Tiêu đề 13. Thermodynamics Easy Ans.pdf ✅

1. (c) 2. (b) Q = U + W  U = Q − W = Q − W (using proper sign) 3. (b) U = Q −W = 35 −15 = 20 J 4. (c) Internal energy depends only on the temperature of the gas. 5. (b) 6. (b) (i) Case → Volume = constant  = 0  PdV (ii) Case → P = constant    = = 1 1 1 1 2 1 2 V V V V PdV P dV PV 7. (d) Q = W + U  35 = −15 + U  U = 50 J 8. (a) JQ = U + W, U = JQ − W U = 4.18 300 − 600 = 654 Joule 9. (d) Work done  = 2 1 PdV , which is state dependent as well as path dependent. 10. (c) Q = U + W  W = 0  Q = U R T f =   2 2 (373 273 ) 2 3 =  R − = 300R. 11. (a) ΔQ = 2kcal = 2 × 10 3 × 4.2J = 8400Jand ΔW = 500J. Hence from ΔQ = ΔU + ΔW, ΔW = ΔQ − ΔU = 8400 – 500 = 7900 J 12. (d) Change in internal energy (U) depends upon initial an find state of the function while Q and W are path dependent also. 13. (a) This is the case of free expansion and in this case W = 0 , U = 0 so temperature remains same i.e. 300 K. 14. (b) Q = U + W  W = Q − U = 100 − 40 = 70 J 15. (b) Work done is not a thermodynamically function. 16. (c) ΔQ = ΔU + ΔW = 167 + 333 = 500 cal 17. (d) Heat always refers to energy in transit from one body to another because of temperature difference. 18. (d) Change in internal energy does not depend upon path so U = Q − W remain constant. 19. (b) Q = U + W ; Q = 200 J and W = −100 J  U = Q − W = 200 − (−100 ) = 300 J 20. (a) During free expansion of a perfect gas no, work is done and also no heat is supplied from outside. Therefore, no change in internal energy. Hence, temperature remain constant. 21. (d) Q = U + W  U = Q − W = 150 − 110 = 40 J 22. (b) From FLOT Q = U + W  Heat supplied to the system so Q → Positive and work is done on the system so W → Negative Hence +Q = U – W 23. (a) 24. (d) State of a thermodynamic state cannot determine by a single variable (P or V or T) 25. (d) R is the universal gas constant. 26. (a) From FLOT  dU = dQ − dW  dU = dQ( 0) ( dW = 0)  dU  0 So temperature will decrease. 27. (b) From FLOT Q = U + W Work done at constant pressure (ΔW)P = (ΔQ)P − ΔU (ΔQ)P − (ΔQ)V (As we know (ΔQ)V = ΔU) Also (ΔQ)P = mcPΔT and (ΔQ)V = mcVΔT  (ΔW)P = m(cP − cV)ΔT  (ΔW)P = 1 × (3.4 × 10 3 − 2.4 × 10 3 ) × 10 = 10 4 cal 28. (d) 29. (a) Ideal gas possess only kinetic energy. 30. (b) The internal energy and entropy depend only on the initial and final states of the system and not on the path followed to attain that state. 31. (c) Q = U + W  Q = 200 cal = 200  4.2 = 840 J and W = 40 J
 U = Q − W = 840 − 40 = 800 J 32. (c) ΔQ = ΔU + ΔW = (Uf − Ui ) + ΔW  30 = (Uf − 40) + 10 Uf = 60J 33. (a) With rise in temperature, internal energy also increases. 34. (a) 35. (b) Heat supplied to a gas raise its internal energy and does some work against expansion, so it is a special case of law of conservation of energy. 36. (c) Change in internal energy is always equal to the heat supplied at constant volume. i.e. ΔU = (ΔQ)V = μCVΔT. For monoatomic gas CV = 3 2 R  ΔU = μ ( 3 2 R)ΔT = 1 × 3 2 × 8.31 × (100 − 0) = 12.48 × 10 2 J 37. (c) ΔU = μCVΔT = n ( R γ−1 )ΔT ⇒ ΔU = PΔV (γ−1) = P(2V−V) γ−1 = PV (γ−1) 38. (b) ΔU = μCVΔT = 2 × 4.96 × (342 − 340) = 19.84 cal 39. (a) 40. (c) According to FLOT ΔQ = ΔU + P(ΔV)  ΔU = ΔQ − P(ΔV) = 1500 − (2.1 × 10 5 )(2.5 × 10 −3 ) =975 Joule 41. (a) Q = U + W  U = Q − W = 64.18 −6 =19.08kJ  19.1 kJ 42. (a) Given Q = −20 J, W = −8J and U J i = 30 Q = U + W  U = (Q − W)  ( ) Uf − Ui = ( − 30) = −20 − (−8) Uf  U J f = 18 43. (c) Change in internal energy U =  CVT it doesn’t depend upon type of process. Actually it is a state function 44. (c) 45. (a) In first process using ΔQ = ΔU + ΔW  8 × 10 5 = ΔU + 6.5 × 10 5 ΔU = 1.5 × 10J Since final and initial states are same in both process So ΔU will be same in both process For second process using ΔQ = ΔU + ΔW  10 5 = 1.5 × 10 5 + ΔW ΔW = −0.5 × 10 5 J 46. (c) W = PV; here V is negative so W will be negative 47. (b) Entropy is related to second law of thermodynamics. 48. (c) In isothermal process temperature remains constant. 49. (a) If isothermal curves cut each other then at equilibrium two temperature will be there which is impossible. 50. (c) In isothermal expansion temperature remains constant, hence no change in internal energy. 51. (d) W = μRT loge V2 V1 = ( m M ) RT loge V2 V1 = 2.3 × m M RT log10 V2 V1 = 2.3 × 96 32 R(273 + 27) log10 140 70 = 2.3 × 900R log10 2 52. (b) 0.8 5 = P (3 + 5) P = 0.5 m 53. (b) Differentiate PV = constant w.r.t V  PV + VP = 0  V V P P  = −  54. (c) 55. (d) W = −μRT loge V2 V1 = −1 × 8.31 × (273 + 0) loge ( 22.4 11.2 ) = −8.31 × 273 × loge 2 = −1572.5J [  loge 2 = 0.693] 56. (a) E = P , if P = constant, E = constant 57. (c) For isothermal process PV = RT ⇒ P = RT V ∴W= PdV = ∫ RT V dV = RT V2 V1 loge V2 V1 58. (a) E = P 59. (b) For such a case, pressure Compressib ility 1 = 60. (a) Eθ = P = 1.013 × 10 5N/m2 61. (a) In isothermal process, compressibility E = .
62. (c) In isothermal process, exchange of energy takes place between system and surrounding to maintain the system temperature constant. 63. (c) No change in the internal energy of ideal gas but for real gas internal energy increases because work is done against intermolecular forces. 64. (a) In isothermal process temperature remains constant. i.e., T = 0 . Hence according to m T Q C  =  Ciso =  65. (c) This is the case of free expansion of gas. In free expansion U = 0  Temp. remains same. 66. (a) An isothermal process takes place at constant temperature, must be carried out in a vessel with conducting wall so that heat generated should go out at once. 67. (c) For isothermal process dU = 0 and work done ( ) = dW = P V2 − V1 2 2 1 2 V V  V = =  2 PV dW = − 68. (b) In isothermal process, temperature remains constant. 69. (b) In isothermal process, heat is released by the gas to maintain the constant temperature. 70. (a) In isothermal compression, there is always an increase of heat. which must flow out the gas. ΔQ = ΔU + ΔW ⇒ ΔQ = ΔW (∵ ΔU = 0)  ΔQ = −1.5 × 10 4 J = 1.5×10 4 4.18 cal = −3.6 × 10 3 cal 71. (a) In isothermal change, temperature remains constant, Hence U = 0. Also from Q = U + W  Q = W 72. (a) It is an isothermal process. Hence work done = P(V2 − V1 ) = 1 × 10 5 × (1.091 − 1) × 10 −6 = 0.0091 J 73. (c) Q = U + W  U = Q − W = 2240 −168 = 2072 J . 74. (b) Amount of heat given = 540 calories Change in volume ΔV = 1670 c. c Atmospheric pressure P = 1.01 × 10 6 dyne/cm2 Work done against atmospheric pressure W = PΔV = 1.01 × 10 6 × 1670 4.2 × 10 7 ≈ 40 cal 75. (b) Wiso = μRT loge V2 V1 = 1 × 8.31 × 300 loge 20 10 = 1728J 76. (b) W = μRT loge ( V2 V1 ) = 0.2 × 8.3 × loge 2 × (27 + 273) = 0.2 × 8.3 × 300 × 0.693 = 345J 77. (a) For isothermal process P1V1 = P2V2  P2 = P1V1 V2 = 72×1000 900 =80 cm Stress ΔP = P2 − P1 = 80 − 72 = 8cm 78. (d) During isothermal change T = constant  U = 0 also from FLOT, Q = W. 79. (c) Gas cylinder suddenly explodes is an irreversible adiabatic change and work done against expansion reduces the temperature. 