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Prochak stupidity
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James McGinn
2018-03-13 20:00:26 UTC
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On Sunday, January 15, 2017 at 6:52:55 PM UTC-8, James McGinn wrote:

Ed Prochaks Dilemma -- Chan, Ed and Daltons Law

James McGinn wrote:
Chan, it's very hard to overcome the instinct to go along with what
everybody else believes.

Chan:
I think you and I earlier failed to understand how Dalton's law comes into
the picture.

James McGinn wrote:
Actually, I don't see it as being applicable. I'm wondering if maybe you
have made a common error. Vapor pressure of H2O and partial pressure of
different GASES are sometimes both called partial pressure. Maybe it is the
ambiguity in the phrase "partial pressure," that is the source of your
confusion.

Chan:
Vapor pressure contributes to total pressure as (partial) pressure.

James McGinn wrote:
Well, true but--to be candid--so what? Of course it does. But that doesn't
necessarily bring Dalton's into the discussion. Remembers, the underlying
cause of your confusion is, most likely, the ambiguity associated with the
fact that vapor pressure of H2O and partial pressure of different GASES are
sometimes both called partial pressure. (Read the previous sentence over and
over again until the distinction is clear in your mind.) The phrase partial
pressure, therefore, being ambiguous, should be avoided. (Or you could--as
many actually do--choose to remain in a state of permanent confusion. [If
that floats your boat . . . ]) Instead use either of these phrases depending
on which is applicable: 1) Vapor pressure of H2O; or 2) Partial pressure of
different GASES.

You have to be very deliberate about compartamentalizing these two different
concepts in your mind. (There is no substitute for being tough minded. It
does, however, get easier with practice.) Obviously the second of these two
is not applicable, that being Dalton's Law. And so, Dalton's law makes no
difference whatsoever.

More simply put: Dalton's law is a gas law. It is not a vapor law.

Chan:
There is no experimental evidence that moist air is lighter than dry air. It
is correct! Noone could provide a reference.

The experiment I proposed in my other thread "Why moist air lighter than dry
air" is one that could be done and it would indisputably establish the
result.

James McGinn wrote:
Yes, I saw it. I don't want to discourage you but, measuring water vapor is
very difficult. And this is mostly because knowing the size of microdroplets
and knowing the effect of other characteristics associated with
microdroplets--most significantly, the effects of H2O's huge heat capacity--
introduces huge unknowns into the equation. (This is the problem that I
think stumped Espy back in the 1840's.) And another complication is
associate with the amount of static electricity--which, for all we know--may
have a significant effect on size of microdroplets and other
characteristics. So, I'm not saying this makes your approach impossible. I'm
just saying that its success depends greatly on knowing things that, franky,
we don't know a lot about yet.

I think my idea to directly measure the weight of moist vs dry air
(controlling for all other factors) using simple items, like mason jars,
would be easier and much more reproducible. I could envision high school
students doing the experiment--or at least parts of it. The only
difficulting would be getting scales sensitive enough to measure small
differences in air with precision.

Ed Prochak:
http://nvlpubs.nist.gov/nistpubs/jres/83/jresv83n5p419_A1b.pdf
Of course JM thinks it is a conspiracy, so he won't read or understand this
paper. And infact he will deny it even exists.

James McGinn wrote:
LOL. I don't need to deny it. The fact that you can't/won't quote it
directly tells me everything I need to know.

Ed Prochak:
The current paradigm works.

James McGinn wrote:
Vagueness always appears to work.

Ed Prochak:
There is no justification to change the current paradigm until you can
produce better results.

James McGinn wrote:
Better than what? We're talking about measuring something that has never
been measured before. And if measuring something that has never been
measured before results in a change to the paradigm it's hard to imagine how
that would not be a good thing in the long run.

Ed Prochak:
I think Chan brought in the death blow to your premise. Given that using
Dalton's law works in calculating pressure and density in humid air, and you
agree it ONLY works for materials in a gaseous state, then the water must be
gaseous.

James McGinn wrote:
Actually, I am not in agreement with your imaginative reassessment of the
applicability of Dalton's Law. But I give you credit for being creative.

Ed Prochak:
Here are the key lines from the NIST paper I quoted above: "The air density
is conveniently calculated using an equation based on the equation of state
of an air-water vapor mixture. A new formulation of the air density equation
is developed below."

James McGinn wrote:
LOL. You just proved my point. Do you not know the difference between a
measurement/experiment and a calculation?

Ed Prochak:
Dalton's law provides the correct calculation.

James McGinn wrote:
LOL. You flipped from talking about Dalton's and started talking about the
NIST calculations and now you are flopping back to talking about Dalton
again?

Ed Prochak:
Using Dalton's law NIST calculates humid air density measured it and finds
the results match.

James McGinn wrote:
Matches what? It's never been measured. And, as has been explained to you
about 6 times now, Dalton's law deals with partial pressure of gases--not
vapors. It's that simple. The notion that Dalton's deals with H2O vapor is
your delusion. (Leave the rest of us out of it.)

Ed Prochak:
Go argue with NIST.

James McGinn wrote:
About what? Something you you just confirmed they don't have.

Ed Prochak:
Don't take my word.

James McGinn wrote:
You needn't be concerned about that.

