Question

1.The population of India in the year 2000 was 1 billion and it increased exponentially at a rate of 1.6% per year. If the growth rate is maintained, what will be the population in the year 2020? If the growth rate is decreased to 1.2% per year from 2020 onwards and is maintained at that level, what will be the population in the year 2050? Assuming the average human exhales 2.3 pounds of carbon dioxide on an average day, what is the total amount of Carbon released in the atmosphere annually by human exhalation (in billion tonnes) around the year 2020?  

2.Calculate the suspended particulate concentration (in μg/m³) in a sample collected through a hi-vol. sampler: Weight of clean filter = 5.00 g, Weight of the filter after exposure for 24 hours = 5.38 g, Average air flow = 2000 m³ in 24 hours.   

Answer 1

(1) The population of India in in 2000(A₀) = 1billion

Rate of exponential increase/year(k) = 1.6% = 1.6/100 = 0.016

Time (t) = 20 years

The growth function is given by , A = A₀ e^(kt) = 1billion *(e^(0.016*20))

Population of india in year 2020 = 1billion * 1.3770 = 1.377 billion.

If the growth rate is reduced to 1.2% from 2020, k= 1.2/100= 0.012

Time (t) = 2020-2050= 30years

Now , the new A₀ (population in 2020) will be 1.377 billion

So the population of India in 2050, A = A₀ (e^(kt)) = 1.377 (e^(0.012*30)) = 1.9736 billion.

Assuming the average human exales 2.3 pounds of CO₂/ day i.e 0.00104326 tonnes

Then annual human exalation around the year 2020 = 1.377*0.00104326*365 = 0.5243 billion tonnes/year

(2) Weight of the clean filter = 5.00g

Weight of the filter after exposure for 24 hours = 5.38g

So, weight of the perticulates = 5.38-5 = 0.38g

Average air flow = 2000 m³

concentration in microgram/m³ = (0.38*1000000)/2000 = 190 microgram/m³