JEE > DC Pandey Solutions for JEE Physics > DC Pandey Solutions: First Law of Thermodynamics- 2

Download, print and study this document offline |

Page 1 27. D D W W 12 13 < can be seen from area under the curve, while D D V V 1 2 = Þ D D Q Q 12 13 < Þ Q Q 2 1 < or Q Q 1 2 > 28. DW p V V p V CA = - = - 0 0 0 0 0 2 ( ) and DU p V CA = - 3 2 0 0 Þ DQ p V CA = - 5 2 0 0 29. DQ AB = 200 kJ = nC T V D ; DU BC = - 100 kJ and DW BC = - 50 kJ DW AB = 0 Þ DU AB = 200 kJ, DQ CA = 0 D D D D U U U U ABC AB BC CA = + + = 0 or 200 100 0 kJ kJ - + = DU CA DU CA = - 100 kJ D D D Q Q Q AB BC CA + + = + - - + 200 100 50 0 kJ kJ kJ ( ) = 50 kJ D D D W W W A B BC CA + + = + + 0 200 kJ DW CA = = DQ A BC 50 kJ \ DW CA = - 150 kJ 30. D D Q W ab = = = ´ ´ ´ ´ - p p 20 10 2 20 10 2 3 3 = 10 2 p J 31. DW nRT V Q AB B AB = æ è ç ö ø ÷ = = ´ ln 1 9 10 4 J Þ 800 9 10 4 T V B ln = ´ J Þ T V D ln = 225 2 D D D W W W ABCD AB BC = + + + D D W W CD DA = æ è ç ç ö ø ÷ ÷ + - - nRT V V p p p p B A C C B B ln 1 g + - + p V V C D C ( ) 0 = ´ + - - + - 9 10 10 1 5 3 10 2 1 4 5 5 nRT B ( ) = ´ - - 19 10 3 2 10 800 4 5 ( ) T B = ´ + 4 10 1200 4 T B = ´ + ´ ´ ´ ´ 4 10 1200 10 1 100 8 4 5 2.4 = ´ 4 10 5 J 31. D D W W p V p V A B C C B B = + - - 1 g + DW C D = ´ + ´ - ´ - - ´ 9 10 2 10 9 10 1 5 3 1 10 4 5 4 5 / = ´ + ´ ´ - ´ 9 10 3 2 11 10 10 10 4 4 4 = - æ è ç ö ø ÷ ´ = ´ 33 2 1 10 10 4 4 15.5 32. D W p d V kV d V k V = = = ò ò 1 2 2 First Law of Thermodynamics | 58 p V B A D 1 C 1 2 2.4 p V B A C 3 m B A D 1 C 1 2 2.4 32 10N/m Page 2 27. D D W W 12 13 < can be seen from area under the curve, while D D V V 1 2 = Þ D D Q Q 12 13 < Þ Q Q 2 1 < or Q Q 1 2 > 28. DW p V V p V CA = - = - 0 0 0 0 0 2 ( ) and DU p V CA = - 3 2 0 0 Þ DQ p V CA = - 5 2 0 0 29. DQ AB = 200 kJ = nC T V D ; DU BC = - 100 kJ and DW BC = - 50 kJ DW AB = 0 Þ DU AB = 200 kJ, DQ CA = 0 D D D D U U U U ABC AB BC CA = + + = 0 or 200 100 0 kJ kJ - + = DU CA DU CA = - 100 kJ D D D Q Q Q AB BC CA + + = + - - + 200 100 50 0 kJ kJ kJ ( ) = 50 kJ D D D W W W A B BC CA + + = + + 0 200 kJ DW CA = = DQ A BC 50 kJ \ DW CA = - 150 kJ 30. D D Q W ab = = = ´ ´ ´ ´ - p p 20 10 2 20 10 2 3 3 = 10 2 p J 31. DW nRT V Q AB B AB = æ è ç ö ø ÷ = = ´ ln 1 9 10 4 J Þ 800 9 10 4 T V B ln = ´ J Þ T V D ln = 225 2 D D D W W W ABCD AB BC = + + + D D W W CD DA = æ è ç ç ö ø ÷ ÷ + - - nRT V V p p p p B A C C B B ln 1 g + - + p V V C D C ( ) 0 = ´ + - - + - 9 10 10 1 5 3 10 2 1 4 5 5 nRT B ( ) = ´ - - 19 10 3 2 10 800 4 5 ( ) T B = ´ + 4 10 1200 4 T B = ´ + ´ ´ ´ ´ 4 10 1200 10 1 100 8 4 5 2.4 = ´ 4 10 5 J 31. D D W W p V p V A B C C B B = + - - 1 g + DW C D = ´ + ´ - ´ - - ´ 9 10 2 10 9 10 1 5 3 1 10 4 5 4 5 / = ´ + ´ ´ - ´ 9 10 3 2 11 10 10 10 4 4 4 = - æ è ç ö ø ÷ ´ = ´ 33 2 1 10 10 4 4 15.5 32. D W p d V kV d V k V = = = ò ò 1 2 2 First Law of Thermodynamics | 58 p V B A D 1 C 1 2 2.4 p V B A C 3 m B A D 1 C 1 2 2.4 32 10N/m = = = 1 2 1 2 1 2 0 0 pV