Page 1 18. First Law of Thermodynamics Introductory Exercise 18.1 1. (a) D D W p V = = - ´ - 1.7 1.2 0.8 10 5 ( ) J = - ´ 6.8 10 4 J (b) DV = ´ 1.1 10 5 J Þ D D D Q U W = + = - ´ 17.8 10 4 J i.e., 1.78 ´ 10 5 J of heat has flown out of the gas. (c) No, it is independent of the type of the gas. 2. (a) In p V - graph of cyclic process, clockwise rotation gives positive work and anticlockwise gives negative work. And as loop 1 has greater area than loop 2, that is why total work done by the system is positive. (b)As in cyclic process change in internal energy is zero, thatâ€™s why for positive work done by the system, heat flows into the system. (c) In loop â€˜1â€™ work done is positive so, heat flows into the system and in loop â€˜2â€™ work done is negative so heat flows out of the system. 3. As the box is insulated i.e., no heat exchange takes place with surrounding and as the gas expands against vacuum i.e., zero pressure thatâ€™s why no work has been done and there is no change in internal energy. Thus, temperature do not change, internal energy and gas does not do any work. 4. U f nRT nRT = = 2 3 2 Þ n U RT = = ´ ´ ´ 2 3 2 100 3 300 8.314 = 0.0267 mole. 5. D D Q ms = = ´ ´ q 1 387 30 J = 11610 J D D D V V m = = ´ g q r a q 3 = ´ ´ ´ ´ ´ - 1 10 3 7 10 30 3 6 8.92 = ´ - 7.1 10 8 m - 3 D D W p V = = ´ ´ ´ - 1.01 7.06 10 10 5 8 = ´ - 7.13 10 3 J D D D U Q W = - = 11609.99 J Introductory Exercise 18.2 1. (a) At constant volume, DU = 0 Þ DW = 0 D D Q nC T V = Þ D D = = ´ ´ = Q nC V 200 1 3 2 8.314 16.04 K \ T T T f i = + = + D 300 16.04 = 316.04 K (b) At constant pressure, D D T Q nC p = = ´ ´ 200 1 5 2 314 8. = 9.62 K Þ T f = + 300 9.62 K = 309.62 K 2. For adiabatic process, pVg = constant = c (say) \ pdV c V dV c V dV V V V V V V i f i f i f ò ò ò = = - g g Page 2 18. First Law of Thermodynamics Introductory Exercise 18.1 1. (a) D D W p V = = - ´ - 1.7 1.2 0.8 10 5 ( ) J = - ´ 6.8 10 4 J (b) DV = ´ 1.1 10 5 J Þ D D D Q U W = + = - ´ 17.8 10 4 J i.e., 1.78 ´ 10 5 J of heat has flown out of the gas. (c) No, it is independent of the type of the gas. 2. (a) In p V - graph of cyclic process, clockwise rotation gives positive work and anticlockwise gives negative work. And as loop 1 has greater area than loop 2, that is why total work done by the system is positive. (b)As in cyclic process change in internal energy is zero, thatâ€™s why for positive work done by the system, heat flows into the system. (c) In loop â€˜1â€™ work done is positive so, heat flows into the system and in loop â€˜2â€™ work done is negative so heat flows out of the system. 3. As the box is insulated i.e., no heat exchange takes place with surrounding and as the gas expands against vacuum i.e., zero pressure thatâ€™s why no work has been done and there is no change in internal energy. Thus, temperature do not change, internal energy and gas does not do any work. 4. U f nRT nRT = = 2 3 2 Þ n U RT = = ´ ´ ´ 2 3 2 100 3 300 8.314 = 0.0267 mole. 5. D D Q ms = = ´ ´ q 1 387 30 J = 11610 J D D D V V m = = ´ g q r a q 3 = ´ ´ ´ ´ ´ - 1 10 3 7 10 30 3 6 8.92 = ´ - 7.1 10 8 m - 3 D D W p V = = ´ ´ ´ - 1.01 7.06 10 10 5 