hbirge
Dec 6, 2012, 10:34 AM
Here is the set up:
Soil in a 150mL jar, with headspace of varying O2 availability; methane is being produced (this is what we're adding). Because we cannot pull a sample from less than atmospheric pressure, we add 8mL N2 gas to the bottle (mix it with some syringe pumps) and remove 8mL of diluted sample. The concentration of methane in that diluted sample (C2) can be corrected for using C1V1=C2V2. Our time 1 concentration (pre 8mL dilution) is C1 (C1=C2*158/150).
But the next day (time 2), we add another 8mL N2, mix, remove and run the sample on the GC. How do we correct this value for both dilutions? We can correct for the 8mL added that at Time 2 using C1V1=C2V2; but if we'd never diluted it on T1, the [methane] would be higher.
How do we account for each day's additional dilution in the value calculated from the GC?
One of my solutions (I think it's wrong, because it doesn't match the blank bottles i.e. bottles without methane being added, just filled to 20mL with some water and ammonium). Basically, what would we expect if no methane was added, and what did we actually see? Add that difference to what we actually saw... please be kind if I am a complete moron.
Here is the model I came up with (x=time; x-1=previous time; CH4x=the value calculated from the GC data for that day)
CH4x (corrected for dil) = abs(CH4(x-1)*150/158-CHx)+CH4x
Repeat for days 3 and 4 using the previous day's data.
Soil in a 150mL jar, with headspace of varying O2 availability; methane is being produced (this is what we're adding). Because we cannot pull a sample from less than atmospheric pressure, we add 8mL N2 gas to the bottle (mix it with some syringe pumps) and remove 8mL of diluted sample. The concentration of methane in that diluted sample (C2) can be corrected for using C1V1=C2V2. Our time 1 concentration (pre 8mL dilution) is C1 (C1=C2*158/150).
But the next day (time 2), we add another 8mL N2, mix, remove and run the sample on the GC. How do we correct this value for both dilutions? We can correct for the 8mL added that at Time 2 using C1V1=C2V2; but if we'd never diluted it on T1, the [methane] would be higher.
How do we account for each day's additional dilution in the value calculated from the GC?
One of my solutions (I think it's wrong, because it doesn't match the blank bottles i.e. bottles without methane being added, just filled to 20mL with some water and ammonium). Basically, what would we expect if no methane was added, and what did we actually see? Add that difference to what we actually saw... please be kind if I am a complete moron.
Here is the model I came up with (x=time; x-1=previous time; CH4x=the value calculated from the GC data for that day)
CH4x (corrected for dil) = abs(CH4(x-1)*150/158-CHx)+CH4x
Repeat for days 3 and 4 using the previous day's data.