When at rest, a typical person\'s ventilation rate is 6000 ml/min at STP. When o
ID: 476795 • Letter: W
Question
When at rest, a typical person's ventilation rate is 6000 ml/min at STP. When one exhales, carbon dioxide is emitted at a rate of 200 ml/min at STP. If there are 30 people in the room for BIOE 2100 lecture, estimate the concentration (mol/L) in the room after 50 minutes (the length of lecture) if the doors and windows are tightly closed and the ventilation system is shut off. Assume that there is no CO_2 in the room at the beginning of class and that everyone breathes at a consistent rate. State your assumption on the size of the room. Carbon dioxide can be lethal at a concentration of 4.46 Times 10^-3 mol/L. Are we at risk after a 50 minute class? Using your calculations, assumptions, and data from above, how long would we need to stay in the room for the CO_2 to reach a lethal concentration? How many people would need to be in the classroom (using the dimensions that you assumed in Part A) for the CO_2 concentration to reach a lethal level during the 50 minute class period? Do you think it is possible to physically fit that many people in the classroom?Explanation / Answer
Assumption for dimension of lecture hall: length = 8.5m, breadth= 4.6m and height = 3m
volume of hall = l X b X h = 117300 dm3 = 117300 ltrs
Per person, CO2 emission = 200ml/min
amount of CO2 in grams = density of CO2 X volume of CO2
= 1.964 g/l X 200/1000 l/min= 0.3928 g /min
concentration of CO2 /person/min. =0.3928/ (44X 117300),
(where 44 is molecular wt. of CO2 , 117300 is volume of room in ltrs)
= 7.6 X 10-8 mol/l/person/min
a) For 30 people staying for 50 mins,
concentration = 7.6 X 10-8 mol/l X 30 X 50 = 1.10X 10-4mol/l
b) No, people are not at risk as the concentration after 50 mins (1.10X 10-4mol/l) is less than the lethal concentration (4.46 X 10-3 mol/l)
c)time after which CO2 concentration becomes lethal = 4.46 X 10-3 / (7.6 X 10-8X 30) = 1956.14 mins. = 32.6 hours
d) No. of people in the classroom = 4.46 X 10-3 / (7.6 X 10-8 X 50) = 1173 people
No it is not possible to physically fit these many people in the room ,
Assuming average dimension to calculate volume occupied by a person = 1.70m X 0.50m X 0.60m = 0.51m3
= 510 dm3
volume occupied by 1173 persons = 592830 dm3 which is more than volume of the room.
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