nRT RT D D U nC T RT V = = × 1 3 2 0 Þ DQ RT RT = + æ è ç ö ø ÷ = 3 2 1 2 2 0 0 33. pT = constant = = p pV nR p V nR 2 Þ p V 2 = constant \ p V p V 0 2 0 0 2 2 = æ è ç ö ø ÷ Þ V V = 4 0 Þ T p V nR p V nR T = × = = 0 0 0 0 0 2 4 2 2 \ D D U nC T R T T V = = ´ - 2 3 2 2 0 0 ( ) = × = 3 2 3 2 0 0 0 0 R p V R p V 35. DW nRT V V BC C B = æ è ç ç ö ø ÷ ÷ 0 ln = æ è ç ç ö ø ÷ ÷ nRT p p B C 0 ln = × æ è ç ç ö ø ÷ ÷ 2 0 nRT V V B A ln = æ è ç ç ö ø ÷ ÷ 2 0 nRT p p A B ln \ ln ln / ln p p p p B C æ è ç ç ö ø ÷ ÷ = æ è ç ç ö ø ÷ ÷ = 0 0 2 2 4 Þ p p B C = 4 Þ p p p C B = = 4 8 0 36. As, D D W W a b > Þ D D W W 1 2 > while, D D U U 1 2 = Þ D D Q Q 1 2 > 37. h = - = - 1 1 300 600 T T sin k sou r c e = - = = 1 1 2 50 0.5 % 38. As the volume is adiabatically decreased, temperature of the gas increases and as the time elapsed, temperature normalizes i.e., decreases and so pressure also decreases. 39. As the compression is quick , the process is adiabatic while leads to heating of the gas. 40. pV g = constant = = - nRT V V nRTV g g 1 Þ TV g - = 1 constant T T V V L L L L 1 2 2 1 1 2 1 5 3 1 2 1 = æ è ç ç ö ø ÷ ÷ = æ è ç ç ö ø ÷ ÷ = æ è ç ç ö ø ÷ ÷ - - g 2 3 41. pV g = constant = æ è ç ö ø ÷ p n T p g g Þ p T 1 - = g g constant Þ p T g g - µ 1 Þ p T µ - g g 1 As g g - = - = 1 7 5 7 5 1 7 2 / for diatom gases. \ p T µ 3.5 Þ a = 3.5 42. pV x = constant , D D W nR T x = - 1 , D D U n R T = × 5 2 C Q n T nR T x nR T n T = = - + D D D D D 1 5 2 = + - < 5 2 1 0 R R x 5 2 1 R R x < - Þ x - < 1 2 5 59 | First Law of Thermodynamics p B C V A Page 3 27. D D W W 12 13 < can be seen from area under the curve, while D D V V 1 2 = Þ D D Q Q 12 13 < Þ Q Q 2 1 < or Q Q 1 2 > 28. DW p V V p V CA = - = - 0 0 0 0 0 2 ( ) and DU p V CA = - 3 2 0 0 Þ DQ p V CA = - 5 2 0 0 29. DQ AB = 200 kJ = nC T V D ; DU BC = - 100 kJ and DW BC = - 50 kJ DW AB = 0 Þ DU AB = 200 kJ, DQ CA = 0 D D D D U U U U ABC AB BC CA = + + = 0 or 200 100 0 kJ kJ - + = DU CA DU CA = - 100 kJ D D D Q Q Q AB BC CA + + = + - - + 200 100 50 0 kJ kJ kJ ( ) = 50 kJ D D D W W W A B BC CA + + = + + 0 200 kJ DW CA = = DQ A BC 50 kJ \ DW CA = - 150 kJ 30. D D Q W ab = = = ´ ´ ´ ´ - p p 20 10 2 20 10 2 3 3 = 10 2 p J 31. DW nRT V Q AB B AB = æ è ç ö ø ÷ = = ´ ln 1 9 10 4 J Þ 800 9 10 4 T V B ln = ´ J Þ T V D ln = 225 2 D D D W W W ABCD AB BC = + + + D D W W CD DA = æ è ç ç ö ø ÷ ÷ + - - nRT V V p p p p B A C C B B ln 1 g + - + p V V C D C ( ) 0 = ´ + - - + - 9 10 10 1 5 3 10 2 1 4 5 5 nRT B ( ) = ´ - - 19 10 3 2 10 800 4 5 ( ) T B = ´ + 4 10 1200 4 T B = ´ + ´ ´ ´ ´ 4 10 1200 10 1 100 8 4 5 2.4 = ´ 4 10 5 J 31. D D W W p V p V A B C C B B = + - - 1 g + DW C D = ´ + ´ - ´ - - ´ 9 10 2 10 9 10 1 5 3 1 10 4 5 4 5 / = ´ + ´ ´ - ´ 9 10 3 2 11 10 10 10 4 4 4 = - æ è ç ö ø ÷ ´ = ´ 33 2 1 10 10 4 4 15.5 32. D W p d V kV d V k V = = = ò ò 1 2 2 First Law of Thermodynamics | 58 p V B A D 1 C 1 2 2.4 p V B A C 3 m B A D 1 C 1 2 2.4 32 