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Page 1 17. Thermometry, Thermal Expansion & Kinetic Theory of Gases Introductory Exercise 17.1 1. (a) C F 5 32 9 = - for F = 0, C = - ´ 5 9 32 = - ° 17 8 . C (b) K F - = - 273. 5 1 5 32 9 for K = 0, F = - ´ + = - ° 9 5 273 15 32 . 459.67 F 2. (a) x x 5 2 32 9 = - Þ x x = - 10 9 17 8 . Þ 17.8 = - æ è ç ö ø ÷ 10 9 1 x Þ x = ´ = ° 17.8 C 9 160 2 . (b) x x 5 2 32 9 = - / Þ x x = - 5 18 17 8 . Þ 17 8 13 18 . = - x Þ x = - ° 24 65 . C 3. C F - - = - - 5 99 5 32 212 32 Þ C F - = - 5 94 32 180 Þ 52 5 94 32 180 - = - F Þ F = + ´ = ° 32 180 94 47 122 F 4. K F - = - 273 15 5 32 9 . Þ x x - = - 273 15 5 9 17 8 . . Þ 4 9 255 35 x = . Þ x = 574 54 . 5. C F 5 32 9 = - Þ 9 5 32 x x = - Þ 4 5 32 x = - Þ x = - ´ = - ° 5 4 32 40 C 6. D D t t = 1 2 a q = ´ ´ ´ ´ - 1 2 10 86400 30 5 1.2 = ´ ´ = 1.5 1.2 8.64 s 15.55 s given. 7. As from 0°C to 4°C, den sity of wa ter in creases so the vol ume of wooden block above wa ter level in creases and as from 4°C to 10°C den sity of wa ter de creases so the vol ume of block above wa ter de cr eases . 8. V g V g 1 1 2 1 r s = ¢ and V g V g 2 2 2 2 r s = ¢ Þ DV V V V 1 1 1 1 1 1 1 1 = - ¢ = - - s r and DV V 2 2 2 2 1 = - s r \ DV V V V 2 2 1 1 - D = - æ è ç ç ö ø ÷ ÷ - - æ è ç ç ö ø ÷ ÷ = - 1 1 2 2 1 1 1 1 2 2 s r s r s r s r = - - - s r s g r g 1 1 1 2 1 1 1 1 ( ) ( ) DT T Page 2 17. Thermometry, Thermal Expansion & Kinetic Theory of Gases Introductory Exercise 17.1 1. (a) C F 5 32 9 = - for F = 0, C = - ´ 5 9 32 = - ° 17 8 . C (b) K F - = - 273. 5 1 5 32 9 for K = 0, F = - ´ + = - ° 9 5 273 15 32 . 459.67 F 2. (a) x x 5 2 32 9 = - Þ x x = - 10 9 17 8 . Þ 17.8 = - æ è ç ö ø ÷ 10 9 1 x Þ x = ´ = ° 17.8 C 9 160 2 . (b) x x 5 2 32 9 = - / Þ x x = - 5 18 17 8 . Þ 17 8 13 18 . = - x Þ x = - ° 24 65 . C 3. C F - - = - - 5 99 5 32 212 32 Þ C F - = - 5 94 32 180 Þ 52 5 94 32 180 - = - F Þ F = + ´ = ° 32 180 94 47 122 F 4. K F - = - 273 15 5 32 9 . Þ x x - = - 273 15 5 9 17 8 . . Þ 4 9 255 35 x = . Þ x = 574 54 . 5. C F 5 32 9 = - Þ 9 5 32 x x = - Þ 4 5 32 x = - Þ x = - ´ = - ° 5 4 32 40 C 6. D D t t = 1 2 a q = ´ ´ ´ ´ - 1 2 10 86400 30 5 1.2 = ´ ´ = 1.5 1.2 8.64 s 15.55 s given. 7. As from 0°C to 4°C, den sity of wa ter in creases so the vol ume of wooden block above wa ter level in creases and as from 4°C to 10°C den sity of wa ter de creases so the vol ume of block above wa ter de cr eases . 