# (p+a/v^2)v=RT,derive the cyclic rule.i.e. show the products of the partial derivatives of p,v&T taken in a order is equal to -1?

Oct 29, 2016

I am pretty sure that you mean to write $\overline{V} = \frac{V}{n}$ in place of $V$. Then, it seems that your equation of state is:

$\left(P + \frac{a}{{\overline{V}}^{2}}\right) \overline{V} = R T$

or

$P \overline{V} + \frac{a}{\overline{V}} = R T$

THE CYCLIC RULE OF PARTIAL DERIVATIVES

The cyclic rule in general says:

${\left(\frac{\partial x}{\partial y}\right)}_{z} {\left(\frac{\partial y}{\partial z}\right)}_{x} {\left(\frac{\partial z}{\partial x}\right)}_{y} = - 1$.

Since $P$ is a function of $\overline{V}$ and $T$, then if we let $x = \overline{V}$, $y = T$, and $z = P$:

((delbarV)/(delbarT))_P((delT)/(delP))_(barV)((delP)/(delbarV))_(T) = ?

where $P$ is implicitly a function of $T$ and $\overline{V}$.

PROVING THE CYCLIC RULE USING P, V/n, T

To prove the cyclic rule, write the total derivative of $P = P \left(T , \overline{V}\right)$:

$\boldsymbol{\mathrm{dP} = {\left(\frac{\partial P}{\partial T}\right)}_{\overline{V}} \mathrm{dT} + {\left(\frac{\partial P}{\partial \overline{V}}\right)}_{T} \mathrm{db} a r V}$

Now if we divide by $\mathrm{db} a r {V}_{P}$ (the differential change in molar volume at a constant pressure), we get:

${\cancel{{\left(\frac{\partial P}{\partial \overline{V}}\right)}_{P}}}^{0} = {\left(\frac{\partial P}{\partial T}\right)}_{\overline{V}} {\left(\frac{\partial T}{\partial \overline{V}}\right)}_{P} + {\left(\frac{\partial P}{\partial \overline{V}}\right)}_{T} {\cancel{{\left(\frac{\partial \overline{V}}{\partial \overline{V}}\right)}_{P}}}^{1}$

$0 = {\left(\frac{\partial P}{\partial T}\right)}_{\overline{V}} {\left(\frac{\partial T}{\partial \overline{V}}\right)}_{P} + {\left(\frac{\partial P}{\partial \overline{V}}\right)}_{T}$

$- {\left(\frac{\partial P}{\partial \overline{V}}\right)}_{T} = {\left(\frac{\partial P}{\partial T}\right)}_{\overline{V}} {\left(\frac{\partial T}{\partial \overline{V}}\right)}_{P}$

Now if you recall that ${\left(\frac{\partial x}{\partial y}\right)}_{z} = \frac{1}{{\left(\frac{\partial y}{\partial x}\right)}_{z}}$, then multiply by the respective reciprocal partial derivatives to get:

$- {\left(\frac{\partial \overline{V}}{\partial T}\right)}_{P} \cdot {\left(\frac{\partial T}{\partial P}\right)}_{\overline{V}} \cdot {\left(\frac{\partial P}{\partial \overline{V}}\right)}_{T} = {\cancel{{\left(\frac{\partial T}{\partial P}\right)}_{\overline{V}} \cdot {\left(\frac{\partial P}{\partial T}\right)}_{\overline{V}} {\left(\frac{\partial \overline{V}}{\partial T}\right)}_{P} \cdot {\left(\frac{\partial T}{\partial \overline{V}}\right)}_{P}}}^{1}$

Thus:

$\textcolor{b l u e}{{\left(\frac{\partial \overline{V}}{\partial T}\right)}_{P} {\left(\frac{\partial T}{\partial P}\right)}_{\overline{V}} {\left(\frac{\partial P}{\partial \overline{V}}\right)}_{T} = - 1}$

APPLYING THE CYCLIC RULE OF PARTIAL DERIVATIVES

To use the cyclic rule for your equation of state, try rewriting it in terms of $P$.

P-form of the equation of state

$\boldsymbol{P = \frac{R T}{\overline{V}} - \frac{a}{{\overline{V}}^{2}}}$

$\implies \textcolor{g r e e n}{{\left(\frac{\partial P}{\partial \overline{V}}\right)}_{T} = - \frac{R T}{{\overline{V}}^{2}} + \frac{2 a}{{\overline{V}}^{3}}}$

We can use the reciprocal property of partial derivatives to not have to rewrite this in terms of $\overline{V}$ as the dependent variable, which would have proven quite difficult.

