Thermodynamic identity (math) question: one component system

[Hiero]

Homework Statement

Consider a one component system with the only generalized force being pressure

Relevant Equations

Gibbs-Duhem relation

$SdT-Vdp+nd\mu = 0$

$v = V/n$ (volume per mole)

I was just wondering what is wrong with the following logic;

From the Gibbs-Duhem relation we get,

$\frac{\partial \mu}{\partial P}\Big\rvert_T = v$

Now consider,

$\frac{\partial v}{\partial \mu}\Big\rvert_T = \frac{\partial }{\partial \mu}\Big (\frac{\partial \mu}{\partial P}\Big\rvert_T \Big )\Big\rvert_T= \frac{\partial }{\partial P}\Big (\frac{\partial \mu}{\partial \mu}\Big\rvert_T \Big )\Big\rvert_T= \frac{\partial }{\partial P}(1)\Big\rvert_T=0$

I’m pretty sure the result is not true, that it is not generally zero, so which step is flawed?

 

[TSny]

$\frac{\partial }{\partial \mu}\Big (\frac{\partial \mu}{\partial P}\Big\rvert_T \Big )\Big\rvert_T= \frac{\partial }{\partial P}\Big (\frac{\partial \mu}{\partial \mu}\Big\rvert_T \Big )\Big\rvert_T$

I think the mistake is in the step shown above.

If $Z$ is a function of independent variables $x$ and $y$, then the equality of mixed partials would be

Compare to your

 

[Hiero]

Yeah @TSny that makes sense.

What if I don’t differentiate wrt to the dependent variable but I still hold fixed something which is not an independent variable, can I still apply equality of mixed partials? To be clear I am asking if the following is true:

$\frac{\partial }{\partial x}\Big (\frac{\partial Z(x,y)}{\partial y}\Big \rvert_{T(x,y)} \Big ) \Big \rvert_{S(x,y)}=\frac{\partial }{\partial y}\Big (\frac{\partial Z(x,y)}{\partial x}\Big \rvert_{S(x,y)} \Big ) \Big \rvert_{T(x,y)}$

(Where we might have T=S)

I have only thought about it briefly but I’m unable to prove it.

Some related thoughts;

We could assume T(x,y) is invertible so that:
$\frac{\partial Z(x,y(x,T))}{\partial x}\Big \rvert_{T} =\frac{\partial Z(x,y)}{\partial x}\Big \rvert_{y} +\Big ( \frac{\partial Z(x,y)}{\partial y}\Big \rvert_{x} \Big ) \Big ( \frac{\partial y(x,T)}{\partial y}\Big \rvert_{T} \Big )$
And likewise using x(y,T) for the y derivative (and likewise using S in place of T)

We could also just say:
$T(x,y) = \text{constant}$
$\implies \frac{\partial T(x,y)}{\partial x}\Big \rvert_{y}dx + \frac{\partial T(x,y)}{\partial y}\Big \rvert_{x}dy = 0$
$\implies \frac{\partial y}{\partial x}\Big \rvert_{T} = -\frac{\frac{\partial T(x,y)}{\partial x}\Big \rvert_{y}}{\frac{\partial T(x,y)}{\partial y}\Big \rvert_{x}}$

But I’m not sure how to use these to prove or disprove the question I just asked in this post. I’m not even sure if everything I’ve said is even sensible. At this point I feel pretty confused. Thanks for any further clarifications.

 

[Hiero]

$\frac{\partial }{\partial x}\Big (\frac{\partial Z(x,y)}{\partial y}\Big \rvert_{T(x,y)} \Big ) \Big \rvert_{S(x,y)}=\frac{\partial }{\partial y}\Big (\frac{\partial Z(x,y)}{\partial x}\Big \rvert_{S(x,y)} \Big ) \Big \rvert_{T(x,y)}$

I think it’s probably not true but people (at least in physics) including myself have a habit of suppressing the arguments of a function, so now I am paranoid about applying equality of mixed partials.

 

[TSny]

$\frac{\partial }{\partial x}\Big (\frac{\partial Z(x,y)}{\partial y}\Big \rvert_{T(x,y)} \Big ) \Big \rvert_{S(x,y)}=\frac{\partial }{\partial y}\Big (\frac{\partial Z(x,y)}{\partial x}\Big \rvert_{S(x,y)} \Big ) \Big \rvert_{T(x,y)}$

I randomly tried the following:

$Z(x,y) = xy$
$T(x,y) = x+y$
$S(x,y) = x^2-y$

Note that $Z(x,y)$ can be expressed in terms of $T$ and $y$ as $Z = Ty-y^2$. So, if $T$ is held constant,

$\frac{\partial Z(x,y)}{\partial y}\Big \rvert_{T(x,y)} = T-2y$

or, substituting $T= x+y$,

$\frac{\partial Z(x,y)}{\partial y}\Big \rvert_{T(x,y)} = x-y$.

This can be expressed in terms of $S$:

$\frac{\partial Z(x,y)}{\partial y}\Big \rvert_{T(x,y)} = x-x^2+S$

So, $\frac{\partial }{\partial x}\Big (\frac{\partial Z(x,y)}{\partial y}\Big \rvert_{T(x,y)} \Big ) \Big \rvert_{S(x,y)} = \boxed{1-2x}$

In a similar manner, I find

$\frac{\partial Z(x,y)}{\partial x}\Big \rvert_{S(x,y)} = 2x^2+y$

and $\frac{\partial }{\partial y}\Big (\frac{\partial Z(x,y)}{\partial x}\Big \rvert_{S(x,y)} \Big ) \Big \rvert_{T(x,y)} = \boxed{1-4x}$

If I didn't make any errors, this shows that

$\frac{\partial }{\partial x}\Big (\frac{\partial Z(x,y)}{\partial y}\Big \rvert_{T(x,y)} \Big ) \Big \rvert_{S(x,y)}\neq\frac{\partial }{\partial y}\Big (\frac{\partial Z(x,y)}{\partial x}\Big \rvert_{S(x,y)} \Big ) \Big \rvert_{T(x,y)}$

 

[Hiero]

If I didn't make any errors, this shows that

$\frac{\partial }{\partial x}\Big (\frac{\partial Z(x,y)}{\partial y}\Big \rvert_{T(x,y)} \Big ) \Big \rvert_{S(x,y)}\neq\frac{\partial }{\partial y}\Big (\frac{\partial Z(x,y)}{\partial x}\Big \rvert_{S(x,y)} \Big ) \Big \rvert_{T(x,y)}$

Right, disproof by counter example. I should’ve tried that as soon as I failed to prove it. Sorry.

Thank you for taking the time!
Feel like I owe you more than just a like

 

[TSny]

Thank you for taking the time!
Feel like I owe you more than just a like

No problem. Thermodynamics is a quicksand of partial derivatives. I feel like I'm going under at times