mol nRTi = 0.0112 m 3 = 11.2 L pi At Ti = 273 K and pi = 1.0 atm = 1.013 105 Pa, the gas has a volume Vi = For an isochoric process (Vi = Vf), Tf p 20 atm = f = = 20 Tf = 20(273 K ) = 5460 K 1 atm Ti pi (b) The work done on the gas is W = - pV = 0 J. (c) The heat input to the gas is Q = nCV (Tf - Ti ) = (0.50 mol)(20.8 J mol K )(5460 K - 273 K ) = 5.39 10 4 J (d) The pressure ratio is pmax p 20 atm = f = = 20 1 atm pmin pi (e) 17.74. Model: The air is assumed to be an ideal gas. Because the air is compressed without time to exchange heat with its surroundings, the compression is an adiabatic process. Solve: The initial pressure of air in the mountains behind Los Angeles is pi = 60 10 3 Pa at Ti = 273 K. The pressure of this air when it is carried down to the elevation near sea level is pf = 100 103 Pa. The adiabatic compression of a gas leads to an increase in temperature according to Equation 17.39 and Equation 17.40, which are Tf V -1 f = Ti Vi -1 T V f = i Ti Vf -1 p V pf V = pi Vi f = i Vf pi f 1.4 -1 1.4 1 Combining these two equations, (Tf Ti ) = ( pf pi ) -1 100 10 3 Pa Tf = Ti 60 10 3 Pa 5 = (273 K ) 3 0.286 = 316 K = 43C = 109F 17.75. Solve: (a) 50 J of work are done on a gas to compress it to one-third of its original volume at a constant temperature of 77C. How many moles of the gas are in the sample? (b) The number of moles is n= 50 J = 0.0156 mol -(8.31 J mol K )(350 K )(ln 1 ) 3 17.76. Solve: (a) A heated 500 g iron slug is dropped into a 200 cm3 pool of mercury at 15C. If the mercury temperature rises to 90C, what was the initial temperature of the iron slug? (b) The initial temperature was 217C . 17.77. Solve: (a) 2.0 10 5 J must be removed from a sample of nitrogen gas at 20C to convert it to liquid nitrogen. What is the mass of the sample? (b) The mass is M = 0.472 kg. 17.78. Solve: (a) A diatomic gas is adiabatically compressed from 1 atm pressure to 10 atm pressure. What is the compression ratio Vmax/Vmin? (b) The ratio is Vmax Vmin = 10 1.4 = 5.18 . 1 17.79. Model: Assume the helium gas to be an ideal gas. The gas is subjected to isothermal, isochoric, and adiabatic processes. Visualize: Please refer to Figure CP17.79. The gas at point 1 has volume V1 = 1000 cm3 = 1.0 10-3 m3 and pressure p1 = 3.0 atm. At point 2, V2 = 3000 cm3 = 3.0 10-3 m3 and p2 = 1.0 atm. These values mean that T2 = T1, so process 1 2 is an isothermal process. The process 2 3 occurs at constant volume and is thus an isochoric process. Finally, because temperature T3 is lower than T1 or T2, the process 3 1 is an adiabatic process. Solve: (a) The number of moles of gas is n= The temperature T1 can be calculated to be 0.120 g = 0.030 mols 4 g mol T1 = 3.0 1.013 10 5 Pa (1.0 10 -3 m 3 ) p1V1 = 1219 K = 946C = nR (0.030 mol)(8.31 J mol K) ( ) For the isothermal process 1 2, T2 = 1219 K. For the adiabatic process 3 1, p3 = p1 (V1 V3 ) 1000 cm 3 = (3.0 atm) 3000 cm 3 Point 1 2 3 1.67 = 0.48 atm p (atm) 3.0 1.0 0.48 T3 = T1 (V1 V3 ) T (C) 946 946 310 -1 = (1219 K )( 1 ) 3 0.67 = 583 K = 310C V (cm3) 1000 3000 3000 Note that the values obtained above are consistent with the isochoric process 2 3, for which p p2 = 3 p2 = (T2 T3 ) p3 = (1219 K 583 K )(0.48 atm) = 1 atm T2 T3 (b) From Equation 17.15, V W12 = - nRT1 ln 2 = -(0.030 mol)(8.31 J mol K )(1219 K ) ln(3) = -334 J V1 The work done in the ischochoric process is W23 = 0 J . The work