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Equilibrium volume of two differential van der Waal gases

  1. Aug 6, 2017 #1
    1. The problem statement, all variables and given/known data
    Two ideal van der Waals fluids are contained in a cylinder, separated by an internal moveable piston. There is one mole of each fluid, and the two fluids have the same values of the van der Waals constants [itex]b[/itex] and [itex]c[/itex]; the respective values of the van der Waals constant [itex]''a''[/itex] are [itex]a_1[/itex] and [itex]a_2[/itex]. The entire system is in contact with a thermal reservoir of temperature [itex]T[/itex]. Calculate the Helmholtz potential of the composite system as a function of [itex]T[/itex] and the total volume [itex]V[/itex]. If the total volume is doubled (while allowing the internal piston to adjust), what is the work done by the system?

    2. Relevant equations
    The van der Waal Helmholtz potential is [tex]
    F = cNRT - \frac{a}{V}N^2 - T\left ( NR \log(V/N-b) + cNR \log ( cRT) + S_0 \right)
    [/tex]

    3. The attempt at a solution
    So the composite Helmholtz potential will just be the sum of the two. The temperatures and pressures are the same. The negative of the change of the van der Waal potential is the work done. The problem is calculating the equilibrium volume. Equating the temperatures we find [tex]
    \frac{RT}{V_1-Nb} - \frac{a_1}{V_1^2} = \frac{RT}{V_2-Nb} - \frac{a_2}{V_2^2}
    [/tex]
    which along with [itex]V = V_1+V_2[/itex] gives you quintic polynomials to solve for [itex]V_1[/itex] in terms of [itex]V[/itex]. I don't see a way of doing this.
     
  2. jcsd
  3. Aug 6, 2017 #2

    Charles Link

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    I would recommend trying an iterative approach to the solution. You know that to first order ## \frac{RT}{V_1}=\frac{RT}{V_2} ## with ## V=V_1+V_2 ##. You should be able to do a Taylor Series type expansion or get an iterative equation for ## V_1 ## around this point: Basically, you should get an equation of the form ## V_1=(1/2)V +... ##, with the ... being terms that include some function ## f(V_1, a_1, a_2, b) ##. I haven't done it myself yet, but I think it might work. ## \\ ## Editing... And yes, I got it to work. When you have ## RT/(V_1-Nb) ##, you can write it as ##RT/((V_1)(1-Nb/V_1)) ##, and then just keep the first term: ==>> ##RT/(V_1-Nb) \approx (RT/V_1)+(RTNb/V_1^2) ##. You can do a similar thing with the ## RT/(V-V_1-Nb) ## term. There is also a little algebra on the terms ## RT/V_1 -RT/(V-V_1) ##. I am not allowed to give you the complete solution, but can lead you to it if you need additional detail.
     
    Last edited: Aug 6, 2017
  4. Aug 6, 2017 #3
    Thanks for the response. First I should say I didn't mean to write the N as it is 1, although it doesn't really matter. I think I understand what you mean, but I am apprehensive about the right hand side expansion, unless you mean something like [tex]

    \frac{RT}{V_1} + \frac{RTb}{V_1^2} - \frac{a_1}{V_1^2} =\frac{RT}{V-V_1} + \frac{RTb}{(V-V_1)^2} - \frac{a_2}{(V-V_1)^2}
    [/tex]
    which is better than before but still a cubic equation. I'm probably not understanding the method you mean.
     
