# Chapter 1-3 Answers - 1 Chapter 1 1 26,400 ft for H=8 km 2...

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1 Chapter 1. 1. 26,400 ft for H =8 km 2. 0.0041 3. 0.819 hPa (using H =8 km, 1 mile =1.6 ˙ km) 4a) 40,212 km 4b) 111.7 km deg - 1 4c) 96.7 km deg - 1 5. 50,000 km 2 8. 11.175 km s - 1 9. 2 2 gh 2 Chapter 2. 1. 1.276, 1.231, 1.189 kg m - 3 2. 0.795 kg m - 3 3. 212.2 hPa 4. 12,52 N 5. 2.687 × 10 25 molecules m - 3 6. R d = 2 . 87 hPa K - 1 m 3 kg - 1 7. Differentiate P ( v ) with respect to v and set it to zero. See Table 2.1. 8. 461.2, 493.0, 412.5, 1842.7 m - 1 9. 630.2 Pa. 1 kg m - 1 . Pressure with moisture=78467.7 Pa. Density of moist mix=1 kg m - 3 . Density of dry air at same temperature and pressure: 1.00148 kg m - 3 . 10. 1.601 × 10 8 J 11. 211672 J 12. n 0 = 2 . 65 × 10 25 molecules m - 3 , H=8000 m, N =2.12 × 10 29 molecules 13. z λ = H = H ln( σ c H ) 14. gp 0 RT 0 R 0 ze - z/H dz = gp 0 RT 0 1

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15. N 2 m 0 v 2 = 3 N 2 k B T , see # 12. for N . 16. f yellow = c/λ = 6 . 0 × 10 14 s - 1 . f coll = n 0 σ c v 3 × 10 25 × 0 . 4 × 10 - 18 × 811 = 9 . 7 × 10 9 s - 1 . An excited atom might suffer tens of collisions before it relaxes to the lower energy level. 3 Chapter 3 1. d) p 0 z H 2 e - z/H 2. a) ρR ; Let 1 /T x , then take the partial with respect to x . 3. κ T = 1 p ; to get κ θ use pV
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Unformatted text preview: γp 4. β = 1 T 6. 195.8kJ 7. T = T ( 1 e =2 . 78 . . ) . 286 = 205 . 1K. Δ T =-67 . 9K. Therefore, Δ U = 48 . 72J= W , since Q = 0. 8. a) 10.05kJ; b) 7.18kJ; c) dU + V dp → dU . 9. H = RT/g = 8777m ,p = p e-z/H ; then p (1000m) = 892hPa. Δ ≈ V Δ p p V 10. 1.2015kgm-3 . Q =-. 003 × 2 . 5 × 10 6 =-7500J. Use the mass of the dry air only as an approximation. Δ T = M c v Q 11. T 600 = 248 . 7K. Δ U = M c v Δ T, Δ H = 0 12. 0.02Ks-1 ; 20Js-1 ; Δ U per unit time= 14.3Js-1 13. a) 10,000kg. b) mass of an ’air’ molecule=29/ N A = 4 . 82 × 10-23 kg. N = 2 . 08 × 10 26 molecules. c) 1.07 × 10 41 14. c s = √ RT for isothermal; √ γRT, γ = 1 . 40 for adiabatic compression waves 15 (16). θ = T ± p p ² . 286 2...
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