hw1s - = constant or(6 marks(c 1013 millibars(2 marks(d It...

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PHYS0629/PHYS1056 Weather and Climate Solution of Assignment 1 Due date: 5:00pm, September 25, 2015 1. Answer: Solving the equation x = 9 x 5 + 32 , (1) we have x=-40. The required temperature is - 40 F = - 40 C = 233K. 2. Answer: 30 C - 6 . 5 × 10 C = - 35 C. 3. Answer: (a) The density of mercury is about 13 . 5gcm 3 , while the density of water is about 1 . 0gcm 3 . (b) Hence, the water height is about 13 . 5 × 76cm = 10 . 3m. 4. Answer: (a) The saturation mixing ratio at 28 C is m sat = 3 . 752 × 10 7 . 5 T 237 . 7+ T = 23 . 15g / kg . (2) The sea level mixing ratio is 0 . 8 × 23 . 15 = 18 . 52g/kg. Hence, the dew point temperature is T dew = 237 . 7log 10 (18 . 52 / 3 . 752) 7 . 5 - log 10 (18 . 52 / 3 . 752) = 24 . 2 C . (3) (b) 757m. (c) The saturation mixing ratio at 15 C is 3 . 752 × 10 7 . 5 × 15 237 . 7+15 = 10 . 46g / kg . (4) The required RH is 10.46/23.15=45%. (d) The RH is still 40% because there will be no condensation inside the dehumidifier. 5. Answer: (a) PV = nRT . (2 marks) (b) TP (1 γ ) = constant or TP 2 / 7 = constant or PT γ/ (1 γ ) = constant
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Unformatted text preview: ) = constant or . ..(6 marks) (c) 1013 millibars. (2 marks) (d) It is the absolute temperature in the ideal gas law, 25 ◦ = 298K. (6 marks) T = p 922 1013 P 2 / 7 298K = 290 . 1K = 17 . 1 ◦ C . (5) (e) 7 . 9 ◦ C / km. (4 marks) 6. Answer: R=8.31 J/(mol K). PV = nRT or V = nRT/P . At 1atm, the pressure in S.I. unit is 101325N/m 2 . Hence, the required volume is 8 . 31 × (273 + 25) / 101325m 3 = 0 . 0244m 3 . 1 7. Answer: If it rotated faster, there would be less time to accumulate heat during the day, and less time to lose heat during the night. Hence, the day-night temperature variation would be smaller. Also, the highest temperature in summer days would lower and the lowest temperature in winter nights would be higher. The diFerence would be smaller. 2...
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