# How do I calculate the pressure of a known liquid in a sealed container heated above boiling point?

If I have water in a sealed container heated to say 150 degrees, how do I determine the amount of pressure being generated in the container? What about for other liquids? I have searched extensively and cannot figure this out.

I was looking for a formula of sorts, but for example reasons let's say a 5000 ml container with 4500 ml of water in it, with the rest of the space air, heated to 150 degrees Celsius.

• First we would need to know how much water per volume in the sealed container. Also we would need to know if any air is in the container or not. If we knew those things, then the next step is to consult steam tables (either equilibrium or superheated) to determine all the thermodynamic properties of the steam at the right temperature and bulk density. Commented Feb 15, 2015 at 17:27

Note: for convenience, "gas" refers to any gas at a temperature beyond its boiling point; "vapor" refers to any gas evaporated from its liquid state, also implying that the liquid itself is below its boiling point, such as the 150 degrees water in your example.

Dalton's law of partial pressure would come in handy here. The law states that the partial pressure generated by each type of gas particles sum up to the total pressure in a sealed container. The partial pressure of "gasses" can directly be calculated using the ideal gas law. Of course the partial pressure of "gasses" is 0Pa if there is no gas but only vapor in your container.

Now the partial pressure of the vapor of liquids. In a sealed container, if left to attain dynamic equilibrium (which is the steady state in this case) the partial pressure of the vapor of any liquid would be equal to its equilibrium vapor pressure, which is a function of temperature only. Though not excessively accurate, the Antoine equation can be used to estimate this:
$\log P = A- {B \over C+T}$
where P is the equilibrium vapor pressure, A, B and C are substance-specific constants and T is the thermodynamic temperature of the substance. The three substance-specific constants can be checked up on the internet, or provided by other sources, whereas conducting experiments can be a last resort (as it is a very tedious experiment that you need the whole curve of the function to approximate the constants).

If the liquid in question is a mixture of more than one liquid substance, then you'd need a third law - Raoult's law to determine the partial vapor pressure of each component of the mixture:
$p_i = p_i^\star x_i$
where $p_i$ is the partial vapor pressure of component $i$ of the mixture, $p_i^\star$ is the equilibrium vapor pressure of a pure sample of component $i$(in itself, not in a mixture), and $x_i$ is the mole fraction of $i$ in the liquid.
At dynamic equilibrium, combining Dalton's law of partial pressure and Raoult's law yields the expression for total vapor pressure:
$p_{total} = \sum p_i^\star x_i$

Since steam (water vapor) is the most common working fluid in external-combustion engines, steam tables, as mentioned by Curt F above, are widely used and will show the relationship more accurately than the ideal gas law, $$PV=nRT$$.

• DrMoishe, what about for other liquids such as organics? Water was just an example of convenience. Will the ideal gas law alone allow me calculate the pressures that will build up in a tank if I heat a liquid beyond its boiling point? I ask because I feel as if the ideal gas law would be come unapplicable when the pressure is so much that more water cannot evaporate. Is there simply not straightforward way/formula to calculate how much pressure would build up in a container at certain temperatures? Commented Feb 15, 2015 at 19:04
• There are additional factors not included in the ideal gas equation, such as van der Waals force, en.wikipedia.org/wiki/Van_der_Waals_force. In theory, if you know all the forces acting on a substance, you can calculate its pressure/volume/temperature relationship, but in practice the formula is derived empirically. Tables are available for many substances such as CO2 and NH3. J. Willard Gibbs, en.wikipedia.org/wiki/Josiah_Willard_Gibbs, was one of the first to explore chemical thermodynamics; you might want to read his work. Commented Feb 15, 2015 at 22:26
• Increased gas pressure does not stop the liquid from evaporating. The ideal gas law works best at lower P and higher T so might be a first approximation. This sounds like you should be considering experimentation. Commented Oct 14, 2022 at 18:43

There are 3 main factors involved with a liquid in a sealed container: If the container contains no head space and is allegedly strong heating the liquid will cause the pressure to increase. This will continue until there is a phase change or the plastic or tensile limit of the container is reached. If the first the most likely change is that the liquid will become super critical and behave as a gas. The pressure profile of just the liquid if the external pressure is just sufficient to prevent vaporization [a piston] will follow the liquid-vapor equilibrium curve. The pressure necessary to prevent expansion will follow the liquid-solid equilibrium curves. Above the critical T the gas will obey the appropriate equation of state.[The common atmosphere gases are liquids above their critical T.]

If there is head space and enough liquid to remain until the critical temperature the pressure inside will follow the vapor pressure curve of the liquid[s] with allowances for deviations from Raoult's Law if more than one liquid.

If there is head space and an inert gas the original pressure is atmospheric, a sum of the vapor pressure and the inert gas pressure. As the temperature is raised the inert gas will follow the appropriate gas law and the vapor will follow a modified vapor pressure curve. The vapor pressure is a function of temperature and inert gas pressure, the pressure affects the liquid phase more than the gas phase increasing the vapor pressure. I do not know the effect of inert gas pressure on the critical parameters.

A vapor is a GAS in equilibrium with its liquid, nothing to do with boiling points etc.

The volume of the headspace above the liquid (here 500 mL) is not important unless all of your liquid boils while bringing it to the set temperature. If it doesn't, then the system pressure will be on the saturation (or coexistence) curve, which is the curve collecting P/T pairs at which the liquid and vapor coexist (see a phase diagram such as provided by Wolfram, link courtesy of Dr. Pippik). At higher T there might not be a coexistence point, but inspection of the phase diagram reveals that this is not an issue at the desired T, here $$\pu{150 ^\circ C}$$.

To check if liquid and vapor coexist, find the pressure corresponding to $$\pu{150 ^\circ C}$$ on the saturation curve. That pressure, according to the Wolfram site, is 450 kPa.

You can now perform a quick check using the ideal gas law to see approximately what amount of liquid water corresponds to 0.5 L of vapor at the saturation pressure. It comes out to 0.064 mol, which is about 1 mL of liquid, or negligible relative to the 4.5 L of liquid in the container.

So you can conclude that the container at that temperature will be at a pressure of 450 kPa, with most of the water remaining a liquid.