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  • Consider the following liquid-vapour equilibrium <img alt="liquid \rightleftharpoons vapour" src="https://entrancecorner.oncodecogs.com/gif.latex?liquid%20%5Crightleftharpoons%20vapour" /

Consider the following liquid-vapour equilibrium

liquid \rightleftharpoons vapour

Which of the following relations are correct?

Option: 1

\frac{dlnG}{dT^{2}}\: =\: \frac{\Delta H_{v}}{RT^{2}}


Option: 2

\frac{dlnP}{dT}=\frac{\Delta H_{v}}{RT}


Option: 3

\frac{dlnP}{dT^{2}}=\frac{-\Delta H_{v}}{T^{2}}


Option: 4

\frac{dlnP}{dT} = \frac{-\Delta H_{v}}{RT^{2}}


Answers (1)

best_answer

\frac{d\:lnP}{dT}=\frac{-\: \Delta H_{v}}{RT^{2}}

This equilibrium states that, when liquid is heated, it converts into vapour but on cooling, it further converts into liquid, which is derived by Clausius clapeyron and the relationship is written as,

\frac{\mathrm{d} \ln \mathrm{P}}{\mathrm{d} \mathrm{T}}=\frac{-\Delta \mathrm{H}_{\mathrm{v}}}{\mathrm{RT}^{2}}

Therefore, Option(4) is correct.

Posted by

Kuldeep Maurya

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