JEE Main 2014 — Chemistry Physical Chemistry

The pressure exerted by the vapors over the liquid surface when liquid and vapour are in equilibrium is called vapour pressure. Vapour pressure increases with an increase with temperature.

Average kinetic energy of the gas molecules is directly proportional to the absolute temperature. At higher temperatures, more molecules from the liquid have enough KE to escape from the surface of the liquid. Which will increase vapor pressure.

For phase equilibrium for liquid & vapour phases

$\frac{ \text{dP} }{ \text{dT} } = \frac{ \Delta \text{S} }{ \Delta \text{V} }$dTdP​=ΔVΔS​$= \frac{ \Delta {\text{H}}_{\text{vap}} }{ \text{T} \text{.} \Delta {\text{V}}_{\text{vap}} }$=T.ΔVvap​ΔHvap​​

$$\frac{ \text{dP} }{ \text{dT} } = \frac{ \Delta {\text{H}}_{\text{vap}} }{ \text{T} \frac{ \text{RT} }{ \text{P} } } = \frac{ \Delta {\text{H}}_{\text{vap}} \times \text{P} }{ {\text{RT}}^{2} } ( \text{for 1 mole} )$$dTdP​=TPRT​ΔHvap​​=RT2ΔHvap​×P​(for 1 mole)

$$\frac{ \text{dP} }{ \text{P} } = \frac{ \Delta {\text{H}}_{\text{vap}} }{ {\text{RT}}^{2} } \text{dT} \Rightarrow$$PdP​=RT2ΔHvap​​dT⇒

The relation between temperature and vapor pressure is given by the Clausius-Clapeyron equation.

$$\text{ln} \frac{ {\text{P}}_{2} }{ {\text{P}}_{1} } = \frac{ \Delta {\text{H}}_{\text{vap}} }{ \text{R} } [ \frac{ {\text{T}}_{ 2 - } {\text{T}}_{1} }{ {\text{T}}_{1} {\text{T}}_{2} } ]$$lnP1​P2​​=RΔHvap​​[T1​T2​T2−​T1​​]

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ie. vapour increases exponentially with rise in temp ie.

Non linear increase.