Proof of second-order phase transition for a Van Der Waals Gas

In summary, Van der Waals gas describes a second order phase transition because it has a stability-criteria that does not hold in the green-shaded area, but also has a point where the first and second derivatives are equal in the critical point.
  • #1
CGandC
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Homework Statement
I was wondering how is it proved that Van Der Waals Gas is a second-order phase transition.
Relevant Equations
Van Der Waals gas's pressure is: $$ P = \frac{NRT}{V-Nb} - \frac{aN^2}{V^2} $$
( This equation does not exist for V<=NB )
How is it proved that Van Der Waals gas is a second order phase transition?
The second order derivative of the pressure (P ) with respect to volume ( V ) don't have a discontinuity ( except at point V = Nb , but the pressure is not existent for V<=Nb ). So how come Van der waals gas describes a second order phase transition?
 
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  • #2
## b ## is the volume of each particle, so that ## Nb ## is the smallest volume this collection of particles can occupy. It is unnecessary to look mathematically at cases where ## V \leq Nb ## for the Van der Waals gas.
 
  • #3
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  • #4
TeethWhitener said:
Why do you think this? Can you give a citation?
Some light reading:
http://www.pmaweb.caltech.edu/~mcc/Ph127/b/Lecture3.pdf

Actually, after further inquiring about the subject ( I asked a professor teaching statistical physics course ), he told me that the Van-Der-Waals gas is a model characterized by both first order and second order phase transitions, explanation:

The gas describes a first order phase transition because if for example we plot P(v) , we'll get something like this:
1577903887218.png

in the green-shaded area, the following stability-criteria does not hold:

## - \frac{\partial p}{\partial V}>0 ##
( This stability criteria appears in "Thermodynamics and introduction to Thermostatistics - Callen" )
Since the stability-criteria does not hold then we say there is a "jump" of P(v) in the green-shaded area and we have to insert Maxwell-construction model to describe the system's process in the green area.

However, The gas also describes a second order phase transition for a specific type of isotherm - this isotherm describes a temperature called "Critical Temperature" ( also denoted as Tc ), and if we plot P(v) for this isotherm, we'll see that P(v) has stability-criteria fulfilled for all V , but there is a single point in which P(v) has an inflection point - this is the critical point in which the first and second derivatives are zeros ( since this is an inflection point ):
## \frac{\partial p}{\partial V} = \frac{\partial^2 p}{\partial V^2} =0 ##

1577904395833.png


Conclusion: VDW gas desribes mostly a first-order phase transition except for a specific isotherm ( Tc) for which the gas describes only a second order phase transition.
 
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  • #5
What if ##T>T_c##?
 
  • #6
TeethWhitener said:
What if ##T>T_c##?

If T>T_c then there is no phase transition since once we have reached T=T_c and went to higher temperature - our VDW model already passed from one phase into another.
So for T>T_c then P(v) describes only one specific phase and no phase transitions since the stability criteria
$$
- \frac{\partial p}{\partial V}>0
$$
Is satisfied - so there are no discontinuities which would indicate a first-order phase transition and also there are no inflection points which would indicate a second-order phase transition, so P(v) only describes the relationship between P and V for the one particular phase we are currently in for T>T_c.
 
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  • #7
Yes. The important point is that it is the Maxwell construction (and not the vdW equation of state by itself) that generates the 1st order phase transition for ##T<T_c##.

One other minor thing: The ##T<T_c## isotherms also have inflection points, where ##\frac{\partial^2 p}{\partial V^2} = 0##, but it's only at ##T=T_c## that we have a point where ##\frac{\partial p}{\partial V} =\frac{\partial^2 p}{\partial V^2} = 0##.

Edit: to be perfectly honest, I'm not sure you can call the point at ##T=T_c## a second order phase transition. ##\frac{\partial^2 p}{\partial V^2}## is zero, not discontinuous. And all higher derivatives are zero as well. Without the Maxwell construction, the function is smooth (infinitely differentiable) everywhere it's defined. I don't know; I'll have to think harder about it.
 
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Related to Proof of second-order phase transition for a Van Der Waals Gas

What is a Van Der Waals gas?

A Van Der Waals gas is a real gas that takes into account the interactions between molecules, unlike the ideal gas law which assumes no interactions between particles. These interactions can cause deviations from the ideal gas behavior, particularly at high pressures and low temperatures.

What is a second-order phase transition?

A second-order phase transition is a type of phase transition where there is a sudden change in the properties of a substance without a change in its state. This means that the substance remains in the same state (solid, liquid, or gas) but its properties, such as density or specific heat, suddenly change at a specific temperature or pressure.

What is the proof of second-order phase transition for a Van Der Waals gas?

The proof of second-order phase transition for a Van Der Waals gas is based on the behavior of the compressibility factor (Z) at the critical point. At the critical point, Z is equal to 3/8, regardless of the values of the volume and temperature. This behavior is a characteristic of second-order phase transitions.

How is the proof of second-order phase transition for a Van Der Waals gas different from other gases?

The proof of second-order phase transition for a Van Der Waals gas is different from other gases because it considers the interactions between molecules. This leads to a more accurate description of the behavior of real gases, particularly at high pressures and low temperatures.

Why is the proof of second-order phase transition for a Van Der Waals gas important?

The proof of second-order phase transition for a Van Der Waals gas is important because it helps explain the behavior of real gases and provides a more accurate description than the ideal gas law. It also has practical applications in fields such as thermodynamics, materials science, and chemical engineering.

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