# What is the size in m³ of a toxic benzene cloud when if 20.000 kg liquid benzene would vaporize?

Benzene is carcinogen, The short term exposure limit for airborne benzene is 5 ppm for 15 minutes see https://en.wikipedia.org/wiki/Benzene#Benzene_exposure_limits

I would like to know which volume of air could contain toxic benzene if an benzene ship with 2000 ton benzene should vaporise vaporization to the air. Benzene is heavier than air so it would be closed to the ground. How close is not a chemical subject but a physical phenomenon which is out of the scope of this question. In this calculation I use a lot of assumptions, just because the lack of info. But at least it gives a idea what might happen. Hopefully, on a later moment, some assumptions could be replace by facts, making the calculation more accurate. For now it is meant to get an idea what would happen in the worst case scenario. Assumption only 1% of the benzene vaporise. The question how much benzene could vaporise is out of the scope of this question. Calculation adjusted with info of @maurice : Benzene is not a ideal gas. The vapor pressure of benzene is about 0.04 bar at 0°C.

• Mol mass benzene: 78,11184 g/mol

• Mol / kg: ~12,8 mol benzene (1000 g / 78,11184 g/mol)

• 0.022413969545014... m3/mol at 0 °C, see https://en.wikipedia.org/wiki/Molar_volume using Avogadro's law

• The vapor pressure of benzene is at 0°C: 0.04 bar

• 1 kg benzene liquid: 0,01148 m³ benzene gas at 1 bar (~12,8 mol benzene * 0.022413969545014... m3/mol * 0,04bar / 1 bar)

• Benzene to vaporize: 1 %

• Mass vaporized benzene: 20.000 kg ( 2000.000 kg * 1%)

• Total 100% benzene gas: 230 m³ ( 20.000 kg liquid benzene * 0,01148 m³/kg)

• Assumption height benzene cloud from the ground:

• 1 meter high: Surface: 230 m²
• 3 meter high: Surface: 77 m² (230 m² / 3m)
• Assumption gas will spread as a circle:

• Radius with height of 3 meter: 8,5 m (230 m³ = 3,14 * r²)

• Diameter cloud of 3 m high: 17 m

Which can move with the speed of wind which can be 0 m/s (0 Beaufort) up to 28 m/s (10 Beaufort) Benzene will probably mix with air. But airborne benzene stay toxic up to 5 pmm.

So the maximum size of the toxic cloud would be:

• The short term exposure limit for airborne benzene is 5 ppm
• Maximum amount of gas which could be carcinogen: ~46 * 10⁶m³ (230 m³ / 5* 10-⁶ ppm)
• Assumption height benzene cloud from the ground:
• 1 meter high: Surface: 46 * 10⁶m²
• 3 meter high: Surface: 9,2 * 10⁸m² ( 46 * 10⁶m³ / 3m)
• Assumption gas will spread as a circle:
• Radius with height of 3 meter: 3,8 km (46 * 10⁶m² = 3,14 * r²)
• Diameter cloud of 3 m high: 7,6 km

If the cloud gets larger it would be less than 5 pmm and is not toxic anymore. This will be a matter of windspeed and time, which both are beyond the scope of this question.

The key question for me is: Is the calculation for a vessel loaded with 2000 ton benzene where 1% leaks generate a cloud with carcinogen air between about 230 m³ and 46 * 10⁶m³ correct? Or do I miss important issues.

• Benzene doesn't become toxic at a particular concentration. The rules about exposure define the acceptable concentration over a period of time to keep the risk low enough not to be a significant risk. And the big risks of a big cloud are not primarily toxicity. Sep 2 '20 at 11:02
• @matt_black Is carcinogen in the category of toxic? If not, although it is not my question, what is than the main risk of a big cloud? I know it is said there should be no exposure to benzene at all. But is so no calculation can be made. Sep 2 '20 at 11:18
• It is not clear if the limit 5 ppm for air is assumed to be V/V or w/w. I imply the former, but just guessing. The difference is almost 3 times. 1 ppm(V/V) is approximately 78/29 ppm (w/w) Sep 2 '20 at 11:39
• Your calculation uses Avogadro's law, which implies that benzene can form a perfect gas at $0$°C and $1$ bar. But benzene cannot form a gas with a pressure of $1$ bar at $0$°C. The vapor pressure of benzene is about $0.04$ bar at $0$°C. So all your calculations are about $25$ times too high. Sep 2 '20 at 11:48
• @maurice I used your information to correct the calculation in my question Sep 2 '20 at 12:24

## 1 Answer

According to vapor tension online calculator, at $$t=\pu{20 ^\circ C}$$, the vapor pressure of benzene equates to $$p=\pu{10.018 kPa \simeq 10 kPa = 0.1 atm = 100 000 ppmv}$$

If we consider benzene vapor as an ideal gas, the saturated vapor density $$\rho$$ is the maximal concentration of benzene vapor at this given temperature:

$$\rho = \frac {pM}{RT} = \pu{\frac {10000 \cdot 0.078}{8.314 \cdot 293.15} kg m^-3}\simeq \pu{0.320 kg m^-3}$$

The volume of air, saturated by $$m=\pu{20 000 kg}$$ of benzene is:

$$V_\mathrm{saturated}=\frac m\rho \simeq \pu{62500 m^3}$$

The air volume for the $$\pu{15 min}$$ safety threshold at $$\pu{5 ppmv}$$ is:

$$V_\mathrm{threshold}=V_\mathrm{saturated} \frac{100000}{5}\simeq \pu{1.25e9 m^3}$$

Assuming an uniform, pie-shaped cloud of height $$h=\pu{3 m}$$, the cloud would have a diameter $$D$$ : $$D=\sqrt{\frac{4 \cdot V_\mathrm{threshold}}{\pi h}}=\sqrt{\frac{4 \cdot \pu{1.25e9 m^3}}{3.14 \cdot \pu{3 m}}}\simeq \pu{23 000 m}$$

The diameter for the analogical cloud of the saturated air would be $$\pu{163 m}$$.

For obvious reasons, the real cloud shape and concentration distribution would be very different and hard to predict.

The 3D shape of the cloud very depends on the wind speed, the vertical air thermal gradient and the wind speed vertical gradient ( See the Richardson's number ). This determines the level of turbulent mixing in low hundreds meters of troposphere.

But the estimate above provides you a ballpark figure.

• @Buttonword Thanks. Ehm, well, while your answer improvement is good, it is like a rewriting the answer. I see my English has weak times more often than I like. :-) Sep 2 '20 at 13:53
• I agree my editing of the text went far; and while working in the text, wasn't recognized by mine as close to a rewrite as from your perspective. Regardless if these edits may be reverted, my future handling of urtexts is going to be more careful. Sep 2 '20 at 14:02
• @Buttonword No hard feelings, I have left the changes there. I was smiling seeing all the green and pink fields. In fact, I appreciate your editor work. Some cases are stupid errors, some cases are language mistakes. I learn new fine features of English every now and then, like the rule for comma-separated adjectives or "ballpark figure". Sep 2 '20 at 14:14