10N/m = = = 1 2 1 2 1 2 0 0 pV nRT RT D D U nC T RT V = = × 1 3 2 0 Þ DQ RT RT = + æ è ç ö ø ÷ = 3 2 1 2 2 0 0 33. pT = constant = = p pV nR p V nR 2 Þ p V 2 = constant \ p V p V 0 2 0 0 2 2 = æ è ç ö ø ÷ Þ V V = 4 0 Þ T p V nR p V nR T = × = = 0 0 0 0 0 2 4 2 2 \ D D U nC T R T T V = = ´ - 2 3 2 2 0 0 ( ) = × = 3 2 3 2 0 0 0 0 R p V R p V 35. DW nRT V V BC C B = æ è ç ç ö ø ÷ ÷ 0 ln = æ è ç ç ö ø ÷ ÷ nRT p p B C 0 ln = × æ è ç ç ö ø ÷ ÷ 2 0 nRT V V B A ln = æ è ç ç ö ø ÷ ÷ 2 0 nRT p p A B ln \ ln ln / ln p p p p B C æ è ç ç ö ø ÷ ÷ = æ è ç ç ö ø ÷ ÷ = 0 0 2 2 4 Þ p p B C = 4 Þ p p p C B = = 4 8 0 36. As, D D W W a b > Þ D D W W 1 2 > while, D D U U 1 2 = Þ D D Q Q 1 2 > 37. h = - = - 1 1 300 600 T T sin k sou r c e = - = = 1 1 2 50 0.5 % 38. As the volume is adiabatically decreased, temperature of the gas increases and as the time elapsed, temperature normalizes i.e., decreases and so pressure also decreases. 39. As the compression is quick , the process is adiabatic while leads to heating of the gas. 40. pV g = constant = = - nRT V V nRTV g g 1 Þ TV g - = 1 constant T T V V L L L L 1 2 2 1 1 2 1 5 3 1 2 1 = æ è ç ç ö ø ÷ ÷ = æ è ç ç ö ø ÷ ÷ = æ è ç ç ö ø ÷ ÷ - - g 2 3 41. pV g = constant = æ è ç ö ø ÷ p n T p g g Þ p T 1 - = g g constant Þ p T g g - µ 1 Þ p T µ - g g 1 As g g - = - = 1 7 5 7 5 1 7 2 / for diatom gases. \ p T µ 3.5 Þ a = 3.5 42. pV x = constant , D D W nR T x = - 1 , D D U n R T = × 5 2 C Q n T nR T x nR T n T = = - + D D D D D 1 5 2 = + - < 5 2 1 0 R R x 5 2 1 R R x < - Þ x - < 1 2 5 59 | First Law of Thermodynamics p B C V A x < 7 5 Þ x < 1.4 but x > 1 as for x < 1, C will become positive. \ 1 < < x 1.4 43. C n C n C n n R V V V = + + = 1 2 1 2 1 2 13 6 (a) 2 5 2 4 5 2 2 4 15 6 ´ + ´ + = R R R (b) 2 5 2 4 3 2 2 4 11 6 ´ + ´ + = R R R (c) 2 3 2 4 5 2 2 4 13 6 ´ + ´ + = R R R and (d) 2 6 2 4 3 2 2 4 12 6 ´ + ´ + = R R R Passage 44 & 45 44. D D W p V pV Q ABCA = ´ ´ = = 1 2 2 net 45. CA ® isobaric and BC ® isochoric, \ C C p v = = g 5 3 46. pV g = constant = æ è ç ö ø ÷ p nRT p g Þ p T 1 - = g g constant Þ T p µ - g g 1 \ T p µ - 5 3 1 5 3 / / Þ T p µ 2 5 / \ T T p p p p B A B A c c = æ è ç ç ö ø ÷ ÷ = æ è ç ç ö ø ÷ ÷ = 2 5 2 5 2 3 / / 0.85 \ T T B A = = 0.85 K 850 47. DW nRT AB = - = ´ ´ - 1 1 25 3 150 5 3 1 g = ´ = 75 25 1875 J J 48. DW BC = 0 , D D Q U BC BC = = ´ - n R T T C B 3 2 ( ) = ´ - æ è ç ö ø ÷ n R p V nR p V nR C B 3 2 = - æ è ç ö ø ÷ 3 2 1 3 2 3 p p V A A = - = - ´ = - × 1 2 1 2 3 2 3 4 p V p V nRT A B B = - ´ ´ ´ = - 3 4 1 25 3 850 5312.5 J 49. DW AB = + ( ) ve, T T A B = p p V V p = - + 0 0 0 2 3 2 Þ nRT V p V V p = - + 0 0 0 2 3 2 or T p nRV V p nR V = - + 0 0 2 0 0 2 3 2 Þ y ax bx = + 2 is parabola . Again, p p V nRT p p = - × + 2 3 2 0 0 Þ is also equation of parabola. While going from A to B temperature first increases ad than decreases. 