Type in your answers below. Then copy and paste questions and answers into an e-mail document and send them to your instructor. It's a good idea to copy and paste your information frequently to back-up of your work.A bit of info for part II
The dynamic behavior of a trace gas.
t1/2 = 0.7 t
The atmospheric life-time t is related to S and Ceq by Ceq=S t. Thus it is very useful for estimating the equilibrium concentration of a trace gas given S.
The half-life= 0.7t is most useful for estimating the dynamic behavior (changes over time) of a trace gas. The half-life is the time required for the gap between the concentration and the equilibrium concentration to cut in half.
If all emission sources are turned off, then the equilibrium concentration is zero. For this case the half-life is the time it takes for the concentration to reduce to half its present value. If the initial concentration is 80 ppm and the half-life is 12 years then the respective concentration at times of 0, 12, 24, and 36 years would be 80, 40, 20, and 10 ppm.
If there is an emission source then Ceq=St. For example ,say S=20 ppm/yr and the life-time (t) is 10 years (half-life of about 7 years). For this case Ceq equals 200 ppm. If the concentration is initially 40ppm, the initial gap between it and Ceq is 160 ppm. The gap decreases by half every 7 years. So at times of 0, 7, 14, and 21 years the gap is 160, 80, 40, and 20 respectively. Thus, the respective concentrations at these time would be 40, 120, 160, 180 ppm.up to questions