Minimum amount of water in pressure cooker

Question

When cooking I would like to know the minimum amount of water I can add to my 6.2 L pressure cooker, worst case, so without any food. At second ring cooking the pressure is 0.8 bar.

I assume I need to solve $pV = nRT$ and know the density of water vapour at 0.8 bar, but that introduces $m = \rho V$, where I suppose $m$ and $V$ are unknowns?

Question

My high school chemistry is really rusty. Can someone help me in the right direction, and perhaps also tell if it at all something that can be solved without knowing the temperature?

Answer 1

In a pressure vessel, the volume $V$ is approximately constant. It is given as $V = 6.2\ \mathrm l$. However, this volume includes

• liquid water
• steam
• air
• food

in unknown amounts.

The gauge pressure is given as $p_\mathrm e = 0.8\ \mathrm{bar}$; i.e. the absolute pressure is approximately $p = 1.8\ \mathrm{bar}$.

We may use so-called steam tables to look up the properties of water at the given pressure (in the following, I use parameter values taken from the REFPROP – NIST Standard Reference Database 23, Version 9.0). We may find that the saturation point (equilibrium of liquid water and steam) at a pressure of $p = 1.8\ \mathrm{bar}$ corresponds to a temperature of $T = 117\ \mathrm{^\circ C}$. At this point, the density of liquid water is $\rho_\mathrm l = 946\ \mathrm{g/l}$ and the density of steam is $\rho_\text{steam} = 1.02\ \mathrm{g/l}$.

If we ignore the volume of the remaining liquid water, air, and food, we may consider the limiting case in which the entire volume $V$ is filled with steam. Thus, the mass $m_\text{steam}$ is given by $$\begin{align} m_\text{steam} &= \rho_\text{steam} \cdot V \\[3pt] &= 1.02\ \mathrm{\frac gl} \times 6.2\ \mathrm l \\[3pt] &= 6.3\ \mathrm g \end{align}$$ Therefore, only $6.3\ \mathrm g$ of water (which corresponds to about $6.3\ \mathrm{ml}$ of cold water at normal pressure) are required to fill the entire volume with steam at the given pressure.

However, a significant amount of excess liquid water should remain in the cooker in order for it to function properly. The liquid water is heated in the lower part of the cooker and boils at about $117\ \mathrm{^\circ C}$. The steam is distributed in the entire volume and condenses on colder surfaces (food and in particular the walls of the cooker). The condensed water runs down and it is collected in the lower part of the cooker. This circulation is essential for the uniform distribution of heat within the cooker; hence, the cooker should not dry out. The additional amount of liquid water should be sufficient to wet all surfaces (food and cooker) thoroughly.

Up to now, we have ignored the air that remains in the cooker. At initial (cold) conditions (pressure $p_0 = 1\ \mathrm{bar}$ and temperature $T_0 =20\ \mathrm{^\circ C}$), the density of dry air is approximately $\rho_{\text{air},0} = 1.189\ \mathrm{g/l}$. If we now ignore the volume of the liquid water, steam, and food, we may consider the limiting case in which the entire volume $V$ is filled with air. The mass $m_\text{air}$ is given by $$\begin{align} m_\text{air} &= \rho_{\text{air},0} \cdot V \\[3pt] &= 1.189\ \mathrm{\frac gl} \times 6.2\ \mathrm l \\[3pt] &= 7.37\ \mathrm g \end{align}$$ The air cannot escape. At the final pressure of $p = 1.8\ \mathrm{bar}$ and temperature of $T = 117\ \mathrm{^\circ C}$, the trapped air is compressed to a density of approximately $\rho_\text{air} = 1.607\ \mathrm{g/l}$. The corresponding new volume is $$\begin{align} V_\text{air} &= \frac{m_\text{air}}{\rho_\text{air}} \\[3pt] &= \frac{7.37\ \mathrm g}{1.607\ \mathrm{g/l}} \\[3pt] &= 4.6\ \mathrm l \end{align}$$ We see that a large part of the volume is taken up by the trapped air. The amount of water that is actually required to create steam at the given pressure is reduced accordingly.

Answer 2

It depends on the temperature at which you want to cook it, because, theoretically, you can also have 1.8 bar absolute pressure with only a few water molecules in there, if you just heat it up enough: $$p = \frac{nRT}{V}$$ As you increase T at fixed values for n and V also p gets bigger.

So lets take 121 °C (394.15 K) for an example.

And expand our formula by $n = \frac{m}{M}$ and $m = \rho V(aq)$:

$$V(aq) = \frac{pVM}{RT\rho} $$

And insert all into it:

$$V(aq) = \frac{1.8E5~\mathrm{Pa} \times 0.0062~\mathrm{m^3} \times 0.018~\mathrm{kg/mol}}{8.314 ~\mathrm{J/(K~mol)} \times 394.15~\mathrm{K} \times 1000~\mathrm{kg/m^3}} $$

And out comes:
$6.1 \times 10^{-6}$ m$^3$ = $6.1$ mL of Water.

So basically thats the minimum amount of water that will produce this pressure when heated up to 121 °C in your pot. Every additional water will condense, resulting in wet steam and furthermore in liquid water on the inner surface of your pot.

Note: This really is only water vapor, without any liquid water, it would look absolutly transparent. And there is also no other gas pressure included, like the chamber would be free from any other gas (air etc.). Under normal conditions you would therefore need even less water vapor.