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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?

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    $\begingroup$ Can't figure out exactly what you are trying to calculate. The minimum "about" (=amount?) of water to do what? What will hapen beneath or above the minimum? $\endgroup$ – AstronAUT Sep 24 '15 at 23:35
  • $\begingroup$ Thanks. Yes, it should have been "amount". When pressure cooking you want as little water as possible, as it only dilutes the food, but some is required in order to get it under pressure. $\endgroup$ – Jasmine Lognnes Sep 24 '15 at 23:58
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    $\begingroup$ You might want to post this question on the cooking.SE site. It sounds like this is more of a practical matter that someone who cooks with a pressure cooker can explain without resorting to gas laws and physical chemistry, subjects which will only ruin your appetite anyway. $\endgroup$ – Todd Minehardt Sep 25 '15 at 0:03
  • $\begingroup$ Find the manual for your pressure cooker; it should tell you. Hopefully it's online somewhere if you don't have a paper copy. $\endgroup$ – Jason Patterson Sep 25 '15 at 3:15
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    $\begingroup$ I think this question is on-topic and fits well here at chem.se. The initial question as worded is slightly unclear. I believe the question is, how much water needs to be added to a 6.2 L pressure cooker to ensure that the gauge pressure can reach 0.8 bar (of wet steam)? $\endgroup$ – Curt F. Sep 25 '15 at 4:17
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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.

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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.

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  • $\begingroup$ 1.8 in the equation should be 0.8. Apart from that, it can it really be correct that I can get it to boil with less than 1mL of water? I would expect the right answer to be around 1dL. $\endgroup$ – Jasmine Lognnes Sep 25 '15 at 8:23
  • $\begingroup$ @JasmineLognnes Normal ambient or atmospheric air pressure is 1 bar. The value 0.8 bar given for your cooker is the excess pressure (so-called gauge pressure) above ambient pressure. Thus, the total or absolute pressure is 1 bar + 0.8 bar = 1.8 bar. $\endgroup$ – Loong Sep 25 '15 at 10:09
  • $\begingroup$ Found the problem: accidently used the specific gas constant :) 6.1 mL is still not much water but sounds legitimate to me. $\endgroup$ – AstronAUT Sep 25 '15 at 13:19

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