PHYS 1101 Practice problem set 6, Chapter 19: 7, 12, 19, 30, 37, 44, 53, 61, 69

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Harold Goodman
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1 PYS 0 Practice problem set 6, hapter 9: 7,, 9, 0, 7, 44,, 6, Solve: (a) he heat extracted from the cold reservoir is calculated as follows: (b) he heat exhausted to the hot reservoir is K J in 0 J + in 00 J + 0 J 0 J 9.. Model: Process A is adiabatic, process B is isochoric, process is isothermal, and process D is isobaric. Visualize: Please refer to Figure Ex9.. Solve: Process A is adiabatic, so 0 J. ork s is positive as the gas expands. Since s + E th 0 J, E th must be negative. he temperature falls during an adiabatic expansion. Process B is isochoric. No work is done ( s 0 J), and is negative as heat energy is removed to lower the temperature ( E th negative). Process is isothermal, so 0 and E th 0 J. he gas is compressed, so s is negative. s for an isothermal process, so is negative. eat energy is withdrawn during the compression to keep the temperature constant. Finally, work s is positive during the isobaric expansion of process D. emperature increases, so E th is also positive. his makes s + E th positive. E th s A + 0 B 0 0 D + + +

2 9.9. Model: he efficiency of a arnot engine ( arnot ) depends only on the temperatures of the hot and cold reservoirs. On the other hand, the thermal efficiency () of a heat engine depends on the heats and. Visualize: Please refer to Figure Ex9.9. Solve: (a) According to the first law of thermodynamics, out +. For engine (a), 0 J, 0 J and out 0 J, so the first law of thermodynamics is obeyed. For engine (b), 0 J, 7 J and out 4 J, so the first law is violated. For engine (c) the first law of thermodynamics is obeyed. (b) For the three heat engines, the maximum or arnot efficiency is Engine (a) has 00 K arnot 600 K J 0 J out 0.60 his is larger than arnot, thus violating the second law of thermodynamics. For engine (b), out 0.40 < 4 J 0 J so the second law is obeyed. Engine (c) has a thermal efficiency that is so the second law of thermodynamics is obeyed. 0 J 0. < 0 J arnot arnot 9.0. Model: he coefficient of performance of a arnot refrigerator depends only on the temperatures of the cold and hot reservoirs. Solve: Using the definition of the coefficient of performance for a arnot refrigerator, K arnot ( + ) 7 K 60 K K 60 K K o increase the coefficient of performance to 0.0, the hot-reservoir temperature changes to. hus, K arnot 60 K K 60 K Since is less than, the hot-reservoir temperature must be decreased by 6 K or Solve: e can also obtain the desired result using the and K equations for the arnot engine and the arnot refrigerator: ence the equation for K becomes K ( ) K

3 9.44. Visualize: Refer to Figure P9.44 in your textbook. Solve: (a) is given as 000 J. Using the energy transfer equation for the heat engine, he thermal efficiency of a arnot engine is + out + out out 00 K K out ( ) 000 J J o determine and 4, we turn our attention to the arnot refrigerator, which is driven by the output of the heat engine with in out. he coefficient of performance is K 400 K K 400 K in out 4 K J 000 J Using now the energy transfer equation in + 4, we have out J J 00 J (b) From part (a) 00 J and 000 J, so >. (c) Although 000 J and 00 J, the two devices together do not violate the second law of thermodynamics. his is because the hot and cold reservoirs are different for the heat engine and the refrigerator.