8 = ´ - 7.13 10 3 J D D D U Q W = - = 11609.99 J Introductory Exercise 18.2 1. (a) At constant volume, DU = 0 Þ DW = 0 D D Q nC T V = Þ D D = = ´ ´ = Q nC V 200 1 3 2 8.314 16.04 K \ T T T f i = + = + D 300 16.04 = 316.04 K (b) At constant pressure, D D T Q nC p = = ´ ´ 200 1 5 2 314 8. = 9.62 K Þ T f = + 300 9.62 K = 309.62 K 2. For adiabatic process, pVg = constant = c (say) \ pdV c V dV c V dV V V V V V V i f i f i f ò ò ò = = - g g = - + - + c V V V i f g g 1 1 = - - - + - + c V V f i g g g 1 1 1 = × - × - - + - + p V V p V V f f f i i f g g g g 1 1 1 = - - = - - p V p V p V p V f f i i i i f f 1 1 g g (Proved) 3. DW AB = + 500 J,DQ AB = + 250 J Þ DU AB = - 250 J DW AC = + 700 J, DQ AC = + 300 J Þ DU AC = - 400 J (a) Path BC is isochoric process, i.e., DW BC = 0 \ D D D D Q U U U BC BC AC AB = = - = - - - = - 400 250 150 J J J ( ) (b) D D D W W W CDA CD DA = + = - + = - 800 0 800 J J (Work is negative as volume is decreasing) D D D U U U CDA AC AC = = - = 400 J Þ D D D Q W U CDA CDA CDA = + = - + = - 800 400 400 J J J 4. (a) T pV nR = = ´ ´ ´ ´ - 1 10 2 10 1 2 5 8.314 = 240.6 K (b) D D W p V = - = ´ ´ ´ - - g 1 2 10 5 10 5 3 1 5 3 = 10 2 3 3 / J 5. (a) DK p m p m p m m i f i f = - = - æ è ç ç ö ø ÷ ÷ 2 2 2 2 2 2 1 1 = ´ ´ ´ - é ë ê ù û ú - - ( ) 10 10 200 2 1 10 10 1 3 2 3 2.01 = - é ë ê ù û ú = 2 100 1 2 199 J (b) D D Q nC T V = Þ D D D T Q nC Q m M C V V = = = ´ = ´ ´ ´ M Q m R D 3 200 199 2010 3 8.314 = ° 0.8 C 6. D D D D D W W W W W A B BC CD DA = + + + = æ è ç ç ö ø ÷ ÷ + - nRT m p p p V V C BC 1 1 2 2 ( ) + æ è ç ç ö ø ÷ ÷ + - nRTm p p p V V 2 1 1 1 2 ( ) = - æ è ç ç ö ø ÷ ÷ + - nR T T p p p V p V ( ) ln 2 1 2 1 1 2 1 1 + - p V p V 1 1 1 2 = - æ è ç ç ö ø ÷ ÷ ( ) ln p V p V p p 2 2 1 1 2 1 7. D W ABCA = (+)ve Þ DW AB = (+)ve, DW BC = 0, DW CA = - ( )ve For BC, DQ = - ( )ve Þ DU BC = - ( )ve and DW BC = 0 For CA, DU = - ( )ve Þ DQ CA = - ( )ve as DW CA = - ( )ve. DU DW DQ AB + + + BC - 0 - CA - - - Total 0 + + For AB, as DU ABCA = 0 and DU BC = - ( )ve, DU CA = - ( ) ve Þ DU AB = - ( ) ve As D D Q W ABCA ABCA = = + ( )ve and DQ BC = - ( )ve DQ CA = - ( )ve Þ DQ AB = + ( )ve In isobaric process, D D D W p V nR T = = = ´ ´ - = 0.2 8.314 166.3 ( ) 300 200 J 9. DW pdV V dV V = = = ò ò a a 2 3 1 3 = ´ ´ ´ ´ - 1 1.01 3 5 10 2 1 5 3 3 ( ) = ´ 1.18 10 6 J 49 | First Law of Thermodynamics Page 3 18. First Law of Thermodynamics Introductory Exercise 18.1 1. (a) D D W p V = = - ´ - 1.7 1.2 0.8 10 5 ( ) J = - ´ 6.8 10 4 J (b) DV = ´ 1.1 10 5 J Þ D D D Q U W = + = - ´ 17.8 10 4 J i.e., 1.78 ´ 10 5 J of heat has flown out of the gas. (c) No, it is independent of the type of the gas. 2. (a) In p V - graph of cyclic process, clockwise rotation gives positive work and anticlockwise gives negative work. And as loop 1 has greater area than loop 2, that is why total work done by the system is positive. (b)As in cyclic process change in internal energy is zero, thatâ€™s why for positive work done by the system, heat flows into the system. (c) In loop â€˜1â€™ work done is positive so, heat flows into the system and in loop â€˜2â€™ work done is negative so heat flows out of the system. 