done in the adiabatic process is W31 = nCV (T1 - T3 ) = (0.030 mol)(12.5 J mol K )(1219 K - 583 K ) = 239 J (c) For the process 1 2, T = 0 K Eth = 0 J Q = - W = 334 J For the process 2 3, W = 0 J, For the process 3 1, Q = 0 J. Eth = Q = nCV (T3 - T2 ) = -239 J 17.80. Model: The gas is an ideal gas, and its thermal energy is the total kinetic energy of the moving molecules. Visualize: Please refer to Figure P17.80. Solve: (a) The piston is floating in static equilibrium, so the downward force of gravity on the piston's mass must exactly balance the upward force of the gas, Fgas = pA where A = r2 is the area of the face of the piston. Since the upper part of the cylinder is evacuated, there is no gas pressure force pushing downward. Thus, M piston g = pA p = M piston g A = Cu Vpiston g A = Cu gh = (8920 kg m 3 )(9.80 m s 2 )(0.040 m ) = 3500 Pa (b) The gas volume is V1 = r 2 L = (0.030) 2 (0.20 m ) = 5.65 10 -4 m 3 . The number of moles is n= The number of molecules is -4 3 p1V1 (3500 Pa )(5.65 10 m ) = 8.12 10 -4 mol = RT1 (8.31 J mol K)(293 K) N = nNA = (8.12 10-4 mol)(6.02 1023 mol-1) = 4.89 1020 (c) The thermal energy is Eth = NK avg = N 1 m vavg 2 ( ) 2 Molecular nitrogen N2 has an atomic mass number of A = 28, so the mass of one molecule is m = 28 (1.661 10-27 kg) = 4.65 10-26 kg If the average speed is vavg = 550 m/s, then the thermal energy is Eth = (4.89 1020)(0.5)(4.65 10-26 kg)(550 m/s)2 = 3.44 J (d) The pressure in the gas is determined simply by the weight of the piston. That will not change as heat is added, so the heating takes place at constant pressure with Q = nCPT. The temperature increase is T = Q 2.0 J = = 85 K nCP (8.12 10 -4 mol)(29.1 J mol K ) This raises the gas temperature to T2 = T1 + T = 378 K = 105C. (e) Noting that the volume of a cylinder is V = r2L and that r doesn't change, the ideal gas relationship for an isobaric process is V2 V1 L L T 378 K = 2 = 1 L2 = 2 L1 = 20 cm = 25.8 cm 293 K T2 T1 T2 T1 T1 (f) The work done by the gas is Wgas = Fgas y . The force exerted on the piston by the gas is Fgas = pA = pr 2 = 9.90 N This force is applied through y = 5.8 cm = 0.058 m, so the work done is Wgas = (9.90 N)(0.058 m) = 0.574 J Thus, 0.574 J is the work done by the gas on the piston. The work done on the gas is -0.574 J. (20C), and the liquid nitrogen. The boiling point of liquid nitrogen is 196C = 77 K. Solve: The piece of iron has mass Miron = 197 g = 0.197 kg and volume Viron = Miron/iron = (0.197 kg)/(7870 kg/m3) = 25 10-6 m3 = 25 cm3 = 25 mL. The heat lost by the iron is Qiron = MironcironT = (0.197 kg)(449 J/kg K)(77 K 293 K) = 1.911 104 J This heat causes mass M of the liquid nitrogen to boil. Energy conservation requires 17.81. Model: There is a thermal interaction between the iron, assumed to be initially at room temperature Qiron + QN2 = Qiron + MLf = 0 M = - The volume of liquid nitrogen boiled away is thus Qiron 1.911 10 4 J = = 0.0960 kg 1.99 10 5 J / kg Lf Vboil = M 0.0960 kg = = 1.19 10 -4 m 3 = 119 mL N2 810 kg / m 3 Now the volume of nitrogen gas (at 77 K) is 1500 mL before the iron is dropped in. The volume of the piece of iron excludes 25 mL of gas, so the initial gas volume, when the lid is sealed and the liquid starts to boil, is V1 = 1475 mL. The pressure is p1 = 1.0 atm and the temperature is T1 = 77 K. Thus the number of moles of nitrogen gas is n1 = p1V1 (101, 300 