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Chemistry Forums for Students => Physical Chemistry Forum => Topic started by: Big-Daddy on June 23, 2013, 12:32:45 PM
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On a phase diagram, the lines are "lines of equilibrium", i.e. at any point on these lines of equilibrium there is an established equilibrium between the two phases on either side of the line. This is according to ChemGuide (http://www.chemguide.co.uk/physical/phaseeqia/phasediags.html).
However the equilibrium cannot just spring up at this boundary which the line represents. Either side of it, the equilibrium must still be occurring, but less appreciably. For instance, liquids always have a liquid-vapour equilibrium, even in conditions (of T and P) not on the line in the phase diagram for that substance; only sometimes they are mostly liquid and we can call this equilibrium negligible.
Thus, what values or range of the equilibrium constant do the lines in standard phase diagrams tend to represent? e.g. looking at the phase diagram for water, http://en.wikipedia.org/wiki/File:Phase-diag2.svg, the blue line in the diagram represents the equilibrium line for the equilibrium H2O (l) <-> H2O (g), i.e. at the points of (T,P) along the line is this equilibrium constant within some appreciable range which means the phases are considered at equilibrium. What is this range of values?
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A rare instance where ChemGuide uses some clumsy language.
The line in a phase diagram is where the standard Gibbs energies of the two phases are exactly equal, so there is no driving force for one phase to turn into the other (or vice-versa). So let's say you are in the liquid phase at a certain temperature. There is always an equilibrium between the liquid and solid state, and the Gibbs energies of the two phases are temperature dependent, so you can calculate a "rate of evaporation".
For instance, let’s look at the standard Gibbs energy change (J/mol) for water going from liquid to vaporous state (calculated from heat capacities) as a function of temperature with the pressure held constant at one atmosphere.
298.15 8558.434
303.15 7965.748
313.15 6790.625
323.15 5628.796
333.15 4479.846
343.15 3343.394
353.15 2219.084
363.15 1106.585
373.15 5.593691
383.15 -1084.17
393.15 -2162.96
You see quite clearly that there is a change in sign of the standard Gibbs energy change at about 373 K, or 100 Celcius, right where you know the boiling point is.
Now let’s take some other point, say 353.15 K. There is still an equilibrium going on at this temperature. The vaporization of liquid water is not thermodynamically spontaneous at this temperature, but even so there is still an equilibrium, so we know that there will always be some water present as water vapor. So if it’s not-spontaneous, why does a puddle of water evaporate to completion even at room temperature? Well that’s because of La Chatellier’s principle – the vapor can drift away, which continually drives the equilibrium toward vaporization, even though it’s not a spontaneous process. Still, the standard Gibbs energy change becomes less positive as the temperature increases, which naturally predicts what we know to be the case: at higher temperature, vaporization will happen faster, because the equilibrium more and more favors the gas state.
As we go from 353.15 K and increase the temperature, at around 373.15 the sign of the Gibbs energy change switches sign from positive to negative. Then the process of vaporization becomes spontaneous. All this really means is that the gas state is favored over the liquid state. Practically, nothing has really changed – there is still and equilibrium, so you will still have liquid hanging around. Of course, boiling is a little more complex than this because of cavitation and so on, but don’t make your life too difficult.
Anyway, point is that that there is a temperature where the standard Gibbs energy change is zero, because the liquid and gas states have the same standard Gibbs energy. At this point the gas and liquid are equally favored and there is 1 to 1 equilibrium (K = 1). The line in the phase diagram is made by plotting out where this temperature resides for every pressure.
Make sense?
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Nice analysis Corribus.
Does this say anything about rates? Say at 95 C will rate of conversion of liq to vapor be only marginally less than at 100 C?
I think not, but if so, where's the discontinuity. The smooth variation of ΔG doesn't reveal that?
