Chemistry question. How to calculate amount of vapor in a sealed tank?

Lets say I have an empty 5000lt tank and I pour 200 liters of gasoline and seal it. How can I calculate how much of that gasoline will turn into vapor?

I think you’d have to also know the temperature and pressure of the tank.

PV=nrT Pressure times volume equals the amount present times a constant (the constant takes care of things like what units you are using, the exact materials involved, etc) times tempeture.
One mole of any gas at STP takes up 22.4 liters of space.

Solve the equation for n, that’s the amount of vapor present after equilibrium is reached.

You can work that around to moles and the amount of moles can be converted to liquid measure.
I think . . . maybe not.

You need to know the vapor pressure of the gasoline at that temperature. Then use PV=nRT. P= vapor pressure, V = the remaining air space in the tank, R is a constant and T is the temperature in K. Solve for n to get moles.

If my assumptions are correct, then I get around 15 grams of gasoline vapors in 4800 liters of space at 20C. Does this number sound right? I don’t want an exact answer, just a rule of thumb.

What values are you using for the vapour pressure of gasoline and the molecular weight? It’s been a long time since I did these kind of problems while I can come up with a fairly close guess for molecular weight, I don’t know the vapour pressure (at 20C) offhand.

Nitpick–the equation and the conversion only apply to an ideal gas. That being said, I would not hesitate to use the ideal gas law for a first approximation.

In any event, as others have stated, all liquids have a characteristic vapor pressure. Assuming that some liquid is still present at equilibrium, the vapor pressure in a sealed container is only dependent on the ambient temperature.

Gasoline is actually a fairly complex mixture of several hundred aliphatic hydrocarbon compounds containing carbon chains ranging from 4-12 carbon atoms in each molecule. One important component in gasoline is 2,2,4-Trimethylpentane (aka iso-octane), which is used as the reference standard for octane ratings. Iso-octane is often used as a representative compound for gasoline.

At 25 deg C (298 K), the vapor pressure of iso-octane is 49.3 mm Hg.

200 L of liquid iso-octane poured into an empty 5000 L tank will have a vapor space of 4800 L.

At equilibrium at 25 deg C, 49.3 mm Hg of iso-octane in a 4800 L space corresponds to 12.73 moles of iso-octane vapor (assuming ideal gas behavior), or 1.45 kg of iso-octane vapor. This corresponds to a 2.1 L of liquid iso-octane having vaporized.

At 21 deg C (294 K), the vapor pressure of iso-octane is 40.6 mm Hg. This corresponds to 10.63 mol of iso-octane, or 1.21 kg of iso-octane vapor in a 4800 L space.

To get a more accurate answer, you would then go through an iterative process because the vapor space has increased due to the evaporation of some of the iso-octane. This does not change the above results to any significant extent, though. You’d also want to use a more accurate equation of state than the ideal gas law.

Thanks. This is more accurate than I had hoped for. So it is very roughly 1.5 kilos of gasoline in a 5000lt tank in mild weather :slight_smile:

I’m asking because I heard a guy saying that the level in a gasoline tank can rise due to condensed vapor. My gut feeling was that even if all vapor was somehow magically condensed, the rise in the liquid level would be minuscule.