# Air and Noise Pollution

 Question 1
A sample of air analyzed at $25^{\circ}C$ and 1 atm pressure is reported to contain 0.04 ppm of $SO_2$. Atomic mass of S = 32, O = 16. The equivalent $SO_2$ concentration (in $\mu g/m^3$) will be__________. (round off to the nearest integer)
 A 105 B 118 C 138 D 162
GATE CE 2022 SET-2   Environmental Engineering
Question 1 Explanation:
Concentration of $SO_2$ in ppm = 0.04
Let's equivalent concentration in $=\mu g/m^3$ is $x$.
$x\mu g$ of $SO_2$ present in $1m^3$ of air.
We know, 1 mole of $SO_2$ (at $0^{\circ}C$, 1 atm) has volume of 22.4 L.
at $25^{\circ}C$ and 1 atm, volume of $SO_2$
$=\frac{22.4}{273+0}\times (273+25)=24.45 lit$
$\frac{x \times 10^{-6}}{64}$ mole of $SO_2$ has volume
$\frac{24.45 \times x \times 10^{-6}}{64} lit \; in \; 1m^3$ of air.
$0.382x \times 10^{-3}m^3$ of $SO_2$ present in $10^{6}m^3$ of air.
$0.382x \times 10^{-3}=0.04$
$x=104.7 \approx105$
0.04 ppm of $SO_2=105\mu g/m^3$ of $SO_2$
 Question 2
A process equipment emits 5 kg/h of volatile organic compounds (VOCs). If a hood placed over the process equipment captures 95% of the VOCs, then the fugitive emission in kg/h is
 A 0.25 B 4.75 C 2.5 D 0.48
GATE CE 2022 SET-2   Environmental Engineering
Question 2 Explanation:
VOC emission = 5 Kg/h
Capturing Efficiency =95%
Fugitive emission = (5/100)*5 Kg/h=0.25 Kg/h
 Question 3
i. Value of the wind profile exponent for the 'very unstable' atmosphere is smaller than the wind profile exponent for the 'neutral' atmosphere.
ii. Downwind concentration of air pollutants due to an elevated point source will be inversely proportional to the wind speed.
iii. Value of the wind profile exponent for the 'neutral' atmosphere is smaller than the wind profile exponent for the 'very unstable' atmosphere.
iv. Downwind concentration of air pollutants due to an elevated point source will be directly proportional to the wind speed.

Select the correct option.
 A i is False and ii is True B i is True and iv is True C ii is False and iii is False D iii is False and iv is False
GATE CE 2021 SET-2   Environmental Engineering
Question 3 Explanation:
$\text{Concentration} \propto \frac{1}{\text{wind speed}}$
 Question 4
Which of the following statement(s) is/are correct?
 A Increased levels of carbon monoxide in the indoor environment result in the formation of carboxyhemoglobin and the long term exposure becomes a cause of cardiovascular diseases B Volatile organic compounds act as one of the precursors to the formation of photochemical smog in the presence of sunlight C Long term exposure to the increased level of photochemical smog becomes a cause of chest constriction and irritation of the mucous membrane D Increased levels of volatile organic compounds in the indoor environment will result in the formation of photochemical smog which is a cause of cardiovascular diseases
GATE CE 2021 SET-2   Environmental Engineering
Question 4 Explanation:
Indoor environment photochemical smog does not form.
 Question 5
A baghouse filter has to treat $12 \mathrm{~m}^{3} / \mathrm{s}$ of waste gas continuously. The baghouse is to be divided into 5 sections of equal cloth area such that one section can be shut down for cleaning and/or repairing, while the other 4 sections continue to operate. An air-to-cloth ratio of $6.0 \mathrm{~m}^{3} / \mathrm{min}-\mathrm{m}^{2}$ cloth will provide sufficient treatment to the gas. The individual bags are of 32 cm in diameter and 5 m in length. The total number of bags (in integer) required in the baghouse is ________________
 A 10 B 20 C 30 D 40
GATE CE 2021 SET-1   Environmental Engineering
Question 5 Explanation:
Given, discharge to be passed = $12 \mathrm{~m}^{3} / \mathrm{s}$
Out of 5 sections, 4 will operate at a time.
Each bag is Thus, surface area of each bag $=\pi \times 0.32 \times 5 \mathrm{~m}^{2}=5.0265 \mathrm{~m}^{2}$
Total surface area required wr.t. given discharge
$=\frac{12 \mathrm{~m}^{3} / \mathrm{s}}{6 \mathrm{~m}^{3} / \mathrm{min}-\mathrm{m}^{2}}=\frac{12 \mathrm{~m}^{3} / \mathrm{s}}{\frac{6}{60} \mathrm{~m}^{3} / \mathrm{s}-\mathrm{m}^{2}}=120 \mathrm{~m}^{2}$
$\therefore$ Working bag filters required $=\frac{120 m^{2}}{5.0265 m^{2}}=$ 23.87 or 24 bag filters
But, since it is asked total (which includes a standby set also), then,Total no. of bag filters $=24 \times \frac{5}{4}=30$
 Question 6
The liquid forms of particulate air pollutants are
 A dust and mist B mist and spray C smoke and spray D fly ash and fumes
GATE CE 2021 SET-1   Environmental Engineering
Question 6 Explanation:
The liquid forms of particulate air pollutants are mist and spray.
Note: Mist is a cloud made of very small drops of water in the air just above the ground which reduces the visibility
 Question 7
Which one of the following statements is correct?
 A Pyrolysis is an endothermic process, which takes place in the absence of oxygen B Pyrolysis is an exothermic process, which takes place in the absence of oxygen C Combustion is an endothermic process, which takes place in the abundance of oxygen D Combustion is an exothermic process, which takes place in the absence of oxygen
GATE CE 2021 SET-1   Environmental Engineering
Question 7 Explanation:
Pyrolysis is an endothermic process as there is a substantial heat input required to raise the biomass to the reaction temperature.
 Question 8
A waste to energy plant burns dry solid waste of composition :