3. As the box is insulated i.e., no heat exchange takes place with surrounding and as the gas expands against vacuum i.e., zero pressure thatâ€™s why no work has been done and there is no change in internal energy. Thus, temperature do not change, internal energy and gas does not do any work. 4. U f nRT nRT = = 2 3 2 Þ n U RT = = ´ ´ ´ 2 3 2 100 3 300 8.314 = 0.0267 mole. 5. D D Q ms = = ´ ´ q 1 387 30 J = 11610 J D D D V V m = = ´ g q r a q 3 = ´ ´ ´ ´ ´ - 1 10 3 7 10 30 3 6 8.92 = ´ - 7.1 10 8 m - 3 D D W p V = = ´ ´ ´ - 1.01 7.06 10 10 5 8 = ´ - 7.13 10 3 J D D D U Q W = - = 11609.99 J Introductory Exercise 18.2 1. (a) At constant volume, DU = 0 Þ DW = 0 D D Q nC T V = Þ D D = = ´ ´ = Q nC V 200 1 3 2 8.314 16.04 K \ T T T f i = + = + D 300 16.04 = 316.04 K (b) At constant pressure, D D T Q nC p = = ´ ´ 200 1 5 2 314 8. = 9.62 K Þ T f = + 300 9.62 K = 309.62 K 2. For adiabatic process, pVg = constant = c (say) \ pdV c V dV c V dV V V V V V V i f i f i f ò ò ò = = - g g = - + - + c V V V i f g g 1 1 = - - - + - + c V V f i g g g 1 1 1 = × - × - - + - + p V V p V V f f f i i f g g g g 1 1 1 = - - = - - p V p V p V p V f f i i i i f f 1 1 g g (Proved) 3. DW AB = + 500 J,DQ AB = + 250 J Þ DU AB = - 250 J DW AC = + 700 J, DQ AC = + 300 J Þ DU AC = - 400 J (a) Path BC is isochoric process, i.e., DW BC = 0 \ D D D D Q U U U BC BC AC AB = = - = - - - = - 400 250 150 J J J ( ) (b) D D D W W W CDA CD DA = + = - + = - 800 0 800 J J (Work is negative as volume is decreasing) D D D U U U CDA AC AC = = - = 400 J Þ D D D Q W U CDA CDA CDA = + = - + = - 800 400 400 J J J 4. (a) T pV nR = = ´ ´ ´ ´ - 1 10 2 10 1 2 5 8.314 = 240.6 K (b) D D W p V = - = ´ ´ ´ - - g 1 2 10 5 10 5 3 1 5 3 = 10 2 3 3 / J 5. (a) DK p m p m p m m i f i f = - = - æ è ç ç ö ø ÷ ÷ 2 2 2 2 2 2 1 1 = ´ ´ ´ - é ë ê ù û ú - - ( ) 10 10 200 2 1 10 10 1 3 2 3 2.01 = - é ë ê ù û ú = 2 100 1 2 199 J (b) D D Q nC T V = Þ D D D T Q nC Q m M C V V = = = ´ = ´ ´ ´ M Q m R D 3 200 199 2010 3 8.314 = ° 0.8 C 6. D D D D D W W W W W A B BC CD DA = + + + = æ è ç ç ö ø ÷ ÷ + - nRT m p p p V V C BC 1 1 2 2 ( ) + æ è ç ç ö ø ÷ ÷ + - nRTm p p p V V 2 1 1 1 2 ( ) = - æ è ç ç ö ø ÷ ÷ + - nR T T p p p V p V ( ) ln 2 1 2 1 1 2 1 1 + - p V p V 1 1 1 2 = - æ è ç ç ö ø ÷ ÷ ( ) ln p V p V p p 2 2 1 1 2 1 7. D W ABCA = (+)ve Þ DW AB = (+)ve, DW BC = 0, DW CA = - ( )ve For BC, DQ = - ( )ve Þ DU BC = - ( )ve and DW BC = 0 For CA, DU = - ( )ve Þ DQ CA = - ( )ve as DW CA = - ( )ve. DU DW DQ AB + + + BC - 0 - CA - - - Total 0 + + For AB, as DU ABCA = 0 and DU BC = - ( )ve, DU CA = - ( ) ve Þ DU AB = - ( ) ve As D D Q W ABCA ABCA = = + ( )ve and DQ BC = - ( )ve DQ CA = - ( )ve Þ DQ AB = + ( )ve In isobaric process, D D D W p V nR T = = = ´ ´ - = 0.2 8.314 166.3 ( ) 300 200 J 9. DW pdV V dV V = = = ò ò a a 2 3 1 3 = ´ ´ ´ ´ - 1 1.01 3 5 10 2 1 5 3 3 ( ) = ´ 1.18 10 6 J 49 | First Law of Thermodynamics Introductory Exercise 18.3 1. DW nRT BB = ln ln V V R B A æ è ç ç ö ø ÷ ÷ = ´ 3 273 5 = 10959 J DW BC = 0 D D D Q U W = + D D D U Q W = - = - 80000 10959 = 69041 T T f i = = ´ 5 5 273 K = 1365 K D D D Q Q Q ABC AB BC = + = + + + D D W U BC BC 0 0 D D Q nC T BC