# Predicting pressure inside a container based on temperature

I'm a mathematician and computer scientist, and for this particular problem I would benefit from some chemical expertise.

Suppose I fill a container up with some liquid propane. I believe this is done under very cold conditions so that the propane remains liquid while the container is being filled. It is then sealed off. Suppose now the inside of the container changes temperature. I would like to calculate the pressure that would be exerted on the walls of the container at any given internal temperature $$T$$.

Intuitively, I would think that the pressure would be a function of temperature and some initial conditions (how full is the tank initially)? If I filled up the tank to 95% capacity with liquid propane, and then heated the tank, I would expect the resulting pressure on the walls of the container to be much higher than if I filled up the tank to only 1% capacity and did the same thing.

However, the only resources I've been able to find so far relate to the vapour pressure, which is a function of only temperature and this seems incomplete.

Can anybody point me in the right direction on how to approach this problem? Some factors that seem to complicate things are:

• The containers always seem to contain some level of liquid (due to internal pressure?), the rest is gas. Does the liquid have an effect on the pressure exerted on the walls? Can I just ignore it?
• If, rather than fill up the tank with liquid at very cold temperatures, the tank is just directly filled with gas, how does this change things? My guess is that given the initial pressure and temperature you can predict the pressure at a different temperature, but I think the fact that some of the gas becomes liquid at some point is tripping me up and I'm not sure how to approach the problem.
• When your tank is 95% full, is the space above the liquid containing air? Commented May 7, 2019 at 10:33
• As the tank is not full then there is always some vapour in equilibrium with the liquid and so you will need to find the saturated vapour pressure vs temperature. If all the liquid evaporated (say if 1% full) then it will behave as a gas and then pressure is proportional to temperature say via the ideal or van der waals gas laws. Commented May 7, 2019 at 10:37
• @porphyrin , for the first case, the liquid would also contribute $\rho gh$ pressure on walls. Commented May 7, 2019 at 10:40
• @William E Ebenezer, yes I has assumed that that was small enough to ignore. Commented May 7, 2019 at 10:42
• Equations of State - of which the ideal gas law is the simplest, relate pressure, temperature and density to one another Commented May 7, 2019 at 12:18

The problem with answering this question is that propane is not remotely close to being a permanent gas at typical room temperature. This means, in practice, that it is easily liquefied at room temperature with a little pressure and you can't, therefore, use the simple gas law equation to work out the pressure exerted by a given amount of the gas in a vessel of fixed volume.

This means that the tank will fairly certainly contain some liquid (as a butane-filled cigarette lighter will at very modest pressures despite butane being a gas at normal temperatures and pressures).

You can, therefore, estimate the pressure inside the vessel as being the vapour pressure of propane at the given temperature (which ignores any other gas included in the vessel though this may be a good approximation as the filling process is likely to sweep out any other gas). The Wikipedia entry on propane has a convenient chart of the vapour pressure and you can look up the value for a given temperature. The amount of the vessel filled won't make much of a difference to this as long as there is some liquid and some gas in the vessel.

Because simple gas laws don't do a good job of predicting the vapour/gas equilibrium, this isn't a gas-law question and the best answer will always be empirical (see the Wikipedia page).

• Well the purpose of solving this problem is to determine initial conditions to keep the pressure of a container safe over forthcoming temperatures, so unfortunately the vapor pressure doesn't help (given that it is only a function of temperature, it provides no information about initial conditions which we can optimize). So I think looking at empirical data may be the best approach. thanks! Commented May 8, 2019 at 13:10
• @CoffeeDonut Also, find out what the typical fill ratio for commercial propane tanks is and stick close to that. Filling the tank completely with any liquid is likely to be very, very bad as they all expand a bit with temperature. As long as there is sufficient remaining space, then empirical vapour pressure is as bad as it can get. Commented May 8, 2019 at 13:18
• So in actuality, the real question may be to determine what constitutes "sufficient remaining space"? Commented May 8, 2019 at 13:29
• @CoffeeDonut exactly, which is why the people who do it have rules about you full the tanks should be (which is usually no more than 80% of the volume). See fill rules. Commented May 8, 2019 at 14:43

As long as head space is available the pressure in the tank is the vapor pressure of the liquid; the tank, bottle, flask must be strong enough to hold the weight of the liquid. [Never put mercury in a flat-bottomed Erlenmeyer flask]. For gas delivery to work the VP must be greater than atmospheric pressure at all possible temperatures of use including temperature drop from heat of vaporization. The head space must be large enough to accommodate any expansion of the liquid. This is done with a safety factor because expanding liquid will condense vapor until the head space vanishes. With no head space the heated liquid will expand and distort and eventually rupture the container. In large amounts this can be catastrophic resulting in BLEVE explosions.

The tank must be strong enough to hold the weight and withstand the vapor pressure at possible temperatures. [spherical shape is better]. Head space must be sufficient that liquid can never fill the container. An inert gas head space will not work. The added pressure increases the internal pressure of the liquid increasing the vapor pressure