Carbon = 35%,
Oxygen = 26%,
Hydrogen = 10%,
Sulphur = 6%,
Nitrogen = 3% and
Inerts = 20%.

Burning rate is 1000 tonnes/d. Oxygen in air by weight is 23%. Assume complete conversion of Carbon to $CO_2$, Hydrogen to $H_2O$, Sulphur to $SO_2$ and Nitrogen to $NO_2$.

Given Atomic weighs : H=1, C=12, N=14, O=16, S=32.

The stoichiometric (theoretical) amount of air (in tonnes/d, round off to the nearest integer) required for complete burning of this waste, is __________.
 A 1602 B 6967 C 1862 D 7216
GATE CE 2020 SET-2   Environmental Engineering
Question 8 Explanation:
$\underbrace{C}_{12}\underbrace{O}_{2}\rightarrow CO_2$
Oxygen required for 350 tonne/day
$\frac{32}{12}\times 350=933.33$
$\underbrace{4H}_{4}+\underbrace{O_2}_{32}\rightarrow 2H_2O$
Oxygen required for 100 tonne/day
$\frac{32}{4}\times 100=800$
$\underbrace{S}_{32}+\underbrace{O_2}_{32}\rightarrow SO_2$
Oxygen required for 60 tonne/day
$\frac{32}{32}\times 60=60$
$\underbrace{N}_{14}+\underbrace{O}_{32}\rightarrow NO_2$
Oxygen required for 30 tonne/day
$\frac{32}{14}\times 30=68.57$
Total $O_2$ = 1861.9 tonne/day
Available $O_2$ in waste = 260 tonne/day
Required =1861.9-260=1601.9 tinne/day
Amount of air required $=\frac{1601.9}{0.23}=6964.78 \; \text{tonne/day} \simeq 6965 \; \text{tonne/day}$
 Question 9
A gas contains two types of suspended particle having average sizes of 2 $\mu m$ and 50 $\mu m$. Amongst the options given below, the most suitable pollution control strategy for removal of these particles is
 A settling chamber followed by bag filter B electrostatic precipitator followed by venturi scrubber C electrostatic precipitator followed by cyclonic separator D bag filter followed by electrostatic precipitator
GATE CE 2020 SET-2   Environmental Engineering
Question 9 Explanation:
Large size particles should be removed first, hence answer should be (A).
Theoretically, gravitational settling chamber should be able to remove particulates down to 5 or 10 $\mu m$, but in actual they are not practical for removal of particles much less than 50 $\mu m$ in size.
Hence, gravitational settling chamber is most suitable choice for removal of suspended particles of size 50 $\mu m$.
After the removal of 50 $\mu m$ size particles, to remove 2 $\mu m$ size particles, we use fabric bag filter.
Note: Fabric bag filter has high collection efficiency for all particle sizes, especially for particles smaller than 10 $\mu m$ in diameter.
 Question 10
A gaseous chemical has a concentration of 41.6 $\mu mol/m^3$ in air at 1 atm pressure and temperature 293 K. The universal gas constant R is $82.05 \times 10^{-6} (m^3 atm)/(mol K)$. Assuming that ideal gas law is valid, the concentration of the gaseous chemical (in ppm, round off to one decimal place), is _______.
 A 41.6 B 1 C 2.8 D 1.6
GATE CE 2020 SET-1   Environmental Engineering
Question 10 Explanation:
\begin{aligned} PV&=nRT\\ V&=\frac{nRT}{P}\\ &=\frac{41.6 \times 10^{-6}\times 32.05 \times 10^{-6}}{1} \times 293\\ &=10^{-6}m^3 \end{aligned}
$41.6\mu$ mole of gas volume of $10^{-6} m^3$
\begin{aligned} 1 ppm&= \frac{\text{1 part of gas}}{10^6 \text{ part of air}} \\ &= \frac{1 m^3\text{ part of gas}}{10^6 m^3\text{ part of air}}\\ 1ppm &=\frac{41.6 \times 10^6 \mu \; moles}{10^6 m^3} \\ 41.6\mu moles/m^3 &=1 ppm \end{aligned}
There are 10 questions to complete.