Best,

James McGinn / Solving Tornadoes
James McGinn
2018-03-30 21:25:21 UTC
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Post by James McGinn
Ed Prochaks Dilemma -- Chan, Ed and Daltons Law
Chan, it's very hard to overcome the instinct to go along with what
everybody else believes.
I think you and I earlier failed to understand how Dalton's law comes into
the picture.
Actually, I don't see it as being applicable. I'm wondering if maybe you
have made a common error. Vapor pressure of H2O and partial pressure of
different GASES are sometimes both called partial pressure. Maybe it is the
ambiguity in the phrase "partial pressure," that is the source of your
confusion.
Vapor pressure contributes to total pressure as (partial) pressure.
Well, true but--to be candid--so what? Of course it does. But that doesn't
necessarily bring Dalton's into the discussion. Remembers, the underlying
cause of your confusion is, most likely, the ambiguity associated with the
fact that vapor pressure of H2O and partial pressure of different GASES are
sometimes both called partial pressure. (Read the previous sentence over and
over again until the distinction is clear in your mind.) The phrase partial
pressure, therefore, being ambiguous, should be avoided. (Or you could--as
many actually do--choose to remain in a state of permanent confusion. [If
that floats your boat . . . ]) Instead use either of these phrases depending
on which is applicable: 1) Vapor pressure of H2O; or 2) Partial pressure of
different GASES.
You have to be very deliberate about compartamentalizing these two different
concepts in your mind. (There is no substitute for being tough minded. It
does, however, get easier with practice.) Obviously the second of these two
is not applicable, that being Dalton's Law. And so, Dalton's law makes no
difference whatsoever.
More simply put: Dalton's law is a gas law. It is not a vapor law.
There is no experimental evidence that moist air is lighter than dry air. It
is correct! Noone could provide a reference.
The experiment I proposed in my other thread "Why moist air lighter than dry
air" is one that could be done and it would indisputably establish the
result.
Yes, I saw it. I don't want to discourage you but, measuring water vapor is
very difficult. And this is mostly because knowing the size of microdroplets
and knowing the effect of other characteristics associated with
microdroplets--most significantly, the effects of H2O's huge heat capacity--
introduces huge unknowns into the equation. (This is the problem that I
think stumped Espy back in the 1840's.) And another complication is
associate with the amount of static electricity--which, for all we know--may
have a significant effect on size of microdroplets and other
characteristics. So, I'm not saying this makes your approach impossible. I'm
just saying that its success depends greatly on knowing things that, franky,
we don't know a lot about yet.
I think my idea to directly measure the weight of moist vs dry air
(controlling for all other factors) using simple items, like mason jars,
would be easier and much more reproducible. I could envision high school
students doing the experiment--or at least parts of it. The only
difficulting would be getting scales sensitive enough to measure small
differences in air with precision.
http://nvlpubs.nist.gov/nistpubs/jres/83/jresv83n5p419_A1b.pdf
Of course JM thinks it is a conspiracy, so he won't read or understand this
paper. And infact he will deny it even exists.
LOL. I don't need to deny it. The fact that you can't/won't quote it
directly tells me everything I need to know.
The current paradigm works.
Vagueness always appears to work.
There is no justification to change the current paradigm until you can
produce better results.
Better than what? We're talking about measuring something that has never
been measured before. And if measuring something that has never been
measured before results in a change to the paradigm it's hard to imagine how
that would not be a good thing in the long run.
I think Chan brought in the death blow to your premise. Given that using
Dalton's law works in calculating pressure and density in humid air, and you
agree it ONLY works for materials in a gaseous state, then the water must be
gaseous.
Actually, I am not in agreement with your imaginative reassessment of the
applicability of Dalton's Law. But I give you credit for being creative.
Here are the key lines from the NIST paper I quoted above: "The air density
is conveniently calculated using an equation based on the equation of state
of an air-water vapor mixture. A new formulation of the air density equation
is developed below."
LOL. You just proved my point. Do you not know the difference between a
measurement/experiment and a calculation?
Dalton's law provides the correct calculation.
LOL. You flipped from talking about Dalton's and started talking about the
NIST calculations and now you are flopping back to talking about Dalton
again?
Using Dalton's law NIST calculates humid air density measured it and finds
the results match.
Matches what? It's never been measured. And, as has been explained to you
about 6 times now, Dalton's law deals with partial pressure of gases--not
vapors. It's that simple. The notion that Dalton's deals with H2O vapor is
your delusion. (Leave the rest of us out of it.)
Go argue with NIST.
About what? Something you you just confirmed they don't have.
Don't take my word.
You needn't be concerned about that.
Best,
James McGinn / Solving Tornadoes
Steve BH
2018-03-31 05:28:45 UTC
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Post by James McGinn
Well, true but--to be candid--so what? Of course it does. But that doesn't
necessarily bring Dalton's into the discussion. Remembers, the underlying
cause of your confusion is, most likely, the ambiguity associated with the
fact that vapor pressure of H2O and partial pressure of different GASES are
sometimes both called partial pressure.
And there is a reason for this. Wait for it:

Wait for it........



It's because they are the same thing. "Vapor pressure of H2O" is the partial pressure of water gas H2O. Water in gas phase. The phrase "water vapor" IS water gas, whenever "partial pressures" are being talked about.

Little droplets of water in air, fogs, suspensions, mists, clouds, whatever you want to call these colloidial dispersions of water in air, do not exert any significant partial pressure on behalf of the H2O in them, which is in a liquid state. In order to exert a partial pressure, a compound must be in gas phase, with separate molecules.

The pressures in gases approximately follow the idea gas law: P = RT [n/V] where n = number of moles, or mass/molecular weight = mass/MW

So do the "vapor pressures" of water at various temperatures follow the ideal gas law (close enough), showing that they are gas partial pressures, and the particles they are composed of, are separate gas molecules. If they were little clumps of 10 molecules, they would give only 10% of the expected vapor pressure for their density, revealing them to be gases composed of gigantic molecules with molecular weights of 180 grams per mole, and densities of 180 grams per 30.6 liters (the molar gas volume) near 100 C. This would be 5.9 grams/L or 5.9 kg/m^3.

But this isn't so. Pure water vapor has a density near 18 grams per 30.6 liters = 0.59 grams per liter (or kg per m^3, same thing) at 100 C.

http://web.mnstate.edu/jasperse/Chem210/Handouts/Ch%2010%20Gas%20Math%20Summary%20(p8).pdf

In fact, it is exactly 0.590. Here n/V = P/RT. For water gas at 1 atm n/V = 1 atm/[0.0821 (L*atm/mol*K)*373 K) = 0.00326 moles/liter, or 0.59 g/liter if you multiply by 18 grams/mol.

https://www.thermexcel.com/english/tables/vap_eau.htm

In fact using the above steam table at 102.32 C, where even McGinn thinks water is in gas form, the density is 0.645 kg/m^3 (the partial pressure here is 1.1 atm). Whereas at 96.7 C where McGinn thinks water has magically converted to little droplets in vacuum, the equilibrium pressure has fallen to only 0.9 atm (and it's all partial pressure of gas), and the density of the gas is 0.535 kg/m^3, which is hardly different from the value over 100 C. There is no "blip" in equilibrium water vapor density and pressure as it passes through 100 C. This is just a very artificial point where the partial pressure happens to be the same as the pressure of the atmosphere at sea level on obscure planet called Earth. But water and its behavior couldn't care less. The equilibrium pressure for gas above liquid water passes through 1 atm at that temperature, but it passes smoothly through. The pressure shows no giant change from clumping of molecules.

James, I'm sure by now you can't follow the chemistry or physics above, or read the steam table, either. But it's a nice illustration for people reading this thread and wondering why you're calling somebody named "Prochak" stupid. It's only because he's trying to point out these facts, and you have no idea what he's talking about, and thing he's confused about water vapor. He isn't. If it causes partial pressure, it's water gas.
James McGinn
2018-03-31 06:02:02 UTC
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Post by Steve BH
Post by James McGinn
Well, true but--to be candid--so what? Of course it does. But that doesn't
necessarily bring Dalton's into the discussion. Remembers, the underlying
cause of your confusion is, most likely, the ambiguity associated with the
fact that vapor pressure of H2O and partial pressure of different GASES are
sometimes both called partial pressure.
Wait for it........
It's because they are the same thing.
Why do you think you are the only one that believes this?


"Vapor pressure of H2O" is the partial pressure of water gas H2O. Water in gas phase. The phrase "water vapor" IS water gas, whenever "partial pressures" are being talked about.

Nonsense.
Post by Steve BH
Little droplets of water in air, fogs, suspensions, mists, clouds, whatever you want to call these colloidial dispersions of water in air, do not exert any significant partial pressure on behalf of the H2O in them,
Why would you even assert something so plainly ridiculous?

which is in a liquid state. In order to exert a partial pressure, a compound must be in gas phase, with separate molecules.

