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Van der Waals equation of state and Maxwell construction

  1. Aug 20, 2009 #1
    Hello people :),

    I have a question because I couldn't understand something written in Schwabl's book of Statistical mechanics.

    Can you please explain why we in the Van der Waals equation of state choose a constant pressure under the critical temperature. I understood the part that we want to have some plausible compressibility, therefore the derivative of the free energy with respect to volume should be >= 0, but I don't understand:

    1) Why we choose the Maxwell construction to be a line, not some curve.
    2) why we have to have equal areas above and below the line.

    What would happen if we don't follow those rules?

    Find attached the corresponding graphs.

    Thank you

    Attached Files:

  2. jcsd
  3. Aug 20, 2009 #2
    I'm not sure if this is what you want but experiment shows that below the critical temperature the PV graph has a a flat portion(constant pressure)where liquid and vapour are in equilibrium.Van der Waals improved upon the simple kinetic model ideal gas equation by taking into account the fact that the molecules have a finite size and that they exert forces on eachother.When his results are plotted they do not give the flat portion but they give the peculiar wiggly line which goes above and below the flat portion
    Clearly V de Ws equation is not perfect but it is an improvement on the equation it replaced.
  4. Aug 20, 2009 #3
    Is it that simple? just because the experiments have shown that? isn't there some good theoretical reason?

  5. Aug 20, 2009 #4
    Again I am not sure if this is what you want but there is a good theoretical reason for the flat portion of the graph.To liquify a gas you have to both compress it and cool it below its critical temperature.If it wasn't cooled enough you would end up with a highly compressed gas where the molecules are moving too fast to get continually, momentarily captured by other molecules, as is the case with liquids.If a gas is compressed below its critical point a pressure is reached where it starts to liquify and a meniscus appears.The gas can now be described as a saturated vapour where the gas and liquid are in equilibrium with eachother.With a saturated vapour the volume can be changed without a change in pressure,reducing the volume results in more liquid and less vapour and vice versa.As the volume is reduced futher a point is reached where all of the vapour has changed to liquid and since liquids are highly incompressible a massive increase in pressure is needed for a small reduction in volume.Of course different gases have different characteristics.A good equation should explain experimental results and although V de Ws doesn't do this exactly it is a step up from the ideal gas equation.
  6. Aug 20, 2009 #5
    Thank you, but what about the equal areas?
  7. Aug 20, 2009 #6
    Are you referring to the area of the V de W curve above the flat portion as compared to the area below the flat portion?If so I wasn't aware that the areas were equal but when I recall the curves they certainly seem to be.Interesting if they are equal but I think that it just pops out of the equation.Perhaps you could most easily prove the equality by integration.
  8. Aug 20, 2009 #7
    Yes, the areas should be equal, this is the deal! This line should be constructed in a way giving equal areas according to the integral, it's something related to the free energy I couldn't understand. It's not a mathematical issue, it's a physical issue, there is some kind of a condition or something with the free energy, homogeneity.

    Any ideas guys?
  9. Aug 20, 2009 #8
    The work done on or by a gas is equal to the integral of PdV which is equal to the area under the PV graph.Assume that the work done in the V de Ws wiggly portion of the graph is the same as the work done during the experimentally found flat portion of the graph,if so the areas under both graphs must be the same.With this in mind we can take V de Ws graph and draw a line through through the wiggly portion, the position of the line being such that the area of the V de Ws graph above the line must be exactly equal to the area of the V de Ws graph below the line.Imagine the gas was being compressed,when it reaches the flat portion, V de Ws graph rises above the line showing that extra work is done.Later,when V de Ws graph goes below the line less work is done the total work done being the same as that which is done when P remains constant.I hope others can come in and justify more fully why V de Ws graph, which is not exactly correct, can be manipulated like this.
  10. Aug 20, 2009 #9
    Thank you for your tries and efforts :)

    More explanation guys, come on!
  11. Aug 21, 2009 #10
    Oh my god! this is really disappointing :(
  12. Aug 23, 2009 #11
    Ok, so you have to draw a line somewhere. In the P-V curve, there is a maximum, and to the left of that dP/dV < 0. Where dP/dV < 0, you have negative compressibility which means the system will spontaneously compress to a liquid, so the system is thermodynamically unstable at that point. This means you could draw a line at the top so in any case you would have a line drawn at the point where dP/dV = 0.

    Then you need to minimize the free energy. Pick any constant pressure line that intersects the P-V isotherm at three points. The left point is the volume of the liquid, V_L and the right point is the volume of the gas V_G. At a given volume on that line, that line, number of particles in gas or liquid phases are

    [tex]N_L + N_G = N[/tex]
    [tex]n_LV_L + n_g V_G = V[/tex]

    where n_L(n_G) is the number fraction of particles in the liquid(gas), ie. N_L/N.

    The free energy along that line will be

    [tex]A = n_G A(N,V_G,T) + n_L A(N,V_L,T)[/tex]
    [tex]A = n_G A(N,V_G,T) + (1 - n_G) A(N,V_L,T)[/tex]

    will be a straight line, since N, V_G, V_L and T are fixed. Picking different constant pressure lines will shift the line in the free energy up and down. Now, what constant pressure can you pick that gives the lowest line in the free energy? The Maxwell construction gives you that, and guarantees convexity in the free energy. A more detailed explanation is avilable here:
    http://www.pma.caltech.edu/~mcc/Ph127/b/Lecture3.pdf [Broken]
    Last edited by a moderator: May 4, 2017
  13. Aug 26, 2009 #12
    Thanks pal :)
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