In: Other
6:00 - 8:00 PM
time interval = 2 hours
No. of cigarettes smoken = 50(no. of smokers)*2(no. of cigarettes smoken by each smoker in 1 hour)*2(time interval taken) = 200
Amount of Formaldehyde released = 4 mg *200 = 800 mg
So we have 800 mg CHO in 1000 m3 of air
Now according to the reaction Between CHO and air to yield CO :
Hence after 2 hours only 359.463 mg of CHO is left.
Hence after 2 hours we have 359.463 mg CHO in 1000 m3 of air
8:00 PM - 9:00 PM
No. of smoker now = 75
No. of cigarettes smoken in 1 hour = 75*2*1 = 150
CHO emitted = 150*4mg = 600 mg
Now the total amount of CHO in the room : 359.463+600 = 959.463 mg
After 1 hour amount of CHO left from this
So now we have 643.147 mg of CHO in air at 9:00 PM.
Hence at 9:00 PM we have 257.258 ppmv. (parts per million volume of air)
I took 2500 m^3 since initially there was 1000 m^3 of air and then 1500 m^3 was added.
Please comment for any doubts :)