Chapter 17 - PHYS 132, WEEK #8 NOV 8-12, 2004 17.4. Model:...

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Unformatted text preview: PHYS 132, WEEK #8 NOV 8-12, 2004 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. 17.14. 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 DE th = W + Q = - pDV + Q = (4.0 105 Pa)(200 600) 10-6 m3 100 J = 60 J Thermal energy increases by 60 J. 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 -114 C - 20 C = -134 C = -134 K. The mass is M = rV = 789 kg / m ( 3 )(200 10 -6 m 3 ) = 0.1578 kg The heat needed for the two steps is Q1 = McalcoholDT = (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. 17.23. 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 = 71.2 C. Thus (0.050 kg)( 750 J / kg K)( 71.2C - 20.0C) + (200 10 = 1920 J + 838 (J / K)(71 .2C - Ti water ) = 0 J -6 m )(1000 kg / m )( 4190 J / kg K)( 71.2C - Ti water ) 3 3 fi Ti water = 73.5 C Assess: The thermometer reads 71.2 C for a real temperature of 73.5 C. This is reasonable. 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= M M mol = 1.0 g 32 g mol = 0.03125 mol For the isobaric process, Q = nCP DT = (0.03125 mol)(29.2 J/mol K)(100C) = 91.2 J (b) For the isochoric process, Q = nCV DT = 91.2 J = (0.03125 mol)(20.9 J/mol K)DT fi DT = 140C 17.30. Model: We assume the gas is an ideal gas and g = 1.40 for a diatomic gas. Solve: Using the ideal-gas law, Vi = nRTi pi = (0.10 mol )(8.31 J mol K )(423 K ) (3 1.013 10 g 5 Pa ) = 1.157 10 -3 m 3 For an adiabatic process, p i Vi = p f Vf g 1 1.40 1g p^ fi Vf = Vi i ~ pf = 1.157 10 ( -3 pi ^ m ~ 0.5 pi 3 ) = 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 ^ = (423 K ) ~ = 346.9 K = 73.9C ~ p i Vi pi 1.157 10 -3 m 3 p f Vf 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 = M ice cice DT = ( 0.100 kg )(2090 J kg K )(10 K ) = 2090 J Q 2 = M ice Lf = (0.100 kg ) 3.33 10 5 J kg = 33, 300 J Q3 = M ice cwater DT = ( 0.100 kg )(4190 J kg K )(20 K ) = 8380 J Q 4 = M Al cAl DT = M Al (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. 17.44. Model: The liquefaction of the nitrogen occurs in two steps: lowering nitrogen's temperature from 20C to -196C, and then liquefying it at -196C. Assume the cooling occurs at a constant pressure of 1 atm. 3 -3 3 Solve: The mass of 1.0 L of liquid nitrogen is M = rV = 810 kg m 10 m = 0.810 kg . This mass corresponds to ( )( ) n= M M mol = 810 g 28 g mol = 28.9 mols At constant atmospheric pressure, the heat to be removed from 28.93 mols of nitrogen is Q = ML v + nC P DT = -(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 = Q ice + Qcoffee = (Q 1 + Q 2 + Q3 ) + ( Q4 ) = 0 J Q1 = M ice cice DT = M ice (2090 J kg K )( 20 K ) = M ice ( 41,800 J kg ) Q 2 = M ice Lf = M ice (330, 000 J kg) Q 3 = M ice c water DT = M ice ( 4190 J kg K )(60 K ) = M ice (251, 400 J kg ) Q4 = Mcoffee c coffee DT = 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 M ice ( 41,800 + 330, 000 + 251, 400 ) J kg - 37,710 J = 0 J fi M ice = 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 = Q reactor + Qwater = 0 J The heat from the reactor in Dt = 1 s is Q reactor = -2000 MJ = -2.0 10 J and the heat absorbed by the water is Q water = m water cwater DT = m water (4190 J kg K )(12 K ) fi -2.0 10 J + m water ( 4190 J kg K )(12 K ) = 0 J fi m water = 3.98 10 kg 9 4 9 Each second, 3.98 104 kg of water is needed to remove heat from the nuclear reactor. Thus, the water flow per minute is 60 s 1m 3 1L 4 kg 3.98 10 -3 3 = 2.39 106 L/min s min 1000 kg 10 m 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 Kavg = 1 m vavg 2 m = 2 u = 2 1.661 10 ( ) 2 is the average kinetic energy per molecule. The mass -27 of a hydrogen molecule is ( -27 kg = 3.32 10 ) kg The number of molecules in a 1 g = 0.001 kg sample is N= M m 1 2 = 0.001 kg 3.32 10 -27 kg -27 = 3.01 10 23 The average kinetic energy per molecule is Kavg = 1 2 m vavg ( ) 2 = (3.32 10 kg (700 m s ) = 8.13 10 ) 2 -22 J Thus, the thermal energy in the 1-gram sample of the gas is E th = 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: DE th = 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 Kavg = E th N = 1.487 10 -21 J = 1 m vavg 2 ( ) 2 Solving for the new average speed gives v avg = 2 K avg m = 2 1.478 10 -21 J 3.32 10 -27 kg ( ) = 944 m s 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 p 1 = 3.0 atm = 304,000 Pa, V 1 = 100 cm3 = 1.0 10-4 m3, and T1 = 100 C = 373 K. The number of moles of gas is n= p 1 V1 RT1 = (304, 000 Pa )(1. 0 10 -4 m 3 ) (8.31 J mol K )(373 K ) = 9.81 10 -3 mol At point 2 we have p2 = p1 = 3.0 atm and V2 = 300 cm3 = 3V1. This is an isobaric process, so V2 T2 = V1 T1 fi T2 = V2 V1 T1 = 3(373 K) = 1119 K The gas is heated to raise the temperature from T1 to T2. The amount of heat required is Q = nC P DT = 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 = 100 C = 373 K. This is an isochoric process, so Q = nC V DT = 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. 17.67. 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 = 30 C = 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, DE th = Q + W = nC V DT fi W = nCVT fi 400 J = (0.024 mol)(20.8 J/mol K)(Tf 303 K) fi Tf = 1100 K For an adiabatic process Equation 17.40 is 1 Tf V g -1 f = T0 V g -1 0 T ^ g -1 -4 3 303 K ^ 1.4-1 -5 3 3 fi Vf = V0 0 ~ = 6.0 10 m ~ = 2.39 10 m = 23 .9 cm 1100 K Tf 1 ( ) 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 ^ ln( 2.5) g = i ~ fi 2.5 = (2.0 ) fi g = = 1.32 p f Vf = p i Vi fi p i Vf ln( 2.0) g g pf g (b) Equation 17.40 for an adiabatic process is Tf V g -1 f = Ti V g -1 i V ^ fi = i~ Ti Vf Tf g -1 = ( 2.0 ) 1.32 -1 = (2.0 ) 0.32 = 1. 25 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 p i = 60 103 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 g -1 f = Ti V g -1 i T ^ V ^ fi f~ = i~ Ti Vf g -1 1 p ^g V pf Vf = p i Vi fi f ~ = i pi Vf g g Combining these two equations, 1.4 -1 (T f Ti ) = ( pf p i ) g -1 g 3 100 10 Pa ^ fi Tf = Ti ~ 60 10 3 Pa 1.4 5^ = ( 273 K ) 3 0.286 = 316 K = 43C = 109F ...
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