50. pV 2 = constant DW pdV k V dV k V = = = - æ è ç ö ø ÷ ò ò 2 1 = - = - pV p V p V i f i i f f = - nR T T i f ( ) = - - = - nR T T f i ( ) ( ) ve as T T f i > as T T i f < Þ U U i f < Þ DU = + ( ) ve First Law of Thermodynamics | 60 p B A V p 0/2 p 0 2V 0 V 0 Page 4 27. D D W W 12 13 < can be seen from area under the curve, while D D V V 1 2 = Þ D D Q Q 12 13 < Þ Q Q 2 1 < or Q Q 1 2 > 28. DW p V V p V CA = - = - 0 0 0 0 0 2 ( ) and DU p V CA = - 3 2 0 0 Þ DQ p V CA = - 5 2 0 0 29. DQ AB = 200 kJ = nC T V D ; DU BC = - 100 kJ and DW BC = - 50 kJ DW AB = 0 Þ DU AB = 200 kJ, DQ CA = 0 D D D D U U U U ABC AB BC CA = + + = 0 or 200 100 0 kJ kJ - + = DU CA DU CA = - 100 kJ D D D Q Q Q AB BC CA + + = + - - + 200 100 50 0 kJ kJ kJ ( ) = 50 kJ D D D W W W A B BC CA + + = + + 0 200 kJ DW CA = = DQ A BC 50 kJ \ DW CA = - 150 kJ 30. D D Q W ab = = = ´ ´ ´ ´ - p p 20 10 2 20 10 2 3 3 = 10 2 p J 31. DW nRT V Q AB B AB = æ è ç ö ø ÷ = = ´ ln 1 9 10 4 J Þ 800 9 10 4 T V B ln = ´ J Þ T V D ln = 225 2 D D D W W W ABCD AB BC = + + + D D W W CD DA = æ è ç ç ö ø ÷ ÷ + - - nRT V V p p p p B A C C B B ln 1 g + - + p V V C D C ( ) 0 = ´ + - - + - 9 10 10 1 5 3 10 2 1 4 5 5 nRT B ( ) = ´ - - 19 10 3 2 10 800 4 5 ( ) T B = ´ + 4 10 1200 4 T B = ´ + ´ ´ ´ ´ 4 10 1200 10 1 100 8 4 5 2.4 = ´ 4 10 5 J 31. D D W W p V p V A B C C B B = + - - 1 g + DW C D = ´ + ´ - ´ - - ´ 9 10 2 10 9 10 1 5 3 1 10 4 5 4 5 / = ´ + ´ ´ - ´ 9 10 3 2 11 10 10 10 4 4 4 = - æ è ç ö ø ÷ ´ = ´ 33 2 1 10 10 4 4 15.5 32. D W p d V kV d V k V = = = ò ò 1 2 2 First Law of Thermodynamics | 58 p V B A D 1 C 1 2 2.4 p V B A C 3 m B A D 1 C 1 2 2.4 32 10N/m = = = 1 2 1 2 1 2 0 0 pV nRT RT D D U nC T RT V = = × 1 3 2 0 Þ DQ RT RT = + æ è ç ö ø ÷ = 3 2 1 2 2 0 0 33. pT = constant = = p pV nR p V nR 2 Þ p V 2 = constant \ p V p V 0 2 0 0 2 2 = æ è ç ö ø ÷ Þ V V = 4 0 Þ T p V nR p V nR T = × = = 0 0 0 0 0 2 4 2 2 \ D D U nC T R T T V = = ´ - 2 3 2 2 0 0 ( ) = × = 3 2 3 2 0 0 0 0 R p V R p V 35. DW nRT V V BC C B = æ è ç ç ö ø ÷ ÷ 0 ln = æ è ç ç ö ø ÷ ÷ nRT p p B C 0 ln = × æ è ç ç ö ø ÷ ÷ 2 0 nRT V V B A ln = æ è ç ç ö ø ÷ ÷ 2 0 nRT p p A B ln \ ln ln / ln p p p p B C æ è ç ç ö ø ÷ ÷ = æ è ç ç ö ø ÷ ÷ = 0 0 2 2 4 Þ p p B C = 4 Þ p p p C B = = 4 8 0 36. As, D D W W a b > Þ D D W W 1 2 > while, D D U U 1 2 = Þ D D Q Q 1 2 > 37. h = - = - 1 1 300 600 T T sin k sou r c e = - = = 1 1 2 50 0.5 % 38. As the volume is adiabatically decreased, temperature of the gas increases and as the time elapsed, temperature normalizes i.e., decreases and so pressure also decreases. 