8. V g V g 1 1 2 1 r s = ¢ and V g V g 2 2 2 2 r s = ¢ Þ DV V V V 1 1 1 1 1 1 1 1 = - ¢ = - - s r and DV V 2 2 2 2 1 = - s r \ DV V V V 2 2 1 1 - D = - æ è ç ç ö ø ÷ ÷ - - æ è ç ç ö ø ÷ ÷ = - 1 1 2 2 1 1 1 1 2 2 s r s r s r s r = - - - s r s g r g 1 1 1 2 1 1 1 1 ( ) ( ) DT T = - - æ è ç ç ö ø ÷ ÷ D s r g g g 1 1 2 1 1 1 T T = - - - s r s g r g 1 1 1 2 1 1 1 1 ( ) ( ) DT T = - - æ è ç ç ö ø ÷ ÷ s r g g g 1 1 2 1 1 1 T T D 9. On cool ing brass con tracts more than iron ( ) a a B > Fe such that brass disk gets loosen from hole of iron. 10. V T µ Þ V kT = Þ ln ln ln V k T = + Þ D D V V T T = Þ D D V V T T = = 1 g Introductory Exercise 17.2 1. For ideal gases, pV nRT = Þ T V nR p = = VM mR p Slope = VM mR As slope µ 1 m Þ m m 2 1 < 2. pV nRT = Þ p p T T 2 1 2 1 = = = 360 300 6 5 Þ p p 2 1 6 5 6 5 10 12 = = ´ = atm atm 3. M mix = ´ + ´ + = + = 1 4 28 1 4 44 1 4 1 4 7 11 1 2 36 pV nRT m M RT = = Þ pM m V RT RT = = r Þ r = pM RT \ r = ´ ´ ´ ´ - 101 10 36 10 8 31 290 5 3 . . = ´ ´ = 101 36 8 31 29 15 . . . . kg/m 3 4. pV nRT N N RT A = = Þ N pVN RT A = = ´ ´ ´ ´ ´ ´ ´ ´ - - 10 10 10 250 10 10 300 6 3 6 23 13.6 6.02 8.31 = ´ ´ ´ ´ = ´ 13.6 6.02 8.31 8.21 5 6 10 10 15 15 5. pV nRT = Þ V nR p T = × Slope µ 1 r Þ p p 1 2 > 6. pV nRT = Þ p nRT V = ( ) 1 Þ y mx = is a straight line passing through origin. Introductory Exercise 17.3 1. Av er age ve loc ity de pends on the di rec tion of mo tion of gas mol e cules and as con tainer do not move such that their net ef fect be comes zero, due to the rea son that some mol e cules are mov ing in one di rec tion while other are mov ing in op po site di rec tion. But in case of av er age speed only mag ni tudes are in use which do not cancel each other. 38 | Thermometry, Thermal Expansion & Kinetic Theory of Gases Page 3 17. Thermometry, Thermal Expansion & Kinetic Theory of Gases Introductory Exercise 17.1 1. (a) C F 5 32 9 = - for F = 0, C = - ´ 5 9 32 = - ° 17 8 . C (b) K F - = - 273. 5 1 5 32 9 for K = 0, F = - ´ + = - ° 9 5 273 15 32 . 459.67 F 2. (a) x x 5 2 32 9 = - Þ x x = - 10 9 17 8 . Þ 17.8 = - æ è ç ö ø ÷ 10 9 1 x Þ x = ´ = ° 17.8 C 9 160 2 . (b) x x 5 2 32 9 = - / Þ x x = - 5 18 17 8 . Þ 17 8 13 18 . = - x Þ x = - ° 24 65 . C 3. C F - - = - - 5 99 5 32 212 32 Þ C F - = - 5 94 32 180 Þ 52 5 94 32 180 - = - F Þ F = + ´ = ° 32 180 94 47 122 F 4. K F - = - 273 15 5 32 9 . Þ x x - = - 273 15 5 9 17 8 . . Þ 4 9 255 35 x = . Þ x = 574 54 . 5. C F 5 32 9 = - Þ 9 5 32 x x = - Þ 4 5 32 x = - Þ x = - ´ = - ° 5 4 32 40 C 6. D D t t = 1 2 a q = ´ ´ ´ ´ - 1 2 10 86400 30 5 1.2 = ´ ´ = 1.5 1.2 8.64 s 15.55 s given. 7. As from 0°C to 4°C, den sity of wa ter in creases so the vol ume of wooden block above wa ter level in creases and as from 4°C to 10°C den sity of wa ter de creases so the vol ume of block above wa ter de cr eases . 