Question 1 |

The theoretical aerobic oxidation of biomass (\mathrm{C} 5 \mathrm{H} 7 \mathrm{O} 2 \mathrm{~N} ) is given below:

\mathrm{C}_{5} \mathrm{H}_{7} \mathrm{O}_{2} \mathrm{~N}+5 \mathrm{O}_{2} \rightarrow 5 \mathrm{CO}_{2}+\mathrm{NH}_{3}+2 \mathrm{H}_{2} \mathrm{O}

The biochemical oxidation of biomass is assumed as a first-order reaction with a rate constant of 0.23 / \mathrm{d} at 20^{\circ} \mathrm{C} (logarithm to base e). Neglecting the second-stage oxygen demand from its biochemical oxidation, the ratio of \mathrm{BOD} 5 at 20^{\circ} \mathrm{C} to total organic carbon (TOC) of biomass is (round off to two decimal places).

[Consider the atomic weights of \mathrm{C}, \mathrm{H}, \mathrm{O} and \mathrm{N} as 12 \mathrm{~g} / \mathrm{mol}, 1 \mathrm{~g} / \mathrm{mol}, 16 \mathrm{~g} / \mathrm{mol} and 14 \mathrm{~g} / \mathrm{mol}, respectively]

\mathrm{C}_{5} \mathrm{H}_{7} \mathrm{O}_{2} \mathrm{~N}+5 \mathrm{O}_{2} \rightarrow 5 \mathrm{CO}_{2}+\mathrm{NH}_{3}+2 \mathrm{H}_{2} \mathrm{O}

The biochemical oxidation of biomass is assumed as a first-order reaction with a rate constant of 0.23 / \mathrm{d} at 20^{\circ} \mathrm{C} (logarithm to base e). Neglecting the second-stage oxygen demand from its biochemical oxidation, the ratio of \mathrm{BOD} 5 at 20^{\circ} \mathrm{C} to total organic carbon (TOC) of biomass is (round off to two decimal places).

[Consider the atomic weights of \mathrm{C}, \mathrm{H}, \mathrm{O} and \mathrm{N} as 12 \mathrm{~g} / \mathrm{mol}, 1 \mathrm{~g} / \mathrm{mol}, 16 \mathrm{~g} / \mathrm{mol} and 14 \mathrm{~g} / \mathrm{mol}, respectively]

1.82 | |

3.25 | |

5.44 | |

6.24 |

Question 1 Explanation:

Calculation of B O D_{s} :

\mathrm{C}_{5} \mathrm{H}_{7} \mathrm{O}_{2} \mathrm{~N}+5 \mathrm{O}_{2} \rightarrow 5 \mathrm{CO}_{2}+\mathrm{NH}_{3}+2 \mathrm{H}_{2} \mathrm{O}

Considering 1 mole of biomass

\mathrm{BOD}_{4}=160 \mathrm{gm} / \mathrm{mol}

\therefore \quad \mathrm{BOD}_{5}=\mathrm{BOD}_{4}\left(1-\mathrm{e}^{-\mathrm{k}_{0} \mathrm{t}}\right)

=160\left(1-e^{-0.23 \times 5}\right)

=160 \times 0.6833

=109.34 \mathrm{gm} / \mathrm{mol}

Calculation of TOC:

\mathrm{TOC}=\frac{12 \times 5}{113} \times 113=60 \mathrm{gm} / \mathrm{mol}

( \because 1 mole biomass is considered)

\therefore Required Ratio =\frac{109.34}{60}=1.82

\mathrm{C}_{5} \mathrm{H}_{7} \mathrm{O}_{2} \mathrm{~N}+5 \mathrm{O}_{2} \rightarrow 5 \mathrm{CO}_{2}+\mathrm{NH}_{3}+2 \mathrm{H}_{2} \mathrm{O}

Considering 1 mole of biomass

\mathrm{BOD}_{4}=160 \mathrm{gm} / \mathrm{mol}

\therefore \quad \mathrm{BOD}_{5}=\mathrm{BOD}_{4}\left(1-\mathrm{e}^{-\mathrm{k}_{0} \mathrm{t}}\right)

=160\left(1-e^{-0.23 \times 5}\right)

=160 \times 0.6833

=109.34 \mathrm{gm} / \mathrm{mol}

Calculation of TOC:

\mathrm{TOC}=\frac{12 \times 5}{113} \times 113=60 \mathrm{gm} / \mathrm{mol}

( \because 1 mole biomass is considered)

\therefore Required Ratio =\frac{109.34}{60}=1.82

Question 2 |

Match the following air pollutants with the most appropriate adverse health effects:

\begin{array}{|l|l|} \hline \textbf{Air pollutant} & \textbf{Health effect to human and/or test animal} \\ \hline \text{(P) Aromatic hydrocarbons} & \text{(I) Reduce the capability of the blood to carry oxygen} \\ \hline \text{(Q) Carbon monoxide} & \text{(II) Bronchitis and pulmonary emphysema}\\ \hline \text{(R) Sulfur oxides}& \text{(III) Damage of chromosomes} \\ \hline \text{(S) Ozone} & \text{(IV) Carcinogenic effect} \\ \hline \end{array}

\begin{array}{|l|l|} \hline \textbf{Air pollutant} & \textbf{Health effect to human and/or test animal} \\ \hline \text{(P) Aromatic hydrocarbons} & \text{(I) Reduce the capability of the blood to carry oxygen} \\ \hline \text{(Q) Carbon monoxide} & \text{(II) Bronchitis and pulmonary emphysema}\\ \hline \text{(R) Sulfur oxides}& \text{(III) Damage of chromosomes} \\ \hline \text{(S) Ozone} & \text{(IV) Carcinogenic effect} \\ \hline \end{array}