So, why do you think they distinguish between "vapor" pressure and gas pressure?
Post by Steve BH
The pressures in gases approximately follow the idea gas law: P = RT [n/V] where n = number of moles, or mass/molecular weight = mass/MW
So do the "vapor pressures" of water at various temperatures follow the ideal gas law (close enough), showing that they are gas partial pressures, and the particles they are composed of, are separate gas molecules. If they were little clumps of 10 molecules, they would give only 10% of the expected vapor pressure for their density, revealing them to be gases composed of gigantic molecules with molecular weights of 180 grams per mole, and densities of 180 grams per 30.6 liters (the molar gas volume) near 100 C. This would be 5.9 grams/L or 5.9 kg/m^3.
But this isn't so. Pure water vapor has a density near 18 grams per 30.6 liters = 0.59 grams per liter (or kg per m^3, same thing) at 100 C.
http://web.mnstate.edu/jasperse/Chem210/Handouts/Ch%2010%20Gas%20Math%20Summary%20(p8).pdf
In fact, it is exactly 0.590. Here n/V = P/RT. For water gas at 1 atm n/V = 1 atm/[0.0821 (L*atm/mol*K)*373 K) = 0.00326 moles/liter, or 0.59 g/liter if you multiply by 18 grams/mol.
https://www.thermexcel.com/english/tables/vap_eau.htm
In fact using the above steam table at 102.32 C, where even McGinn thinks water is in gas form, the density is 0.645 kg/m^3 (the partial pressure here is 1.1 atm).
No dumbass. The BP is dependent on both pressure and temperature. And pressure is dependent on size of container.

Your abject ignorance has resulted in false confidence that you understand what you do not understand.


Whereas at 96.7 C where McGinn thinks water has magically converted to little droplets in vacuum, the equilibrium pressure has fallen to only 0.9 atm (and it's all partial pressure of gas), and the density of the gas is 0.535 kg/m^3, which is hardly different from the value over 100 C. There is no "blip" in equilibrium water vapor density and pressure as it passes through 100 C. This is just a very artificial point where the partial pressure happens to be the same as the pressure of the atmosphere at sea level on obscure planet called Earth. But water and its behavior couldn't care less. The equilibrium pressure for gas above liquid water passes through 1 atm at that temperature, but it passes smoothly through. The pressure shows no giant change from clumping of molecules.

Your description has nothing to do with reality.
Post by Steve BH
James, I'm sure by now you can't follow the chemistry or physics above, or read the steam table, either. But it's a nice illustration for people reading this thread and wondering why you're calling somebody named "Prochak" stupid. It's only because he's trying to point out these facts, and you have no idea what he's talking about, and thing he's confused about water vapor. He isn't. If it causes partial pressure, it's water gas
Steve BH
2018-03-31 06:10:53 UTC
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Post by James McGinn
So, why do you think they distinguish between "vapor" pressure and gas pressure?
"They" don't. In this case, a "vapor" is just our common name for the gas made from a substance that we normally encounter as a liquid or solid. It could be CO2 vapor above dry ice, or gasoline vapor above a pool of gasoline. In all cases, there's nothing special about it. It's just gas where we're more used to something else.

And as for the rest of your comments, again, you do not understand the gas laws, so I'm talking to the all. Take a chemistry class. And pass it.
Steve BH
2018-03-31 06:11:39 UTC
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Post by James McGinn
So, why do you think they distinguish between "vapor" pressure and gas pressure?
"They" don't. In this case, a "vapor" is just our common name for the gas made from a substance that we normally encounter as a liquid or solid. It could be CO2 vapor above dry ice, or gasoline vapor above a pool of gasoline. In all cases, there's nothing special about it. It's just gas where we're more used to something else.

And as for the rest of your comments, again, you do not understand the gas laws, so I'm talking to the wall. Take a chemistry class. And pass it.
Steve BH
2018-03-31 08:07:55 UTC
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Post by James McGinn
So, why do you think they distinguish between "vapor" pressure and gas pressure?
Often vapor pressures are used as terms for the special partial pressures of gases that are in equilibrium with another phase, whereas in the generic term of "partial pressure" of a gas, it can be anything (no other phase or equilibration is implied). So the term "vapor" implies a history, and also another phase that drives it, as an equilibrium partial pressure from phase change.

Thus the partial pressure of water gas above liquid at 96.7 C is 0.9 atm. We call this special partial pressure a "vapor pressure" because it's in equilibrium with water and has (once been) liquid water. BUT IT IS A GAS NOW. Of course you can have partial pressure of H2O gas as anything LESS than this, without condensation. That's not properly a vapor pressure, but just a partial pressure.

So not all partial pressures are vapor pressures, but all vapor pressures are partial pressures. They are equilibrium partial pressures. One is a subset of the other. However, they are both taking about the pressure of gas molecules.

Now you calculate the density of water GAS in equilibrium above liquid at 90 C. Congrats, that IS the vapor pressure. It's a special gas partial pressure from another phase.

Go ahead-- show us your math. Or look it up in a table. Any vapor pressure implies a DENSITY (moles per volume) which then can easily be converted to a molecular weight. Use some vapor pressures (I don't care where you get them) to calculate some molecular weights. Are these the weights of gas molecules or something larger? Show me your numbers.
James McGinn
2018-04-01 01:39:52 UTC
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Post by Steve BH
Post by James McGinn
So, why do you think they distinguish between "vapor" pressure and gas pressure?
Often vapor pressures are used as terms for the special partial pressures of gases that are in equilibrium with another phase, whereas in the generic term of "partial pressure" of a gas, it can be anything (no other phase or equilibration is implied). So the term "vapor" implies a history, and also another phase that drives it, as an equilibrium partial pressure from phase change.
Thus the partial pressure of water gas above liquid at 96.7 C is 0.9 atm. We call this special partial pressure a "vapor pressure" because it's in equilibrium with water and has (once been) liquid water. BUT IT IS A GAS NOW. Of course you can have partial pressure of H2O gas as anything LESS than this, without condensation. That's not properly a vapor pressure, but just a partial pressure.
So not all partial pressures are vapor pressures, but all vapor pressures are partial pressures. They are equilibrium partial pressures. One is a subset of the other. However, they are both taking about the pressure of gas molecules.
Now you calculate the density of water GAS in equilibrium above liquid at 90 C. Congrats, that IS the vapor pressure. It's a special gas partial pressure from another phase.
Go ahead-- show us your math. Or look it up in a table. Any vapor pressure implies a DENSITY (moles per volume) which then can easily be converted to a molecular weight. Use some vapor pressures (I don't care where you get them) to calculate some molecular weights. Are these the weights of gas molecules or something larger? Show me your numbers.
Can you find any direct quotes from the literature that supports any of this?
p***@gmail.com
2018-04-01 04:18:04 UTC
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Post by James McGinn
Can you find any direct quotes from the literature that supports any of this?
Why? Can't you find them for yourself, Jim? Are you too stupid to look things up all by yourself? This inability to navigate the internet is precisely why you don't know what you don't know. You haven't a clue, and no one here is the least bit surprised!

It doesn't matter, of course, because you wouldn't understand it in any case.
James McGinn
2018-04-01 12:25:14 UTC
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Post by p***@gmail.com
Post by James McGinn
Can you find any direct quotes from the literature that supports any of this?
Why?
You got nothing!!!
Post by p***@gmail.com
Can't you find them for yourself, Jim? Are you too stupid to look things up all by yourself? This inability to navigate the internet is precisely why you don't know what you don't know. You haven't a clue, and no one here is the least bit surprised!
LOL. Me neither.

