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Thermal: taylor series van der waals equation

  • #1

Homework Statement


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 + ...


Homework 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]

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!
 
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Answers and Replies

  • #2
TSny
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[itex]P = \frac{RT}{v-b} - \frac{a}{v^{2}}[/itex]
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.
 
  • #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.
 
  • #4
TSny
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[itex]P = \frac{RTx}{(1-bx)} - ax^{2}[/itex]
Good. So, you have [itex]P = RTxf(x) - ax^{2}[/itex] where [itex]f(x) = 1/(1-bx)[/itex]
And then expand the first fraction, so do a Taylor expansion of [itex]\frac{RTx}{(1-bx)}[/itex] around the point 1/v.
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].

[itex]f(x)=f(a)+f'(a)(x-a)+\frac{f''(a)}{2!} (x-a)^{2} + \frac{f'''(a)}{3!} (x-a)^{3}+...[/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.]
 
  • #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?
 
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  • #6
TSny
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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?
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.
 
  • #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.
 
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  • #8
TSny
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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]
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].
 
  • #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.
 
  • #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.
 
  • #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!
 
  • #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.
 
  • #13
Is it that when x << 1 then the b is negligible so it's basically (1+bx) = 1?
 
  • #14
TSny
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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?
x = 1/v = n/V. What does the problem say about V and n?
 
  • #15
x = 1/v = n/V. What does the problem say about V and n?
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!
 
  • #16
TSny
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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.
[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.
 
  • #17
ehild
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Is it that when x << 1 then the b is negligible so it's basically (1+bx) = 1?
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
 
  • #18
Ah, perfect. Thank you so much for all the help with this problem! :)
 
  • #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?
 
  • #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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