# Van der waal's equation and resulting curve

1. Nov 15, 2013

### Urmi Roy

So as per van der waal's equation, below critical temperature, for each pressure and temperature, we have only 3 possible volumes for each pressure-temperature pair.

My questions are as follows:
1. So in the usual vapor dome (considering a P-v surface), we have a straight horizontal line corresponding to each pressure and temperature, with vapor-liquid equilibrium...so is van der waals saying this isn't true?

2. In the vapor dome, we have only liquid-vapor equilibrium usually...so what does it mean to say that half-way along the line, we have meta-stable 'only' liquid states and for the other half meta-stable 'only' vapor states ?

Thanks!
-Urmi

2. Nov 15, 2013

### Staff: Mentor

The van der walls equation is only a mathematical approximation to the behavior of a real substance. All real substances have horizontal lines inside the dome.

3. Nov 16, 2013

### Rap

1. Yes, the simple van der Waals is saying it isn't true, but inside that vapor dome, the van der Waals equation is not realistic, not correct. To make it correct, you have to add the "Maxwell construction" which says that for each \/\ -shaped curve (i.e a specific temperature), you construct a straight horizontal line connecting two points on the extreme left and right parts of the curve such that the area of the lower part on the left equals the area of the upper part on the right. Now it will represent a thermodynamically possible situation. For a given volume and temperature, if the volume falls on the straight line portion, the correct pressure will be given by the straight line, and the system will consist of two phases, the liquid and the solid. Note that the two points on the left and right are the only points on the curve that have equal pressures AND equal chemical potentials. See https://en.wikipedia.org/wiki/Van_der_Waals_equation

4. Nov 16, 2013

### Urmi Roy

Chestermiller and Rap: but in reality these meta stable states can be achieved by controlling the process in the right way...

For example. I read that a super-heated liquid can be achieved by keeping it free from any vapor, and absolutely still....we just keep heating it, and it goes above its boiling point at that pressure.

Now referring to my diagram, which part of the curvy green line would the above mentioned process be at? However, along the curvy green line, even though we are on the van der waals isotherm, we reach a variety of pressures...

Also, in my diagram,is T' (along van der waal's isotherm)=Tsat (the temperature along red line)?

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5. Nov 16, 2013

### Rap

If you have two temperatures, you will have two isotherms, Tc and Th (c is colder, h is hotter). The hot isotherm will lie above the cold one. Assuming the Maxwell line (vapor pressure line) for Tc intersects the Th isotherm, it will do so BELOW the Th vapor pressure line. Call the point where the vapor pressure line intersects the isotherm on the left the "liquid point" for that isotherm, and where it intersects on the right the "gas point". The situation you describe will be on the isotherm, somewhat to the right of the liquid point.

T' is just the temperature of a particular isotherm. Along the vapor pressure line, it is Tsat corresponding to the given Psat.

6. Nov 16, 2013

### Urmi Roy

Ok here are some more direct questions which might help me understand what's going on:

1. What is the state of the substance that exists at the intersection of the curvy green (van der waals) isotherm and the red line (usual isotherm).

2. Does the concept of quality exist for the van der waals curve?

3. Why are the states on the curve bounding region b called super-saturated vapor?

(please see point 3, at the bottom of pg 77 (its just one small paragraph!) on:

7. Nov 17, 2013

### Rap

1. The "red line" corresponds to FDB on figure 3.7 in the above link. There are three intersection points, F, D, and B. In stable equilibrium (K to F, then horizontal line FDB, then B to A). At the F intersection point, you have total liquid, but if you increase the volume by any amount (holding T constant), evaporation will occur and vapor will be produced. At the B intersection, you have total vapor, but if you decrease the volume by any amount (again, T constant), condensation will occur and some liquid will be produced. At the D intersection point, you will have some liquid, some vapor.

2. I don't know what you mean by "quality"

3. Its not the states bounding region B, its only part of that region. The part of the van der Waals curve from point C to B on fig 3.7 is the super-saturated vapor. The curve from C to D to E is totally unrealistic. Maybe a better term for the CB portion of the curve would be super-cooled vapor. The system is total vapor and is in metastable equilibrium. The slightest disturbance will cause condensation and a drop in pressure - the point will drop from being on the CB portion of the curve to being on the FDB line, at the same volume and then there will be both liquid and gas. The system will then be in stable equilibrium.

8. Nov 17, 2013

### Staff: Mentor

As I said earlier, the Van der Waals equation is only an crude approximation to the behavior of a real gas, and gives unrealistic predictions of the equilibrium behavior under the dome. To talk about it making any kind of accurate predictions of metastable behavior is a pure pipe-dream. You may spend a lot of time trying to interpret the predictions of Van der Waals under the dome in terms of metastable behavior, but your deliberations will not be realistic in terms of matching actual behavior (except, perhaps, at the very edges of the dome).