Alternatively, of all the equilibria what makes the 100 C equilibrium special (such that high rate transition happens, T remains constant etc. )? The fact that ΔG = 0? Also, in chemical reaction equilibria why isn't ΔG=0 much noteworthy as it is in boiling?
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I think it's hard to speak of differences in rate between boiling and evaporation because there are two different types of processes. Any phase change system is going to be heterogeneous and thus complex. Evaporation is primarily a surface phenomenon - evaporation can only happen at a surface, so rate is going to be significantly determined by the available surface area. In boiling, the entropy change is favorable for spontaneous change from liquid to gas, so you have phase changes throughout the system - although cavitation and the need to form bubbles complicates any simple calculation of rates. That is, it's possible to form metastable systems.
Liquid to gas is especially complicated because the rate is also going to be influenced by pressure. That is, if you put water in a sealed container, the rate of evaporation is going to be time dependent, because as the water evaporates, the pressure will go up, which will slow down the rate.
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Thanks, that post helped me. But I have a few points to clarify:
1) Is this temperature, where ΔG°=0 thus K=1, the boiling point, i.e. the point in temperature at a certain constant pressure above which K for H2O (l) ::equil:: H2O (g) is >1 (i.e. favours gaseous H2O)?
2) Here's the real question.
You said:
At this point the gas and liquid are equally favored and there is 1 to 1 equilibrium (K = 1).
On its own, this makes sense. It would suggest that, at this temperature and pressure, if we add some H2O (l) now, then Q<1 whereas K=1 so the equilibrium will shift forward until Q=1 is re-established. In other words, adding liquid water or water vapour will simply result in some of that substance being converted to the other phase to meet K=1 again.
But I'm not sure then about this:
The line in a phase diagram is where the standard Gibbs energies of the two phases are exactly equal, so there is no driving force for one phase to turn into the other (or vice-versa).
If you add water in one phase or the other, without changing the temperature or pressure, then as per your above explanation we should get a driving force (ΔG, not ΔG°) pushing the reaction to reach K=1 again (this will happen at any temperature and pressure, just that K won't equal 1 except at these points, at which it is defined to equal 1).
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Above the boiling point, DG < 0, so K > 1. Gas phase is favored, and this is mostly because of entropy.
On its own, this makes sense. It would suggest that, at this temperature and pressure, if we add some H2O (l) now, then Q<1 whereas K=1 so the equilibrium will shift forward until Q=1 is re-established. In other words, adding liquid water or water vapour will simply result in some of that substance being converted to the other phase to meet K=1 again.
No, because the activity of pure liquid water is unchanged regardless of how much you add. Adding more water doesn't change Q - it doesn't impact the equilibrium, at least not directly. Adding more water vapour, however will shift the equilibrium because as you know the activity of a gas is related to its pressure. If the water vapor is added to a closed system, the pressure will rise, which will tend to favor condensation.
Also I again stress that vaporization is highly dependent on the available surface area, so a simple treatment of the problem using only Qs and Ks doesn't really apply.
If you add water in one phase or the other, without changing the temperature or pressure, then as per your above explanation we should get a driving force (ΔG, not ΔG°) pushing the reaction to reach K=1 again (this will happen at any temperature and pressure, just that K won't equal 1 except at these points, at which it is defined to equal 1).
Maybe my answer above helps you. Since the activity of pure liquid water is independent of how much of it is there, adding liquid water will not impact the equilibrium because Q (abd hence DG) will be unchanged*. Adding water vapor will increase the pressure, which will shift the equilibrium because Q will be changed. There will then be a driving force to reform equilibrium.
One last point here, related to the *, is that unlike in many simple chemical reactions, here the pressure dependence cannot be neglected. In a closed system, adding some liquid will reduce the volume available for the vaporized gas, so technically the equilibrium will be shifted - although not for the reason you've mentioned. Likewise, adding water vapor shifts the equilibrium due to a change in the activity of the gas, which is related to the pressure change.
