December 21, 2024, 09:05:45 PM
Forum Rules: Read This Before Posting


Topic: Phase equilibria  (Read 33711 times)

0 Members and 8 Guests are viewing this topic.

Offline Big-Daddy

  • Sr. Member
  • *****
  • Posts: 1177
  • Mole Snacks: +28/-94
Re: Phase equilibria
« Reply #30 on: July 11, 2013, 01:42:11 PM »
If you are referring to everyday water vapor it is mixed with air. The partial pressure of water vapor in that air-water mix will fall on that P-T line.

OTOH, if it is a non-equilibrium situation none of this applies. It can pretty much do whatever it wants.

Ah ok - so then the pressure as found on the (P,T) graph refers to the partial pressure of the component for which the phase diagram is being drawn, rather than the total pressure of the system. Important thing to note.

It leads on to another question - how can we calculate the partial "liquid" pressure exerted by a certain species in liquid phase, or partial "solid" pressure exerted by a certain species in solid phase (given that in the above case we were dealing with "gaseous" partial pressure), which would represent the (P,T) points for parts of the phase diagram in the solid or liquid phase regions? What are these 2 new kinds of pressure to be written as a function of, in other words (assuming we cannot set the pressure ourselves externally) - volume and temperature (what is our substitute for the ideal gas equation)? Because so far I've only worked with gaseous pressures until now.

Offline curiouscat

  • Chemist
  • Sr. Member
  • *
  • Posts: 3006
  • Mole Snacks: +121/-35
Re: Phase equilibria
« Reply #31 on: July 11, 2013, 01:47:22 PM »


Ah ok - so then the pressure as found on the (P,T) graph refers to the partial pressure of the component for which the phase diagram is being drawn, rather than the total pressure of the system. Important thing to note.



The P-T diagram you talk about are strictly single component.

Your example introduced air. So P-T diagrams are not strictly directly applicable.

For a single component system, total Pressure is the P. 

Offline curiouscat

  • Chemist
  • Sr. Member
  • *
  • Posts: 3006
  • Mole Snacks: +121/-35
Re: Phase equilibria
« Reply #32 on: July 11, 2013, 01:48:39 PM »

It leads on to another question - how can we calculate the partial "liquid" pressure exerted by a certain species in liquid phase, or partial "solid" pressure exerted by a certain species in solid phase (given that in the above case we were dealing with "gaseous" partial pressure), which would represent the (P,T) points for parts of the phase diagram in the solid or liquid phase regions? What are these 2 new kinds of pressure to be written as a function of, in other words (assuming we cannot set the pressure ourselves externally) - volume and temperature (what is our substitute for the ideal gas equation)? Because so far I've only worked with gaseous pressures until now.

I read that thrice. Does not make any sense to me. Sorry.

Offline Big-Daddy

  • Sr. Member
  • *****
  • Posts: 1177
  • Mole Snacks: +28/-94
Re: Phase equilibria
« Reply #33 on: July 11, 2013, 01:54:03 PM »
The P-T diagram you talk about are strictly single component.

Your example introduced air. So P-T diagrams are not strictly directly applicable.

For a single component system, total Pressure is the P.

Ok. But the situation with the equilibrium always applies - it can only be reached at certain points of P and T, where P is the partial pressure of the substance? At all other conditions of P and T, the reaction will move forward in one direction more than in the other until the required conditions for equilibrium are reached.

Offline curiouscat

  • Chemist
  • Sr. Member
  • *
  • Posts: 3006
  • Mole Snacks: +121/-35
Re: Phase equilibria
« Reply #34 on: July 11, 2013, 01:59:13 PM »
Ok. But the situation with the equilibrium always applies - it can only be reached at certain points of P and T, where P is the partial pressure of the substance? At all other conditions of P and T, the reaction will move forward in one direction more than in the other until the required conditions for equilibrium are reached.

I think yes. For single component systems. For multi-component it's probably only an approximation.

