# Thermal: taylor series van der waals equation

• accountkiller
In summary, to show that the Van der Waals equation of state for a real gas approaches the equation of state of an ideal gas at constant volume and temperature with decreasing number of particles, we can rearrange the equation and use a Taylor series expansion with x = 1/v as a small parameter. Simplifying and keeping only the lowest order term in x, we can obtain the ideal gas equation PV=nRT. It is important to note that x << 1 in this case.

## Homework Statement

Show that at constant volume V and temperature T but decreasing number N=n*N$_{A}$ 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:
$( P + \frac{a}{v^{2}} ) ( v - b ) = RT$
Taylor series expansion:
$f(x)=f(a)+f'(a)(x-a)+\frac{f"(a)}{2!} (x-a)^{2} + ...$

## The Attempt at a Solution

Rearranging...
$( P + \frac{a}{v^{2}} ) ( v - b ) = RT$
$P + \frac{a}{v^{2}} = \frac{RT}{v-b}$
$P = \frac{RT}{v-b} - \frac{a}{v^{2}}$

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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$P = \frac{RT}{v-b} - \frac{a}{v^{2}}$

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.

So...

$P = \frac{RT}{\frac{1}{x} - b} - \frac{a}{\frac{1}{x^{2}}}$
$P = \frac{RT}{\frac{1}{x} - \frac{bx}{x}} - ax^{2}$
$P = \frac{RT}{\frac{1-bx}{x}} - ax^{2}$
$P = \frac{RTx}{(1-bx)} - ax^{2}$

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

$f(x)=f(a)+f'(a)(x-a)+\frac{f''(a)}{2!} (x-a)^{2} + \frac{f'''(a)}{3!} (x-a)^{3}+...$

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

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

I just wanted to make sure if that is the correct way to go now before I continue.

$P = \frac{RTx}{(1-bx)} - ax^{2}$

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

You'll just need to expand $f(x) = 1/(1-bx)$ about $x =$ 0 and then multiply by $RTx$.

$f(x)=f(a)+f'(a)(x-a)+\frac{f''(a)}{2!} (x-a)^{2} + \frac{f'''(a)}{3!} (x-a)^{3}+...$

You'll want to expand about $x = 0$, so $a = 0$ in the Taylor expansion. [See the hint in the statement of the problem.]

Thank you for replying! :) So my Taylor series is the following...?

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

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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Thank you for replying! :) So my Taylor series is the following...?

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

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 $f'(0)$ 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.

Oh right, I forgot to put x=0 in. So:

$\frac{1}{1-bx} = 1+b(1-bx)^{-2}(x-0) + ... = 1+b(x-0)+...$

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:
$\frac{1}{1-bx} = 1$

So then my original equation $P = RTx \frac{1}{1-bx} - ax^{2}$
becomes $P = RTx - ax^{2}$

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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Oh right, I forgot to put x=0 in. So:

$\frac{1}{1-bx} = 1+b(1-bx)^{-2}(x-0) + ... = 1+b(x-0)+...$

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:
$\frac{1}{1-bx} = 1$

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, $f'(0)*(x-0)$ means that you let x = 0 in expression for $f'(x)$ but not in the expression $(x-0)$.

Okay well if I do that then I get:
$\frac{1}{1-bx} = 1 + bx + b^{2}x^{2} + b^{3}x^{3} + ...$

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.

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.

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!

Wait... I went too quickly, I skipped something in my head. What happens to the b?

$P=RTx(1+bx)-ax^{2}$

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.

Is it that when x << 1 then the b is negligible so it's basically (1+bx) = 1?

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?

TSny said:
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!

Wait... I went too quickly, I skipped something in my head. What happens to the b?

$P=RTx(1+bx)-ax^{2}$

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.

$P=RTx(1+bx)-ax^{2}=RT(x+bx^2)-ax^2$

Both the b and a terms are of order x^2.

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

Ah, perfect. Thank you so much for all the help with this problem! :)

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?

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.

## 1. What is the Taylor series expansion of the van der Waals equation?

The Taylor series expansion of the van der Waals equation is a mathematical method used to approximate the behavior of a gas at high pressures and low temperatures. It is an expansion of the van der Waals equation, which is a modified form of the ideal gas law that takes into account the intermolecular forces between gas molecules. The Taylor series expansion allows for a more accurate prediction of gas behavior at non-ideal conditions.

## 2. How is the Taylor series for the van der Waals equation derived?

The Taylor series for the van der Waals equation is derived by taking the first and second derivatives of the equation with respect to pressure and volume. These derivatives are then evaluated at a specific point, typically the critical point, and used to construct the Taylor series expansion. This series is then truncated to a desired order, with higher orders providing more accurate results.

## 3. What is the significance of the Taylor series expansion for the van der Waals equation?

The Taylor series expansion is significant because it allows for a more accurate prediction of gas behavior at non-ideal conditions. This is especially important for gases that deviate significantly from ideal behavior, such as at high pressures and low temperatures. The expansion also provides insight into the effects of intermolecular forces on gas behavior.

## 4. What are the limitations of the Taylor series expansion for the van der Waals equation?

While the Taylor series expansion provides a more accurate prediction of gas behavior compared to the original van der Waals equation, it is not a perfect model. The expansion is only valid for gases that behave according to the van der Waals equation, and it becomes less accurate at extreme pressures and temperatures. Additionally, the accuracy of the expansion depends on the order of the series used, with higher orders providing more accurate results but also requiring more computation.

## 5. How is the Taylor series expansion used in thermodynamic calculations?

The Taylor series expansion is used in thermodynamic calculations to predict gas behavior at non-ideal conditions. It is commonly used in the study of phase transitions, where the behavior of gases deviates significantly from ideal behavior. The expansion is also used in the development of other thermodynamic models and equations, providing a foundation for more complex calculations.