Triple Product Rule Equivalency

[WhiteWolf98]

Homework Statement

I've been trying to apply the cyclic rule to the ideal gas law, and got a different answer to the one given. I was wondering if the two solutions are equivalent; knowing this would be a great help for me, as the derivation of the Maxwell equations are next ~
I'm rather new to partial derivatives, so apologies if this is a silly question

Relevant Equations

$(\frac {\partial x} {\partial y})_z (\frac {\partial y} {\partial z})_x (\frac {\partial z} {\partial x})_y = -1$

$p=\frac {RT} v;~p=p(T,v)~...1$

$v=\frac {RT} p;~v=v(T,p)~...2$

$T=\frac {pv} R;~T=T(p,v)~...3$

$Considering~eq.~1:$

$p=\frac {RT} v \Rightarrow (\frac {\partial p} {\partial v})_T=-\frac {RT} {v^2}$

$Considering~eq.~2:$

$v=\frac {RT} p \Rightarrow (\frac {\partial v} {\partial t})_p=\frac R p$

$Considering~eq.~3:$

$T=\frac {pv} R \Rightarrow (\frac {\partial T} {\partial p})_v=\frac v R$

$(\frac {\partial p} {\partial v})_T (\frac {\partial v} {\partial T})_p (\frac {\partial T} {\partial p})_v =-\frac {RT} {v^2} \cdot \frac R p \cdot \frac v R= -\frac {RT} {pv} = -1$

That's the first, given solution. I was able to get to this on my own, but there's also another I got to in a similar way:

$(\frac {\partial p} {\partial T})_v (\frac {\partial v} {\partial p})_T (\frac {\partial T} {\partial v})_p = -\frac {RT} {pv} = -1$

I'd like to know if these two final equations are the same; and if not, why aren't they? I'm finding it a little difficult to find a pattern with the triple product rule. I understand that there has to be a partial derivative of each variable, but is the second part (what the partial derivative is with respect to), random? With three variables, the partial derivative of a variable can be with respect to two other variables; how do you choose these? If one wants to solve for a general solution, is there any convention/ procedure to follow?

 

[kuruman]

The two equations are the same. In the first case you considered the partial derivative $(\frac {\partial p} {\partial v})_T$, that is you took the derivative with respect to the second variable in $p=p(T,v)$ keeping the first variable constant. You could just as well have written $p=p(v,T)$ and following the same procedure would have given you $(\frac{\partial p} {\partial T})_v$ which is what you have in the second case. Note that once you write the first derivative, the other two follow from the cyclic rule, so whether you write $v=v(T,p)$ or $v=v(p,T)$ makes no difference to the product. Which variable you choose to take derivatives with respect to depends on what you are trying to show or derive.

 

[WhiteWolf98]

I think that answers my question. So the first derivative, in this case, will decide what the other two are with respect to. Once you've decided the first, there is only one choice for what the second and third variable could be with respect to; and as you add variables, you can have more and more, 'combinations' as such, but which are still all equivalent.