Question

How do I prove that the cyclic rule of partial derivatives applies to P, T, and V?

Answer 1

First we need to know what a cyclic rule defines. Then using the partial derivative of the ideal gas equation. We will solve this problem in three cases, first doing a partial derivative with respect to T, then with respect to V and then in third with respect to P. Finally using cyclic rule connect we can prove this problem.

Complete answer:
As per the problem we need to prove that the cyclic rule of partial derivatives applies to P, T, and V.
As per the cyclic rule, the product of each term that is temperature, pressure and volume partial differencing with others keeps the third parameter as constant and gives the output as $- 1$.
Now we know the equation for ideal gas,
$PV = nRT$
Let us assume that $n = 1$.
Now the equation will be,
$PV = RT$
Case I:
$P = \dfrac{{RT}}{V}$
Now differentiate the above equation with respect to temperature T we will get,
${\left( {\dfrac{{\partial P}}{{\partial T}}} \right)_V} = \dfrac{\partial }{{\partial T}}{\left[ {\dfrac{{RT}}{V}} \right]_V}$
$\Rightarrow {\left( {\dfrac{{\partial P}}{{\partial T}}} \right)_V} = \dfrac{1}{V}\dfrac{\partial }{{\partial T}}\left[ {RT} \right]$
$\Rightarrow {\left( {\dfrac{{\partial P}}{{\partial T}}} \right)_V} = \dfrac{R}{V}$
Case II:
$T = \dfrac{{PV}}{R}$
Now differentiate the above equation with respect to volume V we will get,
${\left( {\dfrac{{\partial T}}{{\partial V}}} \right)_P} = \dfrac{\partial }{{\partial V}}{\left[ {\dfrac{{PV}}{R}} \right]_P}$
$\Rightarrow {\left( {\dfrac{{\partial T}}{{\partial V}}} \right)_P} = P\dfrac{\partial }{{\partial V}}{\left[ {\dfrac{V}{R}} \right]_P}$
$\Rightarrow {\left( {\dfrac{{\partial T}}{{\partial V}}} \right)_P} = \dfrac{P}{R}$
Case III:
$V = \dfrac{{RT}}{P}$
Now differentiate the above equation with respect to pressure P we will get,
${\left( {\dfrac{{\partial V}}{{\partial P}}} \right)_T} = \dfrac{\partial }{{\partial V}}{\left[ {\dfrac{{RT}}{P}} \right]_T}$
$\Rightarrow {\left( {\dfrac{{\partial V}}{{\partial P}}} \right)_T} = \dfrac{\partial }{{\partial P}}{\left[ {\dfrac{{RT}}{P}} \right]_T}$
$\Rightarrow {\left( {\dfrac{{\partial V}}{{\partial P}}} \right)_T} = T\dfrac{\partial }{{\partial P}}\dfrac{R}{P}$
$\Rightarrow {\left( {\dfrac{{\partial V}}{{\partial P}}} \right)_T} = - \dfrac{{RT}}{{{P^2}}}$
Now multiplying all the cases we will get,
$\Rightarrow {\left( {\dfrac{{\partial P}}{{\partial T}}} \right)_V}{\left( {\dfrac{{\partial T}}{{\partial V}}} \right)_P}{\left( {\dfrac{{\partial V}}{{\partial P}}} \right)_T} = - \dfrac{{RT}}{{{P^2}}}\dfrac{P}{R}\dfrac{R}{V}$
We know from ideal gas equation,
$PV = RT$
Now using this in the above equation we will get,
$\Rightarrow {\left( {\dfrac{{\partial P}}{{\partial T}}} \right)_V}{\left( {\dfrac{{\partial T}}{{\partial V}}} \right)_P}{\left( {\dfrac{{\partial V}}{{\partial P}}} \right)_T} = - \dfrac{{PV}}{{{P^2}}}\dfrac{P}{R}\dfrac{R}{V} = \dfrac{{ - V{P^2}R}}{{{P^2}RV}} = - 1$
Hence proved.

Note:
Remember if we want to derive the cyclic rule of partial derivative of pressure, volume and temperature, then we write the total derivative of the pressure as a function of temperate T and molar volume V. Then after you can divide the whole derivative by at a constant P then on further solving we will get ${\left( {\dfrac{{\partial P}}{{\partial T}}} \right)_V}{\left( {\dfrac{{\partial T}}{{\partial V}}} \right)_P}{\left( {\dfrac{{\partial V}}{{\partial P}}} \right)_T} = - 1$. Each column of the partial derivative P, T, T, V, V, P where the remaining variable not involved in the derivative is held constant.