8. V g V g 1 1 2 1 r s = ¢ and V g V g 2 2 2 2 r s = ¢ Þ DV V V V 1 1 1 1 1 1 1 1 = - ¢ = - - s r and DV V 2 2 2 2 1 = - s r \ DV V V V 2 2 1 1 - D = - æ è ç ç ö ø ÷ ÷ - - æ è ç ç ö ø ÷ ÷ = - 1 1 2 2 1 1 1 1 2 2 s r s r s r s r = - - - s r s g r g 1 1 1 2 1 1 1 1 ( ) ( ) DT T = - - æ è ç ç ö ø ÷ ÷ D s r g g g 1 1 2 1 1 1 T T = - - - s r s g r g 1 1 1 2 1 1 1 1 ( ) ( ) DT T = - - æ è ç ç ö ø ÷ ÷ s r g g g 1 1 2 1 1 1 T T D 9. On cool ing brass con tracts more than iron ( ) a a B > Fe such that brass disk gets loosen from hole of iron. 10. V T µ Þ V kT = Þ ln ln ln V k T = + Þ D D V V T T = Þ D D V V T T = = 1 g Introductory Exercise 17.2 1. For ideal gases, pV nRT = Þ T V nR p = = VM mR p Slope = VM mR As slope µ 1 m Þ m m 2 1 < 2. pV nRT = Þ p p T T 2 1 2 1 = = = 360 300 6 5 Þ p p 2 1 6 5 6 5 10 12 = = ´ = atm atm 3. M mix = ´ + ´ + = + = 1 4 28 1 4 44 1 4 1 4 7 11 1 2 36 pV nRT m M RT = = Þ pM m V RT RT = = r Þ r = pM RT \ r = ´ ´ ´ ´ - 101 10 36 10 8 31 290 5 3 . . = ´ ´ = 101 36 8 31 29 15 . . . . kg/m 3 4. pV nRT N N RT A = = Þ N pVN RT A = = ´ ´ ´ ´ ´ ´ ´ ´ - - 10 10 10 250 10 10 300 6 3 6 23 13.6 6.02 8.31 = ´ ´ ´ ´ = ´ 13.6 6.02 8.31 8.21 5 6 10 10 15 15 5. pV nRT = Þ V nR p T = × Slope µ 1 r Þ p p 1 2 > 6. pV nRT = Þ p nRT V = ( ) 1 Þ y mx = is a straight line passing through origin. Introductory Exercise 17.3 1. Av er age ve loc ity de pends on the di rec tion of mo tion of gas mol e cules and as con tainer do not move such that their net ef fect be comes zero, due to the rea son that some mol e cules are mov ing in one di rec tion while other are mov ing in op po site di rec tion. But in case of av er age speed only mag ni tudes are in use which do not cancel each other. 38 | Thermometry, Thermal Expansion & Kinetic Theory of Gases 2. KE = 3 2 kT = ´ ´ ´ 3 2 6 10 300 23 8.31 J = ´ ´ - 3 4 10 21 8.31 J = ´ - 6.21 J 10 21 3. v RT M rms = 3 , v He 8.31 = ´ ´ ´ - 3 300 4 10 3 = ´ 1.37 m/s 10 3 v Ne 8.31 20.2 608.5 = ´ ´ ´ = - 3 300 10 3 m/s KE 6.21 J = = ´ - 3 2 10 21 kT 4. v RT M rms = 3 Þ T Mv R = = ´ ´ ´ - rms 8.31 2 3 6 3 4 10 10 3 = 160.45 K 5. r r r r r = + + = - + - + n n n n n n n n 1 1 2 2 1 2 2 1 2 2 2 2 1 1 ( ) = + - r r r 1 2 2 1 n ( ) Þ n 2 1 2 1 = - - = - - r r r r 1.293 1.429 1.251 1.429 = = = 136 178 0.764 76.4% by mass 6. V V p T p T p h g p 2 1 1 2 2 1 0 0 277 = = + ´ ( ) r = ´ + ´ ´ ´ ´ ´ ( ) 1.01 1.01 10 40 10 10 293 10 277 5 3 5 = ´ ´ = 5.01 1.01 5.25 293 277 Þ V V 2 1 3 105 = = 5.25 cm 7. N nN A = = ´ ´ 1 18 6 10 23 = ´ 1 3 10 23 ; S R = = ´ ´ ´ ´ 4 4 6400 10 10 2 3 2 2 p 3.14 ( ) = ´ 5.14 cm 10 18 2 \ N S = ´ ´ 10 3 10 23 18 5.14 = ´ 6.5 10 3 molecules/cm 2 (a) nC nR V = = 3 2 35 J/K Þ n R = = 70 3 2.8 mole (b) U nRT = = ´ = 3 2 35 273 J/K K 9555 J (c) C C R R p V = + = = 5 2 20.8 J/ K mole 8. (a) n C C nR p V ( ) - = = 29.1 J/K Þ n = 29.1 8.314 mole = 3.5 mole (b) C nc n R V V = = = ´ ´ 3 2 3.5 1.5 8.314 = 43.65 J/K C nc n R C nR p p V = = = + 5 2 = + ´ 43.65 3.5 8.314 = 72.75 J/K (c) C nc V V ¢ = = ´ = n R 5 3 72.75 J/K C nc n R p p ¢ = = ´ 7 2 = + ´ 72.75 3.5 8.314 = 101.85 J/K 10. v RT M rms = 3 and v RT M av = 8 p Here 3 8 > p Þ v v rms av > , i e . ., the statement is true. Thermometry, Thermal Expansion & Kinetic Theory of Gases | 39 Page 4 17. Thermometry, Thermal Expansion & Kinetic Theory of Gases Introductory Exercise 17.1 1. (a) C F 5 32 9 = - for F = 0, C = - ´ 5 9 32 = - ° 17 8 . C (b) K F - = - 273. 