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Topic: Partial pressure and concentration  (Read 9810 times)

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Offline Big-Daddy

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Re: Partial pressure and concentration
« Reply #15 on: May 22, 2013, 12:02:06 PM »
Well I would say so.  If you want to know the total force acting on something, you add up all the different forces (and their vector directions).  I would think pressure little different.

OK. So if we want to know exactly how much force is acting over each unit area of the container we have to add up the magnitudes of the pressure exerted on the container by the liquids ("liquid pressure") and by the gases ("gas pressure"), but liquid pressure is completely independent of gas pressure. (Except in that some liquid molecules become gaseous, according to the vapour equilibrium etc.)

Correct?

What I mean is, for there to be pressure on the walls of the container, there has to be a force that pushes against them.  What is the force in this case?  We can say that it is collisions between the moving gas particles and the container walls, but this begs the question - what is causing gas molecules to collide against the walls?  Why do gasses expand outward?

Maybe it helps to consider that for an ideal gas, the energy does not depend on the volume.  It only depends on the temperature.  When you take a certain volume of ideal gas and cram it into a smaller volume at the same temperature, the internal energy does not change.  Yet the pressure (average force) goes up.  Why?  This would almost seem to be contradictory - if the energy of the gas doesn't increase as the volume decrease, how can the pressure increase?  What factor is missing that causes this "force"?

(Note that the energy of a REAL gas IS dependent on volume.  If you compress a real gas, the internal energy DOES change.  WHY?)

Seems like something to do with the law that PV/T stays the same for ideal gases, but a conceptualization of that. The key is probably in that for real gases, energy is dependent on volume. This might suggest that the reason for the constant PV/T ratio in ideal gases is that the molecules do not collide at all with one another? (Or, better phrased, we do not have to deal with the number of collisions of the molecules with one another increasing as the gas is compressed, only consider the number of collisions of the molecules with the container they are being compressed in.)

Offline Corribus

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Re: Partial pressure and concentration
« Reply #16 on: May 22, 2013, 03:18:18 PM »
OK. So if we want to know exactly how much force is acting over each unit area of the container we have to add up the magnitudes of the pressure exerted on the container by the liquids ("liquid pressure") and by the gases ("gas pressure"), but liquid pressure is completely independent of gas pressure. (Except in that some liquid molecules become gaseous, according to the vapour equilibrium etc.)
Liquid pressure isn't completely independent of gas pressure.  If the gas pressure is sufficiently high, it will push down on the liquid beneath it.  Liquids aren't very compressable, and so this will also impact the pressure of the liquid on the external walls on the container.  (In essence, it makes the effective gravitational force stronger.) At least this would seem to be the case.

Seems like something to do with the law that PV/T stays the same for ideal gases, but a conceptualization of that. The key is probably in that for real gases, energy is dependent on volume. This might suggest that the reason for the constant PV/T ratio in ideal gases is that the molecules do not collide at all with one another? (Or, better phrased, we do not have to deal with the number of collisions of the molecules with one another increasing as the gas is compressed, only consider the number of collisions of the molecules with the container they are being compressed in.)
What happens to the entropy of an ideal cas when it is compressed to a smaller volume at constant temperature?
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline Big-Daddy

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Re: Partial pressure and concentration
« Reply #17 on: May 22, 2013, 04:32:53 PM »
Liquid pressure isn't completely independent of gas pressure.  If the gas pressure is sufficiently high, it will push down on the liquid beneath it.  Liquids aren't very compressable, and so this will also impact the pressure of the liquid on the external walls on the container.  (In essence, it makes the effective gravitational force stronger.) At least this would seem to be the case.

Hmm - so liquid pressure is affected by gas pressure (due to the gas pushing down on it), for very high gas pressures, but gas pressure is entirely independent of liquid pressure. And presumably the effect of high gas pressure on liquid pressure is rarely non-negligible. Correct?

What happens to the entropy of an ideal cas when it is compressed to a smaller volume at constant temperature?

It should decrease? ΔST=R·loge(V2/V1) so if V2 is smaller than V1, then ΔST is negative so the entropy has decreased (S°[V2]<S°[V1]).

Offline Corribus

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Re: Partial pressure and concentration
« Reply #18 on: May 22, 2013, 09:00:45 PM »
Hmm - so liquid pressure is affected by gas pressure (due to the gas pushing down on it), for very high gas pressures, but gas pressure is entirely independent of liquid pressure. And presumably the effect of high gas pressure on liquid pressure is rarely non-negligible. Correct?
Well, in the context of the gravitational force causing the liquid to push against the container, I'd say this is correct - since the gas is above the liquid, the liquid can't very well exert a gravitational force on it, can it?

It should decrease? ΔST=R·loge(V2/V1) so if V2 is smaller than V1, then ΔST is negative so the entropy has decreased (S°[V2]<S°[V1]).
Right.  And is that favorable or unfavorable?
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline Big-Daddy

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Re: Partial pressure and concentration
« Reply #19 on: May 23, 2013, 11:18:17 AM »
Right.  And is that favorable or unfavorable?

ΔS being negative should be unfavourable (as it will raise the value of ΔG). Ah, now I see. It's favourable in terms of Gibbs' energy to increase entropy by expanding. Not sure why this won't hold true for real gases?