80. (c) Work done in adiabatic change 1 ( ) 1 2 − − =  R T T 81. (b) In case of adiabatic expansion W = positive and Q = 0 from FLOT Q = U + W  U = −W i.e., U will be negative. 82. (d) For adiabatic process T γ Pγ−1 =constant  T2 T1 = ( P1 P2 ) 1−γ γ  T2 300 = ( 4 1 ) (1−1.4) 1.4 T2 = 300(4) − 0.4 1.4 83. (c) PV γ = constant ⇒ P2 P1 = ( V1 V2 ) γ ⇒ P2 1 = ( V1 V1/4 ) 3/2 = 8  P2 = 8 atm. 84. (d)  PV =constant  1 1 5 / 3 2 2 1 1 2 P (8) P 32 P V V P P  = =         =  85. (c) Volume of the gas d m V = and using  PV = constant We get (32) 128 ' ' ' 7 / 5  = =       =      =   d d V V P P 86. (b) T2 T1 = ( V1 V2 ) γ−1 ⇒ T2 = 300 ( 27 8 ) 5 3 −1 = 300 ( 27 8 ) 2 3 = 300 {( 27 8 ) 1/3 } 2 = 800 ( 3 2 ) 2 = 675K ⇒ ΔT = 675 − 300 = 375 K
87. (a) In thermodynamic processes. Work done = Area covered by PV diagram with V-axis From graph it is clear that iso adi (Area)  (Area)  Wiso  Wadi 88. (a) Since PV = RT and T = constant; PV = constant. 89. (c) For Isothermal process PV =constant ⇒ ( dP dV) = −P V = Slope of Isothermal curve For adiabaticPV γ =constant ⇒ dP dV = −γP V = Slop of adiabatic curve slope Clearly, ( dP dV) adiabatic = γ ( dP dV) Isothermal 90. (d) PV γ =constant  P ( RT P ) γ =constant  P 1−γT γ =constant. 91. (b) Wadi = R γ−1 (Ti − Tf) = R γ−1 (T − T1) 92. (d) dQ = 0 = −2 + dW  dW = 2 J  Work done by the gas = 2 J  Work done on the gas = −2 J 93. (d) Eφ = γP = 1.4 × (1 × 10 5 ) = 1.4 × 10 5N/m2 94. (b) Slope of adiabatic curve =   (Slope of isothermal curve) 95. (a) Due to compression the temperature of the system increases to a very high value. This causes the flow of heat from system to the surroundings, thus decreasing the temperature. This decrease in temperature results in decrease in pressure. 96. (c) Q = U + W = 0  W = −U if W is positive i.e., gas does work then U should be negative meaning internal energy is used in doing work. 97. (a) ( ) 1 T1 T2 R W − − =  2077 .5 joules 1.4 1 8.31 {(273 27) (273 127 )} = − −  + − + = 98. (c) Pressure is reduced, so the temperature falls. 99. (d) Adiabatic Bulk modulus E = P 100.(c) In adiabatic process, no heat transfers between system and surrounding. 101.(b) P2 P1 = ( V1 V2 ) γ  P2 = P1 ( V1 V2 ) γ = P0 (8) 4/3 = 16P0 . 102.(c) In adiabatic process PV γ =constant  ( RT V ) . V γ =constant  TV γ−1= constant 103.(a) TV γ−1 =constant  T2 T1 = ( V1 V2 ) γ  T2 = T1 ( V1 V2 ) γ ⇒ T2 = 300 ( 1 2 ) 0.4 = 227.36 K 104.(b) In adiabatic change Q = constant  Q = 0 So W = – U ( Q = U + W) 105.(d) For adiabatic process from FLOT W = −U ( Q = 0)  W = −(−100 ) = +100 J 106.(a) ( 1) ( ) 1 2 − −  = − = −  R T T U W 1 ( ) 2 1 − − =  R T T 107.(d) TV γ−1 =constant⇒ T2 = T1 ( V1 V2 ) γ−1 = 927 oC 108.(d) The process is very fast, so the gas fails to gain or lose heat. Hence this process in adiabatic 109.(a) ΔU = μCVΔT = 1 × CV(Tf − Ti ) = −CV(Ti − Tf )  |U| = CV (Ti – Tf) 110.(c) T2 = T1 ( V1 V2 ) γ−1 = 273(2) 0.41 = 273 × 1.328 = 363K W = R(T1−T2) γ−1 = 8.31(273−363) 1.41−1 = −1824  |W|  1815 J 111. (d) 112.(c) TV γ−1 =constant ⇒ T2 = T1 ( V1 V2 ) γ−1 = (273 + 18) ( V V/8 ) 0.4 = 668K 113.(a) Q = mc . Here Q = 0 , hence c = 0 P Isothermal Adiabatic V