Understanding Van Der Waals Critical Points

In summary, the difference between a gas and a liquid is one of density, and the critical point in the Van-der-waals equation marks the temperature at which there is a discontinuous change in density between the gas and liquid states. This is due to the balance of attraction and repulsion forces between molecules, and as temperature increases, thermal fluctuations weaken this balance, resulting in a disappearance of the phase transition at higher temperatures.
  • #1
TMSxPhyFor
53
0
Hi

I was studding Van-der-waals equation for real gases when I got in my mind two questions that i couldn't get a sufficient answer for from my professors, hope you have a better explanations :

1- Why there is critical point?
It was shocking for me to know that there is a particular temperature above which we can't convert gases to liquid, I always thought that what ever the gas temperature is, we always can apply a sufficient pressure on the gas to get liquid, but according to van-der-waals that wrong, and I can't get the physical reason beyond that, why at that point we can't force the connections between molecules get stronger to get liquid, anyway i get answer such as "because the density of the gas becomes as much as the liquid but it's will remains gas" of course it's not answering me, I want to know physically why it remains gas? maybe I should dig here into quantum physics to get answer? or I'm getting something wrong?

2- when we converting gas from gas to liquid, why we should reduce gas volume more when gas temperature is lower?
I mean if we have to isotherms, to convert the gas to liquid, we need to reduce ([tex]\Delta V[/tex]) the volume of the gas to accomplish the conversation quite more if gas temperature is lower! somehow it's controversy with my intuition because at lower temperature, molecules energy is less, so it should be easier to push molecules to new state of mater!
For this one I got the answer: because cool liquid has a smaller volume! somehow it's logical but I think that it's volume will reduce very little comparing to the needed decrease that shows van-der-waals eq. to change the state of matter.

Sorry for the lengthen and thanks in advance for everyone.
 
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  • #2
Firstly VdW is not an exactly equation. It is, however, quite a good model. You should look over your course notes and learn the subtleties. They matter.

As to your questions, I'll take the first one, and leave the 2nd for someone else:

What is the difference between a gas and a liquid? A solid is very obviously different --- you get a neat crystalline ordering of atoms, which breaks the translational symmetry of space. There is no such changes in symmetry between gas and liquid. The only difference is one of density. As such, it's really more remarkable that there *is* a phase transition, as opposed to when there is not.

A phase transition is simply when you have a discontinuous change of some macroscopic parameter. In this case, liquid to gas is a change in density. As you alter the temperature (or pressure) very slightly, there is a jump in density. The reason that this happens is the same as the fact that when you apply a magnetic field to a bar magnet, it causes the bar magnet to align in magnetisation with the external field; and when you reverse the external field, as it goes through zero*, the magnetisation changes discontinuously to be pointing the other way too --- it is simply unfavourable for any of the spins in a magnet to be pointing in a contrary way to the majority.

As temperature is raised, thermal fluctuations become stronger, and that discontinuous jump (in density or magnetisation) gets smaller. At the critical temperature, the jump is zero. Beyond the critical temperature, there is no net magnetisation because the spins are thermally agitated so much that they don't really care what their neighbours are doing.

* Footnote: the magnet example only really applies to a single domain at a time. As you may know, real bar magnets exhibit hysteresis. This is actually due to domain wall motion, and not any intrinsic "stickiness" of spins. In a real bar magnet, the majority domains grow in size at the expense of minority domains, but it costs energy to move the boundaries; the energy needed is the area in the hysteresis loop.
 
  • #3
Dear genneth

Thank you for your replay, actually sorry if I expressed my self wrong as I'm not studding in English language, anyway considering VdW as equation or a model doesn't change anything in my question.

What is the difference between Liquid & Gas? very very good question, because actually I don't know and I found no strict definition, saying that the difference is in "density" (as you do and wikipedia) is very shallow as VdW model it's selfs saying that we can get a gas with higher density than liquid but it still gas! also I found no "mathematical" definition for liquid! VdW doesn't saying that before mixed area of matter state (in VdW isotherms) it will be gas and after it it will be liquid (in the sense of mathematics)!

In my opinion the only possible mathematical definition for liquid is to say that the attraction forces between molecules are higher than repulsion forces, which is totally opposite in gases, put it still confuses me to know "how much" those forces should be stronger to consider a matter phase as a liquid?

