[\dfrac{dInP}{dT} = \dfra...] Consider the following liquid-vapour equilibriumLiquid \rightleftharpoons Vapour W

For Competitive Exams

Consider the following liquid-vapour equilibrium

Liquid $\rightleftharpoons$ Vapour

Which of the following relations is correct ?

$\dfrac{dInP}{dT} = \dfrac{-\triangle H_v}{RT}$

$\dfrac{dInP}{dT^2} = \dfrac{-\triangle H_v}{T^2}$

$\dfrac{dInP}{dT} = \dfrac{-\triangle H_v}{RT^2}$

$\dfrac{dInG}{dT^2} = \dfrac{-\triangle H_v}{RT^2}$

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