Pa)(1.475 10 -3 m 3 ) = = 0.234 mol (8.31 J / mol K)(77 K) RT1 119 mL of liquid boils away, so the gas volume increases to V2 = V1 + Vboil = 1475 mL + 119 mL = 1594 mL. The temperature is still T2 = 77 K, but the number of moles of gas has been increased by the liquid that boiled. The number of moles that boiled away is nboil = 96.0 g = 3.429 mol 28 mol / g Thus the number of moles of nitrogen gas increases to n2 = n1 + nboil = 0.234 mol + 3.429 mol = 3.663 mol. Consequently, the gas pressure increases to p2 = n2 RT2 (3.663 mol)(8.31 J / mol K)(77 K) = = 1.470 10 6 Pa = 14.5 atm 1.594 10 -3 m 3 V2 Assess: Don't try this! The large pressure increase could cause a flask of liquid nitrogen to explode, leading to serious injuries.
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Stevens - PEP - 111
18.1. Solve: We can use the ideal-gas law in the form pV = NkBT to determine the Loschmidt number (N/V):1.013 10 5 Pa N p = 2.69 10 25 m -3 = = V kB T (1.38 10 -23 J K )(273 K )()18.2. Solve: Nitrogen is a diatomic molecule, so r 1.0 10-1
Stevens - PEP - 111
19.1. Model: The heat engine follows a closed cycle, starting and ending in the original state. The cycleconsists of three individual processes. Visualize: Please refer to Figure Ex19.1. Solve: (a) The work done by the heat engine per cycle is the a
Stevens - PEP - 111
20.1. Model: This is a wave traveling at constant speed. The pulse moves 1 m to the right every second.Visualize: Please refer to Figure Ex20.1. The snapshot graph shows the wave at all points on the x-axis at t = 0 s. You can see that nothing is h
Stevens - PEP - 111
21.1. Model: The principle of superposition comes into play whenever the waves overlap.Visualize:The graph at t = 1 s differs from the graph at t = 0 s in that the left wave has moved to the right by 1 m and the right wave has moved to the left by
Stevens - PEP - 111
22.1. Visualize: Please refer to Figure Ex22.1.Solve: (a)(b) The initial light pattern is a double-slit interference pattern. It is centered behind the midpoint of the slits. The slight decrease in intensity going outward from the middle indicates
Stevens - PEP - 111
23.1. Model: Light rays travel in straight lines.Solve: (a) The time ist=x 1.0 m = = 3.33 10 -9 s = 3.33 ns c 3 10 8 m / s(b) The refractive indices for water, glass, and zircon are 1.33, 1.50, and 1.96, respectively. In a time of 3.33 ns, l
Stevens - PEP - 111
24.1. Model: Balmer's formula predicts a series of spectral lines in the hydrogen spectrum.Solve: Substituting into the formula for the Balmer series,=91.18 nm 91.18 nm = = 410.3 nm 1 1 1 1 - 2 - 2 2 22 n 2 6where n = 3, 4, 5, 6, . and wher
Stevens - PEP - 111
ELECTROMAGNETIC AND WAVES FIELDSw.1. Model: The net magnetic flux over a closed surface is zero. Visualize: Please refer to Ex34.1. Solve: Because we can't enclose a &quot;net pole&quot; within a surface, Q, = f B . d i = 0 . Since the magnetic field isunif
FSU - CLA - 2123
CLA 2123: The Roman Way First Exam. February 8, 2007Name _Please read all directions carefully; no credit will be given for doing more than is required in each section. Part I. Identifications (35 points). Choose FIVE of the following and identif
Stevens - PEP - 111
14.1. Solve: The frequency generated by a guitar string is 440 Hz. The period is the inverse of the frequency, henceT= 1 1 = = 2.27 10 -3 s = 2.27 ms f 440 Hz14.2. Solve: Your pulse or heart beat is 75 beats per minute. The frequency of your hear
