Pesticide is applied to a 10 ha (hectare) field at an average rate of 1 kg/ha every month. The soil is regarded as 20 cm deep and well-mixed. The pesticide evaporates at a rate of 2% of the amount present per day and it degrades microbially with a rate constant of 0.05 day-1. What will be the steady state concentration of pesticide in g/m3 and in unit of g/g, assuming the soil density as 2500 kg/m3
To find the steady-state concentration of the pesticide in g/m3, we need to determine the amount of pesticide that is being added each month and the amount that is being removed through both evaporation and microbial degradation.
First, let's calculate the amount of pesticide added each month. The field is 10 hectares in size, and the application rate is 1 kg/ha. Therefore, the total amount of pesticide added each month is:
Amount added = 10 ha * 1 kg/ha = 10 kg
Next, let's calculate the amount of pesticide removed through evaporation. The evaporation rate is 2% of the amount present per day. Since we are looking for the steady-state concentration, we can assume that the evaporation rate equals the degradation rate:
Evaporation rate = 2% of the amount present per day = 0.02 * Amount present per day
Similarly, the amount of pesticide removed through microbial degradation is given by:
Microbial degradation rate = 0.05 * Amount present per day
For steady-state conditions, the amount added must equal the amount removed. Thus, we can set up the following equation:
Amount added = Evaporation rate + Microbial degradation rate
10 kg = 0.02 * Amount present per day + 0.05 * Amount present per day
Simplifying the equation, we get:
10 kg = 0.07 * Amount present per day
Solving for the Amount present per day, we find:
Amount present per day = 10 kg / 0.07 = 142.86 kg
Now, let's convert the Amount present per day from kg to g:
Amount present per day = 142.86 kg * 1000 g/kg = 142,860 g/day
To find the steady-state concentration in g/m3, we need to divide the Amount present per day by the volume of soil. The volume of soil is given by the area of the field multiplied by the depth of the soil:
Volume of soil = 10 ha * 10,000 m2/ha * 0.2 m = 20,000 m3
Steady-state concentration = Amount present per day / Volume of soil
Steady-state concentration = 142,860 g/day / 20,000 m3
Steady-state concentration = 7.14 g/m3
Therefore, the steady-state concentration of the pesticide in the soil is 7.14 g/m3.
To convert this concentration to μg/g, we need to take into account the soil density of 2500 kg/m3. First, let's convert the steady-state concentration from g/m3 to μg/m3:
Steady-state concentration = 7.14 g/m3 * 1,000,000 μg/g = 7,140,000 μg/m3
Now, let's convert from μg/m3 to μg/g by dividing the concentration by the soil density:
Steady-state concentration = 7,140,000 μg/m3 / 2500 kg/m3 = 2,856 μg/g
Therefore, the steady-state concentration of the pesticide in the soil is 2,856 μg/g.