[Stat Mech] Different answers when using log space

[Clever-Name]

Homework Statement

I have come across a rather interesting conundrum. Given the configurational potential energy partition function for a non-ideal gas:

$$Z = Z_{internal}\frac{1}{N!}\left(\frac{2\pi m}{h^{2}\beta}\right)^{\frac{3N}{2}}(V^{N} - B_{2}(T)N^{2}V^{N-1})$$

where $B_{2}(T)$ is the second virial coefficient.

I'm supposed to solve for the pressure.

Homework Equations

$$p = \frac{1}{\beta Z}\frac{\partial Z}{\partial V}$$

or

$$p = \frac{1}{\beta}\frac{\partial ln(Z)}{\partial V}$$

The Attempt at a Solution

If we calculate it using the first expression for p we arrive at

$$p = \frac{1}{\beta}\left(\frac{N}{V} - B_{2}(T)N^{2}(N-1)V^{-2} \right)$$

However if you do it using the second form for p you can arrive at two different answers depending on how you 'prepare' the 3rd term in Z

Ignoring terms that don't involve V, if you start with it as written and evaluate

$$\frac{\partial}{\partial V} ln(V^{N} - B_{2}(T)N^{2}V^{N-1})$$
$$= \frac{NV^{N-1} - B_{2}(T)N^{2}(N-1)V^{N-2}}{(V^{N} - B_{2}(T)N^{2}V^{N-1})}$$

in the denominator (for each case) we assume $B_{2}(T)$ is small, so we can write the denominator just as $V^{N}$

So we arrive at

$$\frac{N}{V} - B_{2}(T)N^{2}(N-1)V^{-2}$$

OK, great that's what it should be, but if you start out with:

$$\frac{\partial}{\partial V}ln(V^{N}(1-B_{2}(T)N^{2}V^{-1}))$$

$$= \frac{\partial}{\partial V}\left(ln(V^{N}) + ln(1-B_{2}(T)N^{2}V^{-1})\right)$$

$$= \frac{N}{V}+ \frac{B_{2}(T)N^{2}V^{-2}}{(1-B_{2}(T)N^{2}V^{-1})}$$

Again, $B_{2}(T)$ is small, so:

$$= \frac{N}{V} + B_{2}(T)N^{2}V^{-2}$$

Now we have a different answer...

Why is this happening??

In my prof's notes he uses the second derivation, where we get a different answer from every other calculation.

I suspect
$$\frac{N}{V} - B_{2}(T)N^{2}(N-1)V^{-2}$$

is the correct form but I can't see what's wrong with the other way.

 

[haruspex]

Simplifying the expressions by writing a = B₂(T)N²:

$$\frac{\partial}{\partial V} ln(V^{N} - aV^{N-1})= \frac{NV^{N-1} - a(N-1)V^{N-2}}{(V^{N} - aV^{N-1})}$$
in the denominator (for each case) we assume a is small, so we can write the denominator just as $V^{N}$. So we arrive at
$$\frac{N}{V} - a(N-1)V^{-2}$$
OK, great that's what it should be, but if you start out with:
$$\frac{\partial}{\partial V}ln(V^{N}(1-aV^{-1}))= \frac{\partial}{\partial V}\left(ln(V^{N}) + ln(1-aV^{-1})\right)= \frac{N}{V}+ \frac{aV^{-2}}{(1-aV^{-1})}$$
Again, a is small, so:
$$= \frac{N}{V} + aV^{-2}$$

Your problem is that the error from your 'a is small' approximation is the same order as the smaller of the terms you retain. Therefore that term is entirely untrustworthy.
More appropriate would be:
$$\frac{NV^{N-1} - a(N-1)V^{N-2}}{(V^{N} - aV^{N-1})}= \frac{N}{V}\frac{1-a(1-1/N)V^{-1}}{1 - aV^{-1}}$$
$$≈\frac{N}{V}(1-a(1-1/N)V^{-1})(1 + aV^{-1})≈\frac{N}{V}(1-a(1-1/N)V^{-1} + aV^{-1})=\frac{N}{V}(1+\frac{a}{NV})$$

 

[Clever-Name]

Ok, I'm not sure I'm following here. You have derived the correct expression, sure, but we can arrive at that just as easily by saying $1 - aV^{-1} = 1$ in the second derivation. How come we're not allowed to do that when we have $V^{N}(1-aV^{-1})$. I'm not sure I understand your comment about the error.

 

[haruspex]

I'm not sure I understand your comment about the error.

In the first derivation, you equate the denominator to V^N while that denominator applies to the whole expression. The ignored term, as a fraction of the retained V^N, is -a/V. Dropping it therefore increases the denominator by a fraction a/V, and thus decreases the entire expression by approximately that fraction: x → x - xa/V.
In the answer derived, the leading term is N/V. Thus the expression as a whole has lost roughly (N/V)*(a/V) = Na/V². That is the size of the discrepancy between your two answers.
In your second derivation you made the approximation after separating off the leading term. As a result, the induced error was a second order small quantity.

 

[Clever-Name]

Oh ok that makes sense now, thanks for the help.