So you see the system is very complex and I think you have to approach it differently than the way you approach reaction equilibria.
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No, because the activity of pure liquid water is unchanged regardless of how much you add. Adding more water doesn't change Q - it doesn't impact the equilibrium, at least not directly. Adding more water vapour, however will shift the equilibrium because as you know the activity of a gas is related to its pressure. If the water vapor is added to a closed system, the pressure will rise, which will tend to favor condensation.
I'm not sure if this explains things properly. Take H2O at temperature 403.15 K and standard pressure of 1 bar. The liquid-gas equilibrium H2O (l) ::equil:: H2O (g) has K=p[H2O (g)]/a[H2O (l)] which you say we can rewrite K=p[H2O (g)] (I would have thought the assumption that pure liquids are at unity activity no longer holds for phase equilibria?), where K>>1 at this T and P so that most of the water is gaseous. Surely if we add H2O (l), there will be a driving force for its conversion to H2O (g)? It will not just sit there - which indicates that Q has been reduced in value from K to something well below K, meaning the forward reaction occurs more than the backward reaction, and thus that Q has been affected by adding liquid water.
One last point here, related to the *, is that unlike in many simple chemical reactions, here the pressure dependence cannot be neglected. In a closed system, adding some liquid will reduce the volume available for the vaporized gas, so technically the equilibrium will be shifted - although not for the reason you've mentioned. Likewise, adding water vapor shifts the equilibrium due to a change in the activity of the gas, which is related to the pressure change.
So you see the system is very complex and I think you have to approach it differently than the way you approach reaction equilibria.
I don't mind dependence on pressure - as I am aware, all equilibrium constants have at least a little bit of dependence on pressure, as well as on temperature.
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BD,
You are overthinking it. Though, I think I may have misled you.
You start with
Take H2O at temperature 403.15 K
At 1 atmosphere, how do you have water at this temperature? It's not possible. What I wrote above (providing delta G values above the boiling point) was something of an abstraction. Once the temperature reaches the boiling point, I don't view it as a simple liquid-gas equilibrium any longer (in an open container) because there is enough energy to overcome atmospheric pressure and induce bubble formation. I think it's a more complex problem at that point and you can't treat it the same way. Under 1 atmosphere of pressure, it's not possible to heat water higher than the boiling point (ignoring metastable states). In a closed container, of course, you can get water this hot, because the total pressure in the container will rise with the temperature - but again this is a completely different problem because the external temperature is also changing.
Let's stick to a temperature below the boiling point.
Put water in a container at 50 Celsius. The vapor pressure is 12.3 kPa. Now double the amount of water at the same temperature. Does the vapor pressure change? No. This is because the activity of the "pure" liquid water is approximately 1 and therefore the equilibrium constant (and Q) doesn't dependent on the amount of water present. The vapor pressure is essentially an intensive property.
Now, at high pressures this approximation of the activity of liquid water being 1 begins to weaken, because the standard state is defined for 1 atmosphere (just as the approximation begins to weaken in concentrated solutions, because the water can no longer be approximated as "pure"). In these cases you would have to do something else... use the liquid fugacity or something. I don't know - chemical engineers probably would have to worry about that kind of stuff so you'll have to ask one of them.
Other than that's it's a pretty simple equilibrium problem and the standard approximations apply:
http://chemwiki.ucdavis.edu/Physical_Chemistry/Physical_Properties_of_Matter/Virtual%3A_Phase_Changes/Vapor_Pressure (mind the silly typo in the first equation - they aren't talking about hydrogen gas)
http://chemed.chem.wisc.edu/chempaths/GenChem-Textbook/The-Equilibrium-Constant-in-Terms-of-Pressure-965.html
Anyway, phase equilibria are complicated and not always intuitive and it'll make your head hurt if you think too much about them. I know it does mine. :)
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In these cases you would have to do something else... use the liquid fugacity or something. I don't know - chemical engineers probably would have to worry about that kind of stuff so you'll have to ask one of them.