Offline Big-Daddy

  • Sr. Member
  • *****
  • Posts: 1177
  • Mole Snacks: +28/-94
Re: Phase equilibria
« Reply #35 on: July 11, 2013, 02:09:40 PM »

It leads on to another question - how can we calculate the partial "liquid" pressure exerted by a certain species in liquid phase, or partial "solid" pressure exerted by a certain species in solid phase (given that in the above case we were dealing with "gaseous" partial pressure), which would represent the (P,T) points for parts of the phase diagram in the solid or liquid phase regions? What are these 2 new kinds of pressure to be written as a function of, in other words (assuming we cannot set the pressure ourselves externally) - volume and temperature (what is our substitute for the ideal gas equation)? Because so far I've only worked with gaseous pressures until now.

I read that thrice. Does not make any sense to me. Sorry.

On a phase diagram which is a (P,T) and phase graph, pressure extends to liquid and solid phase species of the component for which the diagram is written. Concepts of pressure apply not only to gases but also to liquids and solids.

"Gaseous" pressure can be approximated, for instance, as P≈nRT/V. In general P=f(T,V,n) is true for any gas (a power series is the highest level method for approaching exactness), where T is temperature, V is volume and n is number of moles of that gas, and P is the partial pressure exerted by that gas. In this case, P would be the total pressure in the single-component system.

What can we say, similarly, about pressures for liquids and solids?

I remember being told by Corribus that the pressure exerted by liquids and solids does not contribute at all to that experienced by gases (partial pressure for a single species is strictly a function of T, V and n; maybe of T, V taken up by the gaseous phase as a whole and n for each gaseous phase species), though gaseous pressure can affect the pressure exerted on a liquid or solid phase. What are the liquid and solid phase species' partial pressures a function of, then? I ask because I have very little knowledge of how pressure is dealt with when it comes to non-gases.

Offline curiouscat

  • Chemist
  • Sr. Member
  • *
  • Posts: 3006
  • Mole Snacks: +121/-35
Re: Phase equilibria
« Reply #36 on: July 11, 2013, 02:17:28 PM »




"Gaseous" pressure can be approximated, for instance, as P≈nRT/V. In general P=f(T,V,n) is true for any gas (a power series is the highest level method for approaching exactness), where T is temperature, V is volume and n is number of moles of that gas, and P is the partial pressure exerted by that gas. In this case, P would be the total pressure in the single-component system.

What can we say, similarly, about pressures for liquids and solids?


You can write an  equation of state for L and S phases. Most cases, not much use. L and S are substantially incompressible.

Hence, P or T don't have much of an effect on V/n i.e. density = constant.


Quote

I remember being told by Corribus that the pressure exerted by liquids and solids does not contribute at all to that experienced by gases (partial pressure for a single species is strictly a function of T, V and n; maybe of T, V taken up by the gaseous phase as a whole and n for each gaseous phase species), though gaseous pressure can affect the pressure exerted on a liquid or solid phase. What are the liquid and solid phase species' partial pressures a function of, then?

This section was similar. Read it twice. Makes no sense. Sorry.

Quote
I ask because I have very little knowledge of how pressure is dealt with when it comes to non-gases.

Read a book on fluid flow or Fluid Mechanics.

Offline Big-Daddy

  • Sr. Member
  • *****
  • Posts: 1177
  • Mole Snacks: +28/-94
Re: Phase equilibria
« Reply #37 on: July 11, 2013, 03:13:17 PM »
You can write an  equation of state for L and S phases. Most cases, not much use. L and S are substantially incompressible.

Hence, P or T don't have much of an effect on V/n i.e. density = constant.

Ok. Let's say I have a container of pure water of volume V at temperature T. This point does not lie on the line of a phase diagram, in other words it is kept isothermally at T which ensures that there is no water vapour. What is the pressure of the liquid water? (which, if I so wanted, I could use to make sure that water at this temperature T and pressure is indeed liquid, as per the phase diagram.) This surely isn't a fluid dynamics problem, because the container is static, as is the water in it.

