# 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.

## Answer and Explanation:

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.

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