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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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- Spring '07
- NORTH