5 1 5 32 9 for K = 0, F = - ´ + = - ° 9 5 273 15 32 . 459.67 F 2. (a) x x 5 2 32 9 = - Þ x x = - 10 9 17 8 . Þ 17.8 = - æ è ç ö ø ÷ 10 9 1 x Þ x = ´ = ° 17.8 C 9 160 2 . (b) x x 5 2 32 9 = - / Þ x x = - 5 18 17 8 . Þ 17 8 13 18 . = - x Þ x = - ° 24 65 . C 3. C F - - = - - 5 99 5 32 212 32 Þ C F - = - 5 94 32 180 Þ 52 5 94 32 180 - = - F Þ F = + ´ = ° 32 180 94 47 122 F 4. K F - = - 273 15 5 32 9 . Þ x x - = - 273 15 5 9 17 8 . . Þ 4 9 255 35 x = . Þ x = 574 54 . 5. C F 5 32 9 = - Þ 9 5 32 x x = - Þ 4 5 32 x = - Þ x = - ´ = - ° 5 4 32 40 C 6. D D t t = 1 2 a q = ´ ´ ´ ´ - 1 2 10 86400 30 5 1.2 = ´ ´ = 1.5 1.2 8.64 s 15.55 s given. 7. As from 0°C to 4°C, den sity of wa ter in creases so the vol ume of wooden block above wa ter level in creases and as from 4°C to 10°C den sity of wa ter de creases so the vol ume of block above wa ter de cr eases . 8. V g V g 1 1 2 1 r s = ¢ and V g V g 2 2 2 2 r s = ¢ Þ DV V V V 1 1 1 1 1 1 1 1 = - ¢ = - - s r and DV V 2 2 2 2 1 = - s r \ DV V V V 2 2 1 1 - D = - æ è ç ç ö ø ÷ ÷ - - æ è ç ç ö ø ÷ ÷ = - 1 1 2 2 1 1 1 1 2 2 s r s r s r s r = - - - s r s g r g 1 1 1 2 1 1 1 1 ( ) ( ) DT T = - - æ è ç ç ö ø ÷ ÷ D s r g g g 1 1 2 1 1 1 T T = - - - s r s g r g 1 1 1 2 1 1 1 1 ( ) ( ) DT T = - - æ è ç ç ö ø ÷ ÷ s r g g g 1 1 2 1 1 1 T T D 9. On cool ing brass con tracts more than iron ( ) a a B > Fe such that brass disk gets loosen from hole of iron. 10. V T µ Þ V kT = Þ ln ln ln V k T = + Þ D D V V T T = Þ D D V V T T = = 1 g Introductory Exercise 17.2 1. For ideal gases, pV nRT = Þ T V nR p = = VM mR p Slope = VM mR As slope µ 1 m Þ m m 2 1 < 2. pV nRT = Þ p p T T 2 1 2 1 = = = 360 300 6 5 Þ p p 2 1 6 5 6 5 10 12 = = ´ = atm atm 3. M mix = ´ + ´ + = + = 1 4 28 1 4 44 1 4 1 4 7 11 1 2 36 pV nRT m M RT = = Þ pM m V RT RT = = r Þ r = pM RT \ r = ´ ´ ´ ´ - 101 10 36 10 8 31 290 5 3 . . = ´ ´ = 101 36 8 31 29 15 . . . . kg/m 3 4. pV nRT N N RT A = = Þ N pVN RT A = = ´ ´ ´ ´ ´ ´ ´ ´ - - 10 10 10 250 10 10 300 6 3 6 23 13.6 6.02 8.31 = ´ ´ ´ ´ = ´ 13.6 6.02 8.31 8.21 5 6 10 10 15 15 5. pV nRT = Þ V nR p T = × Slope µ 1 r Þ p p 1 2 > 6. pV nRT = Þ p nRT V = ( ) 1 Þ y mx = is a straight line passing through origin. Introductory Exercise 17.3 1. Av er age ve loc ity de pends on the di rec tion of mo tion of gas mol e cules and as con tainer do not move such that their net ef fect be comes zero, due to the rea son that some mol e cules are mov ing in one di rec tion while other are mov ing in op po site di rec tion. But in case of av er age speed only mag ni tudes are in use which do not cancel each other. 