Answer
(a) Using Eq. (7) we prove the cyclic relation Eq. (10):
(10) ${\ \ \ }$ ${\left( {\frac{{\partial z}}{{\partial x}}} \right)_y}{\left( {\frac{{\partial x}}{{\partial y}}} \right)_z}{\left( {\frac{{\partial y}}{{\partial z}}} \right)_x} = - 1$
(b) Eq. (10) is verified for $F\left( {x,y,z} \right) = x + y + z = 0$.
(c) The cyclic relation is verified for the Ideal Gas Law.
Work Step by Step
(a) Recall Eq. (7):
${\left( {\frac{{\partial z}}{{\partial x}}} \right)_y} = - \frac{{{F_x}}}{{{F_z}}}$, ${\ \ \ }$ ${\left( {\frac{{\partial z}}{{\partial y}}} \right)_x} = - \frac{{{F_y}}}{{{F_z}}}$
By symmetry, we obtain
${\left( {\frac{{\partial x}}{{\partial y}}} \right)_z} = - \frac{{{F_y}}}{{{F_x}}}$ ${\ \ }$ and ${\ \ }$ ${\left( {\frac{{\partial y}}{{\partial z}}} \right)_x} = - \frac{{{F_z}}}{{{F_y}}}$
Thus,
${\left( {\frac{{\partial z}}{{\partial x}}} \right)_y}{\left( {\frac{{\partial x}}{{\partial y}}} \right)_z}{\left( {\frac{{\partial y}}{{\partial z}}} \right)_x} = \left( { - \frac{{{F_x}}}{{{F_z}}}} \right)\left( { - \frac{{{F_y}}}{{{F_x}}}} \right)\left( { - \frac{{{F_z}}}{{{F_y}}}} \right) = - 1$
Hence, the cyclic relation Eq. (10):
(10) ${\ \ \ }$ ${\left( {\frac{{\partial z}}{{\partial x}}} \right)_y}{\left( {\frac{{\partial x}}{{\partial y}}} \right)_z}{\left( {\frac{{\partial y}}{{\partial z}}} \right)_x} = - 1$
(b) We have $F\left( {x,y,z} \right) = x + y + z = 0$.
The partial derivatives are
${F_x} = 1$, ${\ \ }$ ${F_y} = 1$, ${\ \ }$ ${F_z} = 1$
From part (a) we obtain
${\left( {\frac{{\partial z}}{{\partial x}}} \right)_y} = - \frac{{{F_x}}}{{{F_z}}}$, ${\ \ }$ ${\left( {\frac{{\partial x}}{{\partial y}}} \right)_z} = - \frac{{{F_y}}}{{{F_x}}}$, ${\ \ }$ ${\left( {\frac{{\partial y}}{{\partial z}}} \right)_x} = - \frac{{{F_z}}}{{{F_y}}}$
So, we have
${\left( {\frac{{\partial z}}{{\partial x}}} \right)_y} = - 1$, ${\ \ }$ ${\left( {\frac{{\partial x}}{{\partial y}}} \right)_z} = - 1$, ${\ \ }$ ${\left( {\frac{{\partial y}}{{\partial z}}} \right)_x} = - 1$
${\left( {\frac{{\partial z}}{{\partial x}}} \right)_y}{\left( {\frac{{\partial x}}{{\partial y}}} \right)_z}{\left( {\frac{{\partial y}}{{\partial z}}} \right)_x} = - 1$
Hence, Eq. (10) is verified.
(c) We have the Ideal Gas Law $PV - nRT = 0$.
Write $F\left( {T,P,V} \right) \equiv PV - nRT = 0$.
The partial derivatives are
$\frac{{\partial F}}{{\partial T}} = {F_T} = - nR$, ${\ }$ $\frac{{\partial F}}{{\partial P}} = {F_P} = 1$, ${\ }$ $\frac{{\partial F}}{{\partial V}} = {F_V} = 1$
Using Eq. (7) we obtain
${\left( {\frac{{\partial V}}{{\partial T}}} \right)_P} = - \frac{{{F_T}}}{{{F_V}}}$, ${\ }$ ${\left( {\frac{{\partial T}}{{\partial P}}} \right)_V} = - \frac{{{F_P}}}{{{F_T}}}$, ${\ }$ ${\left( {\frac{{\partial P}}{{\partial V}}} \right)_T} = - \frac{{{F_V}}}{{{F_P}}}$
So,
${\left( {\frac{{\partial V}}{{\partial T}}} \right)_P} = nR$, ${\ }$ ${\left( {\frac{{\partial T}}{{\partial P}}} \right)_V} = \frac{1}{{nR}}$, ${\ }$ ${\left( {\frac{{\partial P}}{{\partial V}}} \right)_T} = - 1$
${\left( {\frac{{\partial V}}{{\partial T}}} \right)_P}{\left( {\frac{{\partial T}}{{\partial P}}} \right)_V}{\left( {\frac{{\partial P}}{{\partial V}}} \right)_T} = - 1$
Hence, the cyclic relation is verified.