9. Nov 17, 2013

### Rap

@Chestermiller - I have to admit I don't understand the metastable idea with regard to the van der Waals equation. The vdW equation can be derived by assuming hard spheres with an attractive potential. I don't understand what causes the metastable state to be temporarily stable. I have heard of supercooled very pure water in a container and when you tap the container, the water freezes to form ice, but again, I don't understand from a microscopic point of view what is going on. I don't know if the metastable states can be successfully modelled in the vdW equation, or whether it is just hand-waving, and needs a more detailed theory to actually describe the metastable states.

10. Nov 17, 2013

### Urmi Roy

The only reason I'm making such a fuss of this is that I have a h/w problem which has a metastable state and I need to use van der waals equation for it.

I and a friend did find a solution by mathematical manipulation of the first law and van der waals equation, but neither of us know what's actually going on!

I didn't write out the problem coz it would be cheating on my part, but after submission tomorrow, I can write out the statement, so that you can at least explain what is going on.

The description in your previous post was very good, Rap. From point 2, I also feel that the van der waals isotherm FEDCBA (in the link) is supposed to be at the same temperature as FBD (pure substance isotherm), coz otherwise your description of the metastable state wouldn't make sense.

Also, 'quality' of vapor is a term we use commonly in problem solving and refers to the mass fraction of vapor inside vapor dome...i,e quality =(mass of vapor/(mass of vapor +mass of liquid))

11. Nov 17, 2013

### Rap

By your definition of quality, then yes, it will apply for the vdW equation. Also, yes, the KFEDCBA curve (which includes metastable equilibrium states) is at the same temperature as the KFDBA curve (all stable equilibrium states).

12. Nov 18, 2013

### DrDu

It is important to understand that the vdW equation describes a homogeneous phase. It turns out that it is favourable in terms of free enthalpy for the system to be inhomogeneous in some ranges of p and T. Calculation of the volume of an inhomogeneous system is trivial when one is given the equation for the homogeneous phases as the volumes of the different phases are additive.

13. Nov 18, 2013

### Rap

@DrDu - Yes, but I would think that temperature and volume are the independent variables for a given vdW curve, so that Helmholtz free energy would be the thermodynamic potential of choice. Did you mean to say free energy?

14. Nov 18, 2013

### DrDu

Also a vdW gas is nothing but a thermodynamical system. In terms of phase transitions I think it is more convenient to argue in terms of intensive quantities T and p. Whether P or V is considered intependent variable is purely convention. Especially the 3 conditions for a phase equilibrium are:
$p_1=p_2$, $T_1=T_2$ and $\mu_1=\mu_2$.

15. Nov 18, 2013

### Urmi Roy

So here's the statement: (this is just with an aim to understand the situation and hence vdw better)

A piston-cylinder apparatus contains 1kg of water in a metastable state, P1=5e6 Pa, T1=260C.
In this effect, this is a supersaturated vapor, since T1<Tsat (Tsat at the above mentioned pressure for water is 263.2C) The water undergoes a spontaneous transition to metastable state, state 2. What is T2 and P2? (an there is another part which requires mathematical manipulation of vdw and first law).

I'm having the most trouble locating where state 1 is in the pure substance/vdw diagram...

Rap- The fact that T1<Tsat and that it is at supersaturated vapor doesn't seem to match your explanation (in post #7), even though I'm completely convinced by it.

Last edited: Nov 18, 2013
16. Nov 18, 2013

### Rap

State 1 is where the Maxwell line for the 263.2 vdW curve (at 5e6Pa) intersects the 260 vdW curve. I'm not clear on what the conditions of the transition are. It's clearly at neither constant pressure or temperature.

You have to know two thermodynamic parameters in order to know what the final state is, then you can find all the others. We can rule out temperature and pressure, since they are what we are looking for. Its not clear to me what those two parameters are. Maybe constant volume and constant internal energy? Also, spontaneous transitions go from metastable to stable, so that statement that it goes to another metastable state puzzles me. Also, I don't see where in post #7 I mentioned anything about T1<Tsat.

17. Dec 14, 2013

### Urmi Roy

Rap- In the solutions given by the TA (which might be wrong) he equates internal energy at state 1 to that in state 2. There is no justification given for it, however. How would I know which parameters to keep constant? Also, since 263.2 is the saturation temp, the solution says that the transition occurs at the constant pressure of 5e6 Pa..and temp changes from 260 to 263.2.
Also, you didn't say anything about T1<Tsat...I was trying to think of it by deducing things from what you did say!

18. Dec 14, 2013

### Rap

Give the question as it was given to you, maybe I can figure that out.

19. Dec 14, 2013

### Urmi Roy

Question copied from the paper:

A piston-cylinder apparatus contains 1kg of water in a metastable state, P1=5e6 Pa, T1=260C.
In this effect, this is a supersaturated vapor, since T1<Tsat (Tsat at the above mentioned pressure for water is 263.2C) The water undergoes a spontaneous transition to metastable state, state 2.
Part(A)What is T2 and P2?
Pasrt(B)- Using van der waals equation of state to describe metastable state, estimate the chane in volume v2-v1

given that Cp_0=0.046(1000/T), where T is temperature in Kelvin.

(The given Cp has a subscript 0, which I don't know why)

20. Dec 15, 2013

### Urmi Roy

I've attached a snapshot from the question paper...

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