Tài liệu Physics exercises_solution: Chapter 19 pptx

19.1: a) b) .J1033.1)C80)(KmolJ3145.8)(mol00.2( 3  TnRVp 19.2: a) b) If the pressure is reduced to 40.0% of its original value, the final volume is )25( of its original value. From Eq. (19.4), J.1015.9 2 5 ln )K)(400.15KmolJ3145.8)(3(ln 3 1 2         V V nRTW 19.3: nRTpV  T constant, so when p increases, V decrease 19.4: At constant pressure, so,TnRVpW  C.62.1C35.1C0.27so,Tand K1.35 K)molJ(8.3145mol)(6 J1075.1 2CK 3      TT nR W T 19.5: a) b) At constant volume, 0.soand0  WdV 19.6: b) J.1050.4)m0900.0m(0.0600Pa)10(1.50 3335 Vp 19.7: a) b) In the first process, 0 1  VpW . In the second process, J.1000.4)m0.080(Pa)10005( 435 2  .VpW 19.8: a) .0 and)(,0),( 41212243212113  WVVpWWVVpW The total work done by the system is ),)(( 122141243213 VVppWWW W which is the area in the p- V plane enclosed by the loop. b) For the process in reverse, the pressures are the same, but the volume changes are all the negatives of those found in part (a), so the total work is negative of the work found in part (a). 19.9: J73J, 254  WQ (work is done on the system), and so J. 327 WQU 19.10: a) J.1078.3)mPa)(0.2101080.1( 425 Vp b) J.107.72J103.78J1015.1 445  WQU c) The relations WQUVpW  and hold for any system. 19.11: The type of process is not specified. We can use WQU  because this applies to all processes. Q is positive since heat goes into the gas; J1200Q W positive since gas expands; J2100W J900J2100-J1200 U We can also use   TRnU  2 3 since this is true for any process for an ideal gas.       C4.14 K)molJ5mol)(8.3143(5.00 J)900(2 3 2 nR U T C113C14.4C127 12  TTT 19.12: At constant volume, the work done by the system is zero, so .QWQU  Because heat flows into the system, Q is positive, so the internal energy of the system increases. 19.13: a) J1015.1)mPa)(-0.501030.2( 535 Vp . (b J1055.2J)1015.1(J1040.1 555  WUQ (heat flows out of the gas). c) No; the first law of thermodynamics is valid for any system. 19.14: a) The greatest work is done along the path that bounds the largest area above the V-axis in the p- V plane (see Fig. (19.8)), which is path 1. The least work is done along path 3. 0 Wb)  in all three cases; 0so,  QWUQ for all three, with the greatest Q for the greatest work, that along path 1. When ,0Q heat is absorbed. 19.15: a) The energy is kcal,139g)kcalg)(9.0(7.0g)kcalg)(4.00.17()gkcal0.4)(g0.2(  and the time required is  mKv 2b)min.16.4h0.273h)kcal(510kcal)139( h.km501sm139kg)60(cal)J(4.186cal)10139(2 3  19.16: a) The container is said to be well-insulated, so there is no heat transfer. b) Stirring requires work. The stirring needs to be irregular so that the stirring mechanism moves against the water, not with the water. c) The work mentioned in part (b) is work done on the system, so ,0W and since no heat has been transferred, .0 WU 19.17: The work done is positive from a to b and negative from b to a; the net work is the area enclosed and is positive around the clockwise path. For the closed path ,0U so A.0 WQ positive value for Q means heat is absorbed. b) J.7200soand0(a),part fromandJ,7200  WQQQ c) For the counterclockwise path, Q = W < 0. W= J7200soJ,7200  Q and heat is liberated, with |Q|=7200 J. 19.18: a), b) The clockwise loop (I) encloses a larger area in the p-V plane than the counterclockwise loop (II). Clockwise loops represent positive work and counterclockwise loops negative work, so .0and0 III  WW Over one complete cycle, the net work ,0 III WW and the net work done by the system is positive. c) For the complete cycle, .so and0 QWU  From part (a), W > 0 so Q > 0, and heat flows into the system. d) Consider each loop as beginning and ending at the intersection point of the loops. Around each loop, .0and0 then,;so,0 IIIIII  WQWQWQU Heat flows into the system for loop I and out of the system for loop II. 19.19: a) Yes; heat has been transferred form the