# Chain relation/ triple partial derivative rule

1. Sep 4, 2011

### Syrus

1. The problem statement, all variables and given/known data

For the van der Waals equation of state, confirm the following property:

(∂P/∂T)V (∂T/∂V)P (∂V/∂P)T = -1

2. Relevant equations

The van der Waals equation of state is:

P = nRT/(v-nb) - an2/V2

*R, n, a, b are const.

3. The attempt at a solution

I have come up with some partial derivatives, however, I cannot seem to figure out the algebra to make their product equal to -1. Perhaps my derivatives are incorrect?

(∂P/∂T)V = nR/(v-nb)-1

(∂T/∂V)P = P - an2/V2 + 2abn3/V3

(∂V/∂P)T = 1/ (2an2/V3 - nRT/(v-nb)2)

Any hints or ideas?

2. Sep 4, 2011

### Punkyc7

I believe there is formula that is (∂X/∂Y) =-(F$_{Y}$/F$_{X}$)

3. Sep 4, 2011

### Syrus

Can you explain what the right side of the equality represents?

4. Sep 4, 2011

### Punkyc7

I am not entirely sure it was just an equation in my book, I will leave it for some one else to answer because I do not want to tell you wrong

5. Sep 5, 2011

Anyone else?

6. Sep 5, 2011

### SammyS

Staff Emeritus
Show how you arrived at the last two partial derivatives. (I suggest using implicit differentiation.)

7. Sep 6, 2011

### Syrus

Well, i think i figured it out. I used the reciprocal identity:

(dx/dy)z = 1/ (dy/dx)z (should be partial derivatives here)

to make the triple partial derivative product a double partial derivative product, and then showed it to be equal to the resulting partial derivative on the other side of the equality (which occurs when you divide -1 by one of the terms originally on the left). It worked well =)

8. Sep 6, 2011

### evo_vil

Sorry for the hijack but i have a similar question:

for a recent semester test we needed to show:
$\frac{\partial{P}}{\partial{V}} \frac{\partial{V}}{\partial{T}} \frac{\partial{T}}{\partial{P}} = -1$

i simply converted each partial into its implicit version and cancelled terms:

$\frac{\partial{P}}{\partial{V}} = \frac{-F_V}{F_P}$

$\frac{\partial{V}}{\partial{T}} = \frac{-F_T}{F_V}$

$\frac{\partial{T}}{\partial{P}} = \frac{-F_P}{F_T}$

resulting in

$\frac{-F_V}{F_P} \frac{-F_T}{F_V} \frac{-F_P}{F_T} = -1$

yet this was marked very clearly wrong...

Why?