Effects of Air Pollution Questions and Answers

Answer: A

Correct answer: 15.4 per cent

The relative risk applies per 10 micrograms, so 25 micrograms is 2.5 such increments:

\[ RR = 1.06^{2.5} = 1.154 \]

an increase of about 15.4 per cent. The distractor \( 6.0 \) applies the risk once regardless of the increment, and \( 150 \) confuses the relative risk with a percentage. Note that the relation compounds rather than adding, though over this range the difference between compounding and simple multiplication is small.

Answer: B

Correct answer: 960

Find the baseline deaths first, then apply the increase in risk:

\[ 2\,000\,000 \times \frac{8}{1000} = 16\,000\ \mathrm{deaths\ per\ year} \]

\[ 16\,000 \times 0.06 = 960\ \mathrm{additional\ deaths} \]

This is the standard health impact calculation: a relative risk from epidemiology, a baseline rate from vital statistics and a population figure together give an attributable burden. It is how the benefit of a proposed air quality improvement is expressed in lives rather than in micrograms.

Answer: C

Correct answer: 8 per cent

Apply the stated proportionality:

\[ 50\ \mathrm{ppm} \times 0.16 = 8\ \mathrm{per\ cent\ carboxyhaemoglobin} \]

Eight per cent is a substantial internal dose - well above the 2 to 3 per cent at which effects on exercise capacity and cardiovascular function become detectable, and enough to cause headache and impaired judgement. The relation holds at equilibrium, which takes several hours to reach, so a short exposure at 50 ppm would give less.

Answer: D

Correct answer: Half its previous value

Visibility varies inversely with the extinction coefficient, so doubling the extinction halves the visual range:

\[ V \propto \frac{1}{b_{ext}} \quad \Rightarrow \quad \frac{V_2}{V_1} = \frac{1}{2} \]

The relation is the Koschmieder formula, with the visual range roughly \( 3.9/b_{ext} \). Its practical consequence is that visibility responds proportionally to fine particulate, which is why haze is a reasonable qualitative indicator of PM2.5 and why visibility records have been used to reconstruct historical particulate trends.

Answer: A

Correct answer: 14.3 per cent

The loss is expressed relative to the clean-air yield:

\[ \frac{4.2 - 3.6}{4.2} \times 100 = \frac{0.6}{4.2} \times 100 = 14.3\ \mathrm{per\ cent} \]

The distractor \( 16.7 \) divides by the polluted yield instead, which overstates the loss. Ozone is the pollutant responsible for the largest crop losses worldwide, precisely because it enters through open stomata and so does most damage on well-watered plants in bright growing conditions.

Answer: B

Correct answer: 10 kg per hectare per year

Exceedance is the amount by which deposition exceeds the critical load:

\[ 22 - 12 = 10\ \mathrm{kg\ N\ per\ hectare\ per\ year} \]

The critical load is the deposition an ecosystem can tolerate without significant harmful effect, and it differs by ecosystem - a nutrient-poor heathland has a low one, a fertile woodland a higher. Mapping exceedance rather than deposition is what allows emission reduction effort to be directed where the ecological benefit is greatest.

Answer: C

Correct answer: 100

pH is the negative logarithm of hydrogen ion concentration, so a difference of two pH units is a factor of \( 10^2 \):

\[ \frac{[H^+]_{4.5}}{[H^+]_{6.5}} = 10^{6.5-4.5} = 10^2 = 100 \]

The distractor \( 2 \) treats the scale as linear, which is the commonest error and badly understates the change. At pH 4.5 most fish species cannot survive, both from the acidity itself and from the aluminium that low pH mobilises out of catchment soils.

Answer: D

Correct answer: 5.0

Divide the polluted rate by the clean rate:

\[ \frac{6.0}{1.2} = 5.0 \]

Zinc corrodes five times faster in the industrial atmosphere, because sulphur dioxide dissolves in the surface moisture film and attacks the metal, while the hygroscopic sulphate formed keeps the surface wet for longer. Ratios of this order are why galvanised structures in industrial areas need far shorter maintenance intervals than the same structures in clean rural air.

Answer: A

Correct answer: 14 years

Divide the coating thickness by the corrosion rate:

\[ \frac{85}{6.0} = 14.2\ \mathrm{years} \]

In clean air, at \( 1.2 \) micrometres a year, the same coating would last about 70 years - so pollution shortens the maintenance interval fivefold. This is the practical form in which materials damage enters an economic assessment: not as an abstract corrosion rate but as the cost of repainting and replacement brought forward by decades.

Answer: B

Correct answer: 13.3 per cent

Express the aerosol forcing as a fraction of the greenhouse forcing:

\[ \frac{0.4}{3.0} \times 100 = 13.3\ \mathrm{per\ cent} \]

The offset creates an awkward policy interaction: sulphate aerosol has been masking part of the warming already committed, so reducing sulphur emissions - entirely desirable for health and for acid deposition - unmasks it. The two effects differ fundamentally in persistence, since aerosol lasts days while carbon dioxide lasts centuries.