Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: Big-Daddy on May 18, 2013, 10:43:32 AM
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Is it correct to say, for any ideal gas, that
[tex]P_{gas} = M_{gas} \cdot RT \\[/tex]
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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I am curious
Did you derive this yourself using
PV=nRT
where
n=m/M
http://en.wikipedia.org/wiki/Ideal_gas_law
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I am curious
Did you derive this yourself using
PV=nRT
where
n=m/M
http://en.wikipedia.org/wiki/Ideal_gas_law
As I explained, M is not Mr in my expression but rather molarity (molar concentration). I derived the expression from PV=nRT, n=MgasV. As for the activity coefficient thing, I read that somewhere and it seems to make sense.
Is this correct?
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As I explained, M is not Mr in my expression but rather molarity (molar concentration).
The needless perils of non standard notation. Most of us use M for Mol Wt and c for molar concentration.
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As I explained, M is not Mr in my expression but rather molarity (molar concentration). I derived the expression from PV=nRT, n=MgasV. As for the activity coefficient thing, I read that somewhere and it seems to make sense.
Is this correct?
I don't think I've seen activity coefficients used for a gas. I may be wrong.
Fugacity / Fugacity coefficient seems more like what you'd want.
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Fugacity / Fugacity coefficient seems more like what you'd want.
I've seen a few websites use γ, but ok the overall point is the same. It's a constant with its own separate method of determination, usually temperature-dependent, such that PA=γRTcA (using the notation you want.)
The needless perils of non standard notation. Most of us use M for Mol Wt and c for molar concentration.
In general I could sympathize. I will use c generally in the future. But as you told me recently the most important thing is to define your units and be quantitative. Given that I wrote Mgas is the gas' molar concentration, directly under my formula, I don't think there was any ambiguity.
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I was confused thanks for the clarification
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Fugacity / Fugacity coefficient seems more like what you'd want.
I've seen a few websites use γ, but ok the overall point is the same.
Which sites? Can you provide links?
I am curious to see how.
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Which sites? Can you provide links?
http://www.life.illinois.edu/crofts/bioph354/thermo_eq.htm
They do not really explain what the activity coefficient γ is or how it's calculated, so it's quite possible that this is similar to what you meant by fugacity coefficient.
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Is the principle correct? That, for a real gas, we are trying to work out a constant which mainly is dependent on temperature, and then we would write
[tex]P_{gas}=c_{gas} \cdot k(t) \cdot RT
Where k(t) is the constant, which is the function of temperature we need. I suppose this k(t) would be a fugacity coefficient, or could be something else, I'm not sure.
This then modifies the ideal gas conversion to a real gas conversion. k(t) could be a polynomial, then the more terms we add the more accurate the conversion will be (though more terms means more constants are needed too).
Is that right?
Edit: As an aside - and I'm expecting a 1-word answer really - is it ever possible for anything besides a gas (i.e. a liquid or solution) to contribute at all towards pressure in a system, or to have partial pressure? (If by "pressure" we mean what we normally mean when we talk about pressure.) I'm counting things like the vapour released from the surface as gases.
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Edit: As an aside - and I'm expecting a 1-word answer really - is it ever possible for anything besides a gas (i.e. a liquid or solution) to contribute at all towards pressure in a system,
Yes. Liquids exert pressure too. Try Scuba Diving.
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Liquids exert pressure too.
That's a different kind of pressure, isn't it? Not the sort of thing we would add to the pressure exerted by the gases. If we had a container with both gases and liquids inside it, do the liquids contribute at all to what we call "the pressure" inside the container? (Besides just decreasing the volume available to the gases) Or does "the pressure" usually refer to just the gases, when gases are present, with no contribution from anything in any other state?
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That's a different kind of pressure, isn't it? Not the sort of thing we would add to the pressure exerted by the gases. If we had a container with both gases and liquids inside it, do the liquids contribute at all to what we call "the pressure" inside the container? (Besides just decreasing the volume available to the gases) Or does "the pressure" usually refer to just the gases, when gases are present, with no contribution from anything in any other state?
Well for one thing, liquids exist in equilibrium with vapor, which contributes to the gas pressure.*
Pressure is any force divided by the area over which it is applied. Liquids are condensed phases, and thus the intermolecular forces generally exceed the other forces involved in creating gas pressure. Liquids and gasses are both fluids, though - in some ways you can think of a liquid as just a very nonideal gas. However liquids still exert pressure on a container because of gravity, which tends to pull the fluid toward the earth. (This is why there is enormous pressure against the walls of a hydroelectric dam and also why there is incredible pressure at the bottom of the ocean.)
* Speaking of gas pressure, it is often said that the pressure is caused by collisions between moving gas molecules and the container walls. Google "what causes gas pressure" and you'll see 99% of answers refer to such collisions, including previous topics here. However this is actually a simplification of what causes the pressure of an ideal gas. Can you think of another explanation?
Hint: Remember pressure has to be caused by a force. What force causes the pressure in an ideal gas? If it helps, what "force" causes a gas to expand into any available space? (And yes the stock quotes are purposely placed.)
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Pressure is any force divided by the area over which it is applied. Liquids are condensed phases, and thus the intermolecular forces generally exceed the other forces involved in creating gas pressure. Liquids and gasses are both fluids, though - in some ways you can think of a liquid as just a very nonideal gas. However liquids still exert pressure on a container because of gravity, which tends to pull the fluid toward the earth. (This is why there is enormous pressure against the walls of a hydroelectric dam and also why there is incredible pressure at the bottom of the ocean.)
OK. So do we quantify this separately, e.g. as "liquid pressure"? If we have a container with both a liquid or mixture of liquids and gas(es), then do we have a separate "liquid pressure" and "gas pressure", the latter being what we normally refer to when in chemistry we say "pressure", where the gases do not at all affect the liquid pressure and the liquids do not affect the gas pressure? (Except for in the phase exchange equilibria, but even there, it's the gaseous molecules of vapor that affect the gas pressure, not the liquid, which plays no part.)
However this is actually a simplification of what causes the pressure of an ideal gas. Can you think of another explanation?
Hint: Remember pressure has to be caused by a force. What force causes the pressure in an ideal gas? If it helps, what "force" causes a gas to expand into any available space? (And yes the stock quotes are purposely placed.)
Not sure what you mean by force. Maybe the lack of attraction between the particles?
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OK. So do we quantify this separately, e.g. as "liquid pressure"?
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.
Not sure what you mean by force. Maybe the lack of attraction between the particles?
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?)
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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.)
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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?
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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]).
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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?
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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)?
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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.
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...
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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.)