(P) - (II), (Q) - (I), (R) - (IV), (S) - (III) | |

(P) - (IV), (Q) - (I), (R) - (III), (S) - (II) | |

(P) - (III), (Q) - (I), (R) - (II), (S) - (IV) | |

(P) - (IV), (Q) - (I), (R) - (II), (S) - (III) |

Question 2 Explanation:

**Aromatic hydrocarbons**such as 3, 4benzpyrene, other polycyclic organic compounds which are originated due to incomplete combustion of hydrocarbons are considered to be carcinogenic agents. These are responsible for cancer.

**Sulphur dioxide \left(\mathrm{SO}_{2}\right) :**It is an irritant gas which effects mucous membrane when inhaled. It leads to bronchial spasms. Asthma patients are badly affected.

**Carbon Monoxide (CO) :**Carbon monoxide has a strong affinity for combining with the haemoglobin of the blood to form carboxyhemoglobin, \mathrm{COHb}. This reduces the ability of the haemoglobin to carry oxygen to the body tissues. CO has about two hundred time the affinity of oxygen for attaching itself to the haemoglobin, so that low levels of CO can still result in high levels of \mathrm{COHb}. Carbon monoxide also affects the central nervous system.

**Ozone \left(\mathrm{O}_{3}\right)**can damage the tissue of the respiratory tract, causing inflammation and irritation, and result in symptoms such as coughing, chest tightness and worsening of asthma symptoms. In addition, ozone causes substantial damage to crops, forests and native plants.

Question 3 |

For the elevation and temperature data given in the table, the existing lapse rate in the environment is ___ { }^{\circ} \mathrm{C} / 100 \mathrm{~m} (round off to two decimal places).

\begin{array}{|l|l|} \hline \begin{array}{l}\text {Evaluation from} \\ \text {ground level}(\mathrm{m})\end{array} & \text{Temperature }\left({ }^{\circ} \mathrm{C}\right) \\ \hline 5 & 14.2 \\ \hline 325 & 16.9 \\ \hline \end{array}

\begin{array}{|l|l|} \hline \begin{array}{l}\text {Evaluation from} \\ \text {ground level}(\mathrm{m})\end{array} & \text{Temperature }\left({ }^{\circ} \mathrm{C}\right) \\ \hline 5 & 14.2 \\ \hline 325 & 16.9 \\ \hline \end{array}

0.28 | |

0.54 | |

0.84 | |

0.92 |

Question 3 Explanation:

Lapse rate,

\lambda=\frac{\text { Difference in Temperature }}{\text { Difference in height }}=\frac{\Delta \mathrm{T}}{\Delta \mathrm{H}}

\lambda=\frac{\Delta \mathrm{T}}{\Delta \mathrm{H}}=\frac{16.9-14.2}{325-5}

\lambda=0.0084375^{\circ} \mathrm{C} / \mathrm{m}

\Rightarrow \quad \lambda=\frac{0.0084375 \times 100}{100 \mathrm{~m}}

\Rightarrow \quad \lambda=\frac{0.084375^{\circ} \mathrm{C}}{100 \mathrm{~m}}=0.84^{\circ} \mathrm{C} / 100 \mathrm{~m}

\lambda=\frac{\text { Difference in Temperature }}{\text { Difference in height }}=\frac{\Delta \mathrm{T}}{\Delta \mathrm{H}}

\lambda=\frac{\Delta \mathrm{T}}{\Delta \mathrm{H}}=\frac{16.9-14.2}{325-5}

\lambda=0.0084375^{\circ} \mathrm{C} / \mathrm{m}

\Rightarrow \quad \lambda=\frac{0.0084375 \times 100}{100 \mathrm{~m}}

\Rightarrow \quad \lambda=\frac{0.084375^{\circ} \mathrm{C}}{100 \mathrm{~m}}=0.84^{\circ} \mathrm{C} / 100 \mathrm{~m}

Question 4 |

Which of the following statements is/are TRUE in relation to the maximum mixing depth (or Height) ' D_{\max } ' in the atmosphere?

D_{max} is always equal to the height of the layer
of unstable air | |

Ventilation coefficient depends on D_{max} | |

A smaller D_{max} will have a smaller air pollution
potential if other meteorological conditions
remain same | |

Vertical dispersion of pollutants occurs up to
D_{max} |

Question 4 Explanation:

The depth of mixing layer in which vertical movement of pollutants are possible is called maximum mixing depth (MMD).

An air parcel at temperature rises and cools. The level where its temperature becomes equal to surrounding air gives the maximum mixing depth value.

Also, ventilation coefficient = MMD \times Average wind speed

High value of ventilation coefficient leads to low air pollution potential.

An air parcel at temperature rises and cools. The level where its temperature becomes equal to surrounding air gives the maximum mixing depth value.

Also, ventilation coefficient = MMD \times Average wind speed

High value of ventilation coefficient leads to low air pollution potential.

Question 5 |

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 5 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

There are 5 questions to complete.

Thank you very helpful