Pressure in the vessel containing water

[Jahnavi]

Homework Statement

Homework Equations

The Attempt at a Solution

I think the pressure that the question is relating is the vapour pressure but I do not understand how that depends on the volume of the water contained in the vessel .Is there a relationship between the two ?

 

[Charles Link]

Hint: If we know the vapor pressure at temperature ## T ##, and vessel A has reached equilibrium, (they tell you that all the water did not evaporate in the closed container, and some liquid water remains at constant volume), what do we know about the pressure (from water vapor) in vessel A? (The problem assumes there is originally no air above the water in the container, but that really wouldn't change the answer).

 

[Jahnavi]

Hi ,

Thanks for replying .

what do we know about the pressure (from water vapor) in vessel A?

Pressure is equal to vapour pressure .

 

[Jahnavi]

Does vapour pressure of a liquid depend only on two things , type of liquid and temperature ?

 

[Charles Link]

Does vapour pressure of a liquid depend only on two things , type of liquid and temperature ?

Yes, that is correct. And you might find it of interest that the boiling point (temperature) is when the vapor pressure is equal to the atmosphere pressure.

 

[Jahnavi]

Thank you

Could you explain the solution to this problem . Why option 4) is the correct choice and not 1) or 3) ?

 

[Charles Link]

Thank you

Could you explain the solution to this problem . Why option 4) is the correct choice and not 1) or 3) ?

View attachment 225286

This is not a good question, in my opinion, (edit: at least for a multiple choice question), because it doesn't say how large the container is, and if the amount of water is perhaps very small. If the container is so large that the vapor pressure at ## 0^o ##C can not be reached when all the water has evaporated, along with any ice that formed has also evaporated, the final result could simply be water in the vapor state at ## 0^o ##C. (They needed to put a qualifying statement in there that the volume is sufficiently small that this possibility can not occur). Some of these multiple choice questions have answers that give a somewhat incomplete solution, and a more complete solution is really in order. ## \\ ## Also, to be precise, the triple point of water is at ##T=+ .01^o ## See: https://www.quora.com/What-is-the-meaning-of-triple-point-of-water ## \\ ## ([Edit:] An item of interest: According to the diagram, at 1.0 atm of air pressure, water does freeze somewhere around ##T=0^o ## C , and although this can't be then called the triple point, there will be a partial pressure of vapor in the atmosphere (much below 1.0 atm), so the water will coexist in 3 phases=solid, liquid, and gas, even though it is technically somewhat incorrect to call it the "triple point" as they did in answer (4), [Edit:] but then, this is not what they specified, because the container was previously evacuated, so there is no 1.0 atm of air pressure). ## \\ ## [And a calculation for you: Estimate the vapor pressure of water for this case where the atmospheric pressure supplies a pressure of 1 atm, at ## T=0.00^o ## C , so that the liquid water and ice coexist with whatever vapor pressure there is. (In this case, there is no precise location on the liquid-vapor line to get this number, because the operating point on the diagram is up on the solid-liquid line, where it is marked "Earth" at P= 1.0 atm. I believe a good answer would be the vapor pressure would be very nearly the same as the pressure at the triple point...Additional editing: And I think I have this part correct, because there is no point on the phase diagram that has ## T=0.00^o \, C ## that has liquid and vapor coexisting, because the pressure of ##P=1.0 ## atm is essentially applied externally, and doesn't result from the system's vapor pressure. Thereby, the vapor pressure at this ## T=0.00^o \, C ## point that is a modified type of "triple point" would need to be estimated, and one "best estimate" would be that of the actual triple point vapor pressure at ## T=.01^o \, C ##. ] ## \\ ## Edit: Reading the question more carefully, the chamber is "evacuated", which means there is no 1.0 atm of air pressure. That would put the system, if it is at ## T=0.00^o ## C on the solid-vapor part of the P-T diagram, so there would be no water present in the liquid form in the final state at equilibrium. ## \\ ##To summarize: ## \\ ## If there were P=1.0 air pressure present, (but this is not what this problem specified), then the result would be 3 phases would coexist at equilibrium at ## T=0.00^o ## C even though it is not the triple point. ## \\ ## In the evacuated container at ## T=0.00^o ## C, all of the water that was originally in liquid form should in fact freeze or vaporize. One thing they don't specify is how accurately is the temperature controlled. For a problem such as this, it will affect the final result. Alternatively, if ## T=T_{triple \, point}=+0.01^o ##C , which is also ## T=0^o## C when rounded off, this is in fact where the triple point is, and the 3 phases, solid, liquid, and gas, would coexist at equlibrium in the previously evacuated container. A temperature difference ## \Delta T=.01^o ##C would affect the final result. They need to be much more specific=the problem is otherwise poorly defined. And meanwhile at ## T=+0.02^o ## C ,(which also is ## T=0^o ## C when rounded off), the system would then be on the liquid-vapor line, and no ice would be present in the final system at equilibrium, (only liquid water and water vapor). ## \\ ## And it does actually make for a very good question, but in this case, as we have determined, they need to be quite specific in describing the scenario, or it can have very different outcomes. ## \\ ## Editing: And there is a common sense approach that if the temperature is somewhere close to ## T=0^o ## C, what ultimately occurs, if you start out in the liquid phase, some water vapor and ice necessarily form, (not necessarily at perfect equilibrium, because the ## \Delta H ## for vapor is 540 cal/gm, while the ## \Delta H ## for freezing is only -80 cal/gm. Thereby, some vapor necessarily forms, and along with that is a lot of cooling of the remaining liquid to necessarily form some ice. The system is not precisely at the triple point, but temporarily all 3 phases will necessarily coexist, but not necessarily in equilibrium. What ultimately becomes the equilibrium state of the system depends very precisely on what the temperature is, as was previously mentioned. ## \\ ## @Jahnavi In any case, an interesting question, but perhaps it is better suited as an essay question than a multiple choice question. :)