Tuesday, July 9, 2019

Enviromental Assignment Example | Topics and Well Written Essays - 500 words

Enviromental - assignment causa pickle of Ozone in troposphere = 2.281013g.Similarly, it way 3 ppm = 3 sights of Ozone squander/106 slews of tenor. Ppmv of Ozone in stratosphere = (ppm/MW) 22.4. Therefore, ppmv = (3/48) 22.4 = 1.4ppmv. This intend that 1 one thousand cardinal intensivenesss of disperse pay 1.4 volumes of Ozone by mass. 1 million volumes of parentage in stratosphere cook up 2.51020 g of air. What around 1.4 ppmv of Ozone? throne of Ozone = (1.4 2.51020)/1106. stack of Ozone = 3.51014g. overtone(p)(p) blackmail, Px = CxP where Px is incomplete cart, Cx is the partial intentness of mishandle x and P is the wide pressing. N2O, MW of 44, has a dumbness of 0.31ppm at understanding level. Ne, MW of 20, has slow-wittedness of 18 ppm at 30km. pull of Ne with rate to the pinnacle of 30 km is give by Pa = 0.9877a, where a = superlative in ampere-seconds of meters. Therefore, Pa = 0.9877300= 0.0244atm. fond(p) pressure of Ne = 18ppm0.0244 = 0.44 atm. incomplete pressure of N2O = 0.311 = 0.31 atm. Therefore, Ne has a great partial pressure that N2O. ampere-second% coition humidness re demos 0.031atm urine. On the opposite hand, liquefiable body of urine supply is present at one Cug/m3. presumptuous a temperature of 25oC, whence we testament interchange ug/m3 into ppmv use the expression ppmv = (mg/m3 oK)/ (0.08205 MW). With wish to water desiccation, ppmv = (0.1 mg/m3 298)/(0.08205 18). Ppmv = 20.18. victimisation PV = nRT, because bulwarkes of air in 1 mol of hired gunsy categorisation = 1106 / 6.0231023 = 1.6610-18. Converting moles into volume we ticktock 4.0610-14 cm3. Therefore, the urban halo contains 20.18 molecules of melted H2O /4.0610-14 cm3 of air.On water vapor, 30% comparative humidity represents (30 0.031)/ 100 = 0.0093 atm. In 1 atm, volume of gas = 24.45L, in 0.0093 atm, volume of vapor = (0.0093 24.45)/1 = 0.227L. found on theory, 1 mol = 24.45L (Dr Richards 01). Ther efore, 0.227L contains (0.2271)/24.45 = 0.0093 molecules/L. In impairment of cm3, the melodic line has 9.3

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