You got nothing!!!
Post by p***@gmail.com
It doesn't matter, of course, because you wouldn't understand it in any case.
Sergio
2018-04-07 14:42:05 UTC
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Post by James McGinn
Can you find any direct quotes from the literature that supports any of this?
Why? Can't you find them for yourself, Jim? Are you too stupid to look things up all by yourself? This inability to navigate the internet is precisely why you don't know what you don't know. You haven't a clue, and no one here is the least bit surprised!
It doesn't matter, of course, because you wouldn't understand it in any case.
McGinn has reading Aphasia.
Steve BH
2018-04-01 04:26:41 UTC
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Post by Steve BH
Post by James McGinn
So, why do you think they distinguish between "vapor" pressure and gas pressure?
Often vapor pressures are used as terms for the special partial pressures of gases that are in equilibrium with another phase, whereas in the generic term of "partial pressure" of a gas, it can be anything (no other phase or equilibration is implied). So the term "vapor" implies a history, and also another phase that drives it, as an equilibrium partial pressure from phase change.
Thus the partial pressure of water gas above liquid at 96.7 C is 0.9 atm. We call this special partial pressure a "vapor pressure" because it's in equilibrium with water and has (once been) liquid water. BUT IT IS A GAS NOW. Of course you can have partial pressure of H2O gas as anything LESS than this, without condensation. That's not properly a vapor pressure, but just a partial pressure.
So not all partial pressures are vapor pressures, but all vapor pressures are partial pressures. They are equilibrium partial pressures. One is a subset of the other. However, they are both taking about the pressure of gas molecules.
Now you calculate the density of water GAS in equilibrium above liquid at 90 C. Congrats, that IS the vapor pressure. It's a special gas partial pressure from another phase.
Go ahead-- show us your math. Or look it up in a table. Any vapor pressure implies a DENSITY (moles per volume) which then can easily be converted to a molecular weight. Use some vapor pressures (I don't care where you get them) to calculate some molecular weights. Are these the weights of gas molecules or something larger? Show me your numbers.
Can you find any direct quotes from the literature that supports any of this?
https://www.chemguide.co.uk/physical/phaseeqia/phasediags.html
p***@gmail.com
2018-04-01 05:09:09 UTC
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Post by James McGinn
Can you find any direct quotes from the literature that supports any of this?
https://www.chemguide.co.uk/physical/phaseeqia/phasediags.html
The problem, of course, is that even in the face of overwhelming evidence, Jim will never concede that water vapor can exist below the boiling point if water. I don't know if it is because he is too stupid to understand the science or if he is just too stubborn to ever admit that he could be wrong... but wrong he is, and wrong he shall remain.

Run, Jim, run, you have no observations, no experiments, and no evidence whatsoever. Perhaps you should find another hobby, you don't have the chops for science...
James McGinn
2018-04-01 12:32:13 UTC
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Post by Steve BH
Post by James McGinn
Can you find any direct quotes from the literature that supports any of this?
https://www.chemguide.co.uk/physical/phaseeqia/phasediags.html
The problem, of course, is that even in the face of overwhelming evidence,
It must be frustrating to be so sure you are right and so completely unable to say how or why.
James McGinn
2018-04-01 12:25:33 UTC
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Post by James McGinn
Post by Steve BH
Post by James McGinn
So, why do you think they distinguish between "vapor" pressure and gas pressure?
Often vapor pressures are used as terms for the special partial pressures of gases that are in equilibrium with another phase, whereas in the generic term of "partial pressure" of a gas, it can be anything (no other phase or equilibration is implied). So the term "vapor" implies a history, and also another phase that drives it, as an equilibrium partial pressure from phase change.
Thus the partial pressure of water gas above liquid at 96.7 C is 0.9 atm. We call this special partial pressure a "vapor pressure" because it's in equilibrium with water and has (once been) liquid water. BUT IT IS A GAS NOW. Of course you can have partial pressure of H2O gas as anything LESS than this, without condensation. That's not properly a vapor pressure, but just a partial pressure.
So not all partial pressures are vapor pressures, but all vapor pressures are partial pressures. They are equilibrium partial pressures. One is a subset of the other. However, they are both taking about the pressure of gas molecules.
Now you calculate the density of water GAS in equilibrium above liquid at 90 C. Congrats, that IS the vapor pressure. It's a special gas partial pressure from another phase.
Go ahead-- show us your math. Or look it up in a table. Any vapor pressure implies a DENSITY (moles per volume) which then can easily be converted to a molecular weight. Use some vapor pressures (I don't care where you get them) to calculate some molecular weights. Are these the weights of gas molecules or something larger? Show me your numbers.
Can you find any direct quotes from the literature that supports any of this?
https://www.chemguide.co.uk/physical/phaseeqia/phasediags.html
Relevance?
Steve BH
2018-04-03 05:09:57 UTC
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Post by Steve BH
Post by James McGinn
Post by Steve BH
Post by James McGinn
So, why do you think they distinguish between "vapor" pressure and gas pressure?
Often vapor pressures are used as terms for the special partial pressures of gases that are in equilibrium with another phase, whereas in the generic term of "partial pressure" of a gas, it can be anything (no other phase or equilibration is implied). So the term "vapor" implies a history, and also another phase that drives it, as an equilibrium partial pressure from phase change.
Thus the partial pressure of water gas above liquid at 96.7 C is 0.9 atm. We call this special partial pressure a "vapor pressure" because it's in equilibrium with water and has (once been) liquid water. BUT IT IS A GAS NOW. Of course you can have partial pressure of H2O gas as anything LESS than this, without condensation. That's not properly a vapor pressure, but just a partial pressure.
So not all partial pressures are vapor pressures, but all vapor pressures are partial pressures. They are equilibrium partial pressures. One is a subset of the other. However, they are both taking about the pressure of gas molecules.
Now you calculate the density of water GAS in equilibrium above liquid at 90 C. Congrats, that IS the vapor pressure. It's a special gas partial pressure from another phase.
Go ahead-- show us your math. Or look it up in a table. Any vapor pressure implies a DENSITY (moles per volume) which then can easily be converted to a molecular weight. Use some vapor pressures (I don't care where you get them) to calculate some molecular weights. Are these the weights of gas molecules or something larger? Show me your numbers.
Can you find any direct quotes from the literature that supports any of this?
https://www.chemguide.co.uk/physical/phaseeqia/phasediags.html
Relevance?
Try reading it.
James McGinn
2018-04-03 08:38:07 UTC
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Post by James McGinn
Post by Steve BH
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Post by Steve BH
Post by James McGinn
So, why do you think they distinguish between "vapor" pressure and gas pressure?
Often vapor pressures are used as terms for the special partial pressures of gases that are in equilibrium with another phase, whereas in the generic term of "partial pressure" of a gas, it can be anything (no other phase or equilibration is implied). So the term "vapor" implies a history, and also another phase that drives it, as an equilibrium partial pressure from phase change.
Thus the partial pressure of water gas above liquid at 96.7 C is 0.9 atm. We call this special partial pressure a "vapor pressure" because it's in equilibrium with water and has (once been) liquid water. BUT IT IS A GAS NOW. Of course you can have partial pressure of H2O gas as anything LESS than this, without condensation. That's not properly a vapor pressure, but just a partial pressure.
So not all partial pressures are vapor pressures, but all vapor pressures are partial pressures. They are equilibrium partial pressures. One is a subset of the other. However, they are both taking about the pressure of gas molecules.
Now you calculate the density of water GAS in equilibrium above liquid at 90 C. Congrats, that IS the vapor pressure. It's a special gas partial pressure from another phase.
Go ahead-- show us your math. Or look it up in a table. Any vapor pressure implies a DENSITY (moles per volume) which then can easily be converted to a molecular weight. Use some vapor pressures (I don't care where you get them) to calculate some molecular weights. Are these the weights of gas molecules or something larger? Show me your numbers.
Can you find any direct quotes from the literature that supports any of this?
https://www.chemguide.co.uk/physical/phaseeqia/phasediags.html
Relevance?
Try reading it.
The fact that you won't quote it tells me all I need to know about it.
Steve BH
2018-04-04 03:39:18 UTC
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Post by James McGinn
So, why do you think they distinguish between "vapor" pressure and gas pressure?
Often vapor pressures are used as terms for the special partial pressures of gases that are in equilibrium with another phase, whereas in the generic term of "partial pressure" of a gas, it can be anything (no other phase or equilibration is implied). So the term "vapor" implies a history, and also another phase that drives it, as an equilibrium partial pressure from phase change.
Thus the partial pressure of water gas above liquid at 96.7 C is 0.9 atm. We call this special partial pressure a "vapor pressure" because it's in equilibrium with water and has (once been) liquid water. BUT IT IS A GAS NOW. Of course you can have partial pressure of H2O gas as anything LESS than this, without condensation. That's not properly a vapor pressure, but just a partial pressure.
So not all partial pressures are vapor pressures, but all vapor pressures are partial pressures. They are equilibrium partial pressures. One is a subset of the other. However, they are both taking about the pressure of gas molecules.
Now you calculate the density of water GAS in equilibrium above liquid at 90 C. Congrats, that IS the vapor pressure. It's a special gas partial pressure from another phase.
Go ahead-- show us your math. Or look it up in a table. Any vapor pressure implies a DENSITY (moles per volume) which then can easily be converted to a molecular weight. Use some vapor pressures (I don't care where you get them) to calculate some molecular weights. Are these the weights of gas molecules or something larger? Show me your numbers.
Can you find any direct quotes from the literature that supports any of this?
https://www.chemguide.co.uk/physical/phaseeqia/phasediags.html
Relevance?
Try reading it.
The fact that you won't quote it tells me all I need to know about it.
This is all quote:


