A butyl rubber bag of total volume 3.0 m3 is used as a biogas digester. Each day it is fed an input of 0.20 m3 of slurry, of which4.0 kg is volatile solids, and a corresponding
volume of digested slurry is removed. (This input corresponds roughly to the waste from 20
pigs.) Assuming that a typical reaction in the digestion process is
C12H22O11 + H2O → 6CH4+ 6CO2
and that the reaction takes 7 days to complete, calculate
(i) the volume of gas
(ii) the heat obtainable by combustion of this gas for each day of operation of the
digester
(iii) how much kerosene would have the same calorific value as one day’s biogas?
for part i do we have to use PV=nRT
do we have to use mol ratio
Yes, use mole ratios ... You also need to specify atmospheric conditions; i.e., temperature - pressure conditions for calculation of volume values. Then you can use PV = nRT to calculate volumes. One could simply assume STP, but that conglomeration you've described is probably not at zero celcius and 1 atm. You'll also need to provide the 'Heat of Combustion' of Methane & Kerosene; or one could extract table values from the Thermo Props Table.
To calculate the volume of gas produced each day, we need to first determine the moles of methane (CH4) produced based on the given information.
(i) Volume of gas:
- Given that the volume of slurry input is 0.20 m3 each day.
- Assuming complete conversion of slurry into gas, the moles of methane produced from the slurry can be calculated using the stoichiometry of the reaction: 1 mole of C12H22O11 produces 6 moles of CH4.
- 0.20 m3 of slurry is fed each day, containing 4.0 kg of volatile solids.
- The molar mass of C12H22O11 is approximately 342.3 g/mol.
- The moles of C12H22O11 in 4.0 kg can be calculated as: (mass/volume) / molar mass = (4000 g / 342.3 g/mol) = 11.67 mol.
- Therefore, the moles of CH4 produced will be: 6 * 11.67 mol = 70 moles.
Now, we can use the Ideal Gas Law (PV = nRT) to calculate the volume of gas produced. However, we need additional information such as the temperature, pressure, and gas constant. Unfortunately, these values are not provided in the given information. Therefore, we cannot proceed with the calculation for the volume of gas without this missing information.
If you have the necessary additional data, please provide it, and I will be happy to assist you with the remaining parts (ii) and (iii) of the question.
To calculate the volume of gas produced by the biogas digester, we can use the ideal gas law equation PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.
However, in this case, we are not provided with the pressure or temperature information, so we cannot directly use the ideal gas law equation. Instead, we can use the stoichiometry of the reaction to determine the volume of gas produced.
Given that the reaction takes 7 days to complete, we can assume that the reaction is completed every day. Since 1 mol of C12H22O11 produces 6 mol of CH4, we can use the following conversion factors:
1 mol C12H22O11 → 6 mol CH4
1 mol CH4 → 22.4 L (at standard temperature and pressure)
We are given that the input of slurry is 0.20 m3, and 4.0 kg of this is volatile solids. We can calculate the number of moles of C12H22O11 using its molar mass, which is 342.3 g/mol.
Number of moles C12H22O11 = mass / molar mass
= 4.0 kg / 342.3 g/mol
= 11.68 mol
Using the stoichiometry, we can calculate the number of moles of CH4 produced:
Number of moles CH4 = 6 mol CH4/mol C12H22O11 x 11.68 mol C12H22O11
= 70.08 mol
Now we can convert the moles of CH4 to volume:
Volume of CH4 = 70.08 mol CH4 x 22.4 L/mol CH4
= 1568.43 L
Therefore, the volume of gas produced each day by the biogas digester is 1568.43 L or 1.568 m3 (approximately).
Moving on to part (ii), to calculate the heat obtainable by combustion of this gas for each day of operation, we need to consider the calorific value or heating value of methane (CH4). The calorific value of methane is approximately 55.5 MJ/kg.
Using the moles of CH4 calculated earlier (70.08 mol), we can calculate the heat obtainable:
Heat obtainable = moles CH4 x calorific value
= 70.08 mol CH4 x (55.5 MJ/kg / 1000 g/mol)
= 3.88 MJ
Therefore, the heat obtainable by combustion of the gas produced each day is approximately 3.88 MJ.
For part (iii), we need to determine how much kerosene would have the same calorific value as one day's biogas. The calorific value of kerosene is approximately 46 MJ/kg.
Using the heat obtainable from part (ii) (3.88 MJ), we can calculate the mass of kerosene:
Mass of kerosene = heat obtainable / calorific value
= 3.88 MJ / (46 MJ/kg)
= 0.084 kg
Therefore, one day's biogas would have the same calorific value as approximately 0.084 kg of kerosene.
Note: These calculations assume ideal conditions and may vary in practice.