39. As the compression is quick , the process is adiabatic while leads to heating of the gas. 40. pV g = constant = = - nRT V V nRTV g g 1 Þ TV g - = 1 constant T T V V L L L L 1 2 2 1 1 2 1 5 3 1 2 1 = æ è ç ç ö ø ÷ ÷ = æ è ç ç ö ø ÷ ÷ = æ è ç ç ö ø ÷ ÷ - - g 2 3 41. pV g = constant = æ è ç ö ø ÷ p n T p g g Þ p T 1 - = g g constant Þ p T g g - µ 1 Þ p T µ - g g 1 As g g - = - = 1 7 5 7 5 1 7 2 / for diatom gases. \ p T µ 3.5 Þ a = 3.5 42. pV x = constant , D D W nR T x = - 1 , D D U n R T = × 5 2 C Q n T nR T x nR T n T = = - + D D D D D 1 5 2 = + - < 5 2 1 0 R R x 5 2 1 R R x < - Þ x - < 1 2 5 59 | First Law of Thermodynamics p B C V A x < 7 5 Þ x < 1.4 but x > 1 as for x < 1, C will become positive. \ 1 < < x 1.4 43. C n C n C n n R V V V = + + = 1 2 1 2 1 2 13 6 (a) 2 5 2 4 5 2 2 4 15 6 ´ + ´ + = R R R (b) 2 5 2 4 3 2 2 4 11 6 ´ + ´ + = R R R (c) 2 3 2 4 5 2 2 4 13 6 ´ + ´ + = R R R and (d) 2 6 2 4 3 2 2 4 12 6 ´ + ´ + = R R R Passage 44 & 45 44. D D W p V pV Q ABCA = ´ ´ = = 1 2 2 net 45. CA ® isobaric and BC ® isochoric, \ C C p v = = g 5 3 46. pV g = constant = æ è ç ö ø ÷ p nRT p g Þ p T 1 - = g g constant Þ T p µ - g g 1 \ T p µ - 5 3 1 5 3 / / Þ T p µ 2 5 / \ T T p p p p B A B A c c = æ è ç ç ö ø ÷ ÷ = æ è ç ç ö ø ÷ ÷ = 2 5 2 5 2 3 / / 0.85 \ T T B A = = 0.85 K 850 47. DW nRT AB = - = ´ ´ - 1 1 25 3 150 5 3 1 g = ´ = 75 25 1875 J J 48. DW BC = 0 , D D Q U BC BC = = ´ - n R T T C B 3 2 ( ) = ´ - æ è ç ö ø ÷ n R p V nR p V nR C B 3 2 = - æ è ç ö ø ÷ 3 2 1 3 2 3 p p V A A = - = - ´ = - × 1 2 1 2 3 2 3 4 p V p V nRT A B B = - ´ ´ ´ = - 3 4 1 25 3 850 5312.5 J 49. DW AB = + ( ) ve, T T A B = p p V V p = - + 0 0 0 2 3 2 Þ nRT V p V V p = - + 0 0 0 2 3 2 or T p nRV V p nR V = - + 0 0 2 0 0 2 3 2 Þ y ax bx = + 2 is parabola . Again, p p V nRT p p = - × + 2 3 2 0 0 Þ is also equation of parabola. While going from A to B temperature first increases ad than decreases. 50. pV 2 = constant DW pdV k V dV k V = = = - æ è ç ö ø ÷ ò ò 2 1 = - = - pV p V p V i f i i f f = - nR T T i f ( ) = - - = - nR T T f i ( ) ( ) ve as T T f i > as T T i f < Þ U U i f < Þ DU = + ( ) ve First Law of Thermodynamics | 60 p B A V p 0/2 p 0 2V 0 V 0 D D D D Q nC T nR T n C R T V V = - = - ( ) = + ( ) ve as C R V > i.e., heat is given to the system. 51. In cyclic process, DU = 0 DW nR T V V = + æ è ç ç ö ø ÷ ÷ 0 2 2 0 0 0 ln + + æ è ç ç ö ø ÷ ÷ 0 2 0 0 0 nRT V V ln = - 2 2 2 0 0 nRT nRT ln ln = = + nRT 0 2 ln ( ) ve i.e., DW > 0 D D D Q U W ab bc supplied = + = - + æ è ç ç ö ø ÷ ÷ nC T T nR T V V V ( ) ln 2 2 2 0 0 0 0 0 = ´ + 2 3 2 4 2 0 0 RT RT ln = + 3 4 2 0 0 RT RT ln 52. ab ® isochoric, bc ® isobaric and ca ® isothermal. DW ab = 0, DU ca = 0 as in ca density is increasing, so volume is decreasing i.e., DW ca = - ( ) ve, i.e., DW ca < 0 in isochoric process DQ ab is positive for increase in temperature. 53. In isochoric process DW = 0. and in adiabatic process DQ = 0 Þ Q 3 to be minimum Þ Q Q Q 2 1 3 > > JEE Corner ¢ Assertion & Reasons 1. In adiabatic expression, DW= + ( )ve while DQ=0 and as according to first law of thermodynamics, D D D Q U W = + Þ D D U W = - i.e., DU = - ( ) ve this implies decrease in temperature. So, Assertion and reason are both true but not correct explanation. 2. Assertion is false, as work done is a path function and not a state function i.e., it depends on the path through which the gas was taken from initial to find state. 