V = Þ C Q n T V BC = D D = ´ ´ = 69041 3 4 273 21.07 C C R p V = + = 29.39 Þ g = = = C C p V 29.39 21.07 1.4 2. D D D Q U W = + ; D D Q nC T p = Þ 1600 1 72 = × × C p Þ C p = 22.22 C C R V p = - = 13.9 Þ g = = C C p V 1.6 D D D W Q U = - = - 1600 nC T V D = - ´ ´ 1600 1 72 13.9 = - 1600 1000.8 J = 599.2 J and D D U nC T V = = ´ ´ 1 72 13.9 = 1001 = 1 kJ 3. D D D W p V = 1 2 = ´ ´ ´ ´ ´ - 1 2 20 10 1 10 5 3 1.01 = ´ = 10 101 1010 J \ p n W t = = ´ D D 100 1010 60 J s = 1.68 kW AIEEE Corner ¢ Subjectve Questions (Level 1) 1. D D D U Q W = - = + 254 73 J J = 327 J 2. (a) D D D T Q nC Q nR V = = = ´ ´ ´ 2 3 2 200 2 1 8.314 = 16 K Þ T T T f i = + = D 316 K (b) D D D T Q nC Q nR p ¢ = = = ´ ´ ´ 2 5 2 200 5 1 8.314 = 9.6 K Þ T T T f i ¢ = + ¢ = D 309.6 K 3. D D U nC T V = , in adiabatic process, DQ = 0 and D D U W = - First Law of Thermodynamics | 50 p 0/5 p 0 p C A V 0 5V 0 V B Page 4 18. First Law of Thermodynamics Introductory Exercise 18.1 1. (a) D D W p V = = - ´ - 1.7 1.2 0.8 10 5 ( ) J = - ´ 6.8 10 4 J (b) DV = ´ 1.1 10 5 J Þ D D D Q U W = + = - ´ 17.8 10 4 J i.e., 1.78 ´ 10 5 J of heat has flown out of the gas. (c) No, it is independent of the type of the gas. 2. (a) In p V - graph of cyclic process, clockwise rotation gives positive work and anticlockwise gives negative work. And as loop 1 has greater area than loop 2, that is why total work done by the system is positive. (b)As in cyclic process change in internal energy is zero, thatâ€™s why for positive work done by the system, heat flows into the system. (c) In loop â€˜1â€™ work done is positive so, heat flows into the system and in loop â€˜2â€™ work done is negative so heat flows out of the system. 3. As the box is insulated i.e., no heat exchange takes place with surrounding and as the gas expands against vacuum i.e., zero pressure thatâ€™s why no work has been done and there is no change in internal energy. Thus, temperature do not change, internal energy and gas does not do any work. 4. U f nRT nRT = = 2 3 2 Þ n U RT = = ´ ´ ´ 2 3 2 100 3 300 8.314 = 0.0267 mole. 5. D D Q ms = = ´ ´ q 1 387 30 J = 11610 J D D D V V m = = ´ g q r a q 3 = ´ ´ ´ ´ ´ - 1 10 3 7 10 30 3 6 8.92 = ´ - 7.1 10 8 m - 3 D D W p V = = ´ ´ ´ - 1.01 7.06 10 10 5 8 = ´ - 7.13 10 3 J D D D U Q W = - = 11609.99 J Introductory Exercise 18.2 1. (a) At constant volume, DU = 0 Þ DW = 0 D D Q nC T V = Þ D D = = ´ ´ = Q nC V 200 1 3 2 8.314 16.04 K \ T T T f i = + = + D 300 16.04 = 316.04 K (b) At constant pressure, D D T Q nC p = = ´ ´ 200 1 5 2 314 8. = 9.62 K Þ T f = + 300 9.62 K = 309.62 K 2. For adiabatic process, pVg = constant = c (say) \ pdV c V dV c V dV V V V V V V i f i f i f ò ò ò = = - g g = - + - + c V V V i f g g 1 1 = - - - + - + c V V f i g g g 1 1 1 = × - × - - + - + p V V p V V f f f i i f g g g g 1 1 1 = - - = - - p V p V p V p V f f i i i i f f 1 1 g g (Proved) 3. DW AB = + 500 J,DQ AB = + 250 J Þ DU AB = - 250 J DW AC = + 700 J, DQ AC = + 300 J Þ DU AC = - 400 J (a) Path BC is isochoric process, i.e., DW BC = 0 \ D D D D Q U U U BC BC AC AB = = - = - - - = - 400 250 150 J J J ( ) (b) D D D W W W CDA CD DA = + = - + = - 800 0 800 J J (Work is negative as volume is decreasing) D D D U U U CDA AC AC = = - = 400 J Þ D D D Q W U CDA CDA CDA = + = - + = - 800 400 400 J J J 4. (a) T pV nR = = ´ ´ ´ ´ - 1 10 2 10 1 2 5 8.314 = 240.6 K (b) D D W p V = - = ´ ´ ´ - - g 1 2 10 5 10 5 3 1 5 3 = 10 2 3 3 / J 5. (a) DK p m p m p m m i f i f = - = - æ è ç ç ö ø ÷ ÷ 2 2 2 2 2 2 1 1 = ´ ´ ´ - é ë ê ù û ú - - ( ) 10 10 200 2 1 10 10 1 3 2 3 2.01 = - é ë ê ù û ú = 2 100 1 2 199 J (b) D D Q nC T V = Þ D D D T Q nC Q m M C V V = = = ´ = ´ ´ ´ M Q m R D 3 200 199 2010 3 8.314 = ° 0.8 C 6. D D D D D W W W W W A B BC CD DA = + + + = æ è ç ç ö ø ÷ ÷ + - nRT m p p p V V C BC 1 1 2 2 ( ) + æ è ç ç ö ø ÷ ÷ + - nRTm p p p V V 2 1 1 1 2 ( ) = - æ è ç ç ö ø ÷ ÷ + - nR T T p p p V p V ( ) ln 2 1 2 1 1 2 1 1 + - p V p V 1 1 1 2 = - æ è ç ç ö ø ÷ ÷ ( ) ln p V p V p p 2 2 1 1 2 1 7. D W ABCA = (+)ve Þ DW AB = (+)ve, DW BC = 0, DW CA = - ( )ve For BC, DQ = - ( )ve Þ DU BC = - ( )ve and DW BC = 0 For CA, DU = - ( )ve Þ DQ CA = - ( )ve as DW CA = - ( )ve. DU DW DQ AB + + + BC - 0 - CA - - - Total 0 + + For AB, as DU ABCA = 0 and DU BC = - ( )ve, DU CA = - ( ) ve Þ DU AB = - ( ) ve As D D Q W ABCA ABCA = = + ( )ve and DQ BC = - ( )ve DQ CA = - ( )ve Þ DQ AB = + ( )ve In isobaric process, D D D W p V nR T = = = ´ ´ - = 0.2 8.314 166.3 ( ) 300 200 J 9. DW pdV V dV V = = = ò ò a a 2 3 1 3 = ´ ´ ´ ´ - 1 1.01 3 5 10 2 1 5 3 3 ( ) = ´ 1.18 10 6 J 49 | First Law of Thermodynamics Introductory Exercise 18.3 1. DW nRT BB = ln ln V V R B A æ è ç ç ö ø ÷ ÷ = ´ 3 273 5 = 10959 J DW BC = 0 D D D Q U W = + D D D U Q W = - = - 80000 10959 = 69041 T T f i = = ´ 5 5 273 K = 1365 K D D D Q Q Q ABC AB BC = + = + + + D D W U BC BC 0 0 D D Q nC T BC V = Þ C Q n T V BC = D D = ´ ´ = 69041 3 4 273 21.07 C C R p V = + = 29.39 Þ g = = = C C p V 29.39 21.07 1.4 2. D D D Q U W = + ; D D Q nC T p = Þ 1600 1 72 = × × C p Þ C p = 22.22 C C R V p = - = 13.9 Þ g = = C C p V 1.6 D D D W Q U = - = - 1600 nC T V D = - ´ ´ 1600 1 72 13.9 = - 1600 1000.8 J = 599.2 J and D D U nC T V = = ´ ´ 1 72 13.9 = 1001 = 1 kJ 3. D D D W p V = 1 2 = ´ ´ ´ ´ ´ - 1 2 20 10 1 10 5 3 1.01 = ´ = 10 101 1010 J \ p n W t = = ´ D D 100 1010 60 J s = 1.68 kW AIEEE Corner ¢ Subjectve Questions (Level 1) 1. D D D U Q W = - = + 254 73 J J = 327 J 2. (a) D D D T Q nC Q nR V = = = ´ ´ ´ 2 3 2 200 2 1 8.314 = 16 K Þ T T T f i = + = D 316 K (b) D D D T Q nC Q nR p ¢ = = = ´ ´ ´ 2 5 2 200 5 1 8.314 = 9.6 K Þ T T T f i ¢ = + ¢ = D 309.6 K 3. D D U nC T V = , in adiabatic process, DQ = 0 and D D U W = - First Law of Thermodynamics | 50 p 0/5 p 0 p C A V 0 5V 0 V B where, D D W nR T = - 1 g Þ D D U nR T = - g 1 for all process. 4. DV = 0 Þ DW = 0 \ D D D Q U nC T V = = = × n R T 5 2 D = - 5 2 ( ) p V p V f f i i = - 5 2 ( ) p V V f i = ´ - ´ ´ - 5 2 5 10 10 10 10 5 5 3 ( ) = ´ ´ ´ = - 5 2 4 10 10 10 5 2 4 J 5. D D D Q U nC T n R T T V i i 1 1 5 2 3 = = = × - ( ) = 5 nRT i D D Q nC T n R T T p i i 2 5 2 6 3 = = × - ( ) = 7.5nRT i \ c Q n T nRT n T T R i i i = = - = D D 12.5 12.5 ( ) 6 5 = 2.5 R 6. DW AB = 0 , DW p V p V BC = ´ = 0 0 0 0 2 1 2 = × = 1 2 300 0 nRT R D D D D D Q U W U W AB BC = + + + ( ) ( ) = + + D D D U U W AB BC BC = + 0 300 R (As T T A C = ) = ´ 2.49 10 3 J = 2.49 kJ 7. D D D U Q W = - = - 1200 2100 J J = 900 J D D T U nC V = = ´ ´ = 900 5 3 2 8.314 14.43 Þ T T T f i = - = ° - ° D 127 C 14.43 C = ° 112.6 C 8. When gas expands it does positive work on the surrounding and for this purpose heat has to be supplied into the system. 