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How to realize the triple point of water?

My question is more or less the same as the one in the link above but with one added doubt. The answer claims that pressure will adjust itself so as to stay on the liquid-vapor line. But the entire apparatus is constrained to stay at a constant volume, which means that the state variables, namely pressure, and temperature should be constrained to move along an isochoric line (i.e.A straight line passing through the origin) on the graph. So it should not have been possible for the pressure to adjust itself on the equilibrium line between liquid and vapor. How can we reconcile for that?

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You're assuming the ideal gas law $PV = nkT$ when you say

constrained to stay at a constant volume, which means that the state variables, namely pressure, and temperature should be constrained to move along an isochoric line (i.e.A straight line passing through the origin)

But that's not at all true for a far-from-ideal material like water in a mix of liquid, gaseous and eventually solid phases. The liquid-vapor line itself doesn't do that:

enter image description here

The inter-molecular forces make it far from ideal by binding some molecules together and in the process changing the number of colliding gas molecules $n$.

Note that this is a different situation from boiling/condensing at constant pressure, i.e. boiling water in an open pot. There, when heat is added, water turns from liquid to gas, but because it's open to the air, the pressure doesn't change and the temperature stays at 100C until all the water is evaporated.

These cells are more like pressure cookers: They're closed, so adding heat to turn water from liquid to gas raises the pressure. And because the pressure has gone up, the equilibrium temperature between gas and liquid goes up.

In a fixed volume, like a pressure cooker or triple-point cell, so long as there's both liquid and gas in the cell, it has to be at a consistent pressure and temperature: It has to lie along the "liquid-vapor" line above. More energy will increase both temperature and pressure; less energy will decrease both temperature and pressure. By removing energy, cooling, the cell, the mixture works down the line until it hits the triple point and solid ice starts appearing.

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  • $\begingroup$ Is there anything that is staying constant through this process, then? $\endgroup$ – Goyal May 29 '18 at 6:14
  • $\begingroup$ @RadhikaGoyal Not sure what your asking. Besides total volume and mass? $\endgroup$ – Bob Jacobsen May 29 '18 at 13:18
  • $\begingroup$ Never mind my previous comment. Ill try to put it in a better way. Is it all right to say that this graph is followed even when the process is not isochoric? As in, say after taking the temperature to 70 to 50-degree Celcius, I suddenly increase the volume of the entire apparatus isothermally. The pressure would thus reduce but will eventually get restored to the equilibrium again, hence bringing us back on the same curve on this graph. Right? $\endgroup$ – Goyal May 29 '18 at 14:18
  • $\begingroup$ @RadhikaGoyal Added a bit more on the equilibrium. If you change the volume (provided there's still enough water to have both gas and liquid), you'll move to a different point on the liquid-vapor line. Both so long as you have both phases, you're on that line. $\endgroup$ – Bob Jacobsen May 29 '18 at 14:25

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