Ha! We simply use a Moeller chart or a steam table and get on with life. ;D
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At 1 atmosphere, how do you have water at this temperature? It's not possible. What I wrote above (providing delta G values above the boiling point) was something of an abstraction. Once the temperature reaches the boiling point, I don't view it as a simple liquid-gas equilibrium any longer (in an open container) because there is enough energy to overcome atmospheric pressure and induce bubble formation. I think it's a more complex problem at that point and you can't treat it the same way. Under 1 atmosphere of pressure, it's not possible to heat water higher than the boiling point (ignoring metastable states). In a closed container, of course, you can get water this hot, because the total pressure in the container will rise with the temperature - but again this is a completely different problem because the external temperature is also changing.
Thanks, the rest of your post is clear, so this is the bit I'm honing in on.
I'm not sure why it should make a difference to the phase of the water itself whether the container is open or not, so long as the water's temperature can be governed (adiabatically, i.e. we assume we can set a certain temperature for the water and keep it there). Phase diagrams for water all show water as being gaseous at 400 K and 1 Pa (example: http://www.lsbu.ac.uk/water/phase.html) which would suggest these conditions are possible in theory; how then would the liquid water be spontaneously converted to vapour, which it will, to maintain the equilibrium (suggesting Q has been affected)? Perhaps the approximation of the activity of liquid water being unity becomes substantially weaker as the temperature rises as well as the pressure?
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i.e. if you add water in one phase or the other, at any T or P, without changing the temperature or pressure, then because water's activity cannot be approximated as 1 for phase equilibria (impression I'm getting - if not, then please explain what exactly makes Q change as you add more water), then the activity of liquid water changes when you add the liquid water so Q changes and this causes the reaction to move to produce more water vapour (reducing the activity of the liquid water again, as the quantity of liquid water is reduced through conversion to gaseous water) to reach the same K as specified at that particular T,P for this phase equilibrium.
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[...] the equilibrium cannot just spring up at this boundary which the line represents. Either side of it, the equilibrium must still be occurring, but less appreciably. For instance, liquids always have a liquid-vapour equilibrium, even in conditions (of T and P) not on the line in the phase diagram for that substance; only sometimes they are mostly liquid and we can call this equilibrium negligible.
No. The equilibrium is a line. One pressure for each temperature, one temperature for each pressure - over the range where the equilibrium exists.
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No. The equilibrium is a line. One pressure for each temperature, one temperature for each pressure - over the range where the equilibrium exists.
This doesn't agree with what Corribus was saying above ...
Are you saying that the equilibrium is there along the line, and then simply vanishes off the line? Surely this cannot be; the H2O (l) :rarrow: H2O (g) must to some extent be reversible because all reactions are. And furthermore if equilibrium only exists on the line then this fails to explain the presence of vapour pressure at points of T,P not on the line.
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Can we come to some kind of agreement on this? I don't feel comfortable moving on before I've established basically whether or not there is an equilibrium on the line (with the line representing T and P where K=1) and if not, how these problems I mentioned above are explained.
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I think the rest of us have an agreement.
Equilibrium is indeed on the line. @Enthapy seems absolutely right. Where does @Corribus say otherwise?
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No. The equilibrium is a line. One pressure for each temperature, one temperature for each pressure - over the range where the equilibrium exists.
This doesn't agree with what Corribus was saying above ...
Yes it does.
Are you saying that the equilibrium is there along the line, and then simply vanishes off the line?
Yes.
Surely this cannot be; the H2O (l) :rarrow: H2O (g) must to some extent be reversible because all reactions are.
Ok. So?
And furthermore if equilibrium only exists on the line then this fails to explain the presence of vapour pressure at points of T,P not on the line.
Nope. Vapor can be present. Just not at equilibrium. When things equilibriate the parameters will again fall on the P-T curve.
