Environmental Engineering Questions

Question

(1) Benzene is accidentally spilled in a soil block with 100 m length, 50 m width, and 1 m deep and you found the concentration in pore water is 10 mg/L. What is the amount of benzene (g) present in pore water if the soil is 60% saturated with a porosity of 0.5?

I tried solving this one and got 15,000 g which does not seem correct so I wanted to double check.

(2) Microorganisms in soil can degrade benzene (C6H6) to produce carbon dioxide using oxygen. How much oxygen is needed to degrade all benzene in pore water assuming all benzene is converted to carbon dioxide in a redox reaction with oxygen?

I do not know how to solve this one.

Answer 1

Problem 1

To calculate the amount of benzene present in pore water, we can use the formula:

\[ \text{Mass of Benzene (g)} = \text{Concentration of Benzene} \times \text{Volume of Pore Water} \]

First, let's calculate the volume of the pore water. The soil block is 60% saturated, meaning that 60% of the pore space is filled with water. The porosity (\( \phi \)) is given as 0.5, which means 50% of the soil volume is pore space. Therefore, the volume of pore water (\( V_{\text{pore}} \)) is calculated as follows:

\[ V_{\text{pore}} = \text{Porosity} \times \text{Saturated Volume of Soil} \]

\[ V_{\text{pore}} = 0.5 \times (100 \, \text{m} \times 50 \, \text{m} \times 1 \, \text{m}) \]

Now, we can calculate the mass of benzene using the concentration of benzene (\( C \)):

\[ \text{Mass of Benzene} = C \times V_{\text{pore}} \]

Given that \( C = 10 \, \text{mg/L} \), we need to convert this to grams per liter:

\[ C = 10 \, \text{mg/L} \times 0.001 \, \text{g/mg} \]

Now, we can substitute the values into the formula:

\[ \text{Mass of Benzene} = (10 \, \text{mg/L} \times 0.001 \, \text{g/mg}) \times (0.5 \times 100 \, \text{m} \times 50 \, \text{m} \times 1 \, \text{m}) \]

Calculating this expression will give you 25g as the mass of benzene present in the pore water.

Problem 2:

The balanced equation is:

C₆H₆ + 15/2 O₂ --> 6CO₂ + 3H₂O

We know the mass of benzene (\(m_{\text{benzene}}\)) from the previous calculation. Let's denote the molar mass of oxygen (\(O_2\)) as approximately 32 g/mol.

The amount of oxygen needed (\(m_{\text{oxygen}}\)) can be calculated using the molar ratio:

m_oxygen = 15/2 × m_Benzene × (32 g/mol) / (78.11 g/mol) = 187.5 g

Substitute the value of m_Benzene into the equation to find m_oxygen. This will give you the mass of oxygen needed to degrade all the benzene in the pore water.