Stevens - PEP - 111
1.1.Solve:1.2.Solve:Solve: (a) The basic idea of the particle model is that we will treat an object as if all its mass is concentrated into a single point. The size and shape of the object will not be considered. This is a reasonable approxim
Stevens - PEP - 111
2.1.Solve:Model: The car is represented by the particle model as a dot. (a) Time t (s) Position x (m) 0 1200 1 975 2 825 3 750 4 700 5 650 6 600 7 500 8 300 9 0(b)2.2. Solve:Diagram (a) (b) (c)Position Negative Negative PositiveVelocity
Stevens - PEP - 111
3.1. Solve: (a) If one component of the vector is zero, then the other component must not be zero (unless the whole vector is zero). Thus the magnitude of the vector will be the value of the other component. For example, if Ax = 0 m and Ay = 5 m, the
Stevens - PEP - 111
4.1. Solve: A force is basically a push or a pull on an object. There are five basic characteristics of forces. (i) A force has an agent that is the direct and immediate source of the push or pull. (ii) Most forces are contact forces that occur at a
Stevens - PEP - 111
5.1.Model: We can assume that the ring is a single massless particle in static equilibrium. Visualize:Solve:Written in component form, Newton's first law is( Fnet ) x = Fx = T1x + T2 x + T3 x = 0 NT1 x = - T1T1y = 0 N Using Newton's first l
Stevens - PEP - 111
6.1. Model: We will assume motion under constant-acceleration kinematics in a plane.Visualize:Instead of working with the components of position, velocity, and acceleration in the x and y directions, we will use the kinematic equations in vector f
Stevens - PEP - 111
7.1. Solve: (a) From t = 0 s to t = 1 s the particle rotates clockwise from the angular position +4 rad to -2 rad. Therefore, = -2 - ( +4 ) = -6 rad in one sec, or = -6 rad s . From t = 1 s to t = 2 s, = 0 rad/s. From t = 2 s to t = 4 s the partic
Stevens - PEP - 111
8.1. Visualize:Solve: Figure (i) shows a weightlifter (WL) holding a heavy barbell (BB) across his shoulders. He is standing on a rough surface (S) that is a part of the earth (E). We distinguish between the surface (S), which exerts a contact forc
Stevens - PEP - 111
Solve: (a) The momentum p = mv = (1500 kg)(10 m /s) = 1.5 10 4 kg m /s . (b) The momentum p = mv = (0.2 kg)( 40 m /s) = 8.0 kg m /s .9.1. Model: Model the car and the baseball as particles.9.2. Model: Model the bicycle and its rider as a particl
Stevens - PEP - 111
10.1. Model: We will use the particle model for the bullet (B) and the bowling ball (BB).Visualize:Solve:For the bullet,KB =For the bowling ball,1 1 2 mB vB = (0.01 kg)(500 m /s) 2 = 1250 J 2 2 1 1 2 mBB vBB = (10 kg)(10 m / s) 2 = 500 J 2
Stevens - PEP - 111
11.1. Visualize:r Please refer to Figure Ex11.1. rSolve: (b) (c)(a) A B = AB cos = ( 4)(5)cos 40 = 15.3. r r C D = CD cos = (2)( 4)cos120 = -4.0. r r E F = EF cos = (3)( 4)cos 90 = 0.11.2. Visualize:r Please refer to Figure Ex11.2. rSolve
Stevens - PEP - 111
12.1.Solve: (b)Model: Model the sun (s), the earth (e), and the moon (m) as spherical. (a)Fs on e =Gms me (6.67 10 -11 N m 2 / kg 2 )(1.99 10 30 kg)(5.98 10 24 kg) = 3.53 10 22 N = (1.50 1011 m ) 2 rs2 e -Fm on e =GMm Me (6.67 10 -1
Stevens - PEP - 111
13.1. Model: The crankshaft is a rotating rigid body.Solve: The crankshaft at t = 0 s has an angular velocity of 250 rad/s. It gradually slows down to 50 rad/s in 2 s, maintains a constant angular velocity for 2 s until t = 4 s, and then speeds up