And then, let's say there is a vapour pressure acting downwards on the water (i.e. let's say we know that the gases in the container are exerting a total pressure of Pg). What's the pressure of the liquid water?

Offline curiouscat

  • Chemist
  • Sr. Member
  • *
  • Posts: 3006
  • Mole Snacks: +121/-35
Re: Phase equilibria
« Reply #38 on: July 11, 2013, 09:44:26 PM »
Pressure on the water can be anything in your first case. Indeterminate. Water at V and T could be at 1 bar, 10 bar, 100 bar.

So long as incompressibility is assumed.

In your second example pressure on water is Pg. Plus gravity head etc. depending on where you measure it.

Offline curiouscat

  • Chemist
  • Sr. Member
  • *
  • Posts: 3006
  • Mole Snacks: +121/-35
Re: Phase equilibria
« Reply #39 on: July 11, 2013, 09:52:29 PM »
in any case, "vapor pressure acting downwards" is a bit incongrous.

Fluid pressure cannot act in a certain direction. it acts in ll directions.

Offline Big-Daddy

  • Sr. Member
  • *****
  • Posts: 1177
  • Mole Snacks: +28/-94
Re: Phase equilibria
« Reply #40 on: July 12, 2013, 07:13:24 PM »
Pressure on the water can be anything in your first case. Indeterminate. Water at V and T could be at 1 bar, 10 bar, 100 bar.

So what are the other factors, besides V and T, that will help me determine what pressure the water in this static beaker is at?

So long as incompressibility is assumed.

Even if it is not assumed, what does that change about the argument? It doesn't suddenly make it possible to express the pressure purely as a function of

In your second example pressure on water is Pg. Plus gravity head etc. depending on where you measure it.

What? The example is the same as the first, except we've added gas. So if the pressure on the water for the first example is P1, is the pressure on the water for the second example equal to P1+Pg? Or does the water's own collisions with the container then stop having effect, so that Pg alone (plus extra effects depending on how deep in the water you are measuring) is the pressure on the water?

Offline curiouscat

  • Chemist
  • Sr. Member
  • *
  • Posts: 3006
  • Mole Snacks: +121/-35
Re: Phase equilibria
« Reply #41 on: July 12, 2013, 11:26:57 PM »
if compressible use an empirical eq relating density to P and T.

Offline Big-Daddy

  • Sr. Member
  • *****
  • Posts: 1177
  • Mole Snacks: +28/-94
Re: Phase equilibria
« Reply #42 on: July 13, 2013, 09:21:10 AM »
Ok.

So we've established that V, T and total pressure of the vapour above the water in the static beaker are all factors to determine the pressure of the water in the beaker. In the case of gases this would be enough, but as you say for liquids it is not enough. What are the other factor(s)?

Offline curiouscat

  • Chemist
  • Sr. Member
  • *
  • Posts: 3006
  • Mole Snacks: +121/-35
Re: Phase equilibria
« Reply #43 on: July 13, 2013, 02:57:24 PM »
Ok.

So we've established that V, T and total pressure of the vapour above the water in the static beaker are all factors to determine the pressure of the water in the beaker. In the case of gases this would be enough, but as you say for liquids it is not enough. What are the other factor(s)?

Gravity, weights, pistons, flow, gases, things that squeeze, external impulses..

Offline curiouscat

  • Chemist
  • Sr. Member
  • *
  • Posts: 3006
  • Mole Snacks: +121/-35
Re: Phase equilibria
« Reply #44 on: July 13, 2013, 02:58:51 PM »
What? The example is the same as the first, except we've added gas. So if the pressure on the water for the first example is P1, is the pressure on the water for the second example equal to P1+Pg? Or does the water's own collisions with the container then stop having effect, so that Pg alone (plus extra effects depending on how deep in the water you are measuring) is the pressure on the water?

Didn't make much sense.

Sponsored Links