"Note: This is now a good point for a quick comment about the use of the words "gas" and "vapour". To a large extent you just use the term which feels right. You don't usually talk about "ethanol gas", although you would say "ethanol vapour". Equally, you wouldn't talk about oxygen as being a vapour - you always call it a gas.

There are various guide-lines that you can use if you want to. For example, if the substance is commonly a liquid at or around room temperature, you tend to call what comes away from it a vapour. A slightly wider use would be to call it a vapour if the substance is below its critical point, and a gas if it is above it. Certainly it would be unusual to call anything a vapour if it was above its critical point at room temperature - oxygen or nitrogen or hydrogen, for example. These would all be described as gases.

This is absolutely NOT something that is at all worth getting worked up about!"


End of quote. Unless you are J. McGuinn

The meaning of lines in phase diagrams:



"Note: I don't want to make any very big deal over this, but this line is actually exactly the same as the graph for the effect of temperature on the saturated vapour pressure of the liquid. Saturated vapour pressure is dealt with on a separate page. A liquid will boil when its saturated vapour pressure is equal to the external pressure.

Suppose you measured the saturated vapour pressure of a liquid at 50°C, and it turned out to be 75 kPa. You could plot that as one point on a vapour pressure curve, and then go on to measure other saturated vapour pressures at different temperatures and plot those as well.

Now, suppose that you had the liquid exposed to a total external pressure of 75 kPa, and gradually increased the temperature. The liquid would boil when its saturated vapour pressure became equal to the external pressure - in this case at 50°C. If you have the complete vapour pressure curve, you could equally well find the boiling point corresponding to any other external pressure.

That means that the plot of saturated vapour pressure against temperature is exactly the same as the curve relating boiling point and external pressure - they are just two ways of looking at the same thing.

If all you are interested in doing is interpreting one of these phase diagrams, you probably don't have to worry too much about this."
James McGinn
2018-04-04 04:26:26 UTC
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On Tuesday, April 3, 2018 at 8:39:22 PM UTC-7, Steve BH wrote:

All you are really doing here Steve is stringing words together in a creative manner to fit your imagination. None of this is going to add up to any kind of evidence of your magical cold steam.
Steve BH
2018-04-04 04:45:21 UTC
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Post by James McGinn
All you are really doing here Steve is stringing words together in a creative manner to fit your imagination. None of this is going to add up to any kind of evidence of your magical cold steam.
McGinn after seeing Hamlet:

"Ehh. Seemed like nothing but a bunch of old famous quotes strung together."
Sergio
2018-04-07 13:57:09 UTC
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All you are really doing here Steve is stringing words together in a creative manner to fit your imagination. None of this is going to add up to any kind of evidence of your magical cold steam.
"Ehh. Seemed like nothing but a bunch of old famous quotes strung together."
some insite into how McGinn's brain works in his responce there.

McGinn seems not to be able to connect enough words together to form
full thoughts, and this indicates reading Aphasia, which is consistent
with all his postings.

McGinn is still looking for his "cold steam" too, [fog].
James McGinn
2018-04-07 16:55:42 UTC
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The fact that you won't quote it tells me all I need to know about it.
"Note: This is now a good point for a quick comment about the use of the words "gas" and "vapour". To a large extent you just use the term which feels right. You don't usually talk about "ethanol gas", although you would say "ethanol vapour". Equally, you wouldn't talk about oxygen as being a vapour - you always call it a gas.
Uh, okay. If you say so.
Post by Steve BH
There are various guide-lines that you can use if you want to. For example, if the substance is commonly a liquid at or around room temperature, you tend to call what comes away from it a vapour. A slightly wider use would be to call it a vapour if the substance is below its critical point, and a gas if it is above it. Certainly it would be unusual to call anything a vapour if it was above its critical point at room temperature - oxygen or nitrogen or hydrogen, for example. These would all be described as gases.
Uh, okay. If you say so.
Post by Steve BH
This is absolutely NOT something that is at all worth getting worked up about!"
End of quote. Unless you are J. McGuinn
"Note: I don't want to make any very big deal over this, but this line is actually exactly the same as the graph for the effect of temperature on the saturated vapour pressure
Uh, what dis?

of the liquid. Saturated vapour pressure is dealt with on a separate page. A liquid will boil when its saturated vapour pressure is equal to the external pressure.