Also, I had another thought. If you have a system in which there are several gases (e.g. A, B, C and D) then does each gas take up a certain volume and a certain partial pressure of its own? By extension, can we say that, within any set of the gases (e.g. A, B and C, or A, C and D, or B, C and D, any set we choose, including just 1 gas or all 4 if we want) the gas laws will apply (e.g. if the gases are ideal, P(total of A, C and D)*V(total of A, C and D)=n(total of A, C and D)*R*T, which could then be true for every set of the gases in the container that we might choose)?

Offline Corribus

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Re: Partial pressure and concentration
« Reply #20 on: May 23, 2013, 02:21:51 PM »
Quote
ΔS being negative should be unfavourable (as it will raise the value of ΔG). Ah, now I see. It's favourable in terms of Gibbs' energy to increase entropy by expanding. Not sure why this won't hold true for real gases?
Right, unfavorable.  Isothermal expansion (or diffusion) of an ideal gas is completely entropic.  Therefore the "force" that causes it is an entropic force - the container is prohibiting the gas from maximizing its entropy, i.e., going to a state of low potential energy.  This is what causes the pressure of the gas.

http://en.wikipedia.org/wiki/Entropic_force

Actually you will notice that the Gibb Free Energy concept includes both an energetic and entropic component, and so it should not be surprising now that there are both energetic and entropic types of forces.  Most people are familiar with energetic forces - such as the force which causes objects to fall (gravity) - but not so many are familiar with entropic forces.  However molecular diffusion, gas pressure and even the tendency of a stretched rubber band to retract itself are all caused, at least in part, by entropic forces.  (And it's not always just one or the other, obviously - the "force" which causes most chemical reactions are a combination of both energetic and entropic forces.

If you want to know more about this topic, you might find the following article interesting:

http://johncarlosbaez.wordpress.com/2012/02/01/entropic-forces/

This concept also holds for a real gas - expansion is also primarily entropic.  However it is not FULLY entropic.  The internal energy of a nonideal gas DOES change during an isothermal expansion.  This is because in a nonideal gas there are intermolecular "bondings" that become weaker (usually) as the volume increases.  In this case there will be a force that resists entropic expansion, and the magnitude of that force is related to the strength of the intermolecular bonds.  Because the pressure in this case is determined through a sum of both the entropic and energetic (enthalpic) forces, we can no longer say that the only force acting on a real, expanding gas is entropy.  Even so, the fact that real gasses also expand spontaneously at most temperatures tells us that the outward entropic force exceeds the "inward" enthalpic force in most instances.  As you cool the gas down, however, or compress it (reduce volume/increase pressure), the entropic force becomes weaker (or the enthalpic force becomes stronger), until such point as the enthalpic force dominates and the gas will no longer expand spontaneously.  This state where the intermolecular forces are greater than the entropic driving force for expansion you may recognize otherwise as a condensed phase, or liquid. :) The temperature at which this happens (for a given pressure) is one way to characterize the vaporization/condensation or sublimation/deposition point*.  Now if you go the other way and warm the gas up, eventually the gas molecules become so far apart that the enthalpic force is practically zero (at least, in comparison to the entropic force), and thus you approach the ideal gas limit, where expansion is completely dominated by entropy.

Anyway, I think this is a cool alternative way to think about the processes of vaporization and gas expansion.  Just some food for thought.

Quote
Also, I had another thought. If you have a system in which there are several gases (e.g. A, B, C and D) then does each gas take up a certain volume and a certain partial pressure of its own? By extension, can we say that, within any set of the gases (e.g. A, B and C, or A, C and D, or B, C and D, any set we choose, including just 1 gas or all 4 if we want) the gas laws will apply (e.g. if the gases are ideal, P(total of A, C and D)*V(total of A, C and D)=n(total of A, C and D)*R*T, which could then be true for every set of the gases in the container that we might choose)?
Assuming all the gasses behave ideally, this would be true.  We have had this discussion before, and recently.

http://www.chemicalforums.com/index.php?topic=67832.0

* I haven't actually tried this calculation to see how close you get to the real condensation point of a gas.  It would be interesting to try...
« Last Edit: May 23, 2013, 03:07:51 PM by Corribus »
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline Big-Daddy

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Re: Partial pressure and concentration
« Reply #21 on: June 02, 2013, 06:34:17 PM »
Thanks Corribus. That's an interesting way to look at the different states in particular.

I'd like to ask one more broad question here. I wrote in my OP that

[tex]P_{gas} = M_{gas} \cdot γRT \\[/tex]

For a non-ideal gas (i.e. any gas) where γ is the fugacity coefficient that corrects from ideality.

But Pgas is still a partial pressure term. Is that correct? If our gases are non-ideal, can we still work in terms of partial pressure, or do we need to use fugacity AKA "effective pressure" (as it is called on Wikipedia). i.e. is this more correct:

[tex]P_{gas} = M_{gas} \cdot γRT \\[/tex]

Or this:

[tex]Φ_{gas} = M_{gas} \cdot γRT \\[/tex]

(Φ is the fugacity of the gas) for non-ideal gas mixtures? In other words, is partial pressure a concept applying purely to ideal gases, which cannot be corrected for real behaviour.

Is it correct to say, for any ideal gas, that



As this seems to follow from the ideal gas equation.
(Pgas is the partial pressure of the gas, Mgas is the gas' molar concentration, R is the ideal gas constant, T is the absolute temperature)

Is the usual modification for non-ideality to rewrite this:

[tex]P_{gas} = M_{gas} \cdot γRT \\[/tex]

(γ is the activity coefficient.)

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