38 | Thermometry, Thermal Expansion & Kinetic Theory of Gases 2. KE = 3 2 kT = ´ ´ ´ 3 2 6 10 300 23 8.31 J = ´ ´ - 3 4 10 21 8.31 J = ´ - 6.21 J 10 21 3. v RT M rms = 3 , v He 8.31 = ´ ´ ´ - 3 300 4 10 3 = ´ 1.37 m/s 10 3 v Ne 8.31 20.2 608.5 = ´ ´ ´ = - 3 300 10 3 m/s KE 6.21 J = = ´ - 3 2 10 21 kT 4. v RT M rms = 3 Þ T Mv R = = ´ ´ ´ - rms 8.31 2 3 6 3 4 10 10 3 = 160.45 K 5. r r r r r = + + = - + - + n n n n n n n n 1 1 2 2 1 2 2 1 2 2 2 2 1 1 ( ) = + - r r r 1 2 2 1 n ( ) Þ n 2 1 2 1 = - - = - - r r r r 1.293 1.429 1.251 1.429 = = = 136 178 0.764 76.4% by mass 6. V V p T p T p h g p 2 1 1 2 2 1 0 0 277 = = + ´ ( ) r = ´ + ´ ´ ´ ´ ´ ( ) 1.01 1.01 10 40 10 10 293 10 277 5 3 5 = ´ ´ = 5.01 1.01 5.25 293 277 Þ V V 2 1 3 105 = = 5.25 cm 7. N nN A = = ´ ´ 1 18 6 10 23 = ´ 1 3 10 23 ; S R = = ´ ´ ´ ´ 4 4 6400 10 10 2 3 2 2 p 3.14 ( ) = ´ 5.14 cm 10 18 2 \ N S = ´ ´ 10 3 10 23 18 5.14 = ´ 6.5 10 3 molecules/cm 2 (a) nC nR V = = 3 2 35 J/K Þ n R = = 70 3 2.8 mole (b) U nRT = = ´ = 3 2 35 273 J/K K 9555 J (c) C C R R p V = + = = 5 2 20.8 J/ K mole 8. (a) n C C nR p V ( ) - = = 29.1 J/K Þ n = 29.1 8.314 mole = 3.5 mole (b) C nc n R V V = = = ´ ´ 3 2 3.5 1.5 8.314 = 43.65 J/K C nc n R C nR p p V = = = + 5 2 = + ´ 43.65 3.5 8.314 = 72.75 J/K (c) C nc V V ¢ = = ´ = n R 5 3 72.75 J/K C nc n R p p ¢ = = ´ 7 2 = + ´ 72.75 3.5 8.314 = 101.85 J/K 10. v RT M rms = 3 and v RT M av = 8 p Here 3 8 > p Þ v v rms av > , i e . ., the statement is true. Thermometry, Thermal Expansion & Kinetic Theory of Gases | 39 AIEEE Corner ¢ Subjective Questions (Level 1) 1. C¢ = - = = 5 68 32 9 36 9 4 Þ C¢ = ° 20 C ; K¢ - = - = 273 5 68 32 9 4 Þ K¢ = 293 K C¢ = - = - = - 5 5 32 9 27 9 3 Þ C¢ = - ° 15 C; K¢ - = - = - 273 5 5 32 9 3 Þ K¢ =258 K C¢ = - = = 5 176 32 9 144 9 16 Þ C¢ = ° 80 C; K¢ - = 273 5 16 Þ K¢ =353 K 2. 30 5 32 9 = ¢ - F Þ F¢ = + = ° 54 32 86 F = ° 546 R 5 5 32 9 = ¢ - F Þ F¢ = + = ° 9 32 41 F = ° 501 R - = ¢ - 20 5 32 9 F Þ F¢ = - + = - ° 36 32 41 F = ° 456 R 3. x x 5 32 9 = - Þ 32 9 5 4 5 = - = - x x x Þ x = - ´ = - ° 5 4 32 40 Þ - ° = - ° 40 40 C F 4. D D C F 5 9 = Þ D D F C = = ´ = ° 9 5 9 5 40 72 \ F F 2 1 72 140 2 = + ° = ° . F 5. 32 20 80 20 0 100 0 - - = ¢ - - C Þ 12 60 100 = ¢ C Þ C¢ = ´ = ° 12 100 60 20 C 6. T T p p 2 1 2 1 160 80 2 = = = Þ T T 2 1 2 = \ T 2 2 = ´ = 273.15 K 546.30 K 7. R R t = + 0 1 ( ) a q D Þ 3 50 250 1 100 . . ( ) = + a Þ 1 250 = K or a = = ´ - 10 250 4 10 3 . /°C \ 650 250 1 4 10 3 . . ( ) = + ´ - Dq Þ 4 10 2 = ´ - Dq Þ Dq = 400 Þ q 2 400 = °C Þ Dq = 400 Þ q 2 400 = °C i e . ., boiling point of sulphur is 400°C. 8. T T p p 2 1 2 1 75 45 75 5 120 80 3 2 = = + + = = T T 2 1 3 2 3 2 30015 = = ´ . K = = ° 450225 . K 177.08 C 9. D D g g a a q = - ( ) Br Fe Þ D D q g g a a = × - 1 Br Fe = ´ ´ × - - - 0.01 Br Fe 10 6 10 1 3 2 a a = - - 10 6 3 ( ) a a Br Fe \ q q a a 2 1 3 10 6 = + - - ( ) Br Fe = ° + - = ° + ´ - 30 10 6 30 100 6 3 C C 0.63 Br Fe ( ) a a = ° 57.78 C . 