FSU - CLA - 2123
Classics 2123: The Roman Way Lecture 3. The Early History of Rome and Roman Republican Government Readings: BHR 15-41; Shelton 2-4, 7-8, 251-3, 255-259, 262, 264-5Spring 2008Early History of Rome Regal period: dimly known, mostly through legends;
FSU - CLA - 2123
Classics 2123: The Roman Way Lecture 4. Roman Family LifeSpring 2008Readings: BHR 129-31; Shelton nos. 15, 17-23, 25-27, 30-37, 44-45, 50, 54-56, 59-61, 63-67, 72, 75, 119-120, 124-5I. Definition of the family - familia - power of the father (p
FSU - CLA - 2123
Classics 2123: The Roman Way Lecture 4. Roman Family LifeSpring 2008Readings: BHR 129-31; Shelton nos. 15, 17-23, 25-27, 30-37, 44-45, 50, 54-56, 59-61, 63-67, 72, 75, 119-120, 124-5I. Definition of the family - familia - power of the father (p
FSU - CLA - 2123
Classics 2123: The Roman Way Lecture 6. Internal Disorders; Roman Slavery Readings: BHR 82-92; Shelton 207-209, 219, 227-229, 317-318Spring 2008I. Establishment of Roman provincial government Provincia: sphere of action of a magistrate with imper
UMiami - ACC - 212
Chapter 10Standard CostingAccounting 21210 - 1Learning Objective 1 Describe standard costing and indicate why standard costing is important.Accounting 212 10 - 2Why is Standard Costing Used?A standard is a preestablished benchmark for des
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Classics 2123: The Roman Way Outline for Lecture 7. Roman Religion Readings: BHR 41-44; Shelton nos. 402-419, 423-428Spring 2008Problems Studying Ancient Religious Systems - time and culture (modern politics separated from religion) - vocabulary
FSU - CLA - 2123
Classics 2123: The Roman Way Lecture 8. The Strains of Empire (I) Readings: BHR 72-77, 92-110; Shelton 187-189, 266, 317-318Spring 2008I. Continuation of Roman Imperialism Macedonian Wars - Rome fights four wars in Greece, first against the Maced
FSU - CLA - 2123
Classics 2123: The Roman Way Lecture 9. The Strains of Empire (II) Readings: BHR 111-128, 132-140Spring 2008I. New Developments in the Roman State in the Late Republic Breakdown of concordia in the late Republic the result of many things, includi
UMiami - ACC - 212
Chapter 9The Operating Budget2004 Prentice Hall Business Publishing Introduction to Management Accounting , 2/e Werner/Jones9-1Learning Objective 1Describe some of the benefits of the operating budget.2004 Prentice Hall Business Publishing
FSU - CLA - 2123
Classics 2123: The Roman Way Lecture 10. The Fall of the Roman Republic Readings: BHR 124-179.Fall 2008I. Career of Pompey (Gn. Pompeius Magnus) (106-48 BCE) - begins as supporter of Sulla: raises legions from his father's troops (client army); a
UMiami - ACC - 212
Chapter 11Evaluating PerformanceAccounting 21211 - 1Learning Objective 1Describe centralized and decentralized management styles.Accounting 212 11 - 2Centralized ManagementTop management makes most of the decisions.The most experience
FSU - CLA - 2123
Classics 2123: The Roman Way Lecture 11. Augustus' `Restored' Republic Readings: BHR 167-199; Shelton 38-40, 77-78, 267, 271, 274-276, 294-305.Spring 2008I. From Octavian to `Augustus' - 31 BCE victory over Antony and Cleopatra at Actium; after h
FIU - GEG - 211
