Solved: CHEMISTRY Question for NEET

NEET CHEMISTRYMedium

Consider the following liquid-vapor equilibrium: Liquid ⇌ Vapour. Which of the following relations is correct?

\frac{d \ln P}{dT} = \frac{-\Delta H_v}{RT}

\frac{d \ln P}{dT} = \frac{-\Delta H_v}{T^2}

\frac{d \ln P}{dT} = \frac{\Delta H_v}{RT^2}

\frac{d \ln G}{dT^2} = \frac{-\Delta H_v}{RT^2}

Step-by-Step Solution

The relationship between vapor pressure ( $P$ ) and temperature ( $T$ ) for a liquid-vapor equilibrium is given by the Clausius-Clapeyron equation.

Concept: For the equilibrium $Liquid \rightleftharpoons Vapour$ , the equilibrium constant $K_p$ is equal to the vapor pressure $P$ (assuming pure liquid).
Temperature Dependence: The variation of the equilibrium constant with temperature is given by the van't Hoff equation: $\frac{d \ln K}{dT} = \frac{\Delta H^\circ}{RT^2}$ .
Substitution: Substituting $K = P$ and $\Delta H^\circ = \Delta H_{vap}$ (Enthalpy of vaporization, $\Delta H_v$ ), we get: $\frac{d \ln P}{dT} = \frac{\Delta H_v}{RT^2}$

This equation quantitatively describes how vapor pressure increases with temperature for an endothermic process (vaporization).

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