10. (a) D D l l = - ´ ´ ´ - a q ~ 88.42 2.4 10 30 5 = 0.064 cm (b) D D l l = - ( ) a a q Al St = - ´ ´ - 88.42 2.4 1. ( ) 2 10 30 5 = 0.032 cm l l l S = + = + D 88.42 0.032 cm = 88.45 cm 40 | Thermometry, Thermal Expansion & Kinetic Theory of Gases Page 5 17. Thermometry, Thermal Expansion & Kinetic Theory of Gases Introductory Exercise 17.1 1. (a) C F 5 32 9 = - for F = 0, C = - ´ 5 9 32 = - ° 17 8 . C (b) K F - = - 273. 5 1 5 32 9 for K = 0, F = - ´ + = - ° 9 5 273 15 32 . 459.67 F 2. (a) x x 5 2 32 9 = - Þ x x = - 10 9 17 8 . Þ 17.8 = - æ è ç ö ø ÷ 10 9 1 x Þ x = ´ = ° 17.8 C 9 160 2 . (b) x x 5 2 32 9 = - / Þ x x = - 5 18 17 8 . Þ 17 8 13 18 . = - x Þ x = - ° 24 65 . C 3. C F - - = - - 5 99 5 32 212 32 Þ C F - = - 5 94 32 180 Þ 52 5 94 32 180 - = - F Þ F = + ´ = ° 32 180 94 47 122 F 4. K F - = - 273 15 5 32 9 . Þ x x - = - 273 15 5 9 17 8 . . Þ 4 9 255 35 x = . Þ x = 574 54 . 5. C F 5 32 9 = - Þ 9 5 32 x x = - Þ 4 5 32 x = - Þ x = - ´ = - ° 5 4 32 40 C 6. D D t t = 1 2 a q = ´ ´ ´ ´ - 1 2 10 86400 30 5 1.2 = ´ ´ = 1.5 1.2 8.64 s 15.55 s given. 7. As from 0°C to 4°C, den sity of wa ter in creases so the vol ume of wooden block above wa ter level in creases and as from 4°C to 10°C den sity of wa ter de creases so the vol ume of block above wa ter de cr eases . 8. V g V g 1 1 2 1 r s = ¢ and V g V g 2 2 2 2 r s = ¢ Þ DV V V V 1 1 1 1 1 1 1 1 = - ¢ = - - s r and DV V 2 2 2 2 1 = - s r \ DV V V V 2 2 1 1 - D = - æ è ç ç ö ø ÷ ÷ - - æ è ç ç ö ø ÷ ÷ = - 1 1 2 2 1 1 1 1 2 2 s r s r s r s r = - - - s r s g r g 1 1 1 2 1 1 1 1 ( ) ( ) DT T = - - æ è ç ç ö ø ÷ ÷ D s r g g g 1 1 2 1 1 1 T T = - - - s r s g r g 1 1 1 2 1 1 1 1 ( ) ( ) DT T = - - æ è ç ç ö ø ÷ ÷ s r g g g 1 1 2 1 1 1 T T D 9. On cool ing brass con tracts more than iron ( ) a a B > Fe such that brass disk gets loosen from hole of iron. 10. V T µ Þ V kT = Þ ln ln ln V k T = + Þ D D V V T T = Þ D D V V T T = = 1 g Introductory Exercise 17.2 1. For ideal gases, pV nRT = Þ T V nR p = = VM mR p Slope = VM mR As slope µ 1 m Þ m m 2 1 < 2. pV nRT = Þ p p T T 2 1 2 1 = = = 360 300 6 5 Þ p p 2 1 6 5 6 5 10 12 = = ´ = atm atm 3. M mix = ´ + ´ + = + = 1 4 28 1 4 44 1 4 1 4 7 11 1 2 36 pV nRT m M RT = = Þ pM m V RT RT = = r Þ r = pM RT \ r = ´ ´ ´ ´ - 101 10 36 10 8 31 290 5 3 . . = ´ ´ = 101 36 8 31 29 15 . . . . kg/m 3 4. pV nRT N N RT A = = Þ N pVN RT A = = ´ ´ ´ ´ ´ ´ ´ ´ - - 10 10 10 250 10 10 300 6 3 6 23 13.6 6.02 8.31 = ´ ´ ´ ´ = ´ 13.6 6.02 8.31 8.21 5 6 10 10 15 15 5. pV nRT = Þ V nR p T = × Slope µ 1 r Þ p p 