4 7 9.. Model: he heat engine follows a closed cycle. For a diatomic gas, V R and P R. Visualize: Please refer to Figure P9.. Solve: (a) Since 9 K, the number of moles of the gas is n R 6 ( Pa )( 0 0 m ) mol ( 8. J/mol K)( 9 K) At point, V 4 V and p p. he temperature is calculated as follows: 4 9 K 6 K At point, V V 4 V and p p. he temperature is calculated as before: 4 9 K 7 K For process, the work done is the area under the p-versus-v curve. hat is, he change in the thermal energy is ( 0. atm)( 40 cm 0 cm ). atm 0. atm 40 cm 0 cm + s Pa ( 0 0 m )( atm).04 J atm 4 E n.08 0 mol 8. J/mol K 6 K 9 K.9 J th V he heat is s + Eth 6.97 J. For process, the work done is s 0 J and For process, s E n n R th V mol 8. J/mol K 7 K 6 K 0. J 6 4 th V 0. atm 0 cm 40 cm Pa 0 0 m. J E n.08 0 mol 8. J/mol K 9 K 7 K.80 J he heat is Eth + s. J. (b) he efficiency of the engine is (c) he power output of the engine is s (J) (J) E th Net.. 0 net. J 6.97 J % revolutions min 00 min 60 s revolution 60 net 00. J/s.7 Assess: For a closed cycle, as expected, ( s ) net net and ( E th ) net 0 J. 4

5 9.6. Model: he closed cycle in this heat engine includes adiabatic process, isobaric process, 7 7 and isochoric process. For a diatomic gas, V R, P R, and γ.4. Visualize: Please refer to Figure P9.6. Solve: (a) e can find the temperature from the ideal-gas equation as follows: nr γ γ e can use the equation pv pv to find V, ( Pa)(.0 0 m ) ( 0.00 mol)( 8. J/mol K) 407 K / /.4 p Pa (.0 0 m ).69 0 m.0 0 Pa V V p γ he ideal-gas equation can now be used to find, At point, V V so we have nr (.0 0 Pa)(.69 0 m ) nr ( 0.00 mol)( 8. J/mol K) ( 4 0 Pa)(.69 0 m ) (b) For adiabatic process, 0 J, Eth s, and ( 0.00 mol)( 8. J/mol K) 60 K 6479 K ( ) ( 0.00 mol )( 8. J/mol K)( 407 K 60 K) γ nr γ.4 s For isobaric process, 7 ( ) ( 0.00 mol) 7 ( 8. J/mol K)( 6479 K 407 K) n n R P 7.0 J 69 J Eth nv n R 69 J he work done is the area under the p-versus-v graph. ence, For isochoric process, s 0 J and ( ) s Pa.69 0 m.0 0 m 677 J th V E n 0.00 mol 8. J/mol K 60 K 6479 K 09 J (c) he engine s thermal efficiency is E th (J) s (J) (J) Net J % 69 J

6 9.69. Model: he heat engine follows a closed cycle, starting and ending in the original state. Visualize: Please refer to Figure P9.69. he figure indicates the following seven steps. First, the pin is inserted when the heat engine has the initial conditions. Second, heat is turned on and the pressure increases at constant volume from atm to atm. hird, the pin is removed. he flame continues to heat the gas and the volume increases at constant pressure from 0 cm to 00 cm. Fourth, the pin is inserted and some of the weights are removed. Fifth, the container is placed on ice and the gas cools at constant volume to a pressure of atm. Sixth, with the container still on ice, the pin is removed. he gas continues to cool at constant pressure to a volume of 0 cm. Seventh, with no ice or flame, the pin is inserted back in and the weights returned bringing the engine back to the initial conditions and ready to start over. Solve: (a) (b) he work done per cycle is the area inside the curve: out ( p)( V) ( 0,00 Pa)(0 0 6 m ) 0. J (c) eat energy is input during processes and, so +. his is a diatomic gas, with V R and P R. he number of moles of gas is 6 pv (0,00 Pa)(0 0 m ) n mol R (8. J/mol K)(9 K) Process is isochoric, so (p /p ) 879 K. Process is isobaric, so (V /V ) 78 K. hus Similarly, n nr( ) ( mol)(8. J/mol K)(86 K). J V n nr( ) ( mol)(8. J/mol K)(879 K).8 J 7 7 P hus. J +.8 J 78.0 J and the engine s efficiency is out 0. J 78.0 J 0.9.9% 6

HEAT ENGINES AND REFRIGERATORS

HEAT ENGINES AND REFRIGERATORS EA ENGINES AND REFRIGERAORS 9 onceptual uestions 9.. s =. (a) < 0, > 0. ork is done by the system; the area under the curve is positive. s (b) > 0, < 0. ork is done on the system to compress it to a smaller