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Thermal: taylor series van der waals equation

  1. Sep 1, 2012 #1
    1. The problem statement, all variables and given/known data
    Show that at constant volume V and temperature T but decreasing number N=n*N[itex]_{A}[/itex] of particles the Van der Waals equation of state approaches the equation of state of an ideal gas.

    Hint: Rearrange the equation of state into the explicit functional form P=P(v,T) and use x=1/v as a small parameter for a Taylor series P(x)=P(0)+dP/dx x + ...


    2. Relevant equations
    Van der Waals equation of state for a real gas:
    [itex]( P + \frac{a}{v^{2}} ) ( v - b ) = RT[/itex]
    Taylor series expansion:
    [itex]f(x)=f(a)+f'(a)(x-a)+\frac{f"(a)}{2!} (x-a)^{2} + ... [/itex]

    3. The attempt at a solution
    Rearranging...
    [itex]( P + \frac{a}{v^{2}} ) ( v - b ) = RT[/itex]
    [itex]P + \frac{a}{v^{2}} = \frac{RT}{v-b}[/itex]
    [itex]P = \frac{RT}{v-b} - \frac{a}{v^{2}}[/itex]

    Now it's been a while since I've done a Taylor expansion so I don't seem to remember how to go about it. Could someone just point me in the right direction? Thanks!
     
    Last edited by a moderator: Sep 1, 2012
  2. jcsd
  3. Sep 1, 2012 #2

    TSny

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    Go ahead and follow the hint by letting x = 1/v. Then think about simplifying and expanding the first fraction on the RHS as a Taylor series in x.
     
  4. Sep 1, 2012 #3
    So...

    [itex]P = \frac{RT}{\frac{1}{x} - b} - \frac{a}{\frac{1}{x^{2}}}[/itex]
    [itex]P = \frac{RT}{\frac{1}{x} - \frac{bx}{x}} - ax^{2}[/itex]
    [itex]P = \frac{RT}{\frac{1-bx}{x}} - ax^{2}[/itex]
    [itex]P = \frac{RTx}{(1-bx)} - ax^{2}[/itex]

    And then expand the first fraction, so do a Taylor expansion of [itex]\frac{RTx}{(1-bx)}[/itex] around the point 1/v.

    [itex]f(x)=f(a)+f'(a)(x-a)+\frac{f''(a)}{2!} (x-a)^{2} + \frac{f'''(a)}{3!} (x-a)^{3}+...[/itex]

    f(x) = [itex]\frac{RTx}{(1-bx)}[/itex] and a = 1/v , correct?
    So f(a) = [itex]\frac{\frac{RT}{v}}{1-bx}[/itex] = [itex]\frac{RT}{v(1-bx)}[/itex]

    Then for f'(a)... it's been a while since I've taken a simple derivative; is the following correct?
    [itex]f'(a)=f'(\frac{RT}{v} \frac{1}{1-bx})=f'(\frac{RT}{v} (1-bx)^{-1})[/itex]
    [itex]f'(a)=\frac{RT}{v} * -(1-bx)^{-2} * (-b) = \frac{RT}{v} b(1-bx)^{-2}[/itex]

    I just wanted to make sure if that is the correct way to go now before I continue.
     
  5. Sep 1, 2012 #4

    TSny

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    Good. So, you have [itex]P = RTxf(x) - ax^{2}[/itex] where [itex]f(x) = 1/(1-bx)[/itex]
    You'll just need to expand [itex]f(x) = 1/(1-bx)[/itex] about [itex] x = [/itex] 0 and then multiply by [itex]RTx[/itex].

    You'll want to expand about [itex]x = 0[/itex], so [itex]a = 0[/itex] in the Taylor expansion. [See the hint in the statement of the problem.]
     
  6. Sep 1, 2012 #5
    Thank you for replying! :) So my Taylor series is the following...?

    [itex]\frac{1}{1-bx} = f(0) + f'(0)(x-a) + ...[/itex]
    [itex]\frac{1}{1-bx} = 1 + b(1-bx)^{-2} (x-a) + ...[/itex]

    What do I do with the (x-a) [or (x-0) I guess]?
    (I should know this by now but it's just been a while and doing a Taylor series expansion has apparently been completely erased from my memory.)

    Then once I finish some more of the Taylor series, I multiply that whole right hand side by RTx?
     
    Last edited: Sep 1, 2012
  7. Sep 1, 2012 #6

    TSny

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    Yes, that looks pretty good. But, don't forget that [itex]f'(0)[/itex] means to let x = 0 after taking the derivative. And, yes, (x-a) is (x-0). Then, as you say, you'll multiply by RTx.
     