Earth 1. When was Big bang and when did our Earth form? 13.5 billion years ago big bang. 4.53 billion years. 2. Know earth's radius/diameter, thicknesses of crusts and lithosphere 6,370km radius, crust is 40 km, lithosphere is 100 km 3. Know the phys
UMiami - ACC - 212
HAPTER 1MANAGEMENT ACCOUNTING: ITS ENVIRONMENT AND FUTURESOLUTIONS TO CHAPTER 1 QUICK QUIZ 1. 2. 3. 4. 5. C B D C D 6. 7. 8. 9. 10. C B D B D 2004 Prentice Hall, Inc.M1 - 2Chapter 1 Management Accounting: Its Environment and FutureQUICK
FSU - CLA - 2123
Classics 2123: The Roman Way Lecture 12: Virgil's Aeneid, Books I-IVSpring 2008I. Publius Vergilius Maro (Virgil or Vergil) - born ca. 70, died ca. 19 BCE - from northern Italy; ancient tradition that his family lost land in the proscriptions of
FSU - CLA - 2123
Classics 2123: The Roman Way Lecture 12. Virgil's Aeneid, Books II-VIIISpring 2008I. Dido and Aeneas - Dido's story: the wrong done her husband; her exile; her foundation of Carthage - her welcoming of the Trojans (the concern of Jupiter) - the p
FSU - CLA - 2123
Classics 2123: The Roman Way Lecture 14: Virgil's Aeneid, Books VII-XII I. War in the Aeneid war always considered glorious in epic, but Virgil here exploring a civil war war also in the world of the Aeneid not an end, but a means to an endSpring 2
FSU - CLA - 2123
Classics 2123: The Roman Way Ovid's AmoresSpring 2008Readings: Amores (given by Book and poem number): I. 3, 5, 7, 9, 14, 15; II. 1, 4, 7, 13, 14; III. 7, 8, 15 I. Ovid (P. Ovidius Naso) - other great poet of the Augustan age - poet of love par e
FSU - CLA - 2123
Classics 2123: The Roman Way Ovid's Ars Amatoria Readings: Ars Amatoria Books IIIISpring 2008I. Precedents and Heritage - AA meant to be seen as a didactic poem: long history of didactic poetry in Greek and Roman literature - first practitioner H
FSU - CLA - 2123
Classics 2123: The Roman WaySpring 2008Lecture 17. The Julio-Claudians; Imperial Women and the Imperial HouseholdReadings: BHR 201-221; Shelton 78, 429-431 JulioClaudian Emperors: Name given to the first five emperors: Augustus (ruled 27 BCE14 C
FSU - CLA - 2123
Classics 2123: The Roman Way Lecture 18: Social Life in the Early EmpireSpring 2008Readings: BHR 241-5; Shelton 9-13, 41-42, 134-139, 141-157, 160, 220-226, 232-250, 377-81, 383-88, 391-401 I. Education and Schools (Shelton 134-139, 141-157, 160)
FSU - CLA - 2123
Classics 2123: The Roman Way Lecture 19: The Empire at its Height Readings: BHR 224-240; Shelton 52-53, 275-276, 294-316Spring 2008I. Civil War and the Year of the Four Emperors (6869 CE) o at the fall of Nero (68), no relation of Augustus remain
FSU - CLA - 2123
(On-line) Lecture 17. The Julio-Claudians; Imperial Women We define the Julio-Claudians as those emperors related to Augustus or to Livia, his second wife; Julio denotes the `Julian' strand, `Claudian' the Claudian strand since Livia's first husband
FSU - CLA - 2123
On-Line Lecture 18: Social Life in the Early Empire This lecture surveys a number of aspects of social life in the early Empire, roughly the first century CE. Some of the features characteristic of this time period continue on into the middle and lat
FSU - CLA - 2123
1 On-Line Lecture 19. The Empire at its HeightI. Civil War and the Year of the Four Emperors (6869 CE) Nero was the last of the Julio-Claudians, those people who could trace their descent either to Augustus' family (the Julians) or Livia's family (
FSU - CLA - 2123
