Question

ENSC 102-Problem Set #7 Due at the beginning of class on Friday, May 30 Answer each of the following questions in your own words and using your own calculations. For worked problems remember to explain any intermediate steps or assumptions, show all units, and box your final answers. For fu possible credit, make sure your finished work is legible, prepared on loose-leaf paper and shows your name and your lab instructor s name at the top of the first page 1. Concept questions Fully explain the following phenomena in your own words: a. Even though smog had been recognized and relatively well understood for centuries, the 20th century smog problem in Los Angeles was a mystery for quite awhile. b. Secondhand smoke in the home (for example as experienced by the family members of smokers) has been linked to some of the same diseases as those affecting smokers themselves, even though the concentrations of secondhand smoke are relatively low. c. Most deep convection upwards into the atmosphere occurs near the equator, while most deep convection in the ocean occurs near the poles. 2. AQI On a particular day in Riverside, the following pollutant levels are recorded O: 50 ppb PM2s: 60 Hg m3 PMio: 110 ug m3 What is the AQI for each pollutant? What is the overall AQI for the day? 3. Concentration and mixing ratio If the concentration of NH in air is 0.556 ug m at0 °C and 1 atm, what is its mixing ratio in ppb? (Atomic weights of N and H are 14 and 1.01, respectively.) 4. Residence times at steady state The mass of ammonia (NH), nitrous oxide (N20), and methane (CH4) represent around 1 x 10-8%, 3x10-5%, and 7x10-% of the Earth's atmosphere, respectively. (Total mass of the Earth's atmosphere is 5x10s kg.) Calculate the residence times for each of these compounds if their global losses are 5x10, 1x100, and 4x10 kg per year, respectively. 5. Deposition velocity a. Assuming that tropospheric Os over the continents is confined to a layer of the atmosphere extending from the surface of the Earth up to a height of 4 km and that the average deposition velocity of O onto the ground is 0.40 cm s-, how long woukd it take for all the O in the column to be deposited to the ground if all other sources and sinks of O were suddenly switched off? b. Does this match up well to the residence time for O given in Table 5.1? What does this comparison imply about the primary sinks of tropospheric Os?

Answer 1

1. A. Los Angeles smog is a result of ozone formation due to photochemical reactions. This is an event iccurioc during summers as the temperature is hot and sunlight is clear.

Traditional smog involves pollutants like SO2, called primary pollutants, while secondary pollutants involve ozone. These react with primary pollutant and give rise to smog.

Since ozone was believed to be useful to life forms , its reason for smog was the cause of mystery of Los Angles smog.

B. Second hand smoke is the one that is faced by people near to smokers. The adverse effects has wide range of disease. There are two types of smoke coming out from a cigarette- main stream and side stream . Main stream = smoke coming out from mouth

Side stream = smoke coming out from cigarette tip

Side stream smoke contains high levels of harmful chemicals. Like higher concentrations of carbon monoxide and ammonia. These cause harm to nearby people.

C. The equator receive sun rays for the major part of the day. As a result the air near equator becomes hot and rises up. As a result the deep convection current is upward near equator.

Deep sea current near poles are downward. The temperature near poles is very low at the surface. But the temperature in deeper regions of ocean is high. As a result cool water from surface go down to replace hot water. As the hot water rises up it cools down.

2. AQI of ozone = 46 (good)

AQI of PM 2.5 = 153 (unhealthy)

AQI of PM 10 = 78 (moderate)

The overall AQI is 153. The AQI of highest value is considered the overall AQI.

3. mixing ratio is concentration of gas with respect to volume of air. Moles of NH3 IN 0.556 microgram = 0.556/(14+3.03) = 0.0326 micro moles per cubic metre. So 32.6 * 10^(-9) moles per volume. That is 32.6 ppb.

4. Residence time is the amount gas present divided by the global loss.

Mass of ammonia = 1* 10^(-8) * 5 * 10^18 / 100 = 5 * 10^8 kg

Residence time = 5 * 10^8 / ( 5 * 10^10) = 0.01 year

Similarly residence time of N2O = 15*10^11/(1*10^10) = 10 years

Residence time of CH4 = 35*10^11/(4*10^11) = 8.75 years

5 a. Time = length of column / deposition speed

= 4 * 1000 * 100 cm / 0.40 cm per second

= 10^6 seconds

B.