Also I can't see where the "jump" in the density you talking about, because as you know phase transition is not happening suddenly, it takes time and going quite smoothly according to VdW it self! unless I'm missing something, ok it's going faster and faster with increasing temperature (which is also paradoxical to me and my second question is about this point), but I still don't understand the physical (on molecular level) the reason behind disappearing phase transition in higher degrees!
 
  • #4
TMSxPhyFor said:
Hi

I was studding Van-der-waals equation for real gases when I got in my mind two questions that i couldn't get a sufficient answer for from my professors, hope you have a better explanations :

1- Why there is critical point?
It was shocking for me to know that there is a particular temperature above which we can't convert gases to liquid, I always thought that what ever the gas temperature is, we always can apply a sufficient pressure on the gas to get liquid, but according to van-der-waals that wrong, and I can't get the physical reason beyond that, why at that point we can't force the connections between molecules get stronger to get liquid, anyway i get answer such as "because the density of the gas becomes as much as the liquid but it's will remains gas" of course it's not answering me, I want to know physically why it remains gas? maybe I should dig here into quantum physics to get answer? or I'm getting something wrong?

2- when we converting gas from gas to liquid, why we should reduce gas volume more when gas temperature is lower?
I mean if we have to isotherms, to convert the gas to liquid, we need to reduce ([tex]\Delta V[/tex]) the volume of the gas to accomplish the conversation quite more if gas temperature is lower! somehow it's controversy with my intuition because at lower temperature, molecules energy is less, so it should be easier to push molecules to new state of mater!
For this one I got the answer: because cool liquid has a smaller volume! somehow it's logical but I think that it's volume will reduce very little comparing to the needed decrease that shows van-der-waals eq. to change the state of matter.

Sorry for the lengthen and thanks in advance for everyone.

To liquify a gas it must be compressed and at a temperature lower than its critical temperature.In the liquid state the molecules move fairly randomly and they get captured by and escape from the intermolecular forces.In the gas state the molecules are more energetic and are captured less.Compressing a gas above its critical temperature will result in a just a dense gas,its molecules moving too fast to become a liquid.

During liquification the gas will be present with its liquid.The gas is then known as a saturated vapour and the pressure of a saturated vapour is independant of the relative volumes of liquid and vapour present
 
  • #5
Dadface said:
Compressing a gas above its critical temperature will result in a just a dense gas,its molecules moving too fast to become a liquid.

This what i want to know : how came that whatever the applied pressure above critical temperature will be unable to overcome this fast moving molecules, while it's quite possible to be done below critical point? sorry but you just rephrased my question in your post!

Dadface said:
pressure of a saturated vapour is independant of the relative volumes of liquid and vapour present
But we are not speaking here about relative volumes! it's volume of our box or any closed tube, we don't care about there relative volumes at all.
 
  • #6
TMSxPhyFor said:
This what i want to know : how came that whatever the applied pressure above critical temperature will be unable to overcome this fast moving molecules, while it's quite possible to be done below critical point? sorry but you just rephrased my question in your post!


But we are not speaking here about relative volumes! it's volume of our box or any closed tube, we don't care about there relative volumes at all.

1.In a gas the molecules are more energetic and much faster than they are in a liquid.To change a gas to a liquid we have to reduce the speed of the molecules by removing energy.Compressing a gas pushes the molecules closer together but it does not remove energy.On the contrary it adds energy.
 
  • #7
Dadface said:
1.In a gas the molecules are more energetic and much faster than they are in a liquid.To change a gas to a liquid we have to reduce the speed of the molecules by removing energy.Compressing a gas pushes the molecules closer together but it does not remove energy.On the contrary it adds energy.

Thx Dadface, now we getting closer :-), now I just to need to understand why when we applying pressure below critical point, gas became liquid while in this situation we also adding by that energy to our molecules as you mentioned? or it's now became a quantitative question ?
 
  • #8
Maybe this analogy will help:
Imagine the bond between two molecules (be it a hydrogen bond or vdW bond or whatever) as a potential well, a bucket. Put some ping-pong balls in the bucket. Now shake it. If you shake it lightly (low temperature) they'll all stay in - a solid. If you shake it a bit more (moderate temperature) a few will fly out - that's a liquid. Shake it a lot and no ball will stay in for any longer amount of time - that's a gas.

