From the phase diagram of water, we see that water is a liquid at 20 °C, 1 atm. The state of the water at 20 °C, 1 atm is liquid. But there is actually also water vapor in equilibrium with liquid water. What is the definition of “the state of a substance at a specific temperature and pressure?”
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1$\begingroup$ The state of a substance is probably understood as the most important phase describing this substance at a specific temperature and pressure. At $20°$C, and $1$ atm., the concentration of water in the liquid phase is greater than the concentration of water in the vapor phase. So water is considered as liquid under these conditions of $T$ and $p$. $\endgroup$– MauriceCommented Jun 16, 2021 at 8:44
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1$\begingroup$ The water phase diagram means the single component ( but its phases ) system. So at 20 deg C, the pressure 1 atm will finally convert all water vapour to liquid. $\endgroup$– PoutnikCommented Jun 16, 2021 at 11:43
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$\begingroup$ At 20C in a ONE component system the only head pressure being that of water applied pressure greater than the equilibrium vapor pressure will condense all the water vapor. If the applied pressure is less, vacuum chamber, liquid will evaporate until the vapor pressure is reached or all the water is vapor. $\endgroup$– jimchmstCommented Dec 31, 2021 at 5:34
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3 Answers
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I will start by addressing the posted question: the "state of a substance at a specific temperature and pressure" refers to "the most stable phase of the homogeneous substance at the specific p and T". The phase is indicated in the phase diagram by looking up the appropriate T,p point. At $\pu{20 ^\circ C}$ and 1 atm pressure that would be pure liquid water.
Now, addressing the confusion, the system you are referring to (with vapour) is not at 1 atm, at least not 1 atm of water vapour. You are probably thinking of a closed container partly occupied by liquid water and a mixed gas phase with contents under a pressure of 1 atm. Yes, the liquid water in the container will be in equilibrium with vapour (if the liquid has a volume smaller than the container's, and if the temperature is not too high, causing all of the water to form vapour). But the pressure of the water vapour (that is, its partial pressure) will not be 1 atm. It will be much less, closer to the ideal equilibrium vapour pressure of water at that temperature.
Note the question is similar that posed here, from which I borrow the following phase diagram (not of water):
That post states
However, this had me slightly confused because in the situation initially described (at a point somewhere within the liquid region - NOT on the phase boundary) some vapour existed above the liquid in the container ...
I quote the above because it is a related confusion. When water is in equilibrium with vapour the state is necessarily one described by a point on the liquid-gas coexistence line, and most decidedly not an interior point in the region labelled "liquid". An interior point is not one describing equilibrium with vapour (or, there is no coexisting vapour at such an interior point).
answered Jun 16, 2021 at 8:39
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The statement of "But there is actually also a water vapor in equilibrium with liquid water" may have come from a confusion about the difference of a closed system of water and a beaker of water open to air, both under say NTP. When you talk about the state of water on the phase diagram, you are talking about the former case. As Buck answered above, the pressure must be the vapor pressure of water. In the latter case, the pressure comes from the air and it is applied to the liquid part only while the water vapor and air molecules including O2 and N2 etc form approximately ideal gas mixture, and there is a dynamic equilibrium between the vapor and the liquid phase at the vapor/liquid interface. That is the reason for humidity at a certain temperature. The effect of the atmospheric pressure on the liquid part can be seen by that the liquid can be brought into boiling state if the atmospheric pressure is reduced from 1 atm to the value at 25 °C shown on the curve for liquid/vapor equilibrium of the phase diagram. The reason the liquid boils under that condition is that the vapor within the bubble inside the liquid part is equal to external pressure, which is exerted on the liquid and passed to the vapor within the bubble. If you can understand this difference, then there would be no more confusion. Please see another relevant post here: What does the phase diagram mean by "liquid" at NTP? (The liquid and vapor coexist even when the situation is not on the curve).
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The simple answer is that the phase diagram refers to pure water only—either in a single phase, or in some combination of phases.
So let's imagine we have a large piston that contains only water—nothing else—that is maintained at a constant temperature of $\pu{20 ^{\circ}C = 293.15 K}$ And suppose it starts at a pressure of $\pu{1 \times 10^-6 GPa \approx 0.010 atm}$. Checking the phase diagram below, we can see that it will be in the gaseous state. Now let's lower the piston to raise the pressure until we hit the liquid-gas coexistence line which, at $\pu{293.15 K}$, occurs at $\pu{2.34 \times 10^-6 GPa \approx 0.023 atm}$. That's the vapor pressure of water at that temperature. At this point, liquid water will start to form.
As we continue to lower the piston, the pressure doesn't increase. Instead, gaseous water converts to liquid water, but the pressure stays at the equilibrium vapor pressure of water. This proceeds until the piston head reaches the surface of the liquid. With any infinitesimal increase in pressure beyond this, only liquid water will be present.
We can continue to increase the pressure to your desired pressure of $\pu{1 atm}$. At this point, we will of course still have only liquid water. Thus we can see that, when we have only pure water at ${\pu{T = 20 ^{\circ}C and p = 1 atm}}$, it will be entirely in the liquid state, consistent with the phase diagram.
Source of diagram: Pierazzo, E. & Artemieva, Natalia & Ivanov, Boris. (2004). Starting Conditions for Hydrothermal Systems Underneath Martian Craters: Hydrocode Modeling. Special Paper of the Geological Society of America. 384. 10.1130/0-8137-2384-1.443.