gasses to the water (and very likely the can), as indicated by the temperature rise of the water. For the system of the gasses, 0Q . b) The can is given as being constant-volume, so the gasses do no work. Neglecting the thermal expansion of the water, no work is done. c) .0 QWQU 19.20: a) J.1067.1)m1000.1mPa)(0.82410026.2( 53335   Vp J.10032J1067.1)kgJ10kg)(2.2000.1( b) 656 v   . WmLWQU 19.21: a) Using Equation (19.12), orK,9.167d K)molJmol)(20.76(0.185 J645 d V  nC Q T T = 948 K. b) Using Equation (19.14), K.900or K,9.119d .K)molJmol)(29.07(0.185 J645d   TT p nC Q 19.22: a) J.4.99)CK)(40.0molJmol)(12.470100.0( TnC V 19.23:  C30.0mol.00.5 Tn a) For constant p, J3120)CK)(30.0molJmol)(20.7800.5(  TnCQ p 0Q so heat goes into gas. b) For constant V, J1870)CK)(30.0molJmol)(12.4700.5(  TnCQ v Q > 0 so heat goes into gas. c) For constant p, J5540)CK)(30.0molJmol)(36.9400.5(  TnCQ p Q > 0 so heat goes into gas. 19.24: For an ideal gas, ,TCU V  and at constant pressure, RCTnRVp V 2 3 Using.  for a monatomic gas, J.360)m1000.2m10Pa)(8.001000.4( 2 3 2 3 2 3 33334          VpTRnU 19.25: For constant TnCQp p , Since the gas is ideal, nRTpV  and for constant ., TnRVpp  Vp R C nR Vp nCQ p p                   Since the gas expands, 0V and therefore 0 .0  QQ means heat goes into gas. 19.26: For an ideal gas, ,TCU V  and at constant pressure, .TnRVpW  Using RC V 2 3  for a monatomic gas, .)( 2 3 2 3 2 3 WVpTRnU  Then .so, 5 2 2 5  QWWWUQ 19.27: a) For an isothermal process, 4)1K)ln(K)(350.15molJ5mol)(8.314150.0()(ln 12  VVnRTW J.605 b) For an isothermal process for an ideal gas, .0and0  UT c) For a process with ,0U J605 WQ ; 605 J are liberated. 19.28: For an isothermal process, J.335so,0  QWU 19.29: For an ideal gas ,1 VVp CRCCγ  and so  )1( V  RC KmolJ5.65)127.0(K)molJ3145.8(  and K.molJ8.73  RCC Vp 19.30: a) b) )( 1212 TTnRpVpV  J.208K)K)(100.0molJ5mol)(8.314250.0(  c) The work is done on the piston. d) Since Eq. (19.13) holds for any process, J.712K)K)(100.0molJmol)(28.46250.0(  TnCU V e) Either J10924givesor 3  QWUQTnCQ P to three significant figures. f) The lower pressure would mean a correspondingly larger volume, and the net result would be that the work done would be the same as that found in part (b). 19.31:    ,11a) γR C p  and so     J.553 220.111 C0.5KmolJ3145.8mol40.2     TnCQ p     J.454220.1J553b)  γTnCTnC PV (An extra figure was kept for these calculations.) 19.32: a) See also Exercise 19.36;   Pa.1076.4 m0400.0 m0800.0 Pa1050.1 5 3 3 5 2 1 12 3 5                    γ V V pp b) This result may be substituted into Eq. (19.26), or, substituting the above form for 2 p ,    J.1060.1 0400.0 0800.0 1m0800.0Pa1050.1 2 3 1 1 1 435 1 2 1 11 3 2                                    γ V V Vp γ W c) From Eq. (19.22),       ,59.10400.00800.0 321 1212   VVTT and since the final temperature is higher than the initial temperature, the gas is heated (see the note in Section 19.8 regarding “heating” and “cooling.”) 19.33: a) )19.6Examplein as,400.1Use(b) γ From Eq. (19.22),      C495K7681.11K15.293 400.01 2112   VVTT and from Eq. (19.24),      atm.1.291.11atm00.1 400.1 2112   VVpp 19.34: 0gasdiatomicidealfor 4.1  WUQγ for adiabatic process γ ii γ VPPV PdVWU    const     J. 105.1atmL50 L)(30atm)2.1( VPdVU 3 4.11 1.4-1 L)30( 1.4-1 L)10( 1.4 10L1 L30 1 1 V ii V VP 10L L30 i i            The internal energy increases because work is done on the gas 0).( U The temperature increases because the internal energy has increased. 19.35: For an ideal gas TnCU V  . .ofsign theassame theisofsign The TU  decreases.gas theof energy thenegative;issonegativeis.sopositiveis1and )(and so,and 1212 1 1212 1 22 1 11 1 22 1 11 UTTTpp ppTTpTpT pnRTVVTVT      [...]... the spray by mL  mCT  (1 2)mv 2 , or 1 1 L  CT  v 2  ( 4190 J kg  K ) (80 C)  (19 m s) 2  3.4  10 5 J kg 2 2 19. 56: Solving Equations (19. 22) and (19. 