  8. Sep 2, 2012 #7
    Oh right, I forgot to put x=0 in. So:

    [itex]\frac{1}{1-bx} = 1+b(1-bx)^{-2}(x-0) + ... = 1+b(x-0)+...[/itex]

    Again though, what do I do with the (x-0)? Do I put 0 in for that x too? If I do that then all the other terms become zero and I just have:
    [itex]\frac{1}{1-bx} = 1[/itex]

    So then my original equation [itex]P = RTx \frac{1}{1-bx} - ax^{2}[/itex]
    becomes [itex]P = RTx - ax^{2}[/itex]

    I don't see how this will eventually get me to the ideal gas equation of state PV=nRT. I can get the v back but I have an extra a and I don't have an n...
    EDIT: Oh I forgot that v = V/n so I can get my n too, so I just have the extra a.
     
    Last edited: Sep 2, 2012
  9. Sep 2, 2012 #8

    TSny

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    Just reduce (x-0) to x. But don't make x = 0 here. You only set x = 0 for the derivatives in the taylor expansion. Thus, [itex]f'(0)*(x-0)[/itex] means that you let x = 0 in expression for [itex]f'(x)[/itex] but not in the expression [itex](x-0)[/itex].
     
  10. Sep 2, 2012 #9
    Okay well if I do that then I get:
    [itex]\frac{1}{1-bx} = 1 + bx + b^{2}x^{2} + b^{3}x^{3} + ...[/itex]

    Which still doesn't make anything simplified so that I'd get to the ideal gas equation. What am I missing here?

    Thank you for helping me through this, I appreciate it.
     
  11. Sep 2, 2012 #10

    TSny

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    Put it all together and remember you're looking at the case where x << 1. So you can keep the lowest order term in x and drop all the higher order terms.
     
  12. Sep 2, 2012 #11
    Ok so if I take out the higher order terms including the ax^2 and make the appropriate x=1/v and v=V/n substitutions I do get the PV=nRT like I'm supposed so thank you - problem solved! :)

    But I guess the part of it that is still confusing to me is the x << 1 part. Why do I take x << 1? Is that just what you're supposed to do in a Taylor expansion or was I supposed to know that from some wording in my problem?

    Thank you!
     
  13. Sep 2, 2012 #12
    Wait... I went too quickly, I skipped something in my head. What happens to the b?

    [itex]P=RTx(1+bx)-ax^{2}[/itex]

    If I have just P=RTx if I drop the ax^2 then I do get PV=nRT. That's what I wrote down after I read your reply but I don't know why I dropped the bx term too.
     
  14. Sep 2, 2012 #13
    Is it that when x << 1 then the b is negligible so it's basically (1+bx) = 1?
     
  15. Sep 2, 2012 #14

    TSny

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    x = 1/v = n/V. What does the problem say about V and n?
     
  16. Sep 2, 2012 #15
    OH yep there it is in the question. I think I'm just trying to go through it too quickly that I'm missing important things like that. Thank you!
     
  17. Sep 2, 2012 #16

    TSny

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    [itex]P=RTx(1+bx)-ax^{2}=RT(x+bx^2)-ax^2[/itex]

    Both the b and a terms are of order x^2.
     
  18. Sep 2, 2012 #17

    ehild

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    bx becomes negligible with respect to 1 if x tends to zero. b means the own volume of the gas molecules, x is the reciprocal of the specific volume of the gas, so bx=b/v is the relative volume of the molecules with respect to the volume of the gas. As the gas expands, its volume increases while the volume of the molecules stays constant. bx<<1, so it can be neglected.

    ehild
     
  19. Sep 2, 2012 #18
    Ah, perfect. Thank you so much for all the help with this problem! :)
     
  20. Sep 3, 2012 #19
    When a classmate asked me for help with this question and I showed him my work, he brought up a good point: If x<<1 then wouldn't it mean that in my equation P=RTx the x there <<1 too so the RT is negligible as well?
     
  21. Sep 4, 2012 #20

    AGNuke

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    Yes, RTx will be negligible, with respect to what? I don't see any macro quantity anywhere in your equation.

    RTx may be negligible, but it is there. In your previous equation (1+bx), bx was negligible in front of 1. But here, it is a quantity, that's another thing you call it negligible.

    1 in front of million don't makes much sense, but 1 alone has its worth.
     
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