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)

105 | |

118 | |

138 | |

162 |

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

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

0.25 | |

4.75 | |

2.5 | |

0.48 |

Question 2 Explanation:

VOC emission = 5 Kg/h

Capturing Efficiency =95%

Fugitive emission = (5/100)*5 Kg/h=0.25 Kg/h

Capturing Efficiency =95%

Fugitive emission = (5/100)*5 Kg/h=0.25 Kg/h

Question 3 |

Read the statements given below.

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.

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.

i is False and ii is True | |

i is True and iv is True | |

ii is False and iii is False | |

iii is False and iv is False |

Question 3 Explanation:

\text{Concentration} \propto \frac{1}{\text{wind speed}}

Question 4 |

Which of the following statement(s) is/are correct?

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 | |

Volatile organic compounds act as one of the precursors to the formation of photochemical smog in the presence of sunlight | |

Long term exposure to the increased level of photochemical smog becomes a cause of chest constriction and irritation of the mucous membrane | |

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 |

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 ________________

10 | |

20 | |

30 | |

40 |

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

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

dust and mist | |

mist and spray | |

smoke and spray | |

fly ash and fumes |

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

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?

Pyrolysis is an endothermic process, which takes place in the absence of oxygen | |

Pyrolysis is an exothermic process, which takes place in the absence of oxygen | |

Combustion is an endothermic process, which takes place in the abundance of oxygen | |

Combustion is an exothermic process, which takes place in the absence of oxygen |

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 __________.

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 __________.

1602 | |

6967 | |

1862 | |

7216 |

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}

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

settling chamber followed by bag filter | |

electrostatic precipitator followed by venturi scrubber | |

electrostatic precipitator followed by cyclonic separator | |

bag filter followed by electrostatic precipitator |

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.

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 _______.

41.6 | |

1 | |

2.8 | |

1.6 |

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}

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.