9. D D W V V V f i = = - r r( ) = - æ è ç ç ö ø ÷ ÷ = - æ è ç ö ø ÷ r r r r m m f i 1 1 1 1000 1 999.9 = - ´ ´ ´ = - 10 2 1000 5 0.1 999.9 0.02 J (work done is negative as volume decreases) D D Q ms = = ´ ´ q 2 4200 4 = 33600 J D D D U Q W = - = 33600.02 J 10. D D W p V pV p m f = » = r = ´ ´ = - 10 10 10 5 3 0.6 1666.67 J D D Q ms mL = + q = ´ - 10 4200 2 ´ + ´ ´ - 100 10 25 10 2 6 . = + = 4200 25000 29200 J J J D D D U Q W = - = 29200 J - 1666.67 J = 27533.33 J = ´ 2.75 10 4 J 11. D D W p V = = ´ ´ ´ - 1.013 10 1670 10 5 6 = ´ 1.013 167 J = 169.2 J DQ mL = = ´ ´ - 10 10 3 6 2.256 J = 2256 J \ D D D U Q W = - = - ( ) 2256 169.2 J = 2086.8 J » 2087 J 12. D D W p V = = - ´ ´ 2.3 0.5 10 5 = - ´ 1.15 10 5 J DU = - ´ 1.4 10 5 J Þ D D D Q U W = + 51 | First Law of Thermodynamics p /5 0 p 0 p C A V 0 2V 0 V B Page 5 18. First Law of Thermodynamics Introductory Exercise 18.1 1. (a) D D W p V = = - ´ - 1.7 1.2 0.8 10 5 ( ) J = - ´ 6.8 10 4 J (b) DV = ´ 1.1 10 5 J Þ D D D Q U W = + = - ´ 17.8 10 4 J i.e., 1.78 ´ 10 5 J of heat has flown out of the gas. (c) No, it is independent of the type of the gas. 2. (a) In p V - graph of cyclic process, clockwise rotation gives positive work and anticlockwise gives negative work. And as loop 1 has greater area than loop 2, that is why total work done by the system is positive. (b)As in cyclic process change in internal energy is zero, thatâ€™s why for positive work done by the system, heat flows into the system. (c) In loop â€˜1â€™ work done is positive so, heat flows into the system and in loop â€˜2â€™ work done is negative so heat flows out of the system. 3. As the box is insulated i.e., no heat exchange takes place with surrounding and as the gas expands against vacuum i.e., zero pressure thatâ€™s why no work has been done and there is no change in internal energy. Thus, temperature do not change, internal energy and gas does not do any work. 4. U f nRT nRT = = 2 3 2 Þ n U RT = = ´ ´ ´ 2 3 2 100 3 300 8.314 = 0.0267 mole. 5. D D Q ms = = ´ ´ q 1 387 30 J = 11610 J D D D V V m = = ´ g q r a q 3 = ´ ´ ´ ´ ´ - 1 10 3 7 10 30 3 6 8.92 = ´ - 7.1 10 8 m - 3 D D W p V = = ´ ´ ´ - 1.01 7.06 10 10 5 8 = ´ - 7.13 10 3 J D D D U Q W = - = 11609.99 J Introductory Exercise 18.2 1. (a) At constant volume, DU = 0 Þ DW = 0 D D Q nC T V = Þ D D = = ´ ´ = Q nC V 200 1 3 2 8.314 16.04 K \ T T T f i = + = + D 300 16.04 = 316.04 K (b) At constant pressure, D D T Q nC p = = ´ ´ 200 1 5 2 314 8. = 9.62 K Þ T f = + 300 9.62 K = 309.62 K 2. For adiabatic process, pVg = constant = c (say) \ pdV c V dV c V dV V V V V V V i f i f i f ò ò ò = = - g g = - + - + c V V V i f g g 1 1 = - - - + - + c V V f i g g g 1 1 1 = × - × - - + - + p V V p V V f f f i i f g g g g 1 1 1 = - - = - - p V p V p V p V f f i i i i f f 1 1 g g (Proved) 3. DW AB = + 