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[Paper Note] MMReD: A Cross-Modal Benchmark for Dense Context Reasoning

To calculate the amount of methane gas produced, we follow these steps:

1. Identify the given data:

  • Waste water flowrate: 1150 tonnes/day
  • BOD removed: 1200 lb/day
  • Equivalence: 1 mol of BOD = 1 mol of \(O_2\)
  • Conditions: STP (Standard Temperature and Pressure: 0°C, 1 atm)
  • Molar volume of ideal gas at STP: 22.4 L/mol

2. Convert the BOD removed from pounds to moles:

First, convert the weight from pounds to grams (\(1 \text{ lb} \approx 453.59237 \text{ g}\)): $\(\text{Mass of BOD removed} = 1200 \text{ lb/day} \times 453.59237 \text{ g/lb} = 544,310.844 \text{ g/day}\)$

Next, convert grams to moles using the molar mass of \(O_2\) (\(32 \text{ g/mol}\)): $\(\text{Moles of } O_2 \text{ (BOD) removed} = \frac{544,310.844 \text{ g/day}}{32 \text{ g/mol}} \approx 17,009.71 \text{ mol/day}\)$

3. Determine the relationship between BOD and Methane (\(CH_4\)):

In anaerobic digestion, the theoretical methane production is often related to the Chemical Oxygen Demand (COD) or BOD removed. Oxygen demand represents the amount of oxygen required to oxidize methane completely: $\(CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O\)$ From the stoichiometry, 1 mole of \(CH_4\) requires 2 moles of \(O_2\) for complete oxidation. Therefore, 1 mole of \(CH_4\) is equivalent to 2 moles of oxygen demand (BOD).

\[\text{Moles of } CH_4 \text{ produced} = \frac{\text{Moles of BOD removed}}{2}$$ $$\text{Moles of } CH_4 \text{ produced} = \frac{17,009.71}{2} \approx 8,504.86 \text{ mol/day}\]

4. Calculate the volume of methane gas at STP:

Using the molar volume (22.4 L/mol): $\(\text{Volume in Liters} = 8,504.86 \text{ mol/day} \times 22.4 \text{ L/mol} = 190,508.86 \text{ L/day}\)$

Convert Liters to cubic meters (\(1000 \text{ L} = 1 \text{ m}^3\)): $\(\text{Volume in } m^3 = \frac{190,508.86}{1000} \approx 190.51 \text{ m}^3\text{/day}\)$

Final Answer: The amount of methane gas produced is 190.51 \(m^3/\text{day}\).