Question

3) You are using flux chambers to determine rates of soil respiration from the same Brazilian rainforest. Your chamber is a cylinder 20 cm in diameter and 18 cm in height, and you record the following data with an infrared gas analyzer: Time CO2 concentration arts per million, ppm 0 min 410 2 min521 4 min630 6 min 743 8 min859 10 min 973 12 min1081 Measuring soil respiration with a chamber and infrared gas analyzer. What is the flux of CO2 from the soil in moles m2 h1? Assume atmospheric pressure is 1 atm, temperature is 27°C, and the ideal gas constant is 0.0821 (L atm)/(mol-K). Step 1: Calculate the volume of air in the chamber. Step 2: Use the ideal gas equation to calculate the moles of air in the chamber (PV- nRT) Step 3: Plot concentration versus time as a "scatterplot" in Excel or Sheets and fit a linear trendline to get the rate of CO2 concentration change in the chamber (trendline slope). Attach your plot. Step 4: Convert the rate of concentration change into mol/day and divide by the area of soil surface covered by the chamber (hint: 1 ppm 1x10-6 mol/mol)

Answer 1

8 6 4 3 8 0

This is the graph required.

The y-intercept gives the initial concentration of CO₂ = 320 ppm

Slope/gradient of the graph = (change in Y)/(change in X) = (740-600)/(4-2.8) = 140/1.2 = 116.67

So, the change in CO₂ concentration is 116.67 ppm/min = 116.67/0.0167 ppm/hr = 9980.24 ppm/hr

dc PV dt RTA

where F is the flux of CO₂ in the soil,

dc/dt is the change in concentration with respect to time, here it is 9980.24 ppm/hr,

P is air pressure = 1 atm

R is the ideal gas constant = 0.0821 (L·atm)/(mol·K).,

T is the temperature (in Kelvin) = 27ºC = 300.15 K,

V is the volume, and

A is the basal area of the chamber =  2$\pi$r²+ 2$\pi$hr, here r = 10cm = 0.1 m and h = 18cm = 0.18 m

A = 2*3.14(0.01+0.018) = 0.17584 m²

1 ppm = 1/35500 moles per liter

Then 320ppm = 320/35500 moles per liter = 0.00901408451 moles/L

1 mole of gas occupies 22.4L gas by volume so 0.00901408451 moles will occupy 0.00901408451*22.4 = 0.201915493 L of gas by volume

V = 0.201915493 L

10.201915493 -9980.24 0.0465983419 465.062636 , 0.0821 * 300.15% 0.17584

The flux of CO₂ in the soil = 465.062636 moles m^-2h^-1

Step 1&2

PV = nRT, V = nRT/P

P = 1 atm, R = 0.0821 (L·atm)/(mol·K), T = 27^oC = 300.15 K

Let the number of moles of air in the chamber be 1

V = (1*0.0821*300.15)/1 = 24.642315 L

The volume of air in the chamber = 24.642315 L

SO, the number of moles of air in the chamber is 1

9980.24 ppm/hr = 9980.24*0.0416667 ppm/day = 415.8433333 ppm/day = 415.8433333*10^-6 moles/ day

A = 0.17584 m²

The required value = (415.8433333*10^-6)/ 0.17584 moles m^-2day^-1 = 0.00236489612 moles m^-2day^-1