Question

A 90 m³ basement in a residence is found to be contaminated with radon coming from the ground through the floor drains. The concentration of radon in the room is 1.5 Bq/L under steady state conditions. The room behaves as a CSTR and the decay of radon is a first-order reaction with a decay rate constant of 2.09x10-6 per second. If the source of radon is closed off and the room is vented with radon-free air at a rate of 0.14 m3/s, how long will it take to lower the radon concentration to an acceptable level of 0.15 Bq/L.

Answer 1

To find the time required to reduce radon levels to 0.15 Bq/L, apply the integrated first-order rate law, using the given decay rate constant and ventilation rate.

Step-by-step explanation:

The student is asking about the time needed to reduce the radon concentration in a contaminated basement to an acceptable level through ventilation, after the source of radon has been sealed. The decay of radon, which is a first-order reaction, and the steady state removal through ventilation need to be considered to calculate the time required to reach the safe level. Given the decay rate constant (k = 2.09x10-6 s-1) and the ventilation rate (Q = 0.14 m3/s), we can model the system as a continuously stirred tank reactor (CSTR) and apply the first-order rate law to find the time (t).

To calculate the time, we can use the integrated rate law for a first-order reaction, which is ln(Ct/C0) = -(k+Q/V)t, where Ct is the final concentration, C0 is the initial concentration, V is the volume of the room, and t is the time. Solving for the time when Ct=0.15 Bq/L and C0=1.5 Bq/L, we arrange the equation to t = (1/(k+Q/V))ln(C0/Ct).

Plugging the values into the equation, t = (1/(2.09x10-6 s-1 + 0.14 m3/s / 90 m3))ln(1.5 Bq/L/0.15 Bq/L), we can find the time required to decrease the radon levels to the acceptable concentration.