# With RT =PV, what do you expect the product \frac {\partial P}{ \partial R} \frac {\partial R}{...

## Question:

With {eq}RT =PV {/eq}, what do you expect the product {eq}\frac {\partial P}{ \partial R} \frac {\partial R}{ \partial T} \frac {\partial T}{ \partial P} {/eq} to equal? Now prove that {eq}\frac {\partial P}{ \partial R} \frac {\partial R}{ \partial T} \frac {\partial T}{ \partial P} = -1 {/eq}

## Partial Derivative

In calculus, partial derivative is described as a function consisting of a lot variable. This function is the differentiation of one variable when all other variables are considered as constant. Usually, this function is used in vector calculus.

Consider the equation provided in the problem.

{eq}RT = PV \cdots\cdots\rm{(I)} {/eq}

Here, the temperature is {eq}T {/eq}, volume is {eq}V {/eq}, pressure is P and the gas constant is {eq}R {/eq}.

Rewrite the above expression in terms of temperature.

{eq}T = \dfrac{{PV}}{R} {/eq}

Partial differentiate the above expression with respect to pressure.

{eq}\dfrac{{\partial T}}{{\partial P}} = \dfrac{V}{R} \cdots\cdots\rm{(II)} {/eq}

Rewrite Equation (I) in terms of resistance.

{eq}R = \dfrac{{PV}}{T} {/eq}

Partial differentiate the above expression with respect to temperature.

{eq}\dfrac{{\partial R}}{{\partial T}} = \dfrac{{ - PV}}{{{T^2}}} \cdots\cdots\rm{(III)} {/eq}

Rewrite Equation (I) in terms of pressure.

{eq}P = \dfrac{{RT}}{V} {/eq}

Partial differentiate the above expression with respect to resistance.

{eq}\dfrac{{\partial P}}{{\partial R}} = \dfrac{T}{V} \cdots\cdots\rm{(IV)} {/eq}

Multiply Equation (IV), (III) and Equation (II).

{eq}\begin{align*} \dfrac{{\partial P}}{{\partial R}}\dfrac{{\partial R}}{{\partial T}}\dfrac{{\partial T}}{{\partial P}} &= \dfrac{T}{V}\dfrac{{ - PV}}{{{T^2}}}\dfrac{V}{R}\\ &= - \dfrac{{PV}}{{RT}}\\ &= - \dfrac{{PV}}{{PV}}\\ &= - 1 \end{align*} {/eq}

Therefore, the expected product is {eq}- 1 {/eq}.

Thus, proved.