Classics 2123: The Roman Way Study Questions for Lecture 2. Roman Society and Roman ValuesSpring 20081. How did the Romans conceive of virtus? 2. What did a Roman mean by pietas? How did Roman notions of pietas differ from what English speakers m
FSU - CLA - 2123
Classics 2123: The Roman Way Study Questions for Lecture 3. Roman GovernmentSpring 20081. What form of government did the Romans have after the expulsion of the kings? 2. What is a patrician? A plebeian? What is meant by the Struggle of the Order
FSU - CLA - 2123
Classics 2123: The Roman Way Study Questions for Lecture 4. Roman Family ValuesSpring 20081. What is a familia? 2. What is a paterfamilias? What is patria potestas? 3. What is meant by manus? 4. What was the role of the state in Roman marriage? W
FSU - CLA - 2123
Classics 2123: The Roman Way Study Questions for Lecture 5. Roman Imperialism and Expansion 1. What do we mean by `defensive imperialism'?Spring 20082. What kinds of relationships had the Romans forged with the other inhabitants of the Italian pe
FSU - CLA - 2123
Classics 2123: The Roman Way Study Questions for Lecture 6. Internal Disorders; Roman SlaverySpring 20081. What defines a Roman `province'?2. When did the Romans begin to amass provinces?3. What do the following terms mean: imperium, prorogat
FSU - CLA - 2123
Classics 2123: The Roman Way Study Questions for Lecture 7. Roman ReligionSpring 20081. What are some of the difficulties present to us in studying Roman religion?2. What is polytheism? How does it differ from monotheism (aside from the obvious
FSU - CLA - 2123
Classics 2123: The Roman Way Study Questions for Lecture 8. The Strains of Empire (I)Spring 20081. What do the terms dies fasti and dies nefasti mean?2. What was the Lupercalia?3. What was the Saturnalia?4. What is the significance of the y
FSU - CLA - 2123
Classics 2123: The Roman Way Study Questions for Lecture 9. The Strains of Empire (II)Spring 20081. What do we mean by the term concordia? What are the features of concordia? What were some of the factors leading to its demise?2. What do we mea
FSU - CLA - 2123
Classics 2123: The Roman Way Study Questions for Lecture 10: The Fall of the Roman RepublicFall 20081. What is significant about Pompey's command against the pirates and against Mithridates?2. What similarities can be seen in the careers of Sul
FSU - CLA - 2123
Classics 2123: The Roman Way Study Questions for Lecture 11: Augustus' `Restored' RepublicSpring 20081. What benefits did Octavian receive by being named Caesar's heir and adopted son?2. What benefits did Antony receive from his alliance with C
FSU - CLA - 2123
Classics 2123: The Roman WaySpring 2008Study Questions for Lecture 12: Augustus' Reforms / Virgil's Aeneid1. What reforms to the army did Augustus make? How did he deal with the client army?2. Who are the Praetorian Guard and what was their p
FSU - CLA - 2123
Classics 2123: The Roman Way Study Questions for Lecture 13. Virgil's Aeneid, Books I-VISpring 20081. What is our first view of Aeneas?2. What does Jupiter prophesy about Aeneas' future?3. What thematic relevance does the story of Troy's fall
FSU - CLA - 2123
Classics 2123: The Roman Way Study Questions for Lecture 14: Virgil's Aeneid, Books VII-XIISpring 20081. How is the battle between the Trojans and Italians in the Aeneid a civil war? What associations would such a conflict have raised in the mind