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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