24) to eliminate the volumes, 1 1 p  γ γ p1γ 1T1γ  p21T2γ , or T1  T2  1  p   2 2 Using γ  7 for air, T1  (273.15 K)( 1.6010 5 ) 7  449 K, which is 176C 5 2.8010 6 19. 57: a) As the air moves to lower altitude its density increases;.. .19. 36: Equations (19. 22) and (19. 24) may be re-expressed as T2  V1    T1  V2    5 γ 1 γ V  p , 2  1 p1  V2    2 a ) γ  5 , p2  (4.00 atm)(2 3) 3  2.04 atm, T2  (350 K)(2 3) 3  267 K 3 7 2 b) γ  7 , p2  (4.00 atm) (2 3) 5  2.27 atm, T2  (350 K) (2 3) 5  298 K 5 19. 37: a) b) From Eq (19. 25),W  nCV T  (0.450 mol) (12.47 J mol... is no substantial difference between cV and c p 19. 54: a) βTV0  (5.1  10 5 (C) 1 )(70.0 C)(2.00  10 2 ) 3  2.86  10 8 m 3 b) pV  2.88  103 J c) Q  mCT  ρV0CT  (8.9  103 kg m 3 )(8.00  10 6 m 3 )(390 J kg  K)(70.0 C)  194 4 J a) To three figures, U  Q  194 0 J e) Under these conditions, the difference is not substantial 19. 55: For a mass m of ejected spray, the heat of...  601 K iii) Adiabatic: Using Equation (19. 22), T2  T1V1 1 V2 1  (301 K )(V1.67 )  (301 K )( 1 ).67  189 K 2 67 (2V1 ) 19. 39: See Exercise 19. 32 a) p2  p1 (V1 V2 )   (1.10  105 Pa) ((5.00  10 3 m 3 1.100  10 2 m 3 ))1.29  4.50  10 4 Pa b) Using Equation (19. 26), ( p V  p2V2 ) W 1 1  1 [(1.1  105 N m 3 )(5.0  10 3 m 3 )  (4.5  10 4 N m 3 )(1.0  10 2 m 3 )]  , (1.29  1)... Wba  4.789  103 J  6.735  103 J  0  1.95  10 3 J 19. 49: a) Ta = T c 19. 50: a) n  C pQ  T ( 2.5 104 J) ( 29.07 J molK)(40.0 K)  21.5 mol C 20.76 b) U  nCV T  Q CV  (2.5  10 4 J) 29.07  1.79  10 4 J P c) W  Q  U  7.15  10 3 J d) U is the same for both processes, and if dV  0, W  0 and Q  U  1.79  10 4 J 19. 51: U  0, and so Q  W  pV and V  W (2.15  105... signigicant, and modeling the process as adiabatic (no heat loss to the surroundings) is more accurate b) See Problems 19. 59 and 19. 56: The temperature at the higher pressure is T2  (258.15 K)((8.12  104 Pa)/(5.60  104 Pa)) 2 7  287.1 K, which is 13.9C and so the temperature would rise by 11.9 C 19. 58: a) b) The work done is W  p0 (2V0  V0 )  CV ( p0 (2V0 )  p3 (4V0 )) R p3  p0 (2V0 4V0 )  and so... heat flows into the gas 19. 59: a) From constant cross-section area, the volume is proportional to the length, and 1/ γ Eq (19. 24) becomes L2  L1  p1 p 2  and the distance the piston has moved is 1 / 1.400 5   p 1/ γ    1   1    0.250 m 1   1.01  10 Pa     L1  L2  L1   p2     5.21  105 Pa            0.173 m b) Raising both sides of Eq (19. 22) to the power γ... proportional to the Kelvin temperature, 5 4 min  (1.80  10 Pa)(113.7 K 300 K)  6.82  10 Pa 19. 65: a) W  pV  nRT  (0.150 mol)(8.3145 J mol  K)(150 K)  187 J,  nC p T  (0.150 mol)(29.07 J mol  K)(150 K)  654 J, U  Q  W  467 J b) From Eq (19. 24), using the expression for the temperature found in Problem 19. 64, W 1 0.40 (0.150 mol)(8.3145 J mol  K)(150 K)(1  (1 2 )  113 J, 0.40  0 for... adiabatic process, and U  Q  W  W  113 J c) dV  0, so W  0 Using the temperature change as found in Problem 19. 64 and part (b),  nCV T  (0.150 mol)(20.76 J mol  K)(300 K - 113.7 K)  580 J, and U  Q  W  Q  19. 66: a) W  nRT ln    nRT ln(3)  3.29 10 V2 V1 3 J 2 b) See Problem 19. 32(b); nCV T1 (1  (1 3) 3 )  2.33  10 3 J c) V2  3V1 , so pV  2 pV1  2nRT1  6.00  10 3 J d) The most... different, Q must be different for the two paths The heat flow Q is path dependent 19. 46: a) b) W  Wab  Wbc  Wcd  Wda  0  area(bc)  0  area(da )  (7.00 m 3 )(2000 Pa)  (7.00 m 3 )(6000 Pa)  28,000 J; on the gas since W  0 c) Q  U  W  0  (28,000 J)  28,000 J Heat comes out of the gas since Q < 0 19. 47: a) We aren’t told whether the pressure increases or decreases in process bc . or,)21( 2 mvTmCmL  kg.J104.3)sm19( 2 1 )C(80)KkgJ 4190 ( 2 1 522  vTCL 19. 56: Solving Equations (19. 22) and (19. 24) to eliminate the volumes,. heated (see the note in Section 19. 8 regarding “heating” and “cooling.”) 19. 33: a) )19. 6Examplein as,400.1Use(b) γ From Eq. (19. 22),      C495K7681.11K15.293