Who say?
Post by Steve BH
Suppose you measured the saturated vapour pressure of a liquid at 50°C, and it turned out to be 75 kPa. You could plot that as one point on a vapour pressure curve, and then go on to measure other saturated vapour pressures at different temperatures and plot those as well.
Now, suppose that you had the liquid exposed to a total external pressure of 75 kPa, and gradually increased the temperature. The liquid would boil when its saturated vapour pressure became equal to the external pressure - in this case at 50°C. If you have the complete vapour pressure curve, you could equally well find the boiling point corresponding to any other external pressure.
That means that the plot of saturated vapour pressure against temperature is exactly the same as the curve relating boiling point and external pressure - they are just two ways of looking at the same thing.
If all you are interested in doing is interpreting one of these phase diagrams, you probably don't have to worry too much about this."
What? Me worry?
p***@gmail.com
2018-04-07 18:45:12 UTC
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What? Me worry?
So, are you now claiming to be Alfred E Neuman?

https://tinyurl.com/y8x45xyh
Steve BH
2018-04-07 23:58:35 UTC
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The fact that you won't quote it tells me all I need to know about it.
"Note: This is now a good point for a quick comment about the use of the words "gas" and "vapour". To a large extent you just use the term which feels right. You don't usually talk about "ethanol gas", although you would say "ethanol vapour". Equally, you wouldn't talk about oxygen as being a vapour - you always call it a gas.
Uh, okay. If you say so.
Post by Steve BH
There are various guide-lines that you can use if you want to. For example, if the substance is commonly a liquid at or around room temperature, you tend to call what comes away from it a vapour. A slightly wider use would be to call it a vapour if the substance is below its critical point, and a gas if it is above it. Certainly it would be unusual to call anything a vapour if it was above its critical point at room temperature - oxygen or nitrogen or hydrogen, for example. These would all be described as gases.
Uh, okay. If you say so.
Post by Steve BH
This is absolutely NOT something that is at all worth getting worked up about!"
End of quote. Unless you are J. McGuinn
"Note: I don't want to make any very big deal over this, but this line is actually exactly the same as the graph for the effect of temperature on the saturated vapour pressure
Uh, what dis?
of the liquid. Saturated vapour pressure is dealt with on a separate page. A liquid will boil when its saturated vapour pressure is equal to the external pressure.
Who say?
Post by Steve BH
Suppose you measured the saturated vapour pressure of a liquid at 50°C, and it turned out to be 75 kPa. You could plot that as one point on a vapour pressure curve, and then go on to measure other saturated vapour pressures at different temperatures and plot those as well.
Now, suppose that you had the liquid exposed to a total external pressure of 75 kPa, and gradually increased the temperature. The liquid would boil when its saturated vapour pressure became equal to the external pressure - in this case at 50°C. If you have the complete vapour pressure curve, you could equally well find the boiling point corresponding to any other external pressure.
That means that the plot of saturated vapour pressure against temperature is exactly the same as the curve relating boiling point and external pressure - they are just two ways of looking at the same thing.
If all you are interested in doing is interpreting one of these phase diagrams, you probably don't have to worry too much about this."
What? Me worry?
Read this carefully, Alfred boy. You might actually learn something.

Chemguide is the main on line support for Cambridge university chemistry science students.

https://www.chemguide.co.uk/index.html#top

You want physical chemistry:

https://www.chemguide.co.uk/physmenu.html#top

Choose "phase equilibria."

https://www.chemguide.co.uk/physical/phaseeqiamenu.html#top

Right up top is "An introduction to saturated vapor pressure."

Maybe you'd like to read it?

https://www.chemguide.co.uk/physical/phaseeqia/vapourpress.html#top

Quote:

"In this equilibrium, there will be a fixed number of the gaseous particles in the space above the liquid.

When these particles hit the walls of the container, they exert a pressure. This pressure is called the saturated vapour pressure (also known as saturation vapour pressure) of the liquid."


Gaseous particles exert saturation vapour pressure in these circumstances. Can you say "gaseous particles"?

Then the article "phase diagrams for pure substances" is under that:

https://www.chemguide.co.uk/physical/phaseeqia/phasediags.html#top

Oh, but we've been here already. This is the one where you said "Says who?"

Say the professors at Cambridge.
James McGinn
2018-04-08 01:36:22 UTC
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Post by Steve BH
Chemguide is the main on line support for Cambridge university chemistry science students.
https://www.chemguide.co.uk/index.html#top
https://www.chemguide.co.uk/physmenu.html#top
Choose "phase equilibria."
https://www.chemguide.co.uk/physical/phaseeqiamenu.html#top
Right up top is "An introduction to saturated vapor pressure."
Maybe you'd like to read it?
https://www.chemguide.co.uk/physical/phaseeqia/vapourpress.html#top
"In this equilibrium, there will be a fixed number of the gaseous particles in the space above the liquid
Steve, if you weren't a fucking retard I wouldn't have to explain to you that there is nothing on this website that indicates how or if they empirically determined that the particles were genuinely gaseous or just vapor?

So why the fuck are you referring to this POS website since only if they had done so would this website be of any value in this discussion.

We've aleady determined that there are no shortage of retards on this planet that can't distinguish between gaseous H2O and vaporous H2O. So the fact that you've found one more doesn't change anything.
Post by Steve BH
Say the professors at Cambridge.
Who cares?
Steve BH
2018-04-08 03:27:41 UTC
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Post by Steve BH
Chemguide is the main on line support for Cambridge university chemistry science students.
https://www.chemguide.co.uk/index.html#top
https://www.chemguide.co.uk/physmenu.html#top
Choose "phase equilibria."
https://www.chemguide.co.uk/physical/phaseeqiamenu.html#top
Right up top is "An introduction to saturated vapor pressure."
Maybe you'd like to read it?
https://www.chemguide.co.uk/physical/phaseeqia/vapourpress.html#top
"In this equilibrium, there will be a fixed number of the gaseous particles in the space above the liquid
Steve, if you weren't a fucking retard I wouldn't have to explain to you that there is nothing on this website that indicates how or if they empirically determined that the particles were genuinely gaseous or just vapor?
So why the fuck are you referring to this POS website since only if they had done so would this website be of any value in this discussion.
We've aleady determined that there are no shortage of retards on this planet that can't distinguish between gaseous H2O and vaporous H2O. So the fact that you've found one more doesn't change anything.
Post by Steve BH
Say the professors at Cambridge.
Who cares?
Evidently you must, since you asked.

Number of particles per volume has been explained to you. Is it calculated from pressure, temperature, density.

In an ideal gas the number of mols n = PV/RT. Mass m = n[MW] where MW is molecular weight. And density rho = m/V.

So n = m/[MW] = PV/RT

m/V = rho = [MW]P/RT

MW = rho [RT/P]

Try this for air at O C: Temp = 0 C = 273 K. Pressure = 1 atm. R = 0.0821 l*atm/mol*K, rho = 1.3 g/L = kg/m^3.

MW = 1.3 [0.0821 * 273]/1 atm = 29.1 grams/mole.

That's not a bad estimate. If you try it for any pure sample of clear water gas, you'll have densities of ~0.6 g/L at 1 atm and temps of 373 K and come out with
gas molecular weights of about 18 g/mol.