1 2 > 6. pV nRT = Þ p nRT V = ( ) 1 Þ y mx = is a straight line passing through origin. Introductory Exercise 17.3 1. Av er age ve loc ity de pends on the di rec tion of mo tion of gas mol e cules and as con tainer do not move such that their net ef fect be comes zero, due to the rea son that some mol e cules are mov ing in one di rec tion while other are mov ing in op po site di rec tion. But in case of av er age speed only mag ni tudes are in use which do not cancel each other. 38 | Thermometry, Thermal Expansion & Kinetic Theory of Gases 2. KE = 3 2 kT = ´ ´ ´ 3 2 6 10 300 23 8.31 J = ´ ´ - 3 4 10 21 8.31 J = ´ - 6.21 J 10 21 3. v RT M rms = 3 , v He 8.31 = ´ ´ ´ - 3 300 4 10 3 = ´ 1.37 m/s 10 3 v Ne 8.31 20.2 608.5 = ´ ´ ´ = - 3 300 10 3 m/s KE 6.21 J = = ´ - 3 2 10 21 kT 4. v RT M rms = 3 Þ T Mv R = = ´ ´ ´ - rms 8.31 2 3 6 3 4 10 10 3 = 160.45 K 5. r r r r r = + + = - + - + n n n n n n n n 1 1 2 2 1 2 2 1 2 2 2 2 1 1 ( ) = + - r r r 1 2 2 1 n ( ) Þ n 2 1 2 1 = - - = - - r r r r 1.293 1.429 1.251 1.429 = = = 136 178 0.764 76.4% by mass 6. V V p T p T p h g p 2 1 1 2 2 1 0 0 277 = = + ´ ( ) r = ´ + ´ ´ ´ ´ ´ ( ) 1.01 1.01 10 40 10 10 293 10 277 5 3 5 = ´ ´ = 5.01 1.01 5.25 293 277 Þ V V 2 1 3 105 = = 5.25 cm 7. N nN A = = ´ ´ 1 18 6 10 23 = ´ 1 3 10 23 ; S R = = ´ ´ ´ ´ 4 4 6400 10 10 2 3 2 2 p 3.14 ( ) = ´ 5.14 cm 10 18 2 \ N S = ´ ´ 10 3 10 23 18 5.14 = ´ 6.5 10 3 molecules/cm 2 (a) nC nR V = = 3 2 35 J/K Þ n R = = 70 3 2.8 mole (b) U nRT = = ´ = 3 2 35 273 J/K K 9555 J (c) C C R R p V = + = = 5 2 20.8 J/ K mole 8. (a) n C C nR p V ( ) - = = 29.1 J/K Þ n = 29.1 8.314 mole = 3.5 mole (b) C nc n R V V = = = ´ ´ 3 2 3.5 1.5 8.314 = 43.65 J/K C nc n R C nR p p V = = = + 5 2 = + ´ 43.65 3.5 8.314 = 72.75 J/K (c) C nc V V ¢ = = ´ = n R 5 3 72.75 J/K C nc n R p p ¢ = = ´ 7 2 = + ´ 72.75 3.5 8.314 = 101.85 J/K 10. v RT M rms = 3 and v RT M av = 8 p Here 3 8 > p Þ v v rms av > , i e . ., the statement is true. Thermometry, Thermal Expansion & Kinetic Theory of Gases | 39 AIEEE Corner ¢ Subjective Questions (Level 1) 1. C¢ = - = = 5 68 32 9 36 9 4 Þ C¢ = ° 20 C ; K¢ - = - = 273 5 68 32 9 4 Þ K¢ = 293 K C¢ = - = - = - 5 5 32 9 27 9 3 Þ C¢ = - ° 15 C; K¢ - = - = - 273 5 5 32 9 3 Þ K¢ =258 K C¢ = - = = 5 176 32 9 144 9 16 Þ C¢ = ° 80 C; K¢ - = 273 5 16 Þ K¢ =353 K 2. 30 5 32 9 = ¢ - F Þ F¢ = + = ° 54 32 86 F = ° 546 R 5 5 32 9 = ¢ - F Þ F¢ = + = ° 9 32 41 F = ° 501 R - = ¢ - 20 5 32 9 F Þ F¢ = - + = - ° 36 32 41 F = ° 456 R 3. x x 5 32 9 = - Þ 32 9 5 4 5 = - = - x x x Þ x = - ´ = - ° 5 4 32 40 Þ - ° = - ° 40 40 C F 4. D D C F 5 9 = Þ D D F C = = ´ = ° 9 5 9 5 40 72 \ F F 2 1 72 140 2 = + ° = ° . F 5. 