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

Enviromental - Assignment ExampleMass of Ozone in troposphere = 2.281013g.Similarly, it means 3 ppm = 3 saturations of Ozone gas/106 volumes of air. Ppmv of Ozone in stratosphere = (ppm/MW) 22.4. Therefore, ppmv = (3/48) 22.4 = 1.4ppmv. This means that 1 trillion volumes of air have 1.4 volumes of Ozone by mass. 1 million volumes of air in stratosphere represent 2.51020 g of air. What about 1.4 ppmv of Ozone? Mass of Ozone = (1.4 2.51020)/1106. Mass of Ozone = 3.51014g.Partial pressure, Px = CxP where Px is partial pressure, Cx is the partial concentration of gas x and P is the entire pressure. N2O, MW of 44, has a concentration of 0.31ppm at ground level. Ne, MW of 20, has concentration of 18 ppm at 30km. Pressure of Ne with respect to the altitude of 30 km is given by Pa = 0.9877a, where a = altitude in 100s of meters. Therefore, Pa = 0.9877300= 0.0244atm. Partial pressure of Ne = 18ppm0.0244 = 0.44 atm. Partial pressure of N2O = 0.311 = 0.31 atm. Therefore, Ne has a greater pa rtial pressure that N2O.100% relative humidity represents 0.031atm H2O. On the other hand, naiant water supply is present at 100ug/m3. Assuming a temperature of 25oC, then we will convert ug/m3 into ppmv using the convention ppmv = (mg/m3 oK)/ (0.08205 MW). With respect to water vapor, ppmv = (0.1 mg/m3 298)/(0.08205 18). Ppmv = 20.18. Using PV = nRT, then moles of air in 1 mol of aerosolized mixture = 1106 / 6.0231023 = 1.6610-18. Converting moles into volume we get 4.0610-14 cm3. Therefore, the urban atmosphere contains 20.18 molecules of liquid H2O /4.0610-14 cm3 of air.On water vapor, 30% relative 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. Based on theory, 1 mol = 24.45L (Dr Richards 01). Therefore, 0.227L contains (0.2271)/24.45 = 0.0093 molecules/L. In terms of cm3, the atmosphere has 9.3