  5. Aug 6, 2017 #4
    Ok this time I expanded around [itex]V/2[/itex] like you said :P and got [tex]
    V_1 = V/2 + \frac{4(a_2-a_1)}{\frac{16(a_2+a_1)}{V} +\frac{2RTV^2}{(V/2-b)^2}}
    [/tex]
     
  6. Aug 6, 2017 #5

    Charles Link

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    You did sufficient work of your own, and basically got the answer, so I think I am now ok in showing you what I computed. One thing to also note in this calculation is that your Helmholtz potential basically simply sets ## P_1=P_2 ## for the Van der Waals fluids. Anyway, let me show you the algebra I did and you can compare:
    ## P_1=((NRT/(V_1-Nb))-a_1 N^2/V_1^2=P_2=((NRT/(V_2-Nb)-a_2 N^2/V_2^2 ##. Let ## A=Nb ##. And let ## N=1 ##. A little algebra gives: ## RT/(V_1-A)-RT/((V-V_1)-A)=(a_1/V_1^2)-(a_2/(V-V_1)^2) ##. Let the right side of this equation equal ## C ##, and assume ## C ## is small. Also assume ## A ## is small. We see then that ## V_1 \approx V/2 ##. Anyway, let's work with the left side of the equation: As mentioned in my previous post, the result keeping first order terms in ## A ## is ## (RT/V_1)-(RT/(V-V_1))=-(ART/V_1^2)+(ART/(V-V_1)^2)+C ##. We next multiply through on both sides by ## V_1(V-V_1) ## to get ## -2V_1 RT+VRT=V_1(V-V_1) D ## where ## D ## is the right side of the previous equation. This gives ## V_1=V/2-(V_1(V-V_1)D/(2RT)) ##. On the right side of this equation, (for a first iteration), you let ## V_1=V/2 ## everywhere. And then of course, ## V_2=V-V_1 ##. ## \\ ## It's not exactly what you got, but I don't want to go through all the algebra to see whether the two are the same or not=I'll leave that part to you. ## \\ ## Additional editing: When evaluating the ## D ## term on the first iteration gives simply ## C ## with the ## ART ## terms cancelling. I think it is necessary to have an ## A ## term, and that won't occur until a second iteration. I'm still working on comparing with what you have, but you may have it correct. The result I get after one iteration is ## V_1=V/2+(a_2-a_1)/(2RT) ##. ## \\ ## Additional editing: You can ignore the first term that you have in your final answer in the denominator, because it is small compared to the second term in your denominator. Expanding the second term in tour denominator gives pretty close agreement to what I get with a second iteration on the terms containing ## ART ##.
     
    Last edited: Aug 6, 2017
  7. Aug 6, 2017 #6
    Thanks! I notice that if in the denominator the [itex]a[/itex]s and [itex]b[/itex] are ignored, then my final equation matches yours.
     
  8. Aug 6, 2017 #7

    Charles Link

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    My final result after the second iteration is ## V_1=V/2+((a_2-a_1)/(2RT))(1-4b/V) ##. ## \\ ## If you expand the ## (V/2-b)^2 ## in your denominator (factor to get ## (V/2)^2(1-2b/V)^2 =(V/2)^2(1-4b/V+...) ## and you can put it up in the numerator, and ignore the first term (small with the ## a_1+a_2 ## ) in your denominator), your result agrees with mine. ## \\ ## So very good, it looks like your calculation was accurate. :)
     
    Last edited: Aug 6, 2017
  9. Aug 6, 2017 #8

    Charles Link

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    And please see my edited comments of the previous post. I believe you got the right answer. :) ## \\ ## And additional item: It is interesting that the ## b ## terms in the Van der Waals gas are less significant than the ## a ## terms regarding a change in the volume of the system. The reason I believe is because the ## b ## terms, .i.e. ## Nb ##, is the volume of the atoms. This atomic volume is already accounted for and does not change with an expansion or compression of the gas=thereby, it has no separate first order term when computing ## V_1=V/2+ (a_2-a_1)(...)+... ##. Basically, the dynamics of the system works with the volumes ## V_1-Nb ## and ## V_2-Nb ##, and thereby the effect of the ## b ## term in this equation for ## V_1 ## is to convert the volume ## V ## in the ## (a_2-a_1) ## term to the operational part which is ## V-2Nb ##.
     
    Last edited: Aug 6, 2017
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