Try it.
James McGinn
2018-04-08 05:03:15 UTC
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Post by Steve BH
Chemguide is the main on line support for Cambridge university chemistry science students.
https://www.chemguide.co.uk/index.html#top
https://www.chemguide.co.uk/physmenu.html#top
Choose "phase equilibria."
https://www.chemguide.co.uk/physical/phaseeqiamenu.html#top
Right up top is "An introduction to saturated vapor pressure."
Maybe you'd like to read it?
https://www.chemguide.co.uk/physical/phaseeqia/vapourpress.html#top
"In this equilibrium, there will be a fixed number of the gaseous particles in the space above the liquid
Steve, if you weren't a fucking retard I wouldn't have to explain to you that there is nothing on this website that indicates how or if they empirically determined that the particles were genuinely gaseous or just vapor?
So why the fuck are you referring to this POS website since only if they had done so would this website be of any value in this discussion.
We've aleady determined that there are no shortage of retards on this planet that can't distinguish between gaseous H2O and vaporous H2O. So the fact that you've found one more doesn't change anything.
Post by Steve BH
Say the professors at Cambridge.
Who cares?
Evidently you must, since you asked.
Number of particles per volume has been explained to you. Is it calculated from pressure, temperature, density.
In an ideal gas the number of mols n = PV/RT. Mass m = n[MW] where MW is molecular weight. And density rho = m/V.
So n = m/[MW] = PV/RT
m/V = rho = [MW]P/RT
MW = rho [RT/P]
Try this for air at O C: Temp = 0 C = 273 K. Pressure = 1 atm. R = 0.0821 l*atm/mol*K, rho = 1.3 g/L = kg/m^3.
MW = 1.3 [0.0821 * 273]/1 atm = 29.1 grams/mole.
That's not a bad estimate. If you try it for any pure sample of clear water gas, you'll have densities of ~0.6 g/L at 1 atm and temps of 373 K and come out with
gas molecular weights of about 18 g/mol.
Try it.
Explain its relevance you fucking retard.
Steve BH
2018-04-09 00:28:30 UTC
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Post by James McGinn
Post by Steve BH
Post by James McGinn
Post by Steve BH
Chemguide is the main on line support for Cambridge university chemistry science students.
https://www.chemguide.co.uk/index.html#top
https://www.chemguide.co.uk/physmenu.html#top
Choose "phase equilibria."
https://www.chemguide.co.uk/physical/phaseeqiamenu.html#top
Right up top is "An introduction to saturated vapor pressure."
Maybe you'd like to read it?
https://www.chemguide.co.uk/physical/phaseeqia/vapourpress.html#top
"In this equilibrium, there will be a fixed number of the gaseous particles in the space above the liquid
Steve, if you weren't a fucking retard I wouldn't have to explain to you that there is nothing on this website that indicates how or if they empirically determined that the particles were genuinely gaseous or just vapor?
So why the fuck are you referring to this POS website since only if they had done so would this website be of any value in this discussion.
We've aleady determined that there are no shortage of retards on this planet that can't distinguish between gaseous H2O and vaporous H2O. So the fact that you've found one more doesn't change anything.
Post by Steve BH
Say the professors at Cambridge.
Who cares?
Evidently you must, since you asked.
Number of particles per volume has been explained to you. Is it calculated from pressure, temperature, density.
In an ideal gas the number of mols n = PV/RT. Mass m = n[MW] where MW is molecular weight. And density rho = m/V.
So n = m/[MW] = PV/RT
m/V = rho = [MW]P/RT
MW = rho [RT/P]
Try this for air at O C: Temp = 0 C = 273 K. Pressure = 1 atm. R = 0.0821 l*atm/mol*K, rho = 1.3 g/L = kg/m^3.
MW = 1.3 [0.0821 * 273]/1 atm = 29.1 grams/mole.
That's not a bad estimate. If you try it for any pure sample of clear water gas, you'll have densities of ~0.6 g/L at 1 atm and temps of 373 K and come out with
gas molecular weights of about 18 g/mol.
Try it.
Explain its relevance you fucking retard.
The gas laws and the kinetic theory of gases require that all gases have about the same number of particles per volume at the same temperature and pressure. That's how I can tell you offhand that hydrogen is about 2/29 the density of air, and helium is 4/29 the density of air. The way we know the gas laws are true, is that the densities really are about as I have written them. You name a gas and I can tell you how dense it is with regard to air, without having to look anything up. Argon-40 is.... 40/29 as dense as air (same T and P). Don't take any argon balloons.

Water gas (steam) is 18/29 as dense as air at 101 C (meaning compared with air that has a temp of 101 C also).

It is also an **experimental fact** that pure water "vapor" at 99 C is 18/29 as dense as air at 99 C. Nothing has changed below the "boiling point." That means the "gas particles" at 99 C are just the same composition as they are at 101 C. Which is individual water molecules with an atomic weight of 18, giving the gas they form a density of 18 grams per molar volume (which at 100 C is 22.4(373/373) = 30.6 liters/mole).

You, Mr. Atmospheric physicist, should be able to rattle off all the above. The density of steam is given above, directly, no table needed. Instead I find you know none of it.
James McGinn
2018-04-09 04:36:15 UTC
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On Sunday, April 8, 2018 at 5:28:33 PM UTC-7, Steve BH wrote:

(the density of air, and helium is 4/29 the density of air. The way we know the gas laws are true, is that the densities really are about as I have written them. You name a gas and I can tell you how dense it is with regard to air, without having to look anything up. Argon-40 is.... 40/29 as dense as air (same T and

JMcG:
This is all common knowledge.
Face it, Steve. You really don't have a point. Be honest.


P). Don't take any argon balloons.
Post by Steve BH
Water gas (steam) is 18/29 as dense as air at 101 C (meaning compared with air that has a temp of 101 C also).
JMcG:
Irrelevant.
Post by Steve BH
It is also an **experimental fact** that pure water "vapor" at 99 C is 18/29 as dense as air at 99 C. Nothing has changed below the "boiling point." That means the "gas particles" at 99 C are just the same composition as they are at 101 C.
JMcG:
You know this how?

Which is individual water molecules with an atomic weight of 18, giving the gas they form a density of 18 grams per molar volume (which at 100 C is 22.4(373/373) = 30.6 liters/mole).


JMcG:
The difference is only 1%
Do the math:
No Steam in the Atmosphere; H2O Polarity is Variable

Post by Steve BH
You, Mr. Atmospheric physicist, should be able to rattle off all the above. The density of steam is given above, directly, no table needed. Instead I find you know none of it.
You got nothing!!!

James McGinn / Solving Tornadoes
Sergio
2018-04-03 20:51:04 UTC
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So, why do you think they distinguish between "vapor" pressure and gas pressure?
Often vapor pressures are used as terms for the special partial pressures of gases that are in equilibrium with another phase, whereas in the generic term of "partial pressure" of a gas, it can be anything (no other phase or equilibration is implied). So the term "vapor" implies a history, and also another phase that drives it, as an equilibrium partial pressure from phase change.
Thus the partial pressure of water gas above liquid at 96.7 C is 0.9 atm. We call this special partial pressure a "vapor pressure" because it's in equilibrium with water and has (once been) liquid water. BUT IT IS A GAS NOW. Of course you can have partial pressure of H2O gas as anything LESS than this, without condensation. That's not properly a vapor pressure, but just a partial pressure.
So not all partial pressures are vapor pressures, but all vapor pressures are partial pressures. They are equilibrium partial pressures. One is a subset of the other. However, they are both taking about the pressure of gas molecules.
Now you calculate the density of water GAS in equilibrium above liquid at 90 C. Congrats, that IS the vapor pressure. It's a special gas partial pressure from another phase.
Go ahead-- show us your math. Or look it up in a table. Any vapor pressure implies a DENSITY (moles per volume) which then can easily be converted to a molecular weight. Use some vapor pressures (I don't care where you get them) to calculate some molecular weights. Are these the weights of gas molecules or something larger? Show me your numbers.
Can you find any direct quotes from the literature that supports any of this?
https://www.chemguide.co.uk/physical/phaseeqia/phasediags.html
Relevance?
Try reading it.
from the article;

"So, again, what is the significance of this line separating the two areas?