3. Assertion is false, as first law can be applied for both real and ideal gases. 4. During melting of ice its volume decreases, so work done by it is negative and that by atmosphere is positive. So, reason is true explanation of assertion. 5. As D D D Q U W = + Þ D D D U Q W = - , where DU is state function while DQ and DW are path function as for definite 61 | First Law of Thermodynamics p V b a c p d b V a T 0 2T 0 2V 0 V 0 c Page 5 27. D D W W 12 13 < can be seen from area under the curve, while D D V V 1 2 = Þ D D Q Q 12 13 < Þ Q Q 2 1 < or Q Q 1 2 > 28. DW p V V p V CA = - = - 0 0 0 0 0 2 ( ) and DU p V CA = - 3 2 0 0 Þ DQ p V CA = - 5 2 0 0 29. DQ AB = 200 kJ = nC T V D ; DU BC = - 100 kJ and DW BC = - 50 kJ DW AB = 0 Þ DU AB = 200 kJ, DQ CA = 0 D D D D U U U U ABC AB BC CA = + + = 0 or 200 100 0 kJ kJ - + = DU CA DU CA = - 100 kJ D D D Q Q Q AB BC CA + + = + - - + 200 100 50 0 kJ kJ kJ ( ) = 50 kJ D D D W W W A B BC CA + + = + + 0 200 kJ DW CA = = DQ A BC 50 kJ \ DW CA = - 150 kJ 30. D D Q W ab = = = ´ ´ ´ ´ - p p 20 10 2 20 10 2 3 3 = 10 2 p J 31. DW nRT V Q AB B AB = æ è ç ö ø ÷ = = ´ ln 1 9 10 4 J Þ 800 9 10 4 T V B ln = ´ J Þ T V D ln = 225 2 D D D W W W ABCD AB BC = + + + D D W W CD DA = æ è ç ç ö ø ÷ ÷ + - - nRT V V p p p p B A C C B B ln 1 g + - + p V V C D C ( ) 0 = ´ + - - + - 9 10 10 1 5 3 10 2 1 4 5 5 nRT B ( ) = ´ - - 19 10 3 2 10 800 4 5 ( ) T B = ´ + 4 10 1200 4 T B = ´ + ´ ´ ´ ´ 4 10 1200 10 1 100 8 4 5 2.4 = ´ 4 10 5 J 31. D D W W p V p V A B C C B B = + - - 1 g + DW C D = ´ + ´ - ´ - - ´ 9 10 2 10 9 10 1 5 3 1 10 4 5 4 5 / = ´ + ´ ´ - ´ 9 10 3 2 11 10 10 10 4 4 4 = - æ è ç ö ø ÷ ´ = ´ 33 2 1 10 10 4 4 15.5 32. D W p d V kV d V k V = = = ò ò 1 2 2 First Law of Thermodynamics | 58 p V B A D 1 C 1 2 2.4 p V B A C 3 m B A D 1 C 1 2 2.4 32 10N/m = = = 1 2 1 2 1 2 0 0 pV nRT RT D D U nC T RT V = = × 1 3 2 0 Þ DQ RT RT = + æ è ç ö ø ÷ = 3 2 1 2 2 0 0 33. pT = constant = = p pV nR p V nR 2 Þ p V 2 = constant \ p V p V 0 2 0 0 2 2 = æ è ç ö ø ÷ Þ V V = 4 0 Þ T p V nR p V nR T = × = = 0 0 0 0 0 2 4 2 2 \ D D U nC T R T T V = = ´ - 2 3 2 2 0 0 ( ) = × = 3 2 3 2 0 0 0 0 R p V R p V 35. DW nRT V V BC C B = æ è ç ç ö ø ÷ ÷ 0 ln = æ è ç ç ö ø ÷ ÷ nRT p p B C 0 ln = × æ è ç ç ö ø ÷ ÷ 2 0 nRT V V B A ln = æ è ç ç ö ø ÷ ÷ 2 0 nRT p p A B ln \ ln ln / ln p p p p B C æ è ç ç ö ø ÷ ÷ = æ è ç ç ö ø ÷ ÷ = 0 0 2 2 4 Þ p p B C = 4 Þ p p p C B = = 4 8 0 36. As, D D W W a b > Þ D D W W 1 2 > while, D D U U 1 2 = Þ D D Q Q 1 2 > 37. h = - = - 1 1 300 600 T T sin k sou r c e = - = = 1 1 2 50 0.5 % 38. As the volume is adiabatically decreased, temperature of the gas increases and as the time elapsed, temperature normalizes i.e., decreases and so pressure also decreases. 