500 J,DQ AB = + 250 J Þ DU AB = - 250 J DW AC = + 700 J, DQ AC = + 300 J Þ DU AC = - 400 J (a) Path BC is isochoric process, i.e., DW BC = 0 \ D D D D Q U U U BC BC AC AB = = - = - - - = - 400 250 150 J J J ( ) (b) D D D W W W CDA CD DA = + = - + = - 800 0 800 J J (Work is negative as volume is decreasing) D D D U U U CDA AC AC = = - = 400 J Þ D D D Q W U CDA CDA CDA = + = - + = - 800 400 400 J J J 4. (a) T pV nR = = ´ ´ ´ ´ - 1 10 2 10 1 2 5 8.314 = 240.6 K (b) D D W p V = - = ´ ´ ´ - - g 1 2 10 5 10 5 3 1 5 3 = 10 2 3 3 / J 5. (a) DK p m p m p m m i f i f = - = - æ è ç ç ö ø ÷ ÷ 2 2 2 2 2 2 1 1 = ´ ´ ´ - é ë ê ù û ú - - ( ) 10 10 200 2 1 10 10 1 3 2 3 2.01 = - é ë ê ù û ú = 2 100 1 2 199 J (b) D D Q nC T V = Þ D D D T Q nC Q m M C V V = = = ´ = ´ ´ ´ M Q m R D 3 200 199 2010 3 8.314 = ° 0.8 C 6. D D D D D W W W W W A B BC CD DA = + + + = æ è ç ç ö ø ÷ ÷ + - nRT m p p p V V C BC 1 1 2 2 ( ) + æ è ç ç ö ø ÷ ÷ + - nRTm p p p V V 2 1 1 1 2 ( ) = - æ è ç ç ö ø ÷ ÷ + - nR T T p p p V p V ( ) ln 2 1 2 1 1 2 1 1 + - p V p V 1 1 1 2 = - æ è ç ç ö ø ÷ ÷ ( ) ln p V p V p p 2 2 1 1 2 1 7. D W ABCA = (+)ve Þ DW AB = (+)ve, DW BC = 0, DW CA = - ( )ve For BC, DQ = - ( )ve Þ DU BC = - ( )ve and DW BC = 0 For CA, DU = - ( )ve Þ DQ CA = - ( )ve as DW CA = - ( )ve. DU DW DQ AB + + + BC - 0 - CA - - - Total 0 + + For AB, as DU ABCA = 0 and DU BC = - ( )ve, DU CA = - ( ) ve Þ DU AB = - ( ) ve As D D Q W ABCA ABCA = = + ( )ve and DQ BC = - ( )ve DQ CA = - ( )ve Þ DQ AB = + ( )ve In isobaric process, D D D W p V nR T = = = ´ ´ - = 0.2 8.314 166.3 ( ) 300 200 J 9. DW pdV V dV V = = = ò ò a a 2 3 1 3 = ´ ´ ´ ´ - 1 1.01 3 5 10 2 1 5 3 3 ( ) = ´ 1.18 10 6 J 49 | First Law of Thermodynamics Introductory Exercise 18.3 1. DW nRT BB = ln ln V V R B A æ è ç ç ö ø ÷ ÷ = ´ 3 273 5 = 10959 J DW BC = 0 D D D Q U W = + D D D U Q W = - = - 80000 10959 = 69041 T T f i = = ´ 5 5 273 K = 1365 K D D D Q Q Q ABC AB BC = + = + + + D D W U BC BC 0 0 D D Q nC T BC V = Þ C Q n T V BC = D D = ´ ´ = 69041 3 4 273 21.07 C C R p V = + = 29.39 Þ g = = = C C p V 29.39 21.07 1.4 2. D D D Q U W = + ; D D Q nC T p = Þ 1600 1 72 = × × C p Þ C p = 22.22 C C R V p = - = 13.9 Þ g = = C C p V 1.6 D D D W Q U = - = - 1600 nC T V D = - ´ ´ 1600 1 72 13.9 = - 1600 1000.8 J = 599.2 J and D D U nC T V = = ´ ´ 1 72 13.9 = 1001 = 1 kJ 3. D D D W p V = 1 2 = ´ ´ ´ ´ ´ - 1 2 20 10 1 10 5 3 1.01 = ´ = 10 101 1010 J \ p n W t = = ´ D D 100 1010 60 J s = 1.68 kW AIEEE Corner ¢ Subjectve Questions (Level 1) 1. D D D U Q W = - = + 254 73 J J = 327 J 2. (a) D D D T Q nC Q nR V = = = ´ ´ ´ 2 3 2 200 2 1 8.314 = 16 K Þ T T T f i = + = D 316 K (b) D D D T Q nC Q nR p ¢ = = = ´ ´ ´ 2 5 2 200 5 1 8.314 = 9.6 K Þ T T T f i ¢ = + ¢ = D 309.6 K 3. D D U nC T V = , in adiabatic process, DQ = 0 and D D U W = - First Law of Thermodynamics | 50 p 0/5 p 0 p C A V 0 5V 0 V B where, D D W nR T = - 1 g Þ D D U nR T = - g 1 for all process. 