If you shake it violently enough, have a sufficiently high temperature, then any ball that falls in will pretty much fly out instantly. Even if you make the area outside so small they can't help but be inside the bucket (high pressure), they're still bouncing up above the rim. That's your critical temperature!
 
  • #9
I was very serious about the analogy with ferromagnets --- in terms of critical behaviour, the two transitions are identical.

Answer lies in collective behaviour/interactions. Below the critical temperature, there are two densities which are local minima in terms of energy --- corresponding to liquid and gas, respectively. In other words, any change of density away from those would cost energy. As criticality is approached, those minima get shallower and shallower, until they are no longer minima.

For a visual picture, draw yourself the function y = x^4 - a x^2, where a is some constant. Notice that for a > 0, there are two minima. For a < 0, there is only one. At a = 0, you have the critical point.
 
  • #10
If you shake it violently enough, have a sufficiently high temperature, then any ball that falls in will pretty much fly out instantly

I understand from your example that we are dealing here with kind of Energy hole ? is that related some how to Quantum properties of the molecules as I thought from the beginning ?

Answer lies in collective behaviour/interactions. Below the critical temperature, there are two densities which are local minima in terms of energy --- corresponding to liquid and gas, respectively. In other words, any change of density away from those would cost energy. As criticality is approached, those minima get shallower and shallower, until they are no longer minima.

I'm really got your Idia guys, but I'm interested in the reason at it's roots behind having two minima! or to know the reason for :

because the spins are thermally agitated so much that they don't really care what their neighbours are doing
 
  • #11
TMSxPhyFor said:
I understand from your example that we are dealing here with kind of Energy hole ? is that related some how to Quantum properties of the molecules as I thought from the beginning ?

Yes and no.
Yes, because forces between molecules are all dependent on electronic energy, and are thus ultimately quantum-mechanical in nature.
No, because if the molecules are all in their electronic ground state, you have (to a good approximation) a fixed potential-energy curve for the interaction between the molecules, (e.g. a http://en.wikipedia.org/wiki/Lennard-Jones_potential" [Broken]) and so you can treat them as classical particles with this potential-energy function. (which is done all the time to simulate gases and liquids in molecular-dynamics simulations)

In other words, the origin of the potential is quantum-mechanical in nature, but it doesn't need to be - as long as you've got a potential well of some sort, you'll get this behavior purely classically.
 
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  • #12
Thx alxm, now it's became much clearer especially after i read Lennard-Jones potential wiki article.

Now left the second question : Why phase transitions going faster and faster (in the sense of volume increment) with higher temperatures.
 
  • #13
ince the kinetic energy of a molecule is proportional to its temperature, evaporation proceeds more quickly at higher temperatures. As the faster-moving molecules escape, the remaining molecules have lower average kinetic energy, and the temperature of the liquid thus decreases. http://www.farauctions.com" [Broken]
 
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1. What is a Van Der Waals Critical Point?

A Van Der Waals Critical Point is a specific point on a phase diagram that represents the temperature and pressure at which a substance transitions from a liquid to a gas. At this point, the liquid and gas phases have the same density, making it difficult to distinguish between the two.

2. How is a Van Der Waals Critical Point different from a boiling point?

A Van Der Waals Critical Point is different from a boiling point in that it occurs at a specific temperature and pressure, whereas the boiling point can vary depending on external factors such as atmospheric pressure. Additionally, at the critical point, the liquid and gas phases are indistinguishable, whereas at the boiling point, the liquid and gas phases are clearly separated.

3. Why is the Van Der Waals Critical Point important in materials science?

The Van Der Waals Critical Point is important in materials science because it helps to determine the physical properties of substances, such as their melting and boiling points. It also plays a role in understanding the behavior of substances under extreme conditions, such as high pressures and temperatures.

4. How is the Van Der Waals Critical Point measured?

The Van Der Waals Critical Point can be measured experimentally by plotting a phase diagram for a substance and determining the temperature and pressure at which the liquid and gas phases have the same density. This can also be calculated using equations derived from the Van Der Waals equation of state.

5. Can the Van Der Waals Critical Point be manipulated?

No, the Van Der Waals Critical Point is a fundamental property of a substance and cannot be manipulated. However, by understanding and controlling the factors that affect the critical point, such as temperature and pressure, scientists can manipulate the physical properties of a substance, such as its melting and boiling points.

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