Enviromental - Assignment Example

Chemistry 513: Assignment Question From PV = nRT. P= 6×10-7, R = 0.0821, T = 170K. At atm, V = 22.4., at 0.006 atm, V = (0.006 × 22.4)/1 = 0.1344. Since PV = nRT, then n = PV/RT; n = 5.78×10-5. Based on the equation: ppm = (mg/L/MW) ×22.4. mg/L = (5.78×10-5) × 1000 = 5.78×10-2. In this case, ppm of H2O = (5.78×10-2/18) ×22.4 = 7.19×10-2 ppm.Question 2a) 1 ppm = 1 volume of gas/106 volumes of air. Therefore, 0.01 ppm = 0.01 volume of Ozone gas/106 volumes of air. Ppmv = (ppm/MW) × 22.4. Ppmv of ozone = 0.

00467 ppmv O3. This means for every 1 million volume of air, there is 0.00467 volumes of Ozone (Dr Richards 01). Therefore, if 1 million volumes of air = 4.88×1021, then 0.00467 volumes of Ozone = (0.00467 × 4.88×1021)/1×106. Mass of Ozone in troposphere = 2.28×1013g.b) Similarly, it means 3 ppm = 3 volumes 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 million volumes of air have 1.4 volumes of Ozone by mass.

1 million volumes of air in stratosphere represent 2.5×1020 g of air. What about 1.4 ppmv of Ozone? Mass of Ozone = (1.4× 2.5×1020)/1×106. Mass of Ozone = 3.5×1014g.Question 3Partial pressure, Px = Cx×P 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 100’s of meters.

Therefore, Pa = 0.9877300= 0.0244atm. Partial pressure of Ne = 18ppm×0.0244 = 0.44 atm. Partial pressure of N2O = 0.31×1 = 0.31 atm. Therefore, Ne has a greater partial pressure that N2O.Question 4100% relative humidity represents 0.031atm H2O. On the other hand, liquid water is present at 100ug/m3. Assuming a temperature of 25oC, then we will convert ug/m3 into ppmv using the formula 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 gaseous mixture = 1×106 / 6.023×1023 = 1.66×10-18. Converting moles into volume we get 4.06×10-14 cm3. Therefore, the urban atmosphere contains 20.18 molecules of liquid H2O /4.06×10-14 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 (0.227×1)/24.45 = 0.0093 molecules/L. In terms of cm3, the atmosphere has 9.3 molecules of H2O vapor/4.06×10-14 cm3. Therefore liquid water is more abundant as compared to water vapor in the same environment.Work CitedDr Richards. “CHEM 3400 - Mathematics in Environmental Chemistry: Converting Atmospheric Pollutant Concentrations”. Chem3400.blogspot.com. January 16, 2013. Web January 28, 2014. http://chem3400.blogspot.com/2013/01/mathematics-in-environmental-chemistry.html#!