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I think the rest of us have an agreement.
Equilibrium is indeed on the line. @Enthapy seems absolutely right. Where does @Corribus say otherwise?
Here:
So let's say you are in the liquid phase at a certain temperature. There is always an equilibrium between the liquid and solid state, and the Gibbs energies of the two phases are temperature dependent,
And in the general argument I read into that post, it suggests there is always an equilibrium (which is temperature and pressure-dependent) between any two phases at any values of temperature and pressure. And as Corribus goes on to point out:
Anyway, point is that that there is a temperature where the standard Gibbs energy change is zero, because the liquid and gas states have the same standard Gibbs energy. At this point the gas and liquid are equally favored and there is 1 to 1 equilibrium (K = 1). The line in the phase diagram is made by plotting out where this temperature resides for every pressure.
Which says that the line is plotted from the points where the equilibrium constant is equal to 1 - the equilibrium constant being implied to exist at all points of T, P.
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Yes it does.
How? Read my post above ...
Surely this cannot be; the H2O (l) :rarrow: H2O (g) must to some extent be reversible because all reactions are.
Ok. So?
So H2O (l) ::equil:: H2O (g) exists at every point on the P, T curve. By the very definition of this reaction being reversible.
Nope. Vapor can be present. Just not at equilibrium. When things equilibriate the parameters will again fall on the P-T curve.
Ah now I am beginning to understand. But I'm not sure it makes sense - if I'm at a point of P,T not on the line, and I fix the conditions there, there is no vapour pressure? (Either the substance is all vapour, or there is no vapour.)
So the reversible reaction H2O (l) ::equil:: H2O (g) is present under all conditions. Somehow, under conditions of P,T not on the curve, the reaction proceeds in one direction more than in the other, and continues to do so until the P,T conditions are brought back to those on the line, at which point K=1 for the reversible reaction so there will be the same amount (in terms of activity/partial pressure) of the two phases present. This suggests that the equilibrium constant for H2O (l) ::equil:: H2O (g) is 1 but this only becomes relevant when at conditions on the line because off the line, something (dependent on pressure and temperature) ensures that the rate in a certain direction is always higher than the rate in the other. Is this what you're saying? Even then, this requires explanation.
Why does the reaction proceed in one direction more than the other, regardless of the fact that the activity of one side (the reactant in the favoured direction) is decreasing almost to 0? (Because rate in a certain direction is normally a function of the activities on the reacting side of that direction.) And even if the activity of the pure liquid isn't seen as changing, the partial pressure of the vapour is definitely changing, and even then the question remains, what forces the reaction in particular directions and then does not allow them to reach equilibrium under all conditions? (Equilibrium for most processes is achieved over time regardless of the temperature conditions I thought)
What causes the P,T to change as the reaction proceeds in a certain direction, to level off at points on the line when equilibrium can actually be reached?
And also, will there be no vapour and thus no vapour pressure present over a liquid if I hold the conditions to a P, T point not on the phase line (i.e. inside the liquid region)?
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When you have this equilibrium how many phases and components are there.
Ergo how many degrees of freedom.
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When you have this equilibrium how many phases and components are there.
Ergo how many degrees of freedom.
2 phases, 1 component. I don't know how to calculate degrees of freedom, but from a quick online search for degrees of freedom I can't conceivably draw links which would help me answer any of the questions above.
If you explain whether the understanding I suggested in the post above is correct or not that would be a start:
So the reversible reaction H2O (l) ::equil:: H2O (g) is present under all conditions. Somehow, under conditions of P,T not on the curve, the reaction proceeds in one direction more than in the other, and continues to do so until the P,T conditions are brought back to those on the line, at which point K=1 for the reversible reaction so there will be the same amount (in terms of activity/partial pressure) of the two phases present. This suggests that the equilibrium constant for H2O (l) ::equil:: H2O (g) is 1 but this only becomes relevant when at conditions on the line because off the line, something (dependent on pressure and temperature) ensures that the rate in a certain direction is always higher than the rate in the other. Is this what you're saying?
