A room of 500 cubic meters has 50 smokers inside, each smoking two
cigarettes an hour. An individual cigarette emits approximately 1.4 mg
of HCOH (formaldehyde), which can be converted to carbon dioxide as a
first order reaction in atmospheric conditions with a rate constant of
k=0.40/hr. Fresh air enters the room at 1000 cubic meters per hour,
the same as the stale air leaving the room.
 
a). Estimate the steady-state concentration of formaldehyde in the
room air, assuming complete and instant mixing.
b). Estimate the dynamic response of the room air formaldehyde
concentration assuming the room air is clean initially and the fresh
air entering is also clean. How long does it take for the formaldehyde
concentration in the room to exceed 0.05 ppm, which is the threshold
of eye irritation?

Set up the differential equation relating the time rate of change in
concentration of formaldehyde to the rate of production and loss of
this compound:

dC/dt = production rate - decay rate - rate of loss due to air exchange
dC/dt = 100*1.4mg/hr/(500m^3) - 0.4*C - C*1000m^3/hr/(500m^3)
dC/dt = 0.28mg/m^3/hr - 2.4*C/hr

where C is the formaldehyde concentration in mg/m^3.

The steady state concentration is found by setting dC/dt = 0 and solving for C:

0 = 0.28mg/m^3/hr - 2.4*C/hr

C=0.117 mg/m^3


For the transient solution, you need to solve the differential equation:

dC/dt = 0.28mg/m^3/hr - 2.4*C/hr
1/[0.28mg/m^3/hr - 2.4*C/hr] dC = dt
with initial condition C = 0 at t=0

-ln[(0.28-2.4C(t))/0.28]/2.4 = t

0.28mg/m^3/hr - 2.4/hr * C(t) = 0.28mg/m^3/h * exp(-2.4/hr * t)
C(t) = 0.28/2.4 mg/m^3*[1-exp(-2.4/hr * t)]
C(t) = 0.117mg/m^3*[1-exp(-2.4/hr * t)]

where t is in hours, and C is in mg/m^3.

Check to see if this has the correct behavior:

at t=0, C(0) = 0 -- ok
as t-> infinity, C(steady state) = 0.117mg/m^3 -- ok

The question asks when a certain concentration, given in ppm, is
reached.  You'll need to convert mg/m^3 into ppmv.  Note that one mole
of formaldehyde has a mass of ~30 grams, and at 1 atmosphere, one mole
of gas occupies ~ 0.0224 m^3, so to you need to multiply the
concentration in mg/m^3 by a factor of
1/1000(gm/mg)*1/30(mol/gm)*0.0224(m^3/mol)*10^6 = 0.747, to convert to
ppm concentration units.  Conversely, 0.05 ppm = 0.067 mg/m^3.  Use
this value to solve for "t" in the the transient solution.  (If I did
the math right, I get about 0.36 hr, or about 21.4 minutes.