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Say I have 3 closed containers of 1L each. By volume, the 1st one is 50/50 air and water, the 2nd is 20% water and 80% air, and the third is 80% water and 20% air. If I heat all of them to the same temperature of 120 C. Are they all under the same internal pressure?

EDIT: They all start at atmosphere pressure (1 atm) and room temperature (20 C) before I close the lid and start heating them.

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  • $\begingroup$ Do they all have the same mass? Or do they all have the same number of particles? $\endgroup$ Jul 18 '17 at 21:20
  • $\begingroup$ They all start at regular atmosphere pressure before I heat them. $\endgroup$
    – gummibear
    Jul 18 '17 at 21:22
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    $\begingroup$ Aside from the effect of changing the amount of liquid water as a result of different amounts of evaporation and different amounts of volume change from thermal expansion, the three pressures will be about the same. $\endgroup$ Jul 18 '17 at 23:00
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No in general they're not. When you start the experiment you have liquid water. The total number of particles is not the same in the three different containers. If you have to consider the values given in your question , than you basically have three pressure cookers. Even the container containing less water, about 200 g, given that you have 1 l room and stop heating at 120 °C,** provides enough molecules to attain the vapour P. It will be about 2 atm, you may look at the water PT diagram.

(**PV = nRT gives 358 atm)

Let us assume the volume of the containers be much larger and/or the amount of water much less. Then consider that the molecular weight (averaged) of air is almost twice that of water and especially that air in the initial conditions is hundreds time less dense than water. You can then simplify your calculations or at least your reasoning by assuming that your system contains just water, in terms of number of moles (number of particles) and specific heat.

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The containers all start at the same pressure $P_0$, volume $V_0$ and temperature $T_0$. They each consist of two gas species, consisting of a number of particles $N_1$ and $N_2$. These exert partial pressures $P_1$ and $P_2$. By Dalton's law of partial pressures, the total pressure in the container is given by

$$P = P_1+P_2$$

which, when we apply the ideal gas law, becomes

$$P = \frac{N_1kT}{V_0}+\frac{N_2kT}{V_0} = (N_1+N_2)\frac{kT}{V_0}$$

which means that the total pressure in the container as a function of temperature depends only on the total number of particles $N=N_1+N_2$ in the container. Since each gas mixture is at the same temperature, volume, and pressure, we can see that

$$N = \frac{P_0V_0}{kT_0}$$

which is the same in each case. Therefore, the pressure in each container is the same as a function of temperature.

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  • $\begingroup$ But we can see that the different containers start out with different number of molecules, so the pressures must at the end must also be different. You say that the the containers start out with similar conditions, but that is not true, since there are different mixes of fluids in different phases. $\endgroup$ Jul 25 '17 at 12:15
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What you are describing is a pressure cooker.

Probably_someone's answer is right if all water evaporates or boils and you just have a mix of two gases. As he said, pressure will depend on the amount of gas inside of the container.

However, as long as there is liquid water in the container, it will stop boiling when pressure in the container reaches its vapour pressure at its temperature.

According to this steam pressure calculator vapour pressure at 120ºC is 198.665 kPa (about two atmospheres) and that will be the pressure in the container at 120ºC as long as some water remains in it. Since if the container originally had 20%, 50% or 80% of water, some water will remain in liquid state, and therefore in the conditions of the question, the container will be at this pressure.

Only if the container contained very little water (let's say less than 1/20 of the container) or if you heat it more than 120ºC, all water will be converted to steam and probably_someone's answer will apply. Please notice that in this "dry" situation pressure will be lower.

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  • $\begingroup$ What class/subject do you need to take to cover this kind of material? Thanks! $\endgroup$
    – gummibear
    Jul 19 '17 at 0:16
  • $\begingroup$ Three decades ago in Catalonia, most of this was taught in high school courses for 15-16 year old students, in classes of physics and chemistry. At more advanced level I suppose changes of phase are covered in physical chemistry or thermodynamics classes. $\endgroup$
    – Pere
    Jul 19 '17 at 0:26
  • $\begingroup$ Would the same amount of water be converted to gas in all cases? $\endgroup$
    – gummibear
    Jul 19 '17 at 0:45
  • $\begingroup$ No, it doesn't. The same pressure will be reached. If there is a different volume of water, only the remaining of the container can be filled with gas. The more water, the less available space, the less steam. $\endgroup$
    – Pere
    Jul 19 '17 at 7:51
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I'll show you how to do the calculation for 20% water and 80% air, and you can then repeat the calculations for the other mixtures. We will neglect the initial amount of water vapor in the air at 20 C. So there are 200 cc of liquid water and 0.8 l of air. The specific volume of liquid water at 20 C is 1.002 cc/gm, so the mass of water is 199.6 gm and the number of moles of air is determined from the ideal gas law as [(0.9869)(0.8)]/[(0.0821)(293)]=0.0328 moles = 0.952 gm.

From the steam tables, at 120 C, the equilibrium vapor pressure of water vapor is 1.985 bars, the specific volume of liquid water is 1.060 cc/gm and the specific volume of water vapor is 892 cc/gm. So, if x represents the number of grams of water vapor in the gas space, we have (199.6-x) grams of water in the liquid space. So,we have, $$892x+1.060(199.6-x)=1000$$This gives 0.885 gm of water in the gas phase and 199.6-0.885=198.7 gm of liquid water. So the volume of liquid water at 120 C will be (198.7)(1.060)=211 cc and the volume of gas will be 1000-211=789 cc. Neglecting any air dissolved in the water, the partial pressure of the air in the gas space will be $[(1.0)(800)(273+120)]/[(789)(293)]=1.36\ bars$. So the total pressure in the container will be 1.985+1.36 = 3.35 bars.

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  • $\begingroup$ Does gm stand for grams? $\endgroup$
    – gummibear
    Jul 20 '17 at 2:08
  • $\begingroup$ How can the pressure be higher than 1.985 bars at 120 C if that is the equilibrium vapor pressure of water vapor at that temp? $\endgroup$
    – gummibear
    Jul 20 '17 at 3:02
  • $\begingroup$ The partial pressure of the air in the container is not zero (you said that there is air in the container, right?). In this example, the partial pressure of the air is 1.32 bars at 120 C. If there were no air in the container (just pure water), the pressure would be 1.985 bars. $\endgroup$ Jul 20 '17 at 11:51
  • $\begingroup$ Ok, so the pressure in the container IS affected by the ratio of water to air, and more air would mean higher pressure? $\endgroup$
    – gummibear
    Jul 21 '17 at 20:28
  • $\begingroup$ Why don't you work out the other two cases using the template that I provided and see what you get.? I think you will find that the effect of the water to air ratio is very inconsequential for most of the range of possible ratios. Only when there is extremely little water to start with or extremely little air to start with will there be a considerable difference. $\endgroup$ Jul 21 '17 at 21:16

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