I'm a mathematician and computer scientist, and for this particular problem I would benefit from some physics/chemistry expertise. I've already asked this question on chemistry stack exchange, but I figured some physicists might be able to help also.

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 increases in 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. Similarly, I would expect filling the tank to 85% capacity would result in less pressure than filling it to 95% capacity.

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.
  • $\begingroup$ You want to fill the tank with liquid propane at a cold temperature. The liquid will boil producing gas when you open the regulator. The pressure in the tank will be directly proportional to the outside temperature. At very high outside temperatures the regulator may fail. The pressure on the gauge is not a direct measure of the volume of gas in the tank - since part of the volume may be liquid and part volume may be gas . $\endgroup$ May 8, 2019 at 3:39
  • $\begingroup$ The pressure will only depend on the vapor pressure of the propane at the temperature of the container, unless you fill the container so full that liquid expansion due to a temperature increase causes the container to become 100% liquid. $\endgroup$ May 8, 2019 at 3:58

1 Answer 1


In the case there is negligible amount of air captured in the gaseous phase(e.g purged away by propane evaporation, and if there is sufficient space for gas not to be compressed by thermal dilation of liquid propane ( like about fully filled by the liquid).....

.... then pressure is fully determined by the temperature, not being dependent on the amount of liquid propane.

Propane would boil, until the pressure reaches the value of saturated propane vapour pressure for given temperature, which raises about exponentially with temperature, according to Clausius-Clapeyron equation:

$$p=p_\mathrm{0} \cdot \exp\left( \frac {\Delta H_\mathrm{evap}}{R}\cdot \left( \frac{1}{T_\mathrm{0}}-\frac{1}{T}\right)\right)$$ where
R is the gas constant 8.314 J/K/mol
$\Delta H_\mathrm{vap}$ is propane vaporisation enthalpy(+) from Wikipedia propane data page 24.545e3 J/mol
T is absolute temperature (+)Thermal energy of the phase change at constant pressure

The assumption of the constant enthalpy with temperature is a reasonable approximation.

More exact equations could be gained, If there is available function $\Delta H=f(T)$. There could be performed the integration of differentiated CC equation over the T interval.

The linearized approximation of $\Delta H=f(T)$ can be gained by applying Hess law :

$$\Delta H(T)=\Delta H(T0)+(C(l)-C(g))\cdot ( T -T0 )$$

where C(l),C(g) are respective liquid and gas propane molar heat capacities in J/K/mol. It is assuming constant capacities, what is another approximation.

But all that is rather complicated and more pragmatical approach is using the empirical formula based on empirical data

See also the link for the graph and empirical formula of $p=f(T)$.

The empirical formula for propane value pressure from the above link is

$$\log p_\mathrm{mm Hg}=6.82973-\frac{813.2}{248+T}$$

760 mm Hg=101325 Pa

If the amount of air cannot be neglected, then $$p_\mathrm{total}= p_\mathrm{propane} + p_\mathrm{air}$$

$$p_\mathrm{air}= p_\mathrm{air, initial} \cdot \frac{V_\mathrm{gas,initial}}{V_\mathrm{gas, final}}$$

One has also consider $$dV_\mathrm{g}=-\frac{dm_\mathrm{propane,g}}{\rho_\mathrm{propane,l}}$$

If the tank is filled just by a gas, the pressure is very well approximated by the Gay-Lussac law for an ideal gas, saying the ratio $$\frac pT = const$$

where p is pressure and T is absolute temperature.

Better estimation would by the van der Waals gas equation

$$(P+a n^2/V^2)(V-n b)=n R T$$

  • $a=8.779 L^2 bar/mol^2$
  • $b=0.08445 L/mol$

tabelized constants.

  • $\begingroup$ I think the equation you've mentioned is a form of the clausius-clapeyron equation which assumes that the vaporisation enthalpy is constant, which is not true in general - why is this assumption reasonable? Also, how do I determine when the amount of air is "negligible"? I'd like to be able to perform calculations like this with an arbitrary set of initial conditions. For example, comparing the difference between 85% full and 95% full - is the only difference the pressure of air at the start? $\endgroup$ May 8, 2019 at 4:44
  • $\begingroup$ It is science, not math. Negligible values lead to negligible result differences. You decide what is negligible in given context. $\endgroup$
    – Poutnik
    May 8, 2019 at 4:57
  • $\begingroup$ See the answer update. $\endgroup$
    – Poutnik
    May 8, 2019 at 5:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.