# What happens to pressure when a liquid goes supercritical?

Fill a pressure vessel with water, no airspace. Heat it past 374 Degrees C, so that the water goes supercritical but can't expand. What happens to the pressure at that point? Does it continue to increase linearly as the temperature continues to increase, or would the graph form a hockey stick there, as the rate of increase (pressure) suddenly increased, or does it increase exponentially (curved graph) or????

Put another way, is the pressure at 700 Kelvin twice that at 350 K, or???

• The vessel would crack due liquid water thermal dilation long before reaching the critical T. Commented Feb 2 at 7:21
• Huh, one would need a line of constant density drawn on P-V phase diagram to tell, but I don't think you'd notice anything at this temp. Why not? Because there would be no vapor phase and pressure would get over crit. value long before crit. temp. Commented Feb 2 at 14:36
• At such high pressures supercrit. fluids are liquid-like and there is no phase transition at crossing crit. temp. The pressure would increase in fashion that is neither linear nor exponential, or any simple function afaict. Commented Feb 2 at 14:44
• Over my head, guys. I can't copy and paste the chart on page 2, here www-pub.iaea.org/MTCD/publications/PDF/P1500_CD_Web/htm/pdf/…
– J.O.
Commented Feb 3 at 2:32
• continued: --d C, so they never get too close to 374 C. I know that zirconium combines with the oxygen in steam and supercritical water, releasing hydrogen; don't care; I need to know what happens to the pressure if that plant overheats and its coolant goes supercritical. Again, if your plant is running at 325 C, and it overheats to 650 C, does the pressure double, or what?
– J.O.
Commented Feb 3 at 2:50

Typically, isochores in a $$p-T$$ diagram are straight lines. This is evident in the ideal gas law, but also a prediction from the simple van der Waals' equation of state.

We put this to test using a more sophisticated equation of state, the Peng Robinson's equation of state $$$$p = \frac{RT}{V - b} - \frac{a}{[V + (1 - \sqrt{2})b)][V + (1 + \sqrt{2})b]} \tag{1}$$$$ Even though it still has two parameters, the one related to the attraction forces has a dependency with temperature $$$$a = 0.45724\,\frac{\alpha(T_\mathrm{r},\omega)R^2T_c^2}{p_\mathrm{c}} \tag{2}$$$$ in which the accentric factor is taken into account via the correlation established by Peng and Robinson $$$$\alpha(T_\mathrm{r},\omega) = [1 + (0.37464 + 1.54226\omega - 0.26992\omega^2)(1 - \sqrt{T_r})]^2 \tag{3}$$$$

The results are shown below:

This is a $$p - T$$ diagram but in reduced coordinates. There are ten isochores which are also displayed in reduced form. The supercritical region that interests you lies in the rectangle for which $$p_\mathrm{r} > 1$$ and $$T_\mathrm{r} > 1$$. When $$T_\mathrm{r}$$ is rather low, there seems to be some curvature in all the isochores due to that both terms are important in the equation. However, for the supercritical region we have essentially straight lines. This agrees with the explanation given by Oscar Lanzi.

References

• The parameters for water used in the equation of state were extracted from the DIPPR Data Compilation of Pure Compound Properties, ASCII FIles, National Institute of Science and Technology, Standard Reference Data, Gaithersburg, MD, 1458 chemicals, extant 1995.

$$\begin{array}{|c|c|c|c|} \hline T_\mathrm{c} \; \pu{(K)} & p_\mathrm{c} \; \pu{(bar)} & V_\mathrm{c} \; \pu{(dm3/mol)} & \text{Accentric factor} \\ \hline 647.1 & 220.55 & 0.0559 & 0.345 \\ \hline \end{array}$$

Assuming that the vessel does not blow up, the "gas" pressure ultimately becomes proportional to absolute temperature, but the proportionality constant may be greater than that derived from the ideal gas law.

Illustrating with the Redlich-Kwong equation:

$$P=\dfrac{RT}{V-b}-\dfrac{a}{\sqrt{T}V(V+b)},$$

where P is pressure, V is molar volume, T is absolute temperature and $$a,b$$ are deemed constants for a given gas. Under the conditions assumed in the problem $$V$$ is also constant as the temperature is raised and therefore the first term becomes mathematically dominant. Thus

$$P\approx\dfrac{RT}{V-b},$$

which exceeds the ideal gas law through the "excluded volume" from the rigid structure in the gas molecules. The excluded volume effect is also present in ordinary gases, but the relative effect would be much larger in the pressure-vessel situation considered here.

• Well, that's illustrative, but for details one would could rater use some modified Tait_equation, I think. Commented Feb 3 at 19:11

Mostly way over my head--it would really help if you would define your variables--but I gather that pressure is proportional to temperature, and that crossing the supercritical point doesn't change that, so that twice the temperature (in Kelvins) means twice the pressure. Interesting that twice the pressure means half again the velocity in firearms, and twice the temperature seems to equal about half again the efficiency in Rankine-cycle power plants. Must be another law of thermodynamics in there ... . I'm thinking of the differences between pressurized water nuclear power plants and molten salt reactors, and if I got this right I got what I needed, so thanx, guyz.