39. As the compression is quick , the process is adiabatic while leads to heating of the gas. 40. pV g = constant = = - nRT V V nRTV g g 1 Þ TV g - = 1 constant T T V V L L L L 1 2 2 1 1 2 1 5 3 1 2 1 = æ è ç ç ö ø ÷ ÷ = æ è ç ç ö ø ÷ ÷ = æ è ç ç ö ø ÷ ÷ - - g 2 3 41. pV g = constant = æ è ç ö ø ÷ p n T p g g Þ p T 1 - = g g constant Þ p T g g - µ 1 Þ p T µ - g g 1 As g g - = - = 1 7 5 7 5 1 7 2 / for diatom gases. \ p T µ 3.5 Þ a = 3.5 42. pV x = constant , D D W nR T x = - 1 , D D U n R T = × 5 2 C Q n T nR T x nR T n T = = - + D D D D D 1 5 2 = + - < 5 2 1 0 R R x 5 2 1 R R x < - Þ x - < 1 2 5 59 | First Law of Thermodynamics p B C V A x < 7 5 Þ x < 1.4 but x > 1 as for x < 1, C will become positive. \ 1 < < x 1.4 43. C n C n C n n R V V V = + + = 1 2 1 2 1 2 13 6 (a) 2 5 2 4 5 2 2 4 15 6 ´ + ´ + = R R R (b) 2 5 2 4 3 2 2 4 11 6 ´ + ´ + = R R R (c) 2 3 2 4 5 2 2 4 13 6 ´ + ´ + = R R R and (d) 2 6 2 4 3 2 2 4 12 6 ´ + ´ + = R R R Passage 44 & 45 44. D D W p V pV Q ABCA = ´ ´ = = 1 2 2 net 45. CA ® isobaric and BC ® isochoric, \ C C p v = = g 5 3 46. pV g = constant = æ è ç ö ø ÷ p nRT p g Þ p T 1 - = g g constant Þ T p µ - g g 1 \ T p µ - 5 3 1 5 3 / / Þ T p µ 2 5 / \ T T p p p p B A B A c c = æ è ç ç ö ø ÷ ÷ = æ è ç ç ö ø ÷ ÷ = 2 5 2 5 2 3 / / 0.85 \ T T B A = = 0.85 K 850 47. DW nRT AB = - = ´ ´ - 1 1 25 3 150 5 3 1 g = ´ = 75 25 1875 J J 48. DW BC = 0 , D D Q U BC BC = = ´ - n R T T C B 3 2 ( ) = ´ - æ è ç ö ø ÷ n R p V nR p V nR C B 3 2 = - æ è ç ö ø ÷ 3 2 1 3 2 3 p p V A A = - = - ´ = - × 1 2 1 2 3 2 3 4 p V p V nRT A B B = - ´ ´ ´ = - 3 4 1 25 3 850 5312.5 J 49. DW AB = + ( ) ve, T T A B = p p V V p = - + 0 0 0 2 3 2 Þ nRT V p V V p = - + 0 0 0 2 3 2 or T p nRV V p nR V = - + 0 0 2 0 0 2 3 2 Þ y ax bx = + 2 is parabola . Again, p p V nRT p p = - × + 2 3 2 0 0 Þ is also equation of parabola. While going from A to B temperature first increases ad than decreases. 50. pV 2 = constant DW pdV k V dV k V = = = - æ è ç ö ø ÷ ò ò 2 1 = - = - pV p V p V i f i i f f = - nR T T i f ( ) = - - = - nR T T f i ( ) ( ) ve as T T f i > as T T i f < Þ U U i f < Þ DU = + ( ) ve First Law of Thermodynamics | 60 p B A V p 0/2 p 0 2V 0 V 0 D D D D Q nC T nR T n C R T V V = - = - ( ) = + ( ) ve as C R V > i.e., heat is given to the system. 51. In cyclic process, DU = 0 DW nR T V V = + æ è ç ç ö ø ÷ ÷ 0 2 2 0 0 0 ln + + æ è ç ç ö ø ÷ ÷ 0 2 0 0 0 nRT V V ln = - 2 2 2 0 0 nRT nRT ln ln = = + nRT 0 2 ln ( ) ve i.e., DW > 0 D D D Q U W ab bc supplied = + = - + æ è ç ç ö ø ÷ ÷ nC T T nR T V V V ( ) ln 2 2 2 0 0 0 0 0 = ´ + 2 3 2 4 2 0 0 RT RT ln = + 3 4 2 0 0 RT RT ln 52. ab ® isochoric, bc ® isobaric and ca ® isothermal. DW ab = 0, DU ca = 0 as in ca density is increasing, so volume is decreasing i.e., DW ca = - ( ) ve, i.e., DW ca < 0 in isochoric process DQ ab is positive for increase in temperature. 