32 20 80 20 0 100 0 - - = ¢ - - C Þ 12 60 100 = ¢ C Þ C¢ = ´ = ° 12 100 60 20 C 6. T T p p 2 1 2 1 160 80 2 = = = Þ T T 2 1 2 = \ T 2 2 = ´ = 273.15 K 546.30 K 7. R R t = + 0 1 ( ) a q D Þ 3 50 250 1 100 . . ( ) = + a Þ 1 250 = K or a = = ´ - 10 250 4 10 3 . /°C \ 650 250 1 4 10 3 . . ( ) = + ´ - Dq Þ 4 10 2 = ´ - Dq Þ Dq = 400 Þ q 2 400 = °C Þ Dq = 400 Þ q 2 400 = °C i e . ., boiling point of sulphur is 400°C. 8. T T p p 2 1 2 1 75 45 75 5 120 80 3 2 = = + + = = T T 2 1 3 2 3 2 30015 = = ´ . K = = ° 450225 . K 177.08 C 9. D D g g a a q = - ( ) Br Fe Þ D D q g g a a = × - 1 Br Fe = ´ ´ × - - - 0.01 Br Fe 10 6 10 1 3 2 a a = - - 10 6 3 ( ) a a Br Fe \ q q a a 2 1 3 10 6 = + - - ( ) Br Fe = ° + - = ° + ´ - 30 10 6 30 100 6 3 C C 0.63 Br Fe ( ) a a = ° 57.78 C . 10. (a) D D l l = - ´ ´ ´ - a q ~ 88.42 2.4 10 30 5 = 0.064 cm (b) D D l l = - ( ) a a q Al St = - ´ ´ - 88.42 2.4 1. ( ) 2 10 30 5 = 0.032 cm l l l S = + = + D 88.42 0.032 cm = 88.45 cm 40 | Thermometry, Thermal Expansion & Kinetic Theory of Gases 11. D D l l ´ = ´ 100 100 % % a q = - ´ ´ ´ - 1.2 10 35 100 5 % = - 0.042% 12. F YA l l YA = = D D a q = ´ 2 10 11 ´ ´ ´ ´ ´ - - 2 10 10 40 6 5 1.2 = ´ ´ = ´ = 4 40 160 192 1.2 N 1.2 N N 13. V g s = - ´ - ( ) 50 45 10 3 kg = ´ - 5 10 3 kg V g ¢ ¢ = - ´ - s ( . ) 50 451 10 3 kg = ´ - 4.9 kg 10 3 V g s l ( ) 1 1 10 3 + + = ´ - g q s g q D D 4.9 1 1 4 9 5 + + = g q g q s l D D . Þ 5 5 + = + g q g q s e D D 4.9 4.9 g g q q q g l s s = + = + 0.1 4.9 4.9 5 1 49 5 D D D g s = ´ + ´ ´ - 1 49 75 5 4 9 12 10 6 . = ´ + ´ - - 272.1 12.2 10 10 6 6 = ´ ° - 2.84 C 10 4 14. M = + = 14 3 17 g/mole = ´ - 17 10 3 kg/mole Þ M = ´ ´ - - 17 10 6033 10 3 23 . kg/molecule = ´ - 282 10 26 . kg/molecule 15. n pV RT = = ´ ´ ´ = - 1.52 8.314 298.15 6.13 10 10 6 2 r = = = ´ ´ - - m V nM V 6.13 2 10 10 3 2 = 1.23 kg/m 3 r r ¢ = ¢ = ¢ = = m V nM V nM V 16 16 = 19 62 3 . kg/m 16. p p V V 2 1 1 2 1 76 6 = = ´ atm = 12.7 atm 17. V p V T T p p p T T V 2 1 1 1 2 2 1 2 2 1 1 = × = × × = ´ ´ 1 270 300 500 3 0.5 m = 900 3 m 18. p V T p V T 1 1 1 2 2 2 = Þ mg A p A h mg A p Ah i f + æ è ç ö ø ÷ × = + æ è ç ö ø ÷ 0 0 293 273 Þ h h f i = = ´ = 373 293 373 293 4 cm 50.9 cm 19. p p 1 2 = Þ n V n V L A L A 1 1 2 2 1 2 25 28 40 4 = = = / / Þ L L 1 2 25 28 1 10 5 56 = ´ = = 0.089 n n 1 2 25 28 40 4 25 280 5 56 = = = = / / 0.089 20. n n n = + 1 2 Þ p V V p V p V ( ) 1 2 1 1 2 2 + = + Þ p p V p V V V = + + 1 1 2 2 1 2 \ p = ´ + ´ + 1.38 0.11 0.69 0.16 0.11 0.16 MP a = + = 0.1518 0.1104 0.27 0.2622 0.27 = 0.97 MP a 21. pV T pV T p V T p V T 1 2 1 1 1 1 2 2 + = + 1 293 600 3 atm K cm ´ = + æ è ç ç ö ø ÷ ÷ p 1 3 3 400 373 200 273 cm K cm K Þ p 1 600 293 400 373 200 273 = + / at m \ p 1 3 2 9 3 2 3 7 3 1 2 7 3 = + æ è ç ö ø ÷ a t m = + = 3 1 5 7 3 . 1.07 2.64 at m = 1.136 atm Thermometry, Thermal Expansion & Kinetic Theory of Gases | 41Read More
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