4. DV = 0 Þ DW = 0 \ D D D Q U nC T V = = = × n R T 5 2 D = - 5 2 ( ) p V p V f f i i = - 5 2 ( ) p V V f i = ´ - ´ ´ - 5 2 5 10 10 10 10 5 5 3 ( ) = ´ ´ ´ = - 5 2 4 10 10 10 5 2 4 J 5. D D D Q U nC T n R T T V i i 1 1 5 2 3 = = = × - ( ) = 5 nRT i D D Q nC T n R T T p i i 2 5 2 6 3 = = × - ( ) = 7.5nRT i \ c Q n T nRT n T T R i i i = = - = D D 12.5 12.5 ( ) 6 5 = 2.5 R 6. DW AB = 0 , DW p V p V BC = ´ = 0 0 0 0 2 1 2 = × = 1 2 300 0 nRT R D D D D D Q U W U W AB BC = + + + ( ) ( ) = + + D D D U U W AB BC BC = + 0 300 R (As T T A C = ) = ´ 2.49 10 3 J = 2.49 kJ 7. D D D U Q W = - = - 1200 2100 J J = 900 J D D T U nC V = = ´ ´ = 900 5 3 2 8.314 14.43 Þ T T T f i = - = ° - ° D 127 C 14.43 C = ° 112.6 C 8. When gas expands it does positive work on the surrounding and for this purpose heat has to be supplied into the system. 9. D D W V V V f i = = - r r( ) = - æ è ç ç ö ø ÷ ÷ = - æ è ç ö ø ÷ r r r r m m f i 1 1 1 1000 1 999.9 = - ´ ´ ´ = - 10 2 1000 5 0.1 999.9 0.02 J (work done is negative as volume decreases) D D Q ms = = ´ ´ q 2 4200 4 = 33600 J D D D U Q W = - = 33600.02 J 10. D D W p V pV p m f = » = r = ´ ´ = - 10 10 10 5 3 0.6 1666.67 J D D Q ms mL = + q = ´ - 10 4200 2 ´ + ´ ´ - 100 10 25 10 2 6 . = + = 4200 25000 29200 J J J D D D U Q W = - = 29200 J - 1666.67 J = 27533.33 J = ´ 2.75 10 4 J 11. D D W p V = = ´ ´ ´ - 1.013 10 1670 10 5 6 = ´ 1.013 167 J = 169.2 J DQ mL = = ´ ´ - 10 10 3 6 2.256 J = 2256 J \ D D D U Q W = - = - ( ) 2256 169.2 J = 2086.8 J » 2087 J 12. D D W p V = = - ´ ´ 2.3 0.5 10 5 = - ´ 1.15 10 5 J DU = - ´ 1.4 10 5 J Þ D D D Q U W = + 51 | First Law of Thermodynamics p /5 0 p 0 p C A V 0 2V 0 V B = - ´ (1.4 + 1.15) 10 5 J = - ´ 2.55 10 5 J Thus, 2.55 ´ 10 5 J of heat flows out of the system and it is independent of the type of the gas. 13. In a cyclic process, U = 0 Þ D D Q W = (a) \ W Q Q Q Q h = + + + ( ) 1 2 3 4 - + + ( ) W W W 1 2 3 = - - + ( ) 5960 5585 2980 3645 - - - ( ) 2200 825 1100 = - = 1040 275 765 J (b) h = = = work done heat supplied 10.83% 1040 9605 14. (a) DW = ´ ¾® ¾® 1 2 AB AC = ´ ´ 1 2 2 0 0 p V = p V 0 0 (b) T p V nR C = 2 0 0 and T p V nR A = 0 0 Þ D D Q nC T nC p V nR CA p p = = - × 0 0 = - × = - 5 2 5 2 0 0 0 0 R p V R p V T p V nR B = 3 0 0 , D D Q nC T AB V = = ´ - æ è ç ö ø ÷ n R p V nR p V nR 3 2 3 0 0 0 0 = ´ 3 2 2 0 0 p V = 3 0 0 p V (c) D D D D Q Q Q W AB BC CA + + = = + - = 3 5 2 0 0 0 0 0 0 p V Q p V p V BC D Þ DQ p V BC = 0 0 2 (d) Temperature is maximum at a point D lying somewhere between B and C where the product pV is maximum. p p V p = - + 2 5 0 0 0 Þ pV p V V p V = - + æ è ç ç ö ø ÷ ÷ 2 5 0 0 0 = - + 2 5 0 0 2 0 p V V p V For pV = maximum d dV pV ( ) = 0 Þ - × + = 2 2 5 0 0 0 0 V p V p Þ V V = 5 4 0 p p V V p p = - × + = 2 5 4 5 5 2 0 0 0 0 0 \ T pV nR max max ( ) = \ p p V R p V R = × × = 5 2 5 4 1 25 8 0 0 0 0 15. V nRT p R R A A A = = ´ ´ = ´ - 2 300 2 10 3 10 5 3 V R R B = ´ ´ = ´ - 2 400 2 10 4 10 5 3 , V R R C = ´ = ´ - 2 400 10 8 10 5 3 V R R 0 3 2 300 10 6 10 = ´ = ´ - DW R = ´ - ´ - 2 10 4 3 10 5 3 ( ) + ´ æ è ç ö ø ÷ + 2 400 8 4 1 R ln ´ - ´ - 10 6 8 10 5 3 ( ) K + ´ æ è ç ö ø ÷ 2 300 3 6 R ln \ DW R R R = + - 200 800 2 200 ln - 600 2 R ln = = 2000 2 1153 R ln J As D D Q W = = 1153 J and DU = 0 cyclic process. First Law of Thermodynamics | 52 1 atm 2 atm p C A V B DRead More

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