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### Topic: Phase equilibria  (Read 31340 times)

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##### Re: Phase equilibria
« Reply #45 on: July 13, 2013, 03:20:21 PM »
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.

#### curiouscat

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##### Re: Phase equilibria
« Reply #46 on: July 13, 2013, 10:48:40 PM »
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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##### Re: Phase equilibria
« Reply #47 on: July 14, 2013, 09:49:25 AM »
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.

#### curiouscat

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##### Re: Phase equilibria
« Reply #48 on: July 14, 2013, 10:55:44 AM »
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?

Quote
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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##### Re: Phase equilibria
« Reply #49 on: July 14, 2013, 11:27:36 AM »
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?
« Last Edit: July 14, 2013, 01:11:26 PM by Big-Daddy »

#### curiouscat

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##### Re: Phase equilibria
« Reply #50 on: July 14, 2013, 01:41:05 PM »
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.

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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?

Quote
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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##### Re: Phase equilibria
« Reply #51 on: July 14, 2013, 01:52:40 PM »
Only because I assumed density invariant with Temperature.

So then $$P = P_{gas} + \rho (T) \cdot g \cdot h$$?

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.

#### curiouscat

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##### Re: Phase equilibria
« Reply #52 on: July 14, 2013, 01:57:08 PM »

So then $$P = P_{gas} + \rho (T) \cdot g \cdot h$$?

Yep. If you want to be more (possibly needlessly) comprehensive: $$\rho (T, P, g, t....)$$
Basically whatever density dependence may exist in your particular liquid.

Quote
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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##### Re: Phase equilibria
« Reply #53 on: July 14, 2013, 02:35:08 PM »
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?

#### curiouscat

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##### Re: Phase equilibria
« Reply #54 on: July 15, 2013, 12:00:31 AM »
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.