Anywhere along this line, there will be an equilibrium between the
liquid and the vapour."


Perhaps McGinn will figure it out after a year or two.
James McGinn
2018-04-03 21:28:14 UTC
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Post by Steve BH
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Post by Steve BH
Post by James McGinn
Post by Steve BH
Post by James McGinn
So, why do you think they distinguish between "vapor" pressure and gas pressure?
Often vapor pressures are used as terms for the special partial pressures of gases that are in equilibrium with another phase, whereas in the generic term of "partial pressure" of a gas, it can be anything (no other phase or equilibration is implied). So the term "vapor" implies a history, and also another phase that drives it, as an equilibrium partial pressure from phase change.
Thus the partial pressure of water gas above liquid at 96.7 C is 0.9 atm. We call this special partial pressure a "vapor pressure" because it's in equilibrium with water and has (once been) liquid water. BUT IT IS A GAS NOW. Of course you can have partial pressure of H2O gas as anything LESS than this, without condensation. That's not properly a vapor pressure, but just a partial pressure.
So not all partial pressures are vapor pressures, but all vapor pressures are partial pressures. They are equilibrium partial pressures. One is a subset of the other. However, they are both taking about the pressure of gas molecules.
Now you calculate the density of water GAS in equilibrium above liquid at 90 C. Congrats, that IS the vapor pressure. It's a special gas partial pressure from another phase.
Go ahead-- show us your math. Or look it up in a table. Any vapor pressure implies a DENSITY (moles per volume) which then can easily be converted to a molecular weight. Use some vapor pressures (I don't care where you get them) to calculate some molecular weights. Are these the weights of gas molecules or something larger? Show me your numbers.
Can you find any direct quotes from the literature that supports any of this?
https://www.chemguide.co.uk/physical/phaseeqia/phasediags.html
Relevance?
Try reading it.
from the article;
"So, again, what is the significance of this line separating the two areas?
Anywhere along this line, there will be an equilibrium between the
liquid and the vapour."
Perhaps McGinn will figure it out after a year or two.
LOL. And in your mind this sentence is evidence that cold steam exists in the atmosphere? Surreal.

Retards should avoid science.
James McGinn
2018-04-09 04:48:50 UTC
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Post by James McGinn
Post by Steve BH
Post by James McGinn
Post by Steve BH
Post by James McGinn
So, why do you think they distinguish between "vapor" pressure and gas pressure?
Often vapor pressures are used as terms for the special partial pressures of gases that are in equilibrium with another phase, whereas in the generic term of "partial pressure" of a gas, it can be anything (no other phase or equilibration is implied). So the term "vapor" implies a history, and also another phase that drives it, as an equilibrium partial pressure from phase change.
Thus the partial pressure of water gas above liquid at 96.7 C is 0.9 atm. We call this special partial pressure a "vapor pressure" because it's in equilibrium with water and has (once been) liquid water. BUT IT IS A GAS NOW. Of course you can have partial pressure of H2O gas as anything LESS than this, without condensation. That's not properly a vapor pressure, but just a partial pressure.
So not all partial pressures are vapor pressures, but all vapor pressures are partial pressures. They are equilibrium partial pressures. One is a subset of the other. However, they are both taking about the pressure of gas molecules.
Now you calculate the density of water GAS in equilibrium above liquid at 90 C. Congrats, that IS the vapor pressure. It's a special gas partial pressure from another phase.
Go ahead-- show us your math. Or look it up in a table. Any vapor pressure implies a DENSITY (moles per volume) which then can easily be converted to a molecular weight. Use some vapor pressures (I don't care where you get them) to calculate some molecular weights. Are these the weights of gas molecules or something larger? Show me your numbers.
Can you find any direct quotes from the literature that supports any of this?
https://www.chemguide.co.uk/physical/phaseeqia/phasediags.html
Relevance?
Try reading it.
from the article;
"So, again, what is the significance of this line separating the two areas?
Anywhere along this line, there will be an equilibrium between the
liquid and the vapour."
So, you retards believe that this sentence proves the existence of "cold steam?"

So, let me get this straight, your "evidence" is a sentence that refers to a line on the H2O phase diagram?

It's too bad you couldn't provide qny empirical evidence, huh.

Lofty Goat
2018-04-04 01:06:45 UTC
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Post by James McGinn
... https://www.chemguide.co.uk/physical/phaseeqia/phasediags.html
Relevance?
Gee, Jim. It's a diagram, constructed from actual measurements, which
demonstrates how thoroughly lost in La-La Land you are. Relevant.
--
Goat
James McGinn
2018-04-04 01:12:45 UTC
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Post by Lofty Goat
Post by James McGinn
... https://www.chemguide.co.uk/physical/phaseeqia/phasediags.html
Relevance?
Gee, Jim. It's a diagram, constructed from actual measurements, which
demonstrates how thoroughly lost in La-La Land you are. Relevant.
--
LOL. You got nothing!!!
James McGinn
2018-04-03 04:37:52 UTC
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Post by Steve BH
Post by James McGinn
Post by Steve BH
Post by James McGinn
So, why do you think they distinguish between "vapor" pressure and gas pressure?
Often vapor pressures are used as terms for the special partial pressures of gases that are in equilibrium with another phase, whereas in the generic term of "partial pressure" of a gas, it can be anything (no other phase or equilibration is implied). So the term "vapor" implies a history, and also another phase that drives it, as an equilibrium partial pressure from phase change.
Thus the partial pressure of water gas above liquid at 96.7 C is 0.9 atm. We call this special partial pressure a "vapor pressure" because it's in equilibrium with water and has (once been) liquid water. BUT IT IS A GAS NOW. Of course you can have partial pressure of H2O gas as anything LESS than this, without condensation. That's not properly a vapor pressure, but just a partial pressure.
So not all partial pressures are vapor pressures, but all vapor pressures are partial pressures. They are equilibrium partial pressures. One is a subset of the other. However, they are both taking about the pressure of gas molecules.
Now you calculate the density of water GAS in equilibrium above liquid at 90 C. Congrats, that IS the vapor pressure. It's a special gas partial pressure from another phase.
Go ahead-- show us your math. Or look it up in a table. Any vapor pressure implies a DENSITY (moles per volume) which then can easily be converted to a molecular weight. Use some vapor pressures (I don't care where you get them) to calculate some molecular weights. Are these the weights of gas molecules or something larger? Show me your numbers.
Can you find any direct quotes from the literature that supports any of this?
https://www.chemguide.co.uk/physical/phaseeqia/phasediags.html
You got nothing!!!
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