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2 phases, 1 component. I don't know how to calculate degrees of freedom, but from a quick online search for degrees of freedom I can't conceivably draw links which would help me answer any of the questions above.
P + F = C + 2
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Your basic misconception here seems that you can take a two phase system, specify both T and P arbitrarily and then get it to equilibrium under those P and T conditions.
Won't happen.
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Your basic misconception here seems that you can take a two phase system, specify both T and P arbitrarily and then get it to equilibrium under those P and T conditions.
Won't happen.
I know, this is what I said in the post I have referred to twice before:
So the reversible reaction H2O (l) ::equil:: H2O (g) is present under all conditions. Somehow, under conditions of P,T not on the curve, the reaction proceeds in one direction more than in the other, and continues to do so until the P,T conditions are brought back to those on the line, at which point K=1 for the reversible reaction so there will be the same amount (in terms of activity/partial pressure) of the two phases present. This suggests that the equilibrium constant for H2O (l) ::equil:: H2O (g) is 1 but this only becomes relevant when at conditions on the line because off the line, something (dependent on pressure and temperature) ensures that the rate in a certain direction is always higher than the rate in the other. Is this what you're saying?
Several questions remain. Let's start with the main one:
Why does the reaction proceed in one direction more than the other, regardless of the fact that the activity of one side (the reactant in the favoured direction) is decreasing almost to 0? (Because rate in a certain direction is normally a function of the activities on the reacting side of that direction.) And even if the activity of the pure liquid isn't seen as changing (BTW: not at all valid once the liquid is entirely converted to gas, activity is 0 then!), the partial pressure of the vapour is definitely changing. I.E. what force causes the reaction to keep moving forward in a certain direction, without equilibrium ever being reached (at values of T and P not on the line).
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Why does the reaction proceed in one direction more than the other, regardless of the fact that the activity of one side (the reactant in the favoured direction) is decreasing almost to 0?
Why? I can't see any activity going down to zero here.
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Why? I can't see any activity going down to zero here.
At any point in P and T the reversible reaction H2O (l) ::equil:: H2O (g) is occurring. But let's say we're at a point of P, T squarely inside the gaseous region, and we're going to hold things at these conditions. Then the forward rate for the reaction overpowers the backward rate - for a reason you have not explained to me. Most importantly, the forward rate stays fairly constant because the activity of the water stays fairly constant (and the conditions are held constant); but the backward rate should be increasing, as the vapour increases in partial pressure, until it can match the backward rate and equilibrium is reached. Where's the problem with the logic there?
And still there remains the question, what are the forward and backward rates functions of, that one continues to dominate the other regardless of how the activity or partial pressure on one side changes?
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as the vapour increases in partial pressure,
How is that consistent with your "we're at a point of P, T squarely inside the gaseous region, and we're going to hold things at these conditions"
Either the pressure is constant or it isn't.
You should double check your hypotheticals for consistency (I hope I don't have to eat crow!).
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How is that consistent with your "we're at a point of P, T squarely inside the gaseous region, and we're going to hold things at these conditions"
Either the pressure is constant or it isn't.
You should double check your hypotheticals for consistency (I hope I don't have to eat crow!).
Valid point. However, maybe the volume of the container is being changed to keep pressure constant? What then.
I only pose this because I want to see what happens if P and T are constant (and yet, obviously, we need to get a gas being produced more than a liquid if we're at a (P,T) point in the gaseous section).
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Valid point. However, maybe the volume of the container is being changed to keep pressure constant? What then.
What then? Pressure is constant. There fails your argument which said: " but the backward rate should be increasing, as the vapour increases in partial pressure,"
No increase in Partial Pressure. End of story.