$\implies \textcolor{g r e e n}{{\left(\frac{\partial T}{\partial P}\right)}_{\overline{V}}} = \frac{1}{\frac{\partial P}{\partial T}} _ \left(\overline{V}\right)$

$= \frac{1}{\frac{R}{\overline{V}}} = \textcolor{g r e e n}{\frac{\overline{V}}{R}}$

T-form of the equation of state

Now to rewrite in terms of $T$ and evaluate the ${\left(\frac{\partial \overline{V}}{\partial T}\right)}_{P}$ derivative:

$T = \frac{1}{R} \left[P \overline{V} + \frac{a}{\overline{V}}\right]$

$\implies \textcolor{g r e e n}{{\left(\frac{\partial \overline{V}}{\partial T}\right)}_{P}} = \frac{1}{{\left(\frac{\partial T}{\partial \overline{V}}\right)}_{P}}$

$= \frac{1}{\frac{P}{R} - \frac{a}{R {\overline{V}}^{2}}}$

$= \frac{1}{\frac{P {\overline{V}}^{2} - a}{R {\overline{V}}^{2}}}$

$= \textcolor{g r e e n}{\frac{R {\overline{V}}^{2}}{P {\overline{V}}^{2} - a}}$

Showing that these derivatives explicitly reduce to -1

Now we combine these to see if we get $- 1$:

((delbarV)/(delT))_P((delT)/(delP))_(barV)((delP)/(delbarV))_T stackrel(?)(=) -1

$= \frac{R {\overline{V}}^{2}}{P {\overline{V}}^{2} - a} \frac{\overline{V}}{R} \left(- \frac{R T}{{\overline{V}}^{2}} + \frac{2 a}{{\overline{V}}^{3}}\right)$

$= \frac{R {\overline{V}}^{2}}{P {\overline{V}}^{2} - a} \left(- \frac{T}{\overline{V}} + \frac{2 a}{R {\overline{V}}^{2}}\right)$

$= - \frac{T R {\overline{V}}^{2}}{\overline{V} \left(P {\overline{V}}^{2} - a\right)} + \frac{2 a R {\overline{V}}^{2}}{R {\overline{V}}^{2} \left(P {\overline{V}}^{2} - a\right)}$

$= - \frac{T {R}^{2} {\overline{V}}^{3}}{R {\overline{V}}^{2} \left(P {\overline{V}}^{2} - a\right)} + \frac{2 a R {\overline{V}}^{2}}{R {\overline{V}}^{2} \left(P {\overline{V}}^{2} - a\right)}$

Now make sure that you recall that

$- \frac{x}{a + b} + \frac{y}{a + b} = \frac{- \left(x - y\right)}{a + b} = - \frac{x - y}{a + b}$.

Thus:

$\implies \frac{- \left(T {R}^{2} {\overline{V}}^{3} - 2 a R {\overline{V}}^{2}\right)}{R {\overline{V}}^{2} \left(P {\overline{V}}^{2} - a\right)}$

$= \frac{- \cancel{R {\overline{V}}^{2}} \left(T R \overline{V} - 2 a\right)}{\cancel{R {\overline{V}}^{2}} \left(P {\overline{V}}^{2} - a\right)}$

$= \frac{- \left(R T \overline{V} - 2 a\right)}{P {\overline{V}}^{2} - a}$

Now we substitute $R T = P \overline{V} + \frac{a}{\overline{V}}$ from the original equation to get:

$\implies \frac{- \left(\left(P \overline{V} + \frac{a}{\overline{V}}\right) \overline{V} - 2 a\right)}{P {\overline{V}}^{2} - a}$

$= - \frac{P {\overline{V}}^{2} + a - 2 a}{P {\overline{V}}^{2} - a}$

$= - \cancel{\frac{P {\overline{V}}^{2} - a}{P {\overline{V}}^{2} - a}}$

$\implies \textcolor{b l u e}{- 1 = {\left(\frac{\partial \overline{V}}{\partial T}\right)}_{P} {\left(\frac{\partial T}{\partial P}\right)}_{\overline{V}} {\left(\frac{\partial P}{\partial \overline{V}}\right)}_{T}}$

as expected!