53. In isochoric process DW = 0. and in adiabatic process DQ = 0 Þ Q 3 to be minimum Þ Q Q Q 2 1 3 > > JEE Corner ¢ Assertion & Reasons 1. In adiabatic expression, DW= + ( )ve while DQ=0 and as according to first law of thermodynamics, D D D Q U W = + Þ D D U W = - i.e., DU = - ( ) ve this implies decrease in temperature. So, Assertion and reason are both true but not correct explanation. 2. Assertion is false, as work done is a path function and not a state function i.e., it depends on the path through which the gas was taken from initial to find state. 3. Assertion is false, as first law can be applied for both real and ideal gases. 4. During melting of ice its volume decreases, so work done by it is negative and that by atmosphere is positive. So, reason is true explanation of assertion. 5. As D D D Q U W = + Þ D D D U Q W = - , where DU is state function while DQ and DW are path function as for definite 61 | First Law of Thermodynamics p V b a c p d b V a T 0 2T 0 2V 0 V 0 c initial and final state DU is constant and so is Q W - . Thus assertion and reason are both true but not correct explanation. 6. Carnot’s engine is ideal heat engine with maximum efficiency but it is not also 100%. So assertion and reason are both true but not correct explanation. 7. pT = constant = × = p pV R p V nR 2 Þ p V 2 = constant \ DW pdV k dV V k V = = = × ò ò 1 2 1 2 / / = 2 k V = = 2 2 2 kV p V / = = - = 2 2 2 pV nR T T nRT T f i ( ) D \ DW = + ( ) ve for DT = + ( ) ve and nRT V T = constant. Þ T V 2 µ or, V T µ 2 Thus assertion is true but reason is false. 8. In adiabatic changes for free expansion, Q = 0, W = 0 and DU = 0 as in free expansion no work is done against any force. For ideal gases pV = constant as DU = 0 Þ T = constant So, assertion and reason are both true but not correct explanation. 9. Assertion and reason are both true and correct explanation. 10. Assertion and reason are both true and correct explanation. ¢ Match the Columns 1. (a) DW p dV pV nR T T f i = = = - ò ( ) = = nRT RT 2 ¾® r (b) DU nC T R T t V = = ´ - 2 3 2 2 ( ) = 3RT ¾® p (c) DW nR T T RT = - - = - ´ ( ) / 2 1 5 3 3 2 2 = - 3RT ¾® s (d)D D U nC T RT V = = 3 ¾® p 2. (a) In ab slope is more so, pressure is less as V nR p T = × , but is constant and in isobaric process. D D D W p V nR T = = and as DT is same in both process so, DW is same for both ¾® r (b) As D D U nC T V = - is same for both process ¾® r (c) As D D D Q U W = + , it is also same for both process ¾® s (d) Nothing can be said about molar heat capacity ® s 3. (a) D = ò W pdV = = ò ò k V dV k dV V = = = 2 2 2 kV pV nR T D ¾® p (b) D D D U nC T nR T V = = 3 2 ¾® s (c) D D D Q nR T nR T = + 2 3 2 = 7 2 nR T D ¾® s (d) ¾® s 4. (a) D D D W p V nR T = = and D D U nC T V = Þ D D W U < ¾® q (b) DW = 0 Þ D D Q U = ,DU = - ( ) ve ® p r , (c) DW = + ( ) ve, DU = - ( ) ve, DQ = 0 ® p (d) DW = + ( ) ve, DU = 0, DQ = + ( ) ve ® p 5. (a) DW p V p V AB = + 0 0 0 0 1 2 First Law of Thermodynamics | 62Read More

Download as PDF

Signup for Free!

Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.

Related Searches