I only pose this because I want to see what happens if P and T are constant (and yet, obviously, we need to get a gas being produced more than a liquid if we're at a (P,T) point in the gaseous section).
Yes. That happens. We end up with all gas at P,T. Where's the contradiction?
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What then? Pressure is constant. There fails your argument which said: " but the backward rate should be increasing, as the vapour increases in partial pressure,"
No increase in Partial Pressure. End of story.
Point taken. I feel like I'm understanding more now.
Two questions to check then: 1) so when the equilibrium is reached, is the constant equal to 1 (so that when the equilibrium is reached and P levels off with time, there are equal activities/fugacities of the substances present - bearing in mind of course that activity of a liquid or solid is not necessarily proportional to number of moles of that substance)?
2) How does vapour pressure come about at (P,T) points other than those on the line. e.g. we live in 298 K at 105 Pa pressure, a point on the diagram in which water lies squarely within the liquid form, so how does water vapour exist under these conditions?
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2) How does vapour pressure come about at (P,T) points other than those on the line. e.g. we live in 298 K at 105 Pa pressure, a point on the diagram in which water lies squarely within the liquid form, so how does water vapour exist under these conditions?
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.
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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.
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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.
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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.
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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.
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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.
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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.
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"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.
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.
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.
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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?
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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.
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in any case, "vapor pressure acting downwards" is a bit incongrous.
Fluid pressure cannot act in a certain direction. it acts in ll directions.
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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?
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if compressible use an empirical eq relating density to P and T.
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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)?
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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..
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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.
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Gravity, weights, pistons, flow, gases, things that squeeze, external impulses..
Only gravity and the weight of the water are present in this case as we're dealing with nothing but an unchanging beaker of water with some gas pressure over it. I realize that any more than this, and we'd be looking at proper fluid dynamics (I am getting a textbook for that).
It strikes me that once you have specified the gaseous pressure around the beaker, the V of the water and the T it is at, the only other factors which even might be involved for this case are the cross-sectional area of the beaker and the value of the gravitational constant g. In a certain gravitational field, if we specify V, T and the cross-sectional area of the beaker we have specified an exact beaker in itself.
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Gravity, weights, pistons, flow, gases, things that squeeze, external impulses..
Only gravity and the weight of the water are present in this case as we're dealing with nothing but an unchanging beaker of water with some gas pressure over it.
No. And gas pressure. If you got a powerful compressor to inject an inert gas at, say, 10 atm. in the gas space in the beaker, the water is obviously going to be at a high pressure too.
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No. And gas pressure. If you got a powerful compressor to inject an inert gas at, say, 10 atm. in the gas space in the beaker, the water is obviously going to be at a high pressure too.
Noted. But as I said, it comes down to: Pg (total gaseous pressure over the liquid); V[H2O] (volume of the liquid); T (temperature of the liquid phase); g (gravitational field strength affecting the beaker); and possibly the cross-sectional area of the beaker in the liquid phase.
If all of these are specified, we've specified an exact environment for the beaker as far as I can see. In that case there must be a straightforward way to calculate the pressure of the liquid?
Sorry if this all sounds unnecessary, just that I want to understand how we calculate pressure in a liquid. I originally thought when first looking at phase diagrams that the pressure on the liquid or solid phase would just be equal to the gaseous pressure surrounding the phase, but now I see that the liquid and solid will exert pressure of their own on the container.
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No. And gas pressure. If you got a powerful compressor to inject an inert gas at, say, 10 atm. in the gas space in the beaker, the water is obviously going to be at a high pressure too.
Noted. But as I said, it comes down to: Pg (total gaseous pressure over the liquid); V[H2O] (volume of the liquid); T (temperature of the liquid phase); g (gravitational field strength affecting the beaker); and possibly the cross-sectional area of the beaker in the liquid phase.
If all of these are specified, we've specified an exact environment for the beaker as far as I can see. In that case there must be a straightforward way to calculate the pressure of the liquid?
Yes. Pgas + δ g h
What's not straightforward?
Sorry if this all sounds unnecessary,
It does. You're moving in circles and not reading enough / not solving enough numerical examples.
Overdoing silly abstraction is killing you, if you ask me.
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Yes. Pgas + δ g h
Ok. But this suggests that the pressure of the water does not depend on the temperature of the water (Pgas will depend on the temperature of the gas, but still the pressure on the water doesn't depend directly on the temperature of the water)? And that, if instead of having 25 cm3 of water I had 50 cm3 of ethanol (with the same gaseous pressure Pgas over it), the pressure would be the same for both (if I'm at the same height from the surface, in the same gravitational field).
You're moving in circles and not reading enough / not solving enough numerical examples.
It's not very difficult to get numerical answers once you have the equations you need, in this case.
I can't find any questions asking me to find the pressure on a liquid in a static beaker. That suggests to me that it is very straightforward indeed, no need to look for numerical problems for such an obvious case. But things aren't clicking yet. When/if it looks like the equations are getting more complicated, I'll start looking for online papers or in a textbook, but until then it just feels like I'm missing something obvious.
Overdoing silly abstraction is killing you, if you ask me.
Since you're one of my most consistent teachers, the question for me to ask is: what should I be doing differently?
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Yes. Pgas + δ g h
Ok. But this suggests that the pressure of the water does not depend on the temperature of the water
Only because I assumed density invariant with Temperature.
And that, if instead of having 25 cm3 of water I had 50 cm3 of ethanol (with the same gaseous pressure Pgas over it), the pressure would be the same for both (if I'm at the same height from the surface, in the same gravitational field).
Why? Do EtOH and H2O have same density?
Since you're one of my most consistent teachers, the question for me to ask is: what should I be doing differently?
I told you my opinion, but you don't like it: Solve problems, crunch numbers. Get your hands dirty. Sweat it out. Armchair theorizing only gets you so far. This isn't like studying Philosophy. :)
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Only because I assumed density invariant with Temperature.
So then [tex]P = P_{gas} + \rho (T) \cdot g \cdot h[/tex]?
Why? Do EtOH and H2O have same density?
No, good point. I misread your δ as change as opposed to density which I've in the past seen as ρ, but really it should have been obvious since I've seen the equation before.
I told you my opinion, but you don't like it: Solve problems, crunch numbers. Get your hands dirty. Sweat it out. Armchair theorizing only gets you so far. This isn't like studying Philosophy. :)
Hmm well, I've got a mechanical engineering textbook on its way. That should have plenty of numerical problems!
I do solve a lot of problems. It's just that usually the answers are available for them, so I rarely have to bring them here and ask. That leaves only the conceptual difficulties which I can only clear up by asking.
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So then [tex]P = P_{gas} + \rho (T) \cdot g \cdot h[/tex]?
Yep. If you want to be more (possibly needlessly) comprehensive: [tex] \rho (T, P, g, t....) [/tex]
Basically whatever density dependence may exist in your particular liquid.
Hmm well, I've got a mechanical engineering textbook on its way. That should have plenty of numerical problems!
I do solve a lot of problems. It's just that usually the answers are available for them, so I rarely have to bring them here and ask. That leaves only the conceptual difficulties which I can only clear up by asking.
Fair enough.
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Fair enough.
Thanks. Last question, then - can we get something similar for the pressure on a solid? I'm guessing it will be P=Pgas+something, but not sure what ... still ρgh?
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Fair enough.
Thanks. Last question, then - can we get something similar for the pressure on a solid? I'm guessing it will be P=Pgas+something, but not sure what ... still ρgh?
Yes and no. Solids are complicated. Within solids pressures are "not isotropic". You have shear stresses etc. that makes direction important while talking about "pressure". Solids also accept directional point loads and moment couples which liquids cannot.
Under a special case, yes ρgh will hold.