In my Process Safety, HAZMAT, and Emergency Response course, I always walk the students through an exercise to demonstrate just how dangerous chemicals such as Cl2 and NH3 are. So many have come to think that these chemicals are only hazardous when we exceed the OSHA/EPA Thresholds for PSM/RMP.
In a space that is 40 ft × 40 ft with an 8ft ceiling, we need only 8 ml/0.00211338 gallons of liquid Cl2 to make the room an IDLH Atmosphere (e.g., 10 ppm). A far cry from OSHA’s PSM TQ of 1500 pounds or EPA’s RMP TQ of 2500 pounds.
How much liquid chlorine does it take to achieve 10 ppm in a room that is 40′ X 40′?
So I’ll explain the theoretical chemistry, but not how to generate chlorine gas in a real space.
Determine the room volume
Length = 40 ft
Width = 40 ft
Height = 8 ftVolume= 40⋅40⋅8 = 12,800 ft3
Convert to cubic meters (since ppm is usually referenced to m³): 12,800 ft3 × 0.0283168=362 m3
Convert 10 ppm to the mass of chlorine gas
10 ppm by volume means:
10 L of Cl2 per 1,000,000 L of air
Room volume in liters:
362 m3 = 362,000 L
Chlorine gas needed:
362,000 × 101,000,000 = 3.62 L of Cl2
Now convert liters of Cl₂ gas to grams using ideal gas law (approx. 1 mole = 22.4 L at STP):
3.6222.4 = 0.162 mol
Molar mass of Cl₂ = 70.9 g/mol:
0.162⋅70.9 = 11.5 g of Cl2
Convert grams of chlorine gas to liquid chlorine
Liquid chlorine density ≈ 1.47 g/mL
11.5 g1.47 g/mL ≈ 7.8 mL of liquid Cl2
To reach 10 ppm chlorine gas in a 40′ × 